# Quantum Chemistry/Printable version

The current, editable version of this book is available in Wikibooks, the open-content textbooks collection, at
https://en.wikibooks.org/wiki/Quantum_Chemistry

# Complex numbers

write a brief review of complex numbers, including euler's relation.

### Sample Question

Use Euler's relation to simplify the following expression, which is relevant to the phase factor in quantum wavefunctions:
$e^{i\pi /4}\cdot e^{-i\pi /3}$
Express your answer in the form a+bi, where a and b are real numbers.

# Integration by parts

Write a brief review of integrals in spherical polar coordinates

### Example

$\int _{0}^{2\pi }\int _{0}^{\pi }\int _{0}^{\infty }e^{-r}r^{3}\sin \theta ~{\textrm {d}}r~{\textrm {d}}\theta ~{\textrm {d}}\phi$

# Probability and statistics

Write a brief summary of the properties of probability distributions. Give an example of integrating a probability distribution to determine a probability.

# Multivariable differentiation

Provide a brief description of partial differentiation.

### Example

Show the calculation of the derivative of this wavefunction with respect to x and y for the function $\psi (x,y)=A\sin \left({\frac {n_{x}\pi }{L}}\right)\cdot \sin \left({\frac {n_{y}\pi }{L}}\right)$

# Implicit differentiation

Provide a brief description of implicit differentiation.

### Example

Differentiate the function with respect to x
$x^{2}+xy+y^{2}$
Where $y=y(x)$

# Operator algebra

Provide a brief description of operator algebra as it relates to quantum chemistry. Provide a short example

# Example 1

Find ⟨x⟩, ⟨x$^2$⟩, ⟨p$_x$⟩ and ⟨p$_x$$^2$⟩ for a quantum harmonic oscillator in the ground state, then determine the uncertainty on the position and momentum. Is the product of the uncertainty on position and momentum consistent with the Heisenberg's Uncertainty Principle?
                                                                         Heisenberg's Uncertainty Principle $\qquad \delta _{x}\cdot \delta _{p}\geq {\frac {\hbar }{2}}\qquad {\text{where}}\ \hbar ={\frac {h}{2\pi }}$
The wavefunction of a quantum harmonic oscillator in the ground state is:
$\qquad \qquad \Psi _{0}(x)={\Bigl{\frac {\alpha }{\pi }}{\Bigr )}^{\frac {1}{4}}e^{\frac {-\alpha x^{2}}{2}}$ Using this wavefunction the average position and the average of the square of the position can be calculated.
The average position:
$\qquad \qquad \qquad \qquad \langle x\rangle =\int _{-\infty }^{\infty }\Psi _{0}(x)^{*}\cdot x\cdot \Psi _{0}(x)\ dx$
$\qquad \qquad \qquad \qquad \quad \ =\int _{-\infty }^{\infty }{\Bigl{\frac {\alpha }{\pi }}{\Bigr )}^{\frac {1}{4}}e^{\frac {-\alpha x^{2}}{2}}\cdot x\cdot {\Bigl{\frac {\alpha }{\pi }}{\Bigr )}^{\frac {1}{4}}e^{\frac {-\alpha x^{2}}{2}}\ dx$
$\qquad \qquad \qquad \qquad \quad \ ={\Bigl{\frac {\alpha }{\pi }}{\Bigr )}^{\frac {1}{2}}\int _{-\infty }^{\infty }x\cdot e^{\frac {-2\alpha x^{2}}{2}}\ dx$
$\qquad \qquad \qquad \qquad \quad \ ={\Bigl{\frac {\alpha }{\pi }}{\Bigr )}^{\frac {1}{2}}\int _{-\infty }^{\infty }xe^{-\alpha x^{2}}\ dx$
                   use $\int xe^{-\alpha x^{2}}\ dx={\frac {-1}{2\alpha }}e^{-\alpha x^{2}}+C$
$\qquad \qquad \qquad \qquad \quad \ ={\Bigl{\frac {\alpha }{\pi }}{\Bigr )}^{\frac {1}{2}}{\biggl{\frac {-1}{2\alpha }}\cdot e^{-\alpha x^{2}}{\biggl ]}_{-\infty }^{\infty }$
$\qquad \qquad \qquad \qquad \quad \ ={\Bigl{\frac {\alpha }{\pi }}{\Bigr )}^{\frac {1}{2}}{\biggl{\frac {-1}{2\alpha }}(0)-{\frac {1}{2\alpha }}(0){\biggl ]}$
$\qquad \qquad \qquad \qquad \langle x\rangle =0$
The average square of the position:
$\qquad \qquad \qquad \qquad \qquad \qquad \qquad \langle x^{2}\rangle =\int _{-\infty }^{\infty }\Psi _{0}(x)^{*}\cdot x^{2}\cdot \Psi _{0}(x)\ dx$
$\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad =\int _{-\infty }^{\infty }{\Bigl{\frac {\alpha }{\pi }}{\Bigr )}^{\frac {1}{4}}e^{\frac {-\alpha x^{2}}{2}}\cdot x^{2}\cdot {\Bigl{\frac {\alpha }{\pi }}{\Bigr )}^{\frac {1}{4}}e^{\frac {-\alpha x^{2}}{2}}\ dx$
$\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad ={\Bigl{\frac {\alpha }{\pi }}{\Bigr )}^{\frac {1}{2}}\int _{-\infty }^{\infty }x^{2}e^{\frac {-2\alpha x^{2}}{2}}\ dx$
$\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad ={\Bigl{\frac {\alpha }{\pi }}{\Bigr )}^{\frac {1}{2}}\int _{-\infty }^{\infty }x^{2}e^{-\alpha x^{2}}\ dx$
                               use $\int x^{2}e^{-\alpha x^{2}}\ dx={\frac {{\sqrt {\pi }}{\text{erf}}({\sqrt {\alpha }}x)}{4\alpha ^{\frac {3}{2}}}}\ -\ {\frac {xe^{-\alpha x^{2}}}{2\alpha }}+C$
$\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad ={\Bigl{\frac {\alpha }{\pi }}{\Bigr )}^{\frac {1}{2}}{\Biggl{\frac {{\sqrt {\pi }}{\text{erf}}({\sqrt {\alpha }}x)}{4\alpha ^{\frac {3}{2}}}}\ -\ {\frac {xe^{-\alpha x^{2}}}{2\alpha }}{\Biggl ]}_{-\infty }^{\infty }$
$\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad ={\Bigl{\frac {\alpha }{\pi }}{\Bigr )}^{\frac {1}{2}}{\Biggl{\frac {\sqrt {\pi }}{4}}{\frac {{\text{erf}}({\sqrt {\alpha }}\ \infty )}{\alpha ^{\frac {3}{2}}}}\ -\ \lim _{x\to \infty }{\frac {xe^{-\alpha x^{2}}}{2\alpha }}{\Biggl ]}\ -\ {\Biggl{\frac {\sqrt {\pi }}{4}}{\frac {{\text{erf}}({\sqrt {\alpha }}-\infty )}{\alpha ^{\frac {3}{2}}}}\ -\ \lim _{x\to -\infty }{\frac {xe^{-\alpha x^{2}}}{2\alpha }}{\Biggl ]}$
                               use $\lim _{x\to \infty }{\text{erf}}(x)=1$ and $\lim _{x\to -\infty }{\text{erf}}(x)=-1$
$\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad ={\Bigl{\frac {\alpha }{\pi }}{\Bigr )}^{\frac {1}{2}}\ {\frac {\sqrt {\pi }}{4}}\ {\frac {1}{\alpha ^{\frac {3}{2}}}}\ {\biggl1-(-1){\biggl ]}$
$\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad ={\frac {\sqrt {\alpha }}{\sqrt {\pi }}}\ {\frac {\sqrt {\pi }}{4}}\ {\frac {1}{\alpha ^{\frac {3}{2}}}}\ {\bigl2{\bigl ]}$
$\qquad \qquad \qquad \qquad \qquad \qquad \qquad \langle x^{2}\rangle ={\frac {1}{2\alpha }}$
The uncertainty on the position:
$\qquad \qquad \qquad \qquad \qquad \qquad \qquad \delta _{x}={\sqrt {\langle x^{2}\rangle -\langle x\rangle ^{2}}}$
$\qquad \qquad \qquad \qquad \qquad \qquad \qquad \ \ \ \ ={\sqrt {{\frac {1}{2\alpha }}-0}}$
$\qquad \qquad \qquad \qquad \qquad \qquad \qquad \delta _{x}={\sqrt {\frac {1}{2\alpha }}}$
The average momentum:
$\qquad \qquad \qquad \qquad \qquad \qquad \langle p_{x}\rangle =\int _{-\infty }^{\infty }\Psi _{0}(x)^{*}\cdot {\widehat {p}}_{x}\cdot \Psi _{0}(x)\ dx\qquad \qquad \qquad \qquad \qquad \qquad {\text{where}}\ {\widehat {p}}_{x}=-i\hbar {\Bigl{\frac {\partial }{\partial x}}{\Bigr )}$
$\qquad \qquad \qquad \qquad \qquad \qquad \qquad =\int _{-\infty }^{\infty }{\Bigl{\frac {\alpha }{\pi }}{\Bigr )}^{\frac {1}{4}}e^{\frac {-\alpha x^{2}}{2}}\cdot -i\hbar {\Biggl{\frac {\partial }{\partial x}}{\Biggr )}\cdot {\Bigl{\frac {\alpha }{\pi }}{\Bigr )}^{\frac {1}{4}}e^{\frac {-\alpha x^{2}}{2}}\ dx$
$\qquad \qquad \qquad \qquad \qquad \qquad \qquad ={\Bigl{\frac {\alpha }{\pi }}{\Bigr )}^{\frac {1}{2}}-i\hbar \int _{-\infty }^{\infty }e^{\frac {-\alpha x^{2}}{2}}{\Biggl{\frac {\partial }{\partial x}}{\Biggr )}\ e^{\frac {-\alpha x^{2}}{2}}\ dx$
$\qquad \qquad \qquad \qquad \qquad \qquad \qquad ={\Bigl{\frac {\alpha }{\pi }}{\Bigr )}^{\frac {1}{2}}-i\hbar \int _{-\infty }^{\infty }e^{\frac {-\alpha x^{2}}{2}}{\biggl-\alpha x\ e^{\frac {-\alpha x^{2}}{2}}{\biggl )}\ dx$
$\qquad \qquad \qquad \qquad \qquad \qquad \qquad =-\alpha {\Bigl{\frac {\alpha }{\pi }}{\Bigr )}^{\frac {1}{2}}-i\hbar \int _{-\infty }^{\infty }e^{\frac {-\alpha x^{2}}{2}}\cdot x\ e^{\frac {-\alpha x^{2}}{2}}\ dx$
$\qquad \qquad \qquad \qquad \qquad \qquad \qquad =-\alpha {\Bigl{\frac {\alpha }{\pi }}{\Bigr )}^{\frac {1}{2}}-i\hbar \int _{-\infty }^{\infty }x\ e^{-\alpha x^{2}}\ dx$
                            use $\int xe^{-\alpha x^{2}}\ dx={\frac {-1}{2\alpha }}e^{-\alpha x^{2}}+C$
$\qquad \qquad \qquad \qquad \qquad \qquad \qquad ={\frac {-\alpha ^{\frac {3}{2}}}{\sqrt {\pi }}}-i\hbar \ {\biggl{\frac {1}{2\alpha }}\ e^{-2\alpha x^{2}}{\biggl ]}_{-\infty }^{\infty }$
$\qquad \qquad \qquad \qquad \qquad \qquad \qquad ={\frac {-\alpha ^{\frac {3}{2}}}{\sqrt {\pi }}}-i\hbar {\biggl{\frac {1}{2\alpha }}(0)-{\frac {1}{2\alpha }}(0){\biggl ]}$
$\qquad \qquad \qquad \qquad \qquad \qquad \langle p_{x}\rangle =0$
The average square of the momentum:
$\qquad \qquad \qquad \qquad \qquad \qquad \langle p_{x}^{2}\rangle =\int _{-\infty }^{\infty }\Psi _{0}(x)^{*}\cdot {\widehat {p}}_{x}^{2}\cdot \Psi _{0}(x)\ dx$
$\qquad \qquad \qquad \qquad \qquad \qquad \qquad =\int _{-\infty }^{\infty }{\Bigl{\frac {\alpha }{\pi }}{\Bigr )}^{\frac {1}{4}}e^{\frac {-\alpha x^{2}}{2}}\cdot {\Biggl-i\hbar {\frac {\partial }{\partial x}}{\Biggr )}{\Biggl-i\hbar {\frac {\partial }{\partial x}}{\Biggr )}\cdot {\Bigl{\frac {\alpha }{\pi }}{\Bigr )}^{\frac {1}{4}}e^{\frac {-\alpha x^{2}}{2}}\ dx$
$\qquad \qquad \qquad \qquad \qquad \qquad \qquad ={\Bigl{\frac {\alpha }{\pi }}{\Bigr )}^{\frac {1}{2}}i^{2}\hbar ^{2}\int _{-\infty }^{\infty }e^{\frac {-\alpha x^{2}}{2}}{\Biggl{\frac {\partial }{\partial x}}{\Biggr )}{\Biggl{\frac {\partial }{\partial x}}{\Biggr )}\ e^{\frac {-\alpha x^{2}}{2}}\ dx$
$\qquad \qquad \qquad \qquad \qquad \qquad \qquad ={\Bigl{\frac {\alpha }{\pi }}{\Bigr )}^{\frac {1}{2}}(-1)\hbar ^{2}\int _{-\infty }^{\infty }e^{\frac {-\alpha x^{2}}{2}}{\Biggl{\frac {\partial }{\partial x}}{\Biggr )}\ {\biggl-\alpha xe^{\frac {-\alpha x^{2}}{2}}{\biggl )}\ dx$
$\qquad \qquad \qquad \qquad \qquad \qquad \qquad =\alpha {\Bigl{\frac {\alpha }{\pi }}{\Bigr )}^{\frac {1}{2}}\hbar ^{2}\int _{-\infty }^{\infty }e^{\frac {-\alpha x^{2}}{2}}{\biggle^{\frac {-\alpha x^{2}}{2}}+\alpha x^{2}e^{\frac {-2\alpha x^{2}}{2}}{\biggl )}\ dx$
$\qquad \qquad \qquad \qquad \qquad \qquad \qquad ={\frac {\alpha ^{\frac {3}{2}}}{\sqrt {\pi }}}\ \hbar ^{2}\int _{-\infty }^{\infty }e^{-\alpha x^{2}}\ dx\int _{-\infty }^{\infty }e^{\frac {-\alpha x^{2}}{2}}\ \alpha x^{2}e^{\frac {-2\alpha x^{2}}{2}}\ dx$
                                        use $\int _{-\infty }^{\infty }e^{-\alpha x^{2}}\ dx={\sqrt {\frac {\pi }{\alpha }}}$
$\qquad \qquad \qquad \qquad \qquad \qquad \qquad ={\frac {\alpha ^{\frac {3}{2}}}{\sqrt {\pi }}}\ \hbar ^{2}{\biggl{\sqrt {\frac {\pi }{\alpha }}}+\alpha \int _{-\infty }^{\infty }x^{2}e^{-\alpha x^{2}}\ dx{\biggl )}$
                               use $\int x^{2}e^{-\alpha x^{2}}\ dx={\frac {{\sqrt {\pi }}{\text{erf}}({\sqrt {\alpha }}x)}{4\alpha ^{\frac {3}{2}}}}\ -\ {\frac {xe^{-\alpha x^{2}}}{2\alpha }}+C$
$\qquad \qquad \qquad \qquad \qquad \qquad \qquad ={\frac {\alpha ^{\frac {3}{2}}}{\sqrt {\pi }}}\ \hbar ^{2}{\Biggl{\sqrt {\frac {\pi }{\alpha }}}+\alpha {\biggl{\frac {{\sqrt {\pi }}\ {\text{erf}}({\sqrt {\alpha }}\ x)}{4\alpha ^{\frac {3}{2}}}}-\ {\frac {xe^{-\alpha x^{2}}}{2\alpha }}{\biggl ]}_{-\infty }^{\infty }{\Biggl )}$
$\qquad \qquad \qquad \qquad \qquad \qquad \qquad ={\frac {\alpha ^{\frac {3}{2}}}{\sqrt {\pi }}}\ \hbar ^{2}{\Biggl{\sqrt {\frac {\pi }{\alpha }}}+\alpha {\Biggl{\frac {\sqrt {\pi }}{4\alpha ^{\frac {3}{2}}}}\ {\text{erf}}({\sqrt {\alpha }}\infty )\lim _{x\to \infty }-\ {\frac {xe^{-\alpha x^{2}}}{2\alpha }}{\Biggl ]}\ -\ {\Biggl{\frac {\sqrt {\pi }}{4\alpha ^{\frac {3}{2}}}}\ {\text{erf}}({\sqrt {\alpha }}-\infty )\lim _{x\to -\infty }-\ {\frac {xe^{-\alpha x^{2}}}{2\alpha }}{\Biggl ]}{\Biggl )}$
                               use $\lim _{x\to \infty }{\text{erf}}(x)=1$ and $\lim _{x\to -\infty }{\text{erf}}(x)=-1$
$\qquad \qquad \qquad \qquad \qquad \qquad \qquad ={\frac {\alpha ^{\frac {3}{2}}}{\sqrt {\pi }}}\ \hbar ^{2}{\Biggl{\sqrt {\frac {\pi }{\alpha }}}+\alpha {\Biggl{\frac {\sqrt {\pi }}{4\alpha ^{\frac {3}{2}}}}\ (1)-\ (0){\Biggl ]}\ -\ {\Biggl{\frac {\sqrt {\pi }}{4\alpha ^{\frac {3}{2}}}}\ (-1)-\ (0){\Biggl ]}{\Biggl )}$
$\qquad \qquad \qquad \qquad \qquad \qquad \qquad ={\frac {\alpha ^{\frac {3}{2}}}{\sqrt {\pi }}}\ \hbar ^{2}{\Biggl{\sqrt {\frac {\pi }{\alpha }}}+\alpha {\biggl{\frac {2{\sqrt {\pi }}}{4\alpha ^{\frac {3}{2}}}}\ {\biggl )}{\Biggl ]}$
$\qquad \qquad \qquad \qquad \qquad \qquad \qquad ={\frac {\alpha }{\sqrt {\pi }}}\ \hbar ^{2}{\Biggl{\sqrt {\pi }}+{\frac {1}{2}}{\sqrt {\pi }}{\Biggl ]}$
$\qquad \qquad \qquad \qquad \qquad \qquad \langle p_{x}^{2}\rangle ={\frac {1}{2}}\alpha \hbar ^{2}$
The uncertainty on the momentum:
$\qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad \delta p_{x}={\sqrt {\langle p_{x}^{2}\rangle -\langle p_{x}\rangle ^{2}}}$
$\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \ \ ={\sqrt {{\frac {a\hbar ^{2}}{2}}-0^{2}}}$
$\qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad \delta p_{x}={\sqrt {\frac {a^{2}}{2}}}\ \hbar$
The product of the uncertainty on the position and the uncertainty on the momentum is:
$\qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad \delta _{x}\cdot \delta p_{x}={\frac {1}{\sqrt {2\alpha }}}\cdot {\sqrt {\frac {\alpha }{2}}}\ \hbar$
$\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \ ={\frac {1}{2}}\ \hbar$
This is equal to ${\frac {\hbar }{2}}$, therefore, a quantum harmonic oscillator in the ground state is consistent with the Heisenberg Uncertainty Principle.

# Example 2
## Question

Starting with the wave equation of a 1D box:
$\Psi \left(x\right)={\sqrt {\frac {2}{L}}}\sin \left({\frac {n\pi }{L}}x\right)$
The classical probability of a particle existing in the region of the box [0, 1/L] is,
$P\left(1/L\right)_{classical}={\frac {1}{L}}$
Where $L$ is the length of the 1D box, $n$ is the principle quantum number, and $x$ is the position of the particle in a 1D box.
Derive an equation that calculates the probability of the particle being in the [0, 1/L] for the quantum mechanical particle in a 1D box. Is this result compatible with the correspondence principle?

## Solution
# Example 3
## Question 3

Calculate the number of nodes for a particle in a 1D box when n=2 and n=3 when the length of the box is L=5, and give the x-intercept of the node(s).
    Wavefunction for a particle in a 1D-Box $\psi (x)={\sqrt {\frac {2}{L}}}\sin \left({\frac {n\pi x}{L}}\right)$

## Answer
#### Part 1

n$_{nodes}$=n-1
n$_{nodes}$=2-1=1
We're looking for one x-intercept.
$0\div {\sqrt {\frac {2}{5}}}={\sqrt {\frac {2}{5}}}\sin \left({\frac {2\pi x}{5}}\right)\div {\sqrt {\frac {2}{5}}}$
$0={\sqrt {\frac {2}{5}}}\sin \left({\frac {2\pi x}{5}}\right)$
Use the form $\sin(kx)$ where $k=\left({\frac {2\pi }{5}}\right)$
$Period=\left({\frac {2\pi }{k}}\right)$
$Period=\left({\frac {2\pi }{\frac {2\pi }{5}}}\right)=5$
$\left({\frac {Period}{2}}\right)$ is the x-intercept
x-intercepts are 0, $\left({\frac {5}{2}}\right)$, 5
0 and 5 are the edges of the box and $\left({\frac {5}{2}}\right)$ is the only node.

#### Part 2

n$_{nodes}$=n-1
n$_{nodes}$=3-1=2
We're looking for two x-intercepts.
$0\div {\sqrt {\frac {3}{5}}}={\sqrt {\frac {3}{5}}}\sin \left({\frac {3\pi x}{5}}\right)\div {\sqrt {\frac {3}{5}}}$
$0={\sqrt {\frac {3}{5}}}\sin \left({\frac {3\pi x}{5}}\right)$
$k=\left({\frac {3\pi }{5}}\right)$ $Period=\left({\frac {3\pi }{\frac {3\pi }{5}}}\right)={\frac {10}{3}}$
x-intercepts are $\left({\frac {5n}{3}}\right)$
x-intercepts = 0, $\left({\frac {5}{3}}\right),\left({\frac {10}{3}}\right),5$
0 and 5 are the edges of the box, $\left({\frac {5}{3}}\right)$ and $\left({\frac {10}{3}}\right)$ are the nodes.

# Example 4
## Question

An electron in a 1D box emits a photon as the electron transitions to a lower energy level. If the length of the 1D box is equal to 1.0 cm, and the quantum number transition is $n_{{3}\rightarrow {2}}$, what is the electromagnetic radiation frequency of the emitted photon?
Solution:
The energy level of a particle in a 1D box at a specific quantum number (${E_{n}}$) is,
$E_{n}={\frac {h^{2}}{8mL^{2}}}{n^{2}}$
Where $h$ is equal to Planck's constant (6.62607015 x 10$^{-34}$ J$\cdot$s), $n$ is equal to the quantum number ($n$ = 1, 2, 3, ...), $m$ is equal to the mass of the particle, and $L$ is equal to the length of the 1D box. For an electron, the mass is equal to 9.10938356 x 10$^{-31}$ kg.
Since ${E_{n}}\propto {n^{2}}$ (assuming the mass and box length are constant), the energy level increases by a factor of 4 as the quantum number increases by a factor of 2. Therefore, if a particle in a 1D box undergoes an energy level transition, there is a difference between the initial and final quantum number energy levels. The energy level difference ($\Delta {E}$) of a particle in a 1D box that has undergone an energy level transition is,
$\Delta {E}={E_{n_{f}}}-{E_{n_{i}}}$
$=\left[{\frac {h^{2}}{8mL^{2}}}{n_{f}^{2}}\right]-\left[{\frac {h^{2}}{8mL^{2}}}{n_{i}^{2}}\right]$
$\Delta {E}={\frac {h^{2}}{8mL^{2}}}{({n_{f}^{2}}-{n_{i}^{2}}})$
Where $n_{f}$ is equal to the final quantum number, and $n_{i}$ is equal to the initial quantum number.
If $n_{f}>n_{i}$, $\Delta {E}$ is a positive value; photon absorbed.
If $n_{f}<n_{i}$, $\Delta {E}$ is a negative value; photon emitted.
Therefore, the energy level difference of an electron in a 1D box with a length of 1.0 cm, which has undergone a $n_{{3}\rightarrow {2}}$ transition is,
$\Delta {E}=\left[{\frac {(6.6261\times 10^{-34}{\text{ J}}{\text{ s}})^{2}}{8(9.1094\times 10^{-31}{\text{ kg}})(1.0\times 10^{-2}{\text{ m}})^{2}}}\right]{({2}^{2}-{3}^{2}})=\left[{\frac {(6.6261\times 10^{-34}{\text{ m}}^{2}{\text{ kg}}{\text{ s}}^{-1})^{2}}{8(9.1094\times 10^{-31}{\text{ kg}})(1.0\times 10^{-2}{\text{ m}})^{2}}}\right]{({2}^{2}-{3}^{2}})$
$\Delta {E}=-3.01\times 10^{-33}{\text{ J}}$
The energy level difference for the electron which underwent a $n_{{3}\rightarrow {2}}$ transition is equal to -3.01 × 10$^{-33}$ J. Since $n_{f}<n_{i}$, 3.01 × 10$^{-33}$ J was emitted from the electron. If the electron underwent a $n_{{2}\rightarrow {3}}$ transition, the electron would absorb the same amount of energy that was emitted from the $n_{{3}\rightarrow {2}}$ transition which was 3.01 × 10$^{-33}$ J.
$\Delta {E}=\left[{\frac {(6.6261\times 10^{-34}{\text{ m}}^{2}{\text{ kg}}{\text{ s}}^{-1})^{2}}{8(9.1094\times 10^{-31}{\text{ kg}})(1.0\times 10^{-2}{\text{ m}})^{2}}}\right]{({3}^{2}-{2}^{2}})=3.01\times 10^{-33}{\text{ J}}$
Therefore,
$E_{{\text{photon}}_{n_{2\rightarrow 3}}}=E_{{\text{photon}}_{n_{3\rightarrow 2}}}$
The energy of a photon has a specific frequency of electromagnetic (EM) radiation, and the energy is directly proportional to the frequency. The energy of the photon is equal to,
$E=h\cdot {f}$
Where $h$ is equal to Planck's constant (6.62607015 x 10$^{-34}$ J$\cdot$s), and $f$ is equal to EM radiation frequency.
Rearranging this equation allows for the calculation of the photon EM radiation frequency,
$E=h\cdot {f}$
$f={\frac {E}{h}}$
The calculated photon energy was equal to 3.01 × 10$^{-33}$ J, therefore the EM radiation frequency of the emitted photon from the $n_{{3}\rightarrow {2}}$ transition of an electron in a 1D box with a length of 1.0 cm is,
$f={\frac {3.01\times 10^{-33}{\text{ J}}}{6.6261\times 10^{-34}{\text{ J}}{\text{ s}}}}$
$f=4.54{\text{s}}^{-1}=4.54{\text{Hz}}$

# Example 5

For a particle (assume the particle is an electron with 1 quantum number.) in a 1D box of length 5 cm, the equation of energy levels of a particle in a 1D box is,
$E_{n}={\frac {h^{2}n^{2}}{8mL^{2}}}$
a. If the length of the 1D box is increased to 10 cm, what is the change in the energy level of this particle in the box?
b. If the length of the 1D box is decreased to 2 cm, what is the change in the energy level of this particle in the box?
c. Explain the effect of length changes on the energy levels of particles in a 1D box.

## Solutions

The energy levels of a particle in 1D box:
$E_{n}={\frac {h^{2}n^{2}}{8mL^{2}}}$
where h is Planck's constant equal to $6.626\times 10^{-34}{\text{ J s}}$
Because this particle is an electron with 1 quantum number:
m is the mass of this particle equal to $9.109\times 10^{-31}{\text{ kg}}$
n is the quantum number of the particle equal to 1
L is the length of this 1D box equal to $5{\text{ cm}}$
for this question,
$\Delta {E}=E_{nf}-E_{ni}$
$=\left[{\frac {h^{2}n^{2}}{8m}}{\frac {1}{L_{f}^{2}}}\right]-\left[{\frac {h^{2}n^{2}}{8m}}{\frac {1}{L_{i}^{2}}}\right]$
$=\left[{\frac {h^{2}n^{2}}{8m}}\right]\times \left[{\frac {1}{L_{f}^{2}}}-{\frac {1}{L_{i}^{2}}}\right]$
a. for part a, the initial length of this 1D box: $L_{i}=5{\text{ cm}}=5\times 10^{-2}{\text{ m}}$, the final length of this 1D box: $L_{f}=10{\text{ cm}}=10\times 10^{-2}{\text{ m}}$
$\Delta {E}=\left[{\frac {h^{2}n^{2}}{8m}}\right]\times \left[{\frac {1}{L_{f}^{2}}}-{\frac {1}{L_{i}^{2}}}\right]$
$=\left[{\frac {(6.626\times 10^{-34}{\text{ J s}})^{2}(1)^{2}}{(8)(9.109\times 10^{-31}{\text{ kg}})}}\right]\left[{\frac {1}{(10\times 10^{-2}{\text{m}})^{2}}}-{\frac {1}{(5\times 10^{-2}{\text{m}})^{2}}}\right]$
$=-1.807\times 10^{-35}{\text{ J}}$
b. for part b, the initial length of this 1D box: $L_{i}=5{\text{ cm}}=5\times 10^{-2}{\text{ m}}$, the final length of this 1D box: $L_{f}=2{\text{ cm}}=2\times 10^{-2}{\text{ m}}$
$\Delta {E}=\left[{\frac {h^{2}n^{2}}{8m}}\right]\times \left[{\frac {1}{L_{f}^{2}}}-{\frac {1}{L_{i}^{2}}}\right]$
$=\left[{\frac {(6.626\times 10^{-34}{\text{ J s}})^{2}(1)^{2}}{(8)(9.109\times 10^{-31}{\text{ kg}})}}\right]\left[{\frac {1}{(2\times 10^{-2}{\text{m}})^{2}}}-{\frac {1}{(5\times 10^{-2}{\text{m}})^{2}}}\right]$
$=1.265\times 10^{-34}{\text{ J}}$
c. Based on the calculations in part a and b, the energy level of the particles in the 1D box is decreased by the increase of this 1D box length, and the energy level of the particles in the 1D box is increased by the decrease of this 1D box length. Therefore, the energy level of the particles in the 1D box is negatively related to the length of this 1D box.

# Example 6

Using the following equation calculate the ground state energies of H, He$^+$ and Li$^{2+}$, $E=-\left({\frac {Z^{2}m_{e}e^{4}}{32\pi ^{2}\epsilon _{0}^{2}\hbar ^{2}}}\right){\frac {1}{n^{2}}}$, then draw a conclusion on the effect of increasing atomic number on the energy of the ground state.
Solution:
Using the Schrödinger equation (ĤΨ=EΨ) the wave functions of the hydrogen atom have been solved. From the Schrödinger equation, the energy of the quantized states of the hydrogen atom can also be determine, as energy is the eigenvalue of the Schrödinger equation. The equation for energy for each energy level in the hydrogen atom is as follows,
    Energy of Energy Levels $E=-\left({\frac {Z^{2}m_{e}e^{4}}{32\pi ^{2}\epsilon _{0}^{2}\hbar ^{2}}}\right){\frac {1}{n^{2}}}$
In the equation stated above Z is the nuclear charge, m$_e$ (9.10938356 x 10$^{-31}$ kg) is the mass of the electron, e (1.60217662 x 10$^{-19}$ C) is the elementary charge, ε$_0$ (8.8541878128 x 10$^{-12}$ Fm$^{-1}$ ) is the vacuum permittivity, ħ is the reduced Planck's constant and n is the principal quantum number.
The wave functions of the hydrogen atom can be extended to any atom with only one electron. Any atom with only one electron will have the electron in the same energy level as the hydrogen atom as electrons must be placed in the lower energy levels before higher ones. The only thing that will differ between hydrogen and other atoms with one electron is the attraction between the nucleus and the electron. As other atoms such as He$^+$ and Li$^{2+}$ have a greater number of protons in their nucleus resulting in a greater Coulombic attraction of the electron to the nucleus affecting the orbital size. As a result the energy of the ground state will differ between one electron systems with different nuclear charges.
This can be demonstrated by calculating the ground state energies for H, He$^+$ and Li$^{2+}$ and seeing how the energies differ. Starting with calculating the ground state of hydrogen where Z=1 and n=1,
$E=-\left({\frac {Z^{2}m_{e}e^{4}}{32\pi ^{2}\epsilon _{0}^{2}\hbar ^{2}}}\right){\frac {1}{n^{2}}}$
$E=-\left({\frac {(1)^{2}(9.10938356\times 10^{-31}{\text{kg}})(1.60217662\times 10^{-19}{\text{C}})^{4}}{32\pi ^{2}(8.8541878128\times 10^{-12}{\text{Fm}})^{2}(1.054571817\times 10^{-34}{\text{Js}})^{2}}}\right){\frac {1}{1^{2}}}$
$E=-2.17987225\times 10^{-18}{\text{J}}$
$1{\text{eV}}=1.602176634\times 10^{-19}{\text{J}}$
$E=-13.60569243{\text{eV}}$
$1{\text{Hartree}}=4.359744722207185\times 10^{-18}{\text{J}}$
$E=-0.5{\text{Hartree}}$
Then for He$^+$ where Z=2 and n=1,
$E=-\left({\frac {Z^{2}m_{e}e^{4}}{32\pi ^{2}\epsilon _{0}^{2}\hbar ^{2}}}\right){\frac {1}{n^{2}}}$
$E=-\left({\frac {(2)^{2}(9.10938356\times 10^{-31}{\text{kg}})(1.60217662\times 10^{-19}{\text{C}})^{4}}{32\pi ^{2}(8.8541878128\times 10^{-12}{\text{Fm}})^{2}(1.054571817\times 10^{-34}{\text{Js}})^{2}}}\right){\frac {1}{1^{2}}}$
$E=-8.71948901\times 10^{-18}{\text{J}}$
$E=-54.42276978{\text{eV}}$
$E=-2{\text{Hartree}}$
Finally for Li$^{2+}$ where Z=3 and n=1,
$E=-\left({\frac {Z^{2}m_{e}e^{4}}{32\pi ^{2}\epsilon _{0}^{2}\hbar ^{2}}}\right){\frac {1}{n^{2}}}$
$E=-\left({\frac {(3)^{2}(9.10938356\times 10^{-31}{\text{kg}})(1.60217662\times 10^{-19}{\text{C}})^{4}}{32\pi ^{2}(8.8541878128\times 10^{-12}{\text{Fm}})^{2}(1.054571817\times 10^{-34}{\text{Js}})^{2}}}\right){\frac {1}{1^{2}}}$
$E=-1.96188503\times 10^{-17}{\text{J}}$
$E=-122.4512322{\text{eV}}$
$E=-4.5{\text{Hartree}}$
The ground state energy for the hydrogen atom is equal to -0.5 Hartree, for He$^+$ it is -2 Hartree and for Li$^{2+}$ it is -4.5 Hartree. The energies are increasing in a non-linear fashion which can be attributed to the Z$^2$ term in the numerator. This means as Z increases more energy will be required to ionize a one electron atom.

# Example 7
## Example Problem 7

Using calculus, derive the most probable radius of an electron 1s orbital.

### Solution

To determine the equation for the most probable radius of an electron in an 1s orbital we must calculate the value of r at the maximum point on the probability distribution. This is done by setting the derivative of radial probability to zero and solving for radius.
The general formula for the radial probability is.
${\text{Radial Probabilty}}=P(r)=R_{nl}^{*}(r)R_{nl}(r)\cdot r^{2}=(R_{nl}(r))^{2}\cdot r^{2}$
After inserting the wavefunction of a 1s orbital we can simplify the radial probability equation as follows.
    Radial Component of Wave Function of 1s Orbital $R_{10}(r)=2\left({\frac {Z}{a_{o}}}\right)^{\frac {3}{2}}\cdot e^{{\frac {-Z}{a_{o}}}r}$
$P_{1s}(r)=\left(2\left({\frac {Z}{a_{o}}}\right)^{\frac {3}{2}}\cdot e^{{\frac {-Z}{a_{o}}}r}\right)^{2}\cdot r^{2}$
$P_{1s}(r)=4\left({\frac {Z}{a_{o}}}\right)^{3}\cdot e^{{\frac {-2Z}{a_{o}}}r}\cdot r^{2}$
$P_{1s}(r)={\frac {4Z^{3}}{a_{o}^{3}}}\cdot e^{{\frac {-2Z}{a_{o}}}r}\cdot r^{2}$
The derivative of the simplified probability function must then be calculated.
${\frac {\delta }{\delta r}}\left(P_{1s}(r)\right)={\frac {\delta }{\delta r}}\left({\frac {4Z^{3}}{a_{o}^{3}}}\cdot e^{{\frac {-2Z}{a_{o}}}r}\cdot r^{2}\right)$
The constants can be excluded from the rest of the equation by moving them to the front of the derivative.
${\frac {\delta }{\delta r}}\left(P_{1s}(r)\right)={\frac {4Z^{3}}{a_{o}^{3}}}{\frac {\delta }{\delta r}}\left(e^{{\frac {-2Z}{a_{o}}}r}\cdot r^{2}\right)$
We see that the function is composed of the product of two smaller functions of the radius, multiplied by constants. Therefore we can apply the derivative product rule to solve for the equation of the derivative of the probability function.
    Product Rule of Derivatives ${\frac {\delta }{\delta r}}\left(f(x)\cdot g(x)\right)={\frac {\delta }{\delta r}}(f(x))\cdot g(x)+f(x)\cdot {\frac {\delta }{\delta r}}(g(x))$
${\frac {\delta }{\delta r}}\left(P_{1s}(r)\right)={\frac {4Z^{3}}{a_{o}^{3}}}\left({\frac {\delta }{\delta r}}\left(e^{{\frac {-2Z}{a_{o}}}r}\right)\cdot r^{2}+\left(\left(e^{{\frac {-2Z}{a_{o}}}r}\right)\cdot {\frac {\delta }{\delta r}}r^{2}\right)\right)$
${\frac {\delta }{\delta r}}\left(P_{1s}(r)\right)={\frac {4Z^{3}}{a_{o}^{3}}}\left(\left({\frac {-2Z}{a_{o}}}\right)\left(e^{{\frac {-2Z}{a_{o}}}r}\right)\cdot r^{2}+\left(\left(e^{{\frac {-2Z}{a_{o}}}r}\right)\cdot 2r\right)\right)$
Simplifying and moving the constants now will be beneficial to simplifying the final equation
${\frac {\delta }{\delta r}}\left(P_{1s}(r)\right)={\frac {8Z^{3}}{a_{o}^{3}}}\left(e^{{\frac {-2Z}{a_{o}}}r}\right)\left(\left({\frac {-Z}{a_{o}}}\right)\cdot r^{2}+r\right)$
${\frac {\delta }{\delta r}}\left(P_{1s}(r)\right)=0={\frac {8Z^{3}}{a_{o}^{3}}}\left(e^{{\frac {-2Z}{a_{o}}}r}\right)\left(\left({\frac {-Z}{a_{o}}}\right)\cdot r^{2}+r\right)$
After setting the derivative of radial probability to zero as seen above it can be seen that the only way that the equation can be equal to zero is if the polynomial part of the formula is equal to zero. This is the case because $e^{x}\neq 0$ and the first part of the equation is constant which means it cannot be zero. Therefore... $0=\left({\frac {-Z}{a_{o}}}\right)\cdot r^{2}+r$
The equation for the exact value of the most probable radius in a 1s orbital is then calculated using the quadradic equation.
    Quadratic Equation $x={\frac {-b\pm \left(b^{2}-4ac\right)^{\frac {1}{2}}}{2a}}$
Therefore let $x=r$, $a={\frac {-Z}{a_{o}}}$, $b=1$ and $c=0$ $r={\frac {-b\pm \left(b^{2}-4ac\right)^{\frac {1}{2}}}{2a}}$
$r={\frac {-(1)\pm \left((1)^{2}-4\left({\frac {-Z}{a_{o}}}\right)(0)\right)^{\frac {1}{2}}}{2\left({\frac {-Z}{a_{o}}}\right)}}$
$r={\frac {-1\pm \left(1-0\right)^{\frac {1}{2}}}{\left({\frac {-2Z}{a_{o}}}\right)}}$
$r={\frac {-1\pm 1}{\frac {-2Z}{a_{o}}}}$
The most probable radius cannot be a negative distance or zero meaning the numerator must be negative so that it is cancelled out by the negative value of the denominator. This means that the plus or minus operator in the numerator must be minus. $r={\frac {-1-1}{\frac {-2Z}{a_{o}}}}$
$r={\frac {1}{\frac {Z}{a_{o}}}}$
$r=r_{\text{most probable}}={\frac {a_{o}}{Z}}$
In conclusion the most probable radius for an electron in an 1s orbital is $r_{\text{most probable}}={\frac {a_{o}}{Z}}$ where that radius is proportional to the bohr radius $\left(a_{o}={\frac {4\pi \epsilon _{o}\hbar ^{2}}{m_{e}e^{2}}}=5.2917721090\cdot 10^{-11}m\right)$ and inversely proportional to the nuclear charge of the nucleus$\left(Z={\text{number of protons in nucleus of atom}}\right)$. This means that for an atom with one proton like hydrogen [H] the most probable radius is $1\cdot a_{o}=5.2917721090\cdot 10^{-11}m$ while for the helium ion [He$^+$] which also only has one electron in the 1s orbital the most probable radius is $0.5\cdot a_{o}=2.645886055\cdot 10^{-11}m$.

# Example 8

Write a question and its solution that quantitatively demonstrates the Heisenberg uncertainty principle holds for the J=0 state of a quantum rigid rotator

## Background Information
### Rigid rotor model

In quantum chemistry, the rigid rotor model is used to describe the rotations in molecules, such as HCl. The assumptions used in the rigid rotor model is that the rotating molecules is rigid, and the changes in bond length that naturally occur in a molecule (such as vibrational) is insignificant compared to the bond length r$_e$, and thus, negligible.
The total energy of the system is the summation of the potential energy and the kinetic energy. The potential energy of a rigid rotor is 0 given the assumption that the rigid rotor bond length is constant. As such, the total energy of the system is equal to KE, which is equal to the angular momentum. The energy level (E$_J$) of a linear rigid rotor model (such as HCl) is given by the equation:
$E_{J}={\frac {\hbar }{2I}}J(J+1)$
In which I is the inertia, based on the reduced mass of the diatomic and bond length, and J is the quantum energy level.
The rigid rotor model is 3 dimensional, and for ease of calculations, rather than 2 sets of masses that are used (m$_1$ and m$_2$), 1 reduced mass is used instead (μ).
And in a spherical model of the rigid rotor, there are 2 variables that are used to determine position, the angels θ and ϕ, given that r is the constant bond length r$_e$.
Thus, the position of the reduced mass is given by the wave function:
$Y(\theta ,\varphi )$

### Heisenberg Uncertainty Principle

The Heisenberg uncertainty principle states that the exact position and momentum of a particle cannot be determines at one given point, and the more precisely either is determined, the less certain the other would be.
However, even thou the exact position and momentum cannot be calculated at a given moment, they can be related, and that relation is:
$\bigtriangleup \phi \bigtriangleup L_{z}\geq {\frac {\hbar }{2}}$
In which $\hbar$ is the reduced Planck constant, $\bigtriangleup \phi$ is the uncertainty of position, and $\bigtriangleup L_{z}$ is the uncertainty of momentum.
However, in the rigid rotor model, this equation does not work. The Heisenberg inequality is recalculated:
$\bigtriangleup \phi \bigtriangleup L_{z}\geq {\frac {1}{2}}|<[L_{z},\phi ]>|={\frac {\hbar }{2}}[1-2\pi |\psi (0)^{2}]$
$\bigtriangleup \phi \bigtriangleup L_{z}\geq {\frac {\hbar }{2}}[1-2\pi |\psi (0)^{2}]$

## Example
#### Question: using the rigid rotor model at J=0, what is the moment of angular momentum and position? Does the rigid rotor follow Heisenberg’s principle at J=0?

In the ground state zero-point, J = 0, and thus, the energy level of the rigid rotor is 0.
Given that the energy is 0, the angular momentum is known (L$_z$ = 0).
The probability density of the position can be obtained using the wavefunction:
Average of position = 0, as such, the variance is calculated:
$\delta \phi ={\sqrt {<\varphi ^{2}>-<\varphi >^{2}}}$
$<\varphi >=\int _{0}^{2\pi }(Y^{*})(\varphi )(Y)d\varphi$
Y = $i^{m+|m|}{\sqrt {{\frac {2J+1}{2}}{\frac {(J-|m|)!}{(J+|m|)!}}}}.P_{0}^{0}cos(\theta )$
at m=0 J=0
Y = ${\frac {1}{\sqrt {4\pi }}}$
$\langle \varphi \rangle =\int _{0}^{2\pi }({\frac {1}{\sqrt {4\pi }}})(\varphi )({\frac {1}{\sqrt {4\pi }}})d\varphi$
$\langle \varphi \rangle =({\frac {1}{\sqrt {4\pi }}})^{2}\int _{0}^{2\pi }(\varphi )d\varphi$
$\langle \varphi \rangle =({\frac {1}{4\pi }})2\pi ^{2}={\frac {\pi }{2}}$
$\langle \varphi \rangle ^{2}={\frac {\pi ^{2}}{4}}$
$\langle \varphi ^{2}\rangle =\int _{0}^{2\pi }(Y^{*})(\varphi )^{2}(Y)d\varphi$
$\langle \varphi ^{2}\rangle =\int _{0}^{2\pi }({\frac {1}{\sqrt {4\pi }}})(\varphi ^{2})({\frac {1}{\sqrt {4\pi }}})d\varphi$
$\langle \varphi ^{2}\rangle =({\frac {1}{\sqrt {4\pi }}})^{2}\int _{0}^{2\pi }(\varphi ^{2})d\varphi$
$\langle \varphi ^{2}\rangle =({\frac {1}{4\pi }}){\frac {8\pi ^{3}}{3}}={\frac {2\pi ^{2}}{3}}$
$\delta \phi ={\sqrt {<\varphi ^{2}>-<\varphi >^{2}}}={\sqrt {{\frac {2\pi ^{2}}{3}}-{\frac {\pi ^{2}}{4}}}}=\pm {\frac {\sqrt {15}}{6}}\pi$
if $\delta \phi$ is plotted on a sphere
Where m = integers between J and – J.
| = number of longitudinal node
J = number of latitudinal nodes
At ground state, J = 0, |m| = 0. Thus, probability of position is equally spread across the sphere.
at J=0,
${\frac {\hbar }{2}}[1-2\pi |\psi (0)^{2}]=0$
thus
$\Delta \phi \cdot \Delta L_{z}\geq {\frac {\hbar }{2}}[1-2\pi |\psi (0)^{2}]$
$(\pm {\frac {\sqrt {15}}{6}}\pi )(0)\geq 0$
is satisfied
Thus, this shows that at ground state, J=0, the rigid rotor model follows the Heisenberg uncertainty.

## References

Anderson, J.M. Introduction to Quantum Chemistry, 1969, W.A. Benjamin, Inc, pg.91-100.
Fayngold & Fayngold, Quantum Mechanics and Quantum Information, pp 384-388.
* ↑ Engel, Thomas, and Philip Reid. “Chapter 20: The Hydrogen Atom.” Physical Chemistry, 3rd ed., Pearson, Upper Saddle River, 2018, p. 478.
* ↑ Engel, Thomas, and Philip Reid. “Chapter 20: The Hydrogen Atom.” Physical Chemistry, 3rd ed., Pearson, Upper Saddle River, 2018, p. 468.
* 1 2 3 4 5 Anderson, J.M. Introduction to Quantum Chemistry, 1969, W.A. Benjamin, Inc, pg.91-100.
* 1 2 3 4 Fayngold & Fayngold, Quantum Mechanics and Quantum Information, pp 384-388.

# Example 9

Write a question and its solution that calculates the locations of the nodes of an electron in a 2s orbital.

## The Question

The question is to find the location of the radial node in a 2s electron for a hydrogen atom. To find the node one can start by analyzing, generally, how many nodes one should expect to see in a 2s electron system. There are two equations that give the number of nodes present in an orbital, the angular node equation and radial node equations:
1. Radial Nodes:
$n-\ell -1=n_{r}$
2. Angular Nodes:
$\ell =n_{a}$
Therefore ℓ must be determined, and based on the table 1 data one can determine ℓ is equal to 0. And the n is equal to the $n$ which is 2, this comes from the number before the orbital type which tells you the principle quantum number.
Table 1: Orbitals and Quantum Number :
| Orbital | Angular Momentum Quantum Number (ℓ) |
| s | 0 |
| p | 1 |
| d | 2 |
| f | 3 |
So how many nodes are there?
First analyze the number of angular nodes:
$\ell =n_{a}$
$0=n_{a}$
Therefore, the number of angular nodes is 0.
Radial nodes:
$n_{r}=n-\ell -1$
$n_{r}=2-0-1$
$n_{r}=1$
Therefore there is one radial node present in a 2s orbital, resulting in the question becoming where is the location of that radial node?
The Wavefunction
Now the next main step is to determine what wavefunction describes the wavefunction this scenario of the 2s electron. The wavefunction can be found online which is$:^{1}$
$\psi _{2s}={\frac {1}{4(2\pi )^{\frac {1}{2}}}}\left({\frac {1}{a_{0}}}\right)^{3/2}\left(2-{\frac {r}{a_{0}}}\right){\text{e}}^{-r/(2a_{0})}$
In the equation the $\psi _{2s}$ is the wavefunction for the 2s electron, 𝒓 is the radius, and $a_{0}$ is the Bohr radius.
Based on the equation we can then solve for the position of the electron. The best way to do this is to find where is the equation going to be equal to zero and what term that contains the position causes this. The first part ${\frac {1}{4(2\pi )^{\frac {1}{2}}}}$ is a constant thus won't change with the radius, the position, of the electron. So the only place that will change with the position of the electron are the 𝒓 terms. With ${\text{e}}^{-r/(2a_{0})}$ the 𝒓 term can be any number and the term won't be zero unless the 𝒓 is approaching infinity, while the $\left(2-{\frac {r}{a_{0}}}\right)$ can potentially be equal to zero since it has 2 subtracted by the position term. Therefore one can set this term equal to zero and solve for 𝒓.
$0=2-{\frac {r}{a_{0}}}$
$2={\frac {r}{a_{0}}}$
$2a_{0}=r$
Therefore, we get the solution to the position of the radial node which is 𝒓 $=2a_{0}$, so when 𝒓 $=2a_{0}$ the probability of the electron being there is 0 all around the nucleus creating a node. The $a_{0}$ has a length of 52.9 pm which means the node is 105.8 pm in radius away from the nucleus.
In Conclusion
In conclusion the approach to the problem of finding nodes for an electron in an orbital boils down to first finding the number of theoretical nodes, then determining the wavefunction, analyzing the wavefunction's variables and solving for 0. After this is all done you will have the solution to the radial node locations. Further problems can be solved as well, because they are follow up questions that are made easier to solve after finding the nodes, like the position of the electron in it's most probable state. This problems follows the solution of arranging the P=$\psi _{2s}$$\left(\psi _{2s}^{*}\right)$$\left(r^{2}\right)$ then finding the derivative of the wavefunction and then simplifying it. The final step is to find the zero points, where the 𝒓 is equal to zero which will give the most probable locations of the electron. The practical applications of finding the nodal locations can help with understanding how orbitals work which can help with making molecular orbital diagrams and SALCs that can be used to determine the way atoms and molecules bond. Other applications include understanding the energy levels of the bonds and orbitals to predict possible interactions between molecules and atoms, for research purposes and chemical engineering.
Reference
1) Branson, J. The Radial Wavefunction Solutions. https://quantummechanics.ucsd.edu/ph130a/130_notes/node233.html (accessed November 16, 2021).
By Dmitry Ivanov

# Example 10
## Question

Use Planck's radiation law to find the surface temperature of a star with a maximum intensity of EM radiation is emitted at 504 nm. Report your answer in Kelvin.

### Getting the correct function

A blackbody material is defined by its ability to absorb ALL radiation that falls onto it. When the blackbody is at constant temperature, the distribution of its emission frequency can be determined by assuming its only direct relation is to temperature. Furthermore, the frequency of the electromagnetic (EM) radiation can also be measured in units of wavenumber ${\text{cm}}^{-1}$.
${\begin{aligned}\rho (\nu ,T)&={\frac {2h\nu ^{2}}{c^{2}}}\left\langle E(\nu ,T)\right\rangle &&\rho (\lambda ,T)={\frac {2hc^{2}}{\lambda ^{5}}}\left\langle E(\nu ,T)\right\rangle \end{aligned}}$
Planck's approach to defining ⟨E(v, T)⟩{\textstyle \langle E(v, T)\rangle } was deriving for a closed form Harmonic Oscillator was based on Boltzmann's distribution. The resulting form of Planck's law can then be applied to the question now that all parameters are known except for the parameter in question. The model below shows the radiation intensity distribution (i.e., area of the curve) of EM with respect to frequency (in Hertz; Hz) and temperature (in Kelvins; K). ${\begin{aligned}d\rho (v,T)&=\rho (\nu ,T)d\nu {\text{, where }}\rho (\nu ,T)=\left({\frac {2h\nu ^{3}}{c^{2}}}\right)\left[\exp \left({\frac {h\nu }{k_{B}T}}\right)\right]^{-1}\\&{\text{Note that }}\left|{\frac {d\nu }{d\lambda }}\right|={\frac {c}{\lambda ^{2}}}\\d\rho (\lambda ,T)&=\rho (\lambda ,T)d\lambda {\text{, where }}\rho (\lambda ,T)=\left({\frac {2hc^{2}}{\lambda ^{5}}}\right)\left[\exp \left({\frac {hc}{k_{B}T\lambda }}\right)\right]^{-1}\end{aligned}}$Planck's law for a blackbody material predicts the behaviour of its quantitative properties (such as frequency, wavelength, and temperature) when the environment are in extreme conditions.
    Planck's Law of BlackBody Radiation $\rho (\lambda ,T)={\frac {2hc^{2}}{\lambda ^{5}}}\times {\frac {1}{\left[\exp({\frac {hc}{k_{B}\lambda T}})\right]-1}}$

### Solving for the temperature

The maximum value of the independent variable of any function can be found by equating the function's derivative to zero. Only $T$ is unknown and is with respect to the independent variable, $\lambda$. Thus, Planck's law will be derived with respect to $\lambda$.
First, for simplicity, let x=hc/k$_B$λT {\textstyle x={\frac {hc}{k_{B}\lambda T}}}.
${d\rho \over d\lambda }=(2hc^{2})\left[{\frac {x}{\lambda ^{6}}}{\frac {e^{x}}{\left(e^{x}-1\right)^{2}}}-{\frac {5}{\lambda ^{6}\left(e^{x}-1\right)}}\right]=0$
The first term consists only of non-zero constants so that leaves:
${\frac {x}{\lambda ^{6}}}{\frac {e^{x}}{\left(e^{x}-1\right)^{2}}}-{\frac {5}{\lambda ^{6}\left(e^{x}-1\right)}}=0$
${\begin{aligned}{\frac {x}{\lambda ^{6}}}{\frac {e^{x}}{\left(e^{x}-1\right)^{2}}}&={\frac {5}{\lambda ^{6}\left(e^{x}-1\right)}}\\{\frac {xe^{x}}{e^{x}-1}}&=5\\{\frac {xe^{x}}{e^{x}-1}}-5&=0\\(x-5)e^{x}+5&=0\end{aligned}}$
Subbing x=hc/k$_B$λT {\textstyle x={\frac {hc}{k_{B}\lambda T}}} back in gives:
${\begin{aligned}\left({\frac {hc}{k_{B}\lambda T}}-5\right)e^{x}&=0&&\rightarrow &&&{\frac {hc}{k_{B}\lambda T}}-5=0\end{aligned}}$
At max intensity, λ$_{max}$=504nm{\textstyle \lambda _{max}=504{\text{nm}}}. Subbing in all known values and constants (in SI units) then rearranging to solve for T will determine the surface temperature of the Sun under these specific conditions.
Known Values :
| Variable | Value | Units |
| $h$ | $6.62607\times 10^{-34}$ | $m^{2}kg \over s$ |
| $\lambda _{max}$ | $504\times 10^{-9}$ | $m$ |
| $k_{B}$ | $1.38065\times 10^{-23}$ | ${\frac {m^{2}kg}{s^{2}K}}$ |
| $c$ | $2.99792\times 10^{8}$ | $m \over s$ |
${\begin{aligned}T={\frac {hc}{5k_{B}\lambda _{max}}}={\frac {(6.62607\times 10^{-34})(2.99792\times 10^{8})}{5(1.38065\times 10^{-23})(504\times 10^{-9})}}=5709.432~{\text{K}}=5710~{\text{K}}\end{aligned}}$
Therefore, the surface temperature of the star with a maximum blackbody radiation emission wavelength of 504 nm is just above 5709 Kelvins.

## References

* 1 2 Quantum Hypothesis Used for Blackbody Radiation Law https://chem.libretexts.org/@go/page/210776 (accessed Nov 20, 2021).
* ↑ "Rotations and Vibrations of Polyatomic Molecules", Molecular Physics, Weinheim, Germany: Wiley-VCH Verlag GmbH, pp. 203–236, retrieved 2021-11-20
* 1 2 3 "1.1: Blackbody Radiation Cannot Be Explained Classically". Chemistry LibreTexts. 2020-03-18. Retrieved 2021-11-20.
* ↑ "Boltzmann constant | Value, Dimensions, Symbol, & Facts | Britannica". www.britannica.com. Retrieved 2021-11-21.

# Example 11
## Example 11

Use the 1D particle in a box model to estimate the wavelength of light required to excite an electron from a pi to pi* MO in ethene.
Solution:
The energy levels of a particle in a 1D box with a specific quantum number $n$, are as follows.
$E_{n}={\frac {h^{2}n^{2}}{8mL^{2}}}$
In this equation $h$ represents Planck's constant, $m$ is the mass of the particle, and $L$ is the length of the box.
The pi electron in the double bond between the carbon atoms in ethene can be approximated to the particle in a 1D box model. This means that the mass of the particle in this question will be the mass of an electron, and the length of the box corresponds to the bond length between the carbon atoms in the molecule ethene.
Additionally, the energy equation above needs to be transformed into a equation for $\Delta E$ since the electron is moving from one energy level to another.
$\Delta E=E_{f}-E_{i}={\frac {h^{2}n_{i}^{2}}{8mL^{2}}}-{\frac {h^{2}n_{f}^{2}}{8mL^{2}}}={\frac {h^{2}}{8mL^{2}}}(n_{f}^{2}-n_{i}^{2})$
The change in energy between the pi and pi* MO in ethene can now be calculated knowing that the bond length between doubly bonded carbon atoms is 133pm and the mass of an electron is 9.1093856x10$^{-31}$ kg. Moving from the ground state n=1 to an excited state of n=2 :
$\Delta E={\frac {h^{2}}{8mL^{2}}}(n_{f}^{2}-n_{i}^{2})$
$\Delta E={\frac {(6.626{\text{x}}10^{-34}{\frac {{\text{m}}^{2}{\text{kg}}}{\text{s}}})^{2}}{8(9.109\times 10^{-31}{\text{kg}})(133\times 10^{-12}{\text{m}})^{2}}}(2^{2}-1^{2})$
$\Delta E=1.0217\times 10^{-17}{\text{J}}$
Now that the energy required to excite the electron to the pi* orbital is known, the wavelength of light can be calculated through the following equation, where c is the speed of light in a vacuum and $\lambda$ is the wavelength of light.
$\Delta E={\frac {hc}{\lambda }}$
The equation can then be re-arranged to solve for the wavelength of light.
$\lambda ={\frac {hc}{\Delta E}}$
By plugging in the known constants and the value for $\Delta E$ that has been calculated above, the wavelength can be found.
$\lambda ={\frac {hc}{\Delta E}}$
$\lambda ={\frac {(6.626\times 10^{-34}{\frac {{\text{m}}^{2}{\text{kg}}}{\text{s}}})(2.998\times 10^{8}{\frac {\text{m}}{\text{s}}})}{1.0217\times 10^{-17}{\text{J}}}}$
$\lambda =1.9\times 10^{-7}{\text{m}}$
$\lambda =19{\text{nm}}$
Therefore the wavelength of light required to excite an electron from a pi to pi* molecular orbital in ethene is 19nm.

# Example 12

Write a question and its solution that shows the specific selection rule for a quantum rigid rotor.
Consider an N$_2$ molecule with a bond length of 1.09 Å.
(a) Calculate the energy at the quantum number 3 using the specific selection rule for a rigid rotor.
Solution: This bond length is given (1.09 Å), and the reduced mass (μ) and the inertia (I) must be calculated to determine the energy at the angular momentum quantum number $J=3$. The bond length and reduced mass must also be changed to SI units.
$\mu ={\frac {m_{1}m_{2}}{m_{1}+m_{2}}}$
$\mu ={\frac {14.0067u\cdot 14.0067u}{14.0067u+14.0067u}}\cdot ({\frac {1.6605\times 10^{-27}kg}{1u}})$
$\mu =1.1629\times 10^{-26}kg$
Now, calculating the moment of inertia ($I$):
$I=\mu \cdot (r_{e})^{2}$
$=(1.1629\times 10^{-26}kg)\cdot (1.09\mathrm {\AA} \cdot 1.00\times 10^{-10}m)^{2}$
$=1.3816\times 10^{-46}kg\cdot m^{2}$
Since the molecule is linear (N$_2$), the energy level ($E_{J}$) can be calculated with the linear rigid rotor equation:
$E_{J}=J(J+1)\cdot ({\frac {\hbar ^{2}}{2I}})$
$E_{J}=3(3+1)\cdot ({\frac {(1.05457\times 10^{-34}J\cdot s)^{2}}{2(1.3816\times 10^{-46}kg\cdot m^{2})}})$
$E_{J}=12\cdot (4.0247\times 10^{-23}J)$
$E_{J}=4.8296\times 10^{-22}J$
(b) Calculate the quantum number if the transition energy is 2.4176 × 10$^{-22}$J. In reference to part (a), does this value adhere to the specific selection rule? Why or why not?
The linear rigid rotor equation must be rearranged into linear form to solve for $J$.
$E_{J}=J(J+1)\cdot ({\frac {\hbar ^{2}}{2I}})$
$J^{2}+J={\frac {2\cdot E\cdot I}{\hbar ^{2}}}$
$J^{2}+J-{\frac {2\cdot E\cdot I}{\hbar ^{2}}}=0$
$J^{2}+J-{\frac {(2)\cdot (2.4176\times 10^{-22}J)\cdot (1.3816\times 10^{-46}kg\cdot m^{2})}{(1.05457\times 10^{-34}J\cdot s)^{2}}}=0$
$J^{2}+J-6.0069=0$
To solve this relationship, the quadratic formula must be utilized:
$J={\frac {-b\pm {\sqrt {b^{2}-4\cdot a\cdot c}}}{2\cdot a}}$
$J={\frac {-1\pm {\sqrt {1^{2}-4\cdot 1\cdot -6.0069}}}{2\cdot 1}}$
$J={\frac {-1\pm {\sqrt {25.0275}}}{2}}$
$J={\frac {-1\pm {4.9973}}{2}}$
$J=2.0014,-3.0014$
Rigid rotor quantum numbers cannot be negative, ∴ $J=2$.
This transition adheres to the specific selection rule for a quantum rigid rotor because the change in rotational quantum number is $\Delta J=\pm 1$.
(c) Calculate the quantum number if the transition energy is 1.2073 × 10$^{-21}$J. In reference to part (a), does this value adhere to the specific selection rule? Why or why not?
The linear rigid rotor equation must be rearranged into linear form to solve for $J$.
$J^{2}+J-{\frac {2\cdot E\cdot I}{\hbar ^{2}}}=0$
$J^{2}+J-{\frac {(2)\cdot (1.2073\times 10^{-21}J)\cdot (1.3816\times 10^{-46}kg\cdot m^{2})}{(1.05457\times 10^{-34}J\cdot s)^{2}}}=0$
$J^{2}+J-29.9970=0$
To solve this relationship, the quadratic formula must be utilized:
$J={\frac {-b\pm {\sqrt {b^{2}-4\cdot a\cdot c}}}{2\cdot a}}$
$J={\frac {-1\pm {\sqrt {1^{2}-4\cdot 1\cdot -29.9970}}}{2\cdot 1}}$
$J={\frac {-1\pm {\sqrt {120.98813}}}{2}}$
$J={\frac {-1\pm {10.9994}}{2}}$
$J=4.9997,-5.9997$
Rigid rotor quantum numbers cannot be negative, ∴ $J=5$.This transition does not adhere to the specific selection rule for a quantum rigid rotor, because the change in rotational quantum number is not within the $\Delta J=\pm 1$. Compared to part (a), it has a change of $\Delta J=\pm 2$.

## References

* ↑ https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Exercises%3A_Physical_and_Theoretical_Chemistry/Exercises%3A_Aktins_et_al./12.E%3A_Rotational_and_Vibrational_Spectra_(Exercises)
* ↑ https://phys.libretexts.org/Bookshelves/University_Physics/Book%3A_University_Physics_(OpenStax)/Book%3A_University_Physics_I_-_Mechanics_Sound_Oscillations_and_Waves_(OpenStax)/10%3A_Fixed-Axis_Rotation__Introduction/10.06%3A_Calculating_Moments_of_Inertia
* ↑ Anderson, J.M. Introduction to Quantum Chemistry, 1969, W.A. Benjamin, Inc, pg.91-100.

# Example 13

Write a question and its solution that shows the specific selection rule for a quantum harmonic oscillator
Calculate the energy for the vibrational transitions $v\rightarrow v+1$ and $v+6\rightarrow v+7$. If they have the same energy gaps comment on why?
$\bigtriangleup Ev\rightarrow v+1$$=hv_{0}{\Bigl(v+1)+{\tfrac {1}{2}}{\Bigr )}-hv_{0}{\Biglv+{\tfrac {1}{2}}{\Bigr )}$
$=hv_{0}{\biggl(v+1)+{\tfrac {1}{2}}-{\Biglv+{\tfrac {1}{2}}{\Bigr )}{\biggr )}$
$=hv_{0}{\bigglv+1+{\tfrac {1}{2}}-v-{\tfrac {1}{2}}{\biggr )}$
$=hv_{0}$
$\bigtriangleup Ev+1\rightarrow v+2$$=hv_{0}{\Bigl(v+7)+{\tfrac {1}{2}}{\Bigr )}-hv_{0}{\Bigl(v+6)+{\tfrac {1}{2}}{\Bigr )}$
$=hv_{0}{\biggl(v+7)+{\tfrac {1}{2}}-{\Bigl(v+6)+{\tfrac {1}{2}}{\Bigr )}{\biggr )}$
$=hv_{0}{\bigglv+7+{\tfrac {1}{2}}-v-6-{\tfrac {1}{2}}{\biggr )}$
$=hv_{0}$
The energy of the vibrational transitions from from $v\rightarrow v+1$ and $v+6\rightarrow v+7$ have the same energy gap $(hv_{0})$ . This is because of the specific selection rule for the quantum harmonic oscillator. The rule states that the molecule is only allowed to move up or down one vibrational energy level for the transition to occur. If the molecule diverges from the rule $\bigtriangleup V\pm 1$ then it is considered an overtone, and these are unlikely to occur.

# Example 14

Show using calculus the most probable position of a quantum harmonic oscillator in the ground state (n=0)
Question:
What is the most probable position of a quantum harmonic oscillator at the ground state? Calculate this using the probability density equation to find the most probable position at n=0.
    Probability distribution $P_{n}(x)=N_{n}^{2}H_{n}({\sqrt {\alpha }}x)^{2}e^{-\alpha x^{2}}$
Solution:
The Hermite polynomial at n=0 is:
$H_{0}(x)=1$
The normalization factor at n=0 is:
$N_{0}={\frac {1}{\sqrt {2^{1}1!}}}\left({\frac {\alpha }{\pi }}\right)^{1/4}=\left({\frac {\alpha }{\pi }}\right)^{1/4}$
α is a constant and is equal to:
$\alpha ={\sqrt {\frac {k\mu }{\hbar }}}$
The probability distribution at n=0:
$P_{n=0}(x)=\left({\frac {\alpha }{\pi }}\right)^{1/2}e^{-\alpha x^{2}}$
The most probable position is when the maximum probability distribution is:
${\frac {\partial P}{\partial x}}=0$
Applying this partial derivative to the probability distribution gives:
${\frac {\partial }{\partial x}}\left({\frac {\alpha }{\pi }}\right)^{1/2}e^{-\alpha x^{2}}=0$
The constants can be taken out of the derivative:
$\left({\frac {\alpha }{\pi }}\right)^{1/2}{\frac {\partial }{\partial x}}e^{-\alpha x^{2}}=0$
The derivative gives:
$\left({\frac {\alpha }{\pi }}\right)^{1/2}[-2\alpha xe^{-\alpha x^{2}}]=0$
Since it is equal to zero the constants can be divided out leaving:
$[-2\alpha xe^{-\alpha x^{2}}]=0$
Since all of the parts are multiplied they can be divided out leaving:
$x=0$
The point where the probability distribution is at a maximum for the ground state of n=0 for the quantum harmonic oscillator is 0.

# Example 15

Write an example question showing the determination of the bond length of CO using microwave spectroscopy

### Example 15:
## Deriving the Required Equations

When a photon is absorbed by a polar diatomic molecule, such as carbon monoxide, the molecule can be excited rotationally. The energy levels of these excited states are quantized to be evenly spaced. The distance between each rotational absorption lines is defined as twice the rotational constant $({\tilde {\beta }})$ which can be measured via the following equation:
${\tilde {\beta }}=\left({\frac {h}{8\pi ^{2}c{\text{I}}}}\right)$
h = plank's constant = 6.626 ᛫10$^{−34}$ J ᛫ s
c = speed of light = 2.998᛫10$^8$ m ᛫ s$^{-1}$
I = moment of inertia
The energy to required to rotate a molecule around its axis is the moment of inertia $({\text{I}})$. It can be calculated as the sum of the products the masses of the component atoms and their distance from the axis of rotation squared:
${\text{I}}=\sum _{i}m_{i}\cdot r_{i}^{2}$
Working it out for an heterogeneous diatomic molecule:
${\text{I}}=\left(m_{a}\cdot r_{a}^{2}\right)+\left(m_{b}\cdot r_{b}^{2}\right)$
${\text{I}}=r_{a}\cdot r_{b}\left(m_{a}+m_{b}\right)$
The distance from the atom to the center of mass $\left(r_{a}{\text{and}}r_{b}\right)$ cannot easily be measure, however, by setting the origin at the center of mass equation can be derive for the two values that uses the bond length $\left(r_{e}\right)$ as a variable:
$\left(m_{a}\cdot r_{a}\right)=\left(m_{b}\cdot r_{b}\right)=\left(m_{a}\cdot \left(r_{a}-r_{e}\right)\right)=\left(m_{b}\cdot \left(r_{b}-r_{e}\right)\right)$
$r_{a}={\frac {m_{b}\cdot r_{e}}{m_{a}+m_{b}}}$
$r_{b}={\frac {m_{a}\cdot r_{e}}{m_{a}+m_{b}}}$
Substituting these equations into the moment of inertia equation:
${\text{I}}={\biggl{\frac {m_{b}\cdot r_{e}}{m_{a}+m_{b}}}{\biggr )}\cdot {\biggl{\frac {m_{a}\cdot r_{e}}{m_{a}+m_{b}}}{\biggr )}\cdot \left(m_{a}+m_{b}\right)$
${\text{I}}={\biggl{\frac {m_{a}\cdot m_{b}\cdot r_{e}^{2}}{m_{a}+m_{b}}}{\biggr )}$
This equation can be simplified further if we imagine the rigid rotor as a single particle rotating around a fixed point a bond length away. The mass of this particle is the reduced mass $\left(\mu \right)$ of the two atoms that make up the diatomic molecule:
$\mu =\left({\frac {m_{1}\cdot m_{2}}{m_{1}+m_{2}}}\right)$
Simplifying the previous moment of inertia equation, we get:
${\text{I}}=\mu r_{e}^{2}$

## Solving for Bond Length

From here we have everything we need to be able to determine the bond length of a polar diatomic molecule such as carbon monoxide.
First, we must solve for the moment of inertia using the rotation constant:
${\tilde {\beta }}=\left({\frac {h}{8\pi ^{2}c{\text{I}}}}\right)$
${\text{I}}=\left({\frac {h}{8\pi ^{2}c{\tilde {\beta }}}}\right)$
As explained earlier, the rotational constant can be determined by measuring the distance between the rotational absorption lines and halfling it. In the case of ${\ce {^{12}C^{16}O}}$ the rotational constant is $193.1281$ m$^{-1}$. Plugging this value in we can determine the moment of inertia:
${\text{I}}=\left({\frac {6.626\cdot 10^{-34}}{8\pi ^{2}\cdot \left(2.998\cdot 10^{8}\right)\cdot \left({\displaystyle 193.1281}\right)}}\right)$
${\text{I}}=1.44939\cdot 10^{-46}$ kg ᛫ m$^2$
Now that we know inertia, we can rearrange the equation we derived earlier in order to determine the bond length:
${\text{I}}=\mu r_{e}^{2}$
$r_{e}={\sqrt {\frac {\text{I}}{\mu }}}$
$r_{e}={\sqrt {\frac {1.44939\cdot 10^{-46}}{\mu }}}$
The exact atomic mass of ${\ce {^{12}C}}$ is 12.011 amu and ${\ce {^{16}O}}$ is 15.9994 amu. As such the reduce mass is calculated to be:
$\mu =\left({\frac {m_{1}\cdot m_{2}}{m_{1}+m_{2}}}\right)$
$\mu =\left({\frac {12.011\cdot 15.9994}{12.011+15.9994}}\right)$
$\mu =6.86062$ amu
$\mu =6.86062$ amu ᛫ $1.67377\times 10^{-27}\left({\frac {kg}{amu}}\right)$
$\mu =1.140\times 10^{-26}kg$
Plugging in the reduce mass back into our equation we can finally solve for the bond length of a carbon monoxide molecule:
$r_{e}={\sqrt {\frac {1.44939\cdot 10^{-46}}{1.140\times 10^{-26}}}}$
$r_{e}={\sqrt {1.271\cdot 10^{-20}}}$
$r_{e}=1.12739\cdot 10^{-10}m$
$r_{e}=1.12739\mathrm {\AA}$

## References

* ↑ https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Exercises%3A_Physical_and_Theoretical_Chemistry/Exercises%3A_Aktins_et_al./12.E%3A_Rotational_and_Vibrational_Spectra_(Exercises)
* ↑ https://phys.libretexts.org/Bookshelves/University_Physics/Book%3A_University_Physics_(OpenStax)/Book%3A_University_Physics_I_-_Mechanics_Sound_Oscillations_and_Waves_(OpenStax)/10%3A_Fixed-Axis_Rotation__Introduction/10.06%3A_Calculating_Moments_of_Inertia
* ↑ https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Exercises%3A_Physical_and_Theoretical_Chemistry/Exercises%3A_Aktins_et_al./12.E%3A_Rotational_and_Vibrational_Spectra_(Exercises)
* ↑ https://webbook.nist.gov/cgi/cbook.cgi?ID=C630080&Mask=1000
* ↑ https://www.angelo.edu/faculty/kboudrea/periodic/structure_mass.htm

# Example 16

Write an example question showing the calculation of the frequency of EM radiation emitted when a HCl molecule transitions from the J=1–>0 rotational state.
Example 16: Giving the bond length r$_e$=1.27Å, find the frequency of the EM radiation emitted when a HCl molecule transitions from the J=1→0 rotational state
Given the reduced mass (μ) and inertia (I), the energy can be found and subsequently be used to find frequency of the transition.
Note that for rotational states for J=1→0, ΔE is not required as the transition states are in the ground state.
Inertia can be found by using reduced mass of the molecule and their bond length(r$_e$ = 1.27Å):
$I=\mu \cdot r_{e}^{2}$
where reduced mass can be calculated by the following:
$\mu ={\frac {m_{1}\cdot m_{2}}{m_{1}+m_{2}}}$, where $m_{1}=1.00784u$ and $m_{2}=35.453u$
$\mu ={\frac {(1.00784)(35.453)}{1.00784+35.453}}=0.9770u$
Convert to SI units:
$1u=1.66054\times 10^{-27}kg$, thus
$\mu =0.9799u\times {\frac {1.66054\times 10^{-27}kg}{1u}}=1.627\times 10^{-27}kg$
Note that the bond length must also be in SI units,
1Å= 1×10$^{-10}$m
Bond length = 1.27×10$^{-10}$m
Inertia can now be calculated:
$I=(1.627\times 10^{-27}kg)(1.27\times 10^{-10}m)^{2}=2.624\times 10^{-47}kg\cdot m^{2}$
After calculating the inertia, energy can be found:
$E={\frac {\hbar ^{2}}{2I}}J(J+1)=hv$
and $\hbar ={\frac {h}{2\pi }}$, where h is the Planck's constant.
$E={\Biggl{\frac {6.626\times 10^{-34}m^{2}kgs^{-1}}{2\pi }}{\Biggr )}^{2}\times {\frac {1}{2\times 2.624\times 10^{-47}kg\cdot m^{2}}}\times 1(1+1)$
$E=4.240\times 10^{-22}J$
Frequency of the transmission can therefore be found:
$v={\frac {E}{h}}$
$v={\frac {4.240\times 10^{-22}J}{6.626\times 10^{-34}m^{2}kgs^{-1}}}=6.40\times 10^{11}s^{-1}$

# Example 17

The force constant of HCl is determined computationally to be 480 N/m. Given this information find the frequency of EM radiation required to excite the HCl molecule from its ground state to its first excited state.

## Solution

Given the bond constant (k) of HCl , we use the relationship between fundamental frequency and bond constant to find the bond constant.
$\nu _{0}={\frac {1}{2\pi }}{\left({\frac {k}{\mu }}\right)}^{1/2}$
where $K$ is the bond constant.
          $\mu$ is the reduced mass of HCl.
To find the reduced mass of HCl the masses of H and Cl are multiplied and divided by the sum of the masses.
$\mu ={\frac {m_{1}\cdot m_{2}}{m_{1}+m_{2}}}$
For HCl the reduced mass is calculated as
$\mu ={\frac {1.007842u\cdot 35.453u}{1.007842u+35.453u}}=0.979982u$
convert to the SI unit of Kg
$1u=1.66054\cdot 10^{-27}Kg$
$0.979983u=0.979983u\cdot 1.66054\cdot 10^{-27}Kg$
$\mu =1.6273\cdot 10^{-27}Kg$
To find the fundamental frequency
$\nu _{0}={\frac {1}{2\pi }}\left({\frac {K}{\mu }}\right)^{1/2}$
    $={\frac {1}{2\pi }}\left({\frac {480N/m}{1.6273\cdot 10^{-27}Kg}}\right)^{1/2}$
    $=8.6438\cdot 10^{13}hz$
After finding the fundamental frequency, the Energy at different quantum levels can be found by
$E_{v}=h\nu _{0}\left(v+{\frac {1}{2}}\right)$
For the ground state i.e. $v=0$
$E_{0}=h\nu _{0}\left(0+{\frac {1}{2}}\right)$
$E_{0}=h\nu _{0}\left({\frac {1}{2}}\right)$
For the first excited state i.e. $v=1$
$E_{1}=h\nu _{0}\left(1+{\frac {1}{2}}\right)$
$E_{1}=h\nu _{0}\left({\frac {3}{2}}\right)$
The difference in energy between the two states is
$\Delta E=E_{1}-E_{0}$
$\Delta E=h\nu _{0}\left({\frac {3}{2}}\right)-h\nu _{0}\left({\frac {1}{2}}\right)$
$\Delta E=h\nu _{0}$
and Energy is defined as Planck's constant multiplied by frequency
$\Delta E=hv$
$hv=h\nu _{0}$
$v=\nu _{0}=8.6438\cdot 10^{13}hz$

# Example 18

Write an example question showing the determination of the force constant of CO using IR spectroscopy.

## Question

Using IR spectroscopy, calculate the force constant of carbon monoxide (CO) and report the value in kN/m.

## Solution

To solve this question, the IR spectrum of CO is required. The website, webbook.nist.gov/chemistry/ is a database that is available to the public for access to IR spectra. If you are unfamiliar on acquiring an IR spectrum, follow these steps: with the nist website open, go to Search Option -> Name -> Enter: Carbon Monoxide -> Select: IR Spectrum -> Change Transmittance to Absorbance -> Finally, take a screen shot.
The CO IR spectrum earlier acquired is a low resolution rotational-vibrational spectrum. In IR spectrum's, the fundamental frequency can be determined by utilizing the P-branch, Q-branch, and R-branch. Remember that the Q-branch is pure vibrational, which is forbidden, therefore the peak does not exist and is located between the P-branch and R-branch.
Using the website, apps.automeris.io/wpd/ will allow for an accurate collection of data points from the image of the CO IR spectrum taken earlier. For low resolution spectrums, locate the R-branch and P-branch, use the peak maxima as the data point. If you are unfamiliar on acquiring the data points from a image, follow these steps: Open the website -> Load Image -> Choose File -> Open -> 2D(X-Y) Plot -> Align Axes -> Plot Known $X_{1},X_{2}.Y_{1},Y_{2}$ -> Complete -> Insert Known Values -> Ok -> Add Points -> View Data.
Equation 1: Using the data points from the image, the fundamental frequency can be determined by the following relationship.
$v_{0}=\left({\frac {num_{H}-num_{L}}{2}}\right)+num_{L}\equiv num_{H}-\left({\frac {num_{H}-num_{L}}{2}}\right)$
Where,
$v_{0}$ is the fundamental frequency $({\text{cm}}^{-1})$
$num_{H}$ is the highest wavenumber $({\text{cm}}^{-1})$
$num_{L}$ is the lowest wavenumber $({\text{cm}}^{-1})$
Equation 2: The following equation is used to relate the fundamental frequency to the force constant.
$v_{0}={\frac {1}{2\pi c}}\left({\frac {k}{\mu }}\right)^{1/2}$
Where,
$v_{0}$ is the fundamental frequency $({\text{m}}^{-1})$
$c$ is the speed of light $({\text{m/s}})$
$k$ is the force constant $({\text{N/m}})$
$\mu$ is the reduced mass $({\text{kg}})$
Equation 3: The reduced mass equation is required for relating the fundamental frequency to the force constant. Remember to use exact masses, and since the isotopes were not specified, that means we assume the most abundant version is to be used.
For oxygen: $^{16}{\text{O}}:15.994915{\text{ amu}}$
For carbon: $^{12}{\text{C}}:12.000000{\text{ amu}}$
$\mu ={\frac {m_{1}\cdot m_{2}}{m_{1}+m_{2}}}$
Where,
$\mu$ is the reduced mass $({\text{kg}})$
$m_{1}$ is the exact mass 1 $({\text{amu}})$
$m_{2}$ is the exact mass 2 $({\text{amu}})$
${\text{amu}}=1.66054{\text{ x}}10^{-27}{\text{kg}}$
Equation 4: Rearrange equation 2 to solve for the force constant.
$k=4\pi ^{2}v_{0}^{2}c^{2}\mu$
The fundamental frequency of CO from IR spectra is determined using equation 1:
$v_{0}=\left({\frac {2176.206896{\text{ cm}}^{-1}-2122.41379{\text{ cm}}^{-1}}{2}}\right)+2122.41379{\text{ cm}}^{-1}=2149.310343{\text{ cm}}^{-1}$
The reduced mass of CO is calculated using equation 3:
$\mu ={\frac {15.994915{\text{ amu}}\cdot 12.000000{\text{ amu}}}{15.994915{\text{ amu}}+12.000000{\text{ amu}}}}=6.856208708{\text{ amu}}\cdot (1.66054{\text{ x}}10^{-27}{\text{kg/amu}})=1.13850088{\text{ x}}10^{-26}{\text{ kg}}$
The force constant of CO is calculated using equation 4:
$k=4\pi ^{2}\left(2149.310343{\text{ cm}}^{-1}\cdot \left({\frac {100{\text{ cm}}}{1{\text{ m}}}}\right)\right)^{2}(2.998{\text{ x}}10^{8}{\text{ m/s}})^{2}(1.13850088{\text{ x}}10^{-26}{\text{ kg}})=1886.184688{\text{ N/m}}$
$k=1.886184688{\text{ kN/m}}$

# Example 19

Write an example question showing the calculation of the frequency of EM radiation required to excite an H atom from ground to n=2 electronic state.

# Example 20

An electronic transition of H atom from an energy level to the ground state is observed with a corresponding wavelength of 102.5 nm. Determine the initial state of the electron from which transition has occurred.
The Rydberg's phenomenological equation will be used to solve the energy transition problem
    Rydberg's phenomenological equation for hydrogen $\lambda =\mathbb {R} _{H}\left({\frac {1}{n_{1}^{2}}}\right)-\left({\frac {1}{n_{2}^{2}}}\right)$
where the $R_{H}$ is Rydberg constant for hydrogen and is equal to 109737 $cm^{-1}$. $n_{1}$ is the final energy level, $n_{2}$ is the initial energy level for hydrogen transition. Both $n_{1}$ and $n_{2}$ are integers and $n_{2}>n_{1}$.
The wavelength is given as 103nm
Concert to SI units:
$\lambda =103nm*1^{-7}{\frac {cm}{nm}}=1.03^{-5}cm$
The H atom's transition is from unknown energy level to the ground state.
So we can know the final state is the ground state which means $n_{1}=1$
The $n_{2}$ now can be calculated:
${\frac {1}{n_{2}^{2}}}={\frac {1}{n_{1}^{2}}}-{\frac {1}{\lambda *R_{H}}}$
${\frac {1}{n_{2}^{2}}}={\frac {1}{1}}-{\frac {1}{0.00103cm*109737cm^{-1}}}$
$n_{2}^{2}=8.67\approx {9}$
Because the energy level always $>0$
$n_{2}=3$
Therefore, the initial state of the electron from which transition has occurred is energy level 3.

# Example 21

Write a question and its solution that shows the calculation of the vibrational force constant of H19F. Show what the force constant would have to be for HF to absorb visible red light instead of IR light.
If H19F has a vibrational fundamental wavenumber of 4141.3 cm, What is the bond force constant? What would the force constant need to be to absorb visible red light instead of IR light? H has a n exact mass of 1.00782 u, and 19F has an exact mass of 18.998403 u.
Solution:
First, calculate the reduced mass for H19F.
$\mu ={\frac {m_{1}m_{2}}{m_{1}+m_{2}}}$
$\mu ={\frac {(1.00782)(18.998403)}{1.00782+18.998403}}$
$\mu =0.95705\ {\text{u}}\times 1.66\times 10^{-27}{\text{kg}}\ {\text{u}}^{-1}$
$\displaystyle \mu =1.589\times {10^{-27}}\ {\text{kg}}$
Now, solve for the force constant:
$\nu ={\frac {1}{2\pi }}\left({\frac {k}{\mu }}\right)^{\frac {1}{2}}$
$2\pi \cdot c\cdot {\tilde {\nu }}={\left({\frac {k}{\mu }}\right)^{\frac {1}{2}}}$
${\frac {k}{\mu }}=4\pi ^{2}c^{2}{\tilde {\nu }}^{2}$
$k=4\pi ^{2}c^{2}{\tilde {\nu }}^{2}\mu$
$k=4\pi ^{2}(2.998\times 10^{10}{\text{cm}}\cdot {\text{s}}^{-1})^{2}(4141.3\ {\text{cm}})^{2}(1.589\times 10^{-27}\ {\text{kg}})$
$k=966.99{\text{N}}\cdot {\text{m}}^{-1}$
Therefore, the bond force constant is 967 N m$^{-1}$.
Red light is around 700 nm, or 700×10$^{-9}$ m. Converted to wavenumber, that is 14285.71 cm$^{-1}$. Substituting this in,
$k=4\pi ^{2}(2.998\times 10^{10}{\text{cm}}\cdot {\text{s}}^{-1})^{2}(14285.71\ {\text{cm}})^{2}(1.589\times 10^{-27}\ {\text{kg}})$
$k=11506.7\ {\text{N}}\cdot {\text{m}}^{-1}$
Therefore, the bond force constant would have to be 11506 N m$^{-1}$ to absorb visible light.

# Example 22

Make a table showing all the absorption transitions for the first 5 rotational states of HCl. Include a column that shows which are allowed and which are forbidden.

# Example 23

Create a table showing all the possible absorption transitions between the first 5 vibrational states of HCl. Include columns for the energies and frequencies. The last column should indicate if the transition is allowed or forbidden.
Solution:
The vibrational frequency of a linear diatomic molecule depends on its reduced mass and vibrational spring constant (k). The spring constant for $^1$H$^{35}$Cl was measured and reported as 515.825 N/m.
The reduced mass was calculated as follows:
$\mu _{HCl}={\frac {m_{1}\cdot m_{2}}{m_{1}+m_{2}}}={\frac {1.00783{\text{ amu}}\cdot 34.9689{\text{ amu}}}{1.00783{\text{ amu}}+34.9689{\text{ amu}}}}=0.979597{\text{ amu}}$
The reduced mass was converted to SI units as follows:
${\bigl0.979597{\text{ amu}}{\bigr )}{\bigl1.66054\cdot 10^{-27}{\frac {\text{ kg}}{\text{amu}}}{\bigr )}=1.62666\cdot 10^{-27}{\text{ kg}}$
The fundamental vibrational frequency was calculated as follows:
$v_{0}={\frac {1}{2\pi }}{\biggl{\frac {k}{\mu }}{\biggr )}^{\frac {1}{2}}={\frac {1}{2\pi }}{\biggl{\frac {515.825{\frac {\text{N}}{\text{m}}}}{1.62666\cdot 10^{-27}{\text{ kg}}}}{\biggr )}^{\frac {1}{2}}=8.961237\cdot 10^{13}{\text{ s}}^{-1}$
The fundamental vibrational frequency was converted to wavenumbers as follows:
${\tilde {v}}_{0}={\frac {v_{0}}{c}}={\frac {8.961237\cdot 10^{13}{\text{ s}}^{-1}}{2.99792458\cdot 10^{10}{\frac {\text{cm}}{\text{s}}}}}=2989.52{\text{ cm}}^{-1}$
The energy for a given vibrational state can be calculated as follows:
$E_{\vee }=hv_{0}{\bigl\vee +{\frac {1}{2}}{\bigr )}=hc{\tilde {v}}_{0}{\bigl\vee +{\frac {1}{2}}{\bigr )}$
For example, the energy of the vibrational ground state:
$E_{0}=hv_{0}{\bigl0+{\frac {1}{2}}{\bigr )}={\frac {1}{2}}hv_{0}={\frac {1}{2}}hc{\tilde {v}}_{0}$
The transition energy between the ground state and the first excited vibrational state was calculated as follows:
$\Delta E_{1\leftarrow 0}=E_{1}-E_{0}={\frac {3}{2}}hc{\tilde {v}}_{0}-{\frac {1}{2}}hc{\tilde {v}}_{0}=hc{\tilde {v}}_{0}$
$\Delta E_{1\leftarrow 0}={\bigl6.62607015\cdot 10^{-34}{\text{ J}}\cdot {\text{s}}{\bigr )}{\bigl2.99792458\cdot 10^{10}{\frac {\text{cm}}{\text{s}}}{\bigr )}{\bigl2989.52{\text{ cm}}^{-1}{\bigr )}=5.93853\cdot 10^{-20}{\text{J}}$
According to the specific selection rule, vibrational transitions are only allowed between adjacent energy levels (where $\Delta \vee =\pm 1$). Consequently, due to the quantization of energy levels, vibrational transitions allowed by the specific selection rule are equivalent in energy and frequency, such that $\Delta E=hv_{0}=hc{\tilde {v}}_{0}$. Conversely, overtone transitions are those for which $\Delta \vee \neq \pm 1$. As such, overtone frequencies can be calculated as integer multiples of the fundamental frequency. For example, the vibrational frequency for the transition between the ground state and the second excited vibrational state was calculated as follows:
$v_{2\leftarrow 0}=2{\biglv_{0}{\bigr )}=2{\bigl8.961237\cdot 10^{13}{\text{ s}}^{-1}{\bigr )}=1.79247\cdot 10^{14}{\text{ s}}^{-1}$
Similarly, the wavenumber for the transition between the ground state and the third excited vibrational state was calculated as follows:
${\tilde {v}}_{3\leftarrow 0}=3{\bigl{\tilde {v}}_{0}{\bigr )}=8968.57{\text{ cm}}^{-1}$
Finally, the energy for the overtone transition between the first and fourth excited vibrational states was calculated as follows:
$\Delta E_{4\leftarrow 1}=E_{4}-E_{1}={\frac {9}{2}}hc{\tilde {v}}_{0}-{\frac {3}{2}}hc{\tilde {v}}_{0}=3hc{\tilde {v}}_{0}$
$\Delta E_{4\leftarrow 1}=3{\bigl6.62607015\cdot 10^{-34}{\text{ J}}\cdot {\text{s}}{\bigr )}{\bigl2.99792458\cdot 10^{10}{\frac {\text{cm}}{\text{s}}}{\bigr )}{\bigl2989.52{\text{ cm}}^{-1}{\bigr )}=5.34468\cdot 10^{-19}{\text{J}}$
However, because $\Delta \vee \neq \pm 1$, this transition is forbidden .
Frequency, wavenumber, and energy values for transitions between the first five vibrational energy levels of $^1$H$^{35}$Cl. :
| Transition | Frequency (s$^{-1}$) | Wavenumber (cm$^{-1}$) | Energy (J) | Allowed/Forbidden |
| 1←0 | 8.96237·10$^{13}$ | 2989.52 | 5.93853·10$^{-20}$ | Allowed |
| 2←0 | 1.79247·10$^{14}$ | 5979.05 | 2.37541·10$^{-19}$ | Forbidden |
| 2←1 | 8.96237·10$^{13}$ | 2989.52 | 5.93853·10$^{-20}$ | Allowed |
| 3←0 | 2.68871·10$^{14}$ | 8968.57 | 5.34468·10$^{-19}$ | Forbidden |
| 3←1 | 1.79247·10$^{14}$ | 5979.05 | 2.37541·10$^{-19}$ | Forbidden |
| 3←2 | 8.96237·10$^{13}$ | 2989.52 | 5.93853·10$^{-20}$ | Allowed |
| 4←0 | 3.58495·10$^{14}$ | 11958.1 | 1.18771·10$^{-18}$ | Forbidden |
| 4←1 | 2.68871·10$^{14}$ | 8968.57 | 5.34468·10$^{-19}$ | Forbidden |
| 4←2 | 1.79247·10$^{14}$ | 5979.05 | 2.37541·10$^{-19}$ | Forbidden |
| 4←3 | 8.96237·10$^{13}$ | 2989.52 | 5.93853·10$^{-20}$ | Allowed |
| 5←0 | 4.48119·10$^{14}$ | 14947.6 | 1.48463·10$^{-18}$ | Forbidden |
| 5←1 | 3.58495·10$^{14}$ | 11958.1 | 1.18771·10$^{-18}$ | Forbidden |
| 5←2 | 2.68871·10$^{14}$ | 8968.57 | 5.34468·10$^{-19}$ | Forbidden |
| 5←3 | 1.79247·10$^{14}$ | 5979.05 | 2.37541·10$^{-19}$ | Forbidden |
| 5←4 | 8.96237·10$^{13}$ | 2989.52 | 5.93853·10$^{-20}$ | Allowed |
References:

# Example 24

Write a question and its solution showing the probability of a particle being located in the bottom left quadrant of the ground state of the particle in a square.

## Background Information
### Particle In a Square (2D)

The particle in a square model is where a particle in confined to a restricted
region inside a 2D box. For this model we consider a particle which is free to move
anywhere inside the x,y dimensions but cannot exist outside of the 2D box where it would
have infinite potential energy.
The conditions for the potential energy are $\nu ={\begin{Bmatrix}\infty ,&x<0,y<0\\0,&0\leq x,y\leq L\\\infty ,&x,y>L\end{Bmatrix}}$ meaning that the particle spends 100% of its time within the walls of the box.
Since the particle in a square is a 2D model the wave function must take into account the x and y direction ; $(\Psi (x,y))$
    The wave function for a particle in a square $\Psi (x,y)={\frac {2}{L}}sin\left({\frac {n_{x}\pi }{L}}x\right)\left({\frac {n_{y}\pi }{L}}y\right)\cdot {\frac {2}{L}}sin\left({\frac {n_{x}\pi }{L}}x\right)\left({\frac {n_{y}\pi }{L}}y\right)$
The states the quantum system is in are defined by the quantum numbers $n_{x},n_{y}$
In quantum energy levels only discrete energy levels are possible. In this model the particle cannot be motionless meaning the lowest possible energy level must be non-zero (n=1). Since the particle is always moving even in the ground state the particle has a a non- zero energy level. Therefore the ground state of a particle in a box is $n_{x}=1,n_{y}=1$. In quantum mechanical systems there is a presence of nodes where in states above the ground state for the 2D model there are nodal lines that divide the square. As the quantum numbers increase so does the amount of nodal lines as $Nnodallines=n_{x},_{y}-1$. The sign of the wavefunction changes between the nodes and has a value of zero on the nodes.

### Probability Distributions

The probability (P) of a particle being found in an interval can be found by integrating the probability distribution (P(x))
The probability of finding a particle in the range $\left[a_{1},a_{2}\right]$,$\left[b_{1},b_{2}\right]$ is given by; $P=\int \limits _{a_{1}}^{a_{2}}\int \limits _{b_{1}}^{b_{2}}P(x,y)dxdy$
The probability distribution of a a particle is the wave function squared.
For one particle in 2D (particle in a box):
$P(x,y)=\left\vert \Psi (x,y)^{2}\right\vert =\Psi (x,y)^{*}\cdot \Psi (x,y)$ ,where * is the complex conjugate of the wave-function
Therefore the probability of finding a particle over a range can be re-written as:
$P=\int \limits _{a_{1}}^{a_{2}}\int \limits _{b_{1}}^{b_{2}}\Psi ^{*}{\biglx,y)\Psi {\biglx,y)dxdy$

## Example
#### Question: For a particle in a square that is length L, what is the probability of finding the particle in the bottom left quadrant if in the ground state?
##### Solution

If the particle is in the bottom left quadrant it is in the interval; $x=\left[0,{\frac {L}{2}}\right]$ $y=\left[0,{\frac {L}{2}}\right]$
Therefore the probability density is:
$P=\int \limits _{0}^{\tfrac {L}{2}}\int \limits _{0}^{\tfrac {L}{2}}\Psi ^{*}{\biglx,y)\Psi {\biglx,y)dxdy$
The wave-function for a particle in a square is:
$\Psi (x,y)={\frac {2}{L}}\sin \left({\frac {n_{x}\pi }{L}}x\right)\left({\frac {n_{y}\pi }{L}}y\right)\cdot {\frac {2}{L}}\sin \left({\frac {n_{x}\pi }{L}}x\right)\left({\frac {n_{y}\pi }{L}}y\right)$
In the ground state $n_{x}=1$ $n_{y}=1$ , therefore the probability distribution can be written as:
$P=\int \limits _{0}^{\frac {L}{2}}\int \limits _{0}^{\frac {L}{2}}{\biggl{2 \over L}{\biggr )}\sin \left({\frac {\pi }{L}}x\right)\sin \left({\frac {\pi }{L}}y\right)\centerdot {\biggl{2 \over L}{\biggr )}\sin \left({\frac {\pi }{L}}x\right)\sin \left({\frac {\pi }{L}}y\right)dxdy$
Then solving the integral,
$P=\left({\frac {4}{L^{2}}}\right)\int _{0}^{\frac {L}{2}}\int _{0}^{\frac {L}{2}}\sin ^{2}\left({\frac {\pi }{L}}x\right)\sin ^{2}\left({\frac {\pi }{L}}y\right)dxdy$
$P=\left({\frac {4}{L^{2}}}\right)\int _{0}^{\frac {L}{2}}\sin ^{2}\left({\frac {\pi }{L}}x\right)dx\cdot \int _{0}^{\frac {L}{2}}\sin ^{2}\left({\frac {\pi }{L}}y\right)dy$
    From Table of Integrals $\int \sin ^{2}(ax)dx={\frac {x}{2}}-{\frac {\sin(2ax)}{4a}}$
$P=\left({\frac {4}{L^{2}}}\right)\left[{\frac {y}{2}}-{\frac {\sin {\bigl2{\frac {\pi }{L}}y{\bigr )}}{4\left({\frac {\pi }{L}}\right)}}\right]_{0}^{L}\left[{\frac {x}{2}}-{\frac {\sin {\bigl2{\frac {\pi }{L}}x{\bigr )}}{4\left({\frac {\pi }{L}}\right)}}\right]_{0}^{L}$
$P=\left({\frac {4}{L^{2}}}\right){\Biggl{\frac {\frac {L}{2}}{2}}-{\frac {L\sin {\bigl2{\frac {\pi }{L}}{\frac {L}{2}}{\bigr )}}{4\left(\pi \right)}}{\Biggr )}{\Biggl{\frac {\frac {L}{2}}{2}}-{\frac {L\sin {\bigl2{\frac {\pi }{L}}{\frac {L}{2}}{\bigr )}}{4\left(\pi \right)}}{\Biggr )}$
$P=\left({\frac {4}{L^{2}}}\right){\Biggl{\frac {L}{4}}-{\frac {L\sin {\bigl\pi {\bigr )}}{4\left(\pi \right)}}{\Biggr )}{\Biggl{\frac {L}{4}}-{\frac {L\sin {\bigl\pi {\bigr )}}{4\left(\pi \right)}}{\Biggr )}$
Since $\sin \pi$ is zero the whole term goes to zero
$P=\left({\frac {4}{L^{2}}}\right){\Biggl{\frac {L}{4}}-0{\biggr )}{\Biggl{\frac {L}{4}}-0{\Biggr )}$
$P={\frac {4}{8}}$
$P={\frac {1}{4}}$
Therefore the the probability of finding a particle in the bottom left quadrant of a square in the ground state is 0.25 of 25%.

# Example 25
## Introduction

The particle in a one-dimensional box model describes the movement of a particle in one dimension. It is confined to a "box" where it can only move horizontally along the x-axis within the confines of the box. The particle cannot escape the box due to the areas outside of the box having an infinite potential energy, thus trapping the particle inside an infinitely deep well. If the leftmost boundary of the box is position 0 and the rightmost position is position L, the potential energy of the particle inside the box is zero between position 0 and L. This is denoted mathematically as $V(x)=0,0\leq x\leq L$ where $V(x)$ is potential energy as a function of position. Outside the box, the potential energy is $V(x)=\infty$ for $x<0$ and $x>L$. The problem below proposes a situation where the potential energy is zero everywhere along the x-axis, meaning the particle is no longer bound by areas of infinite potential energy like in the 1D box model. Rather, $V(x)=0$ across the interval $-\infty \leq x\leq \infty$.

## Example Problem and Solution
### Problem

Suppose an electron is free to move along the x-axis and its potential energy everywhere is zero. Derive a wavefunction for this system.

### Solution
#### Schrödinger Equation and Definition of the Hamiltonian

First the Hamiltonian operator ${\hat {H}}$ must be defined in order to solve the Schrödinger equation for this system.
    Time-independent Schrödinger equation ${\hat {H}}\psi =E\psi$
The Hamiltonian is simply the total energy of the system. It is expressed as the sum of the total kinetic energy (${\hat {T}}$) and total potential energy (${\hat {V}}$) of the system. Therefore the Hamiltonian is defined as,
${\hat {H}}={\hat {T}}+{\hat {V}}$
However, the problem states that the potential energy everywhere is zero, so the Hamiltonian is simply equal to the total kinetic energy of the system.
${\hat {H}}={\hat {T}}+0$
${\hat {H}}={\hat {T}}$

#### Definition of the Kinetic Energy Operator and Substitution into the Schrödinger Equation

In one dimension, the kinetic energy operator is defined as,
    ${\hat {T}}=-{\frac {\hbar ^{2}}{2m}}{\frac {d^{2}}{dx^{2}}}$
So the Hamiltonian becomes,
${\hat {H}}=-{\frac {\hbar ^{2}}{2m_{e}}}{\frac {d^{2}}{dx^{2}}}$
Where $m_{e}$ is the mass of an electron and $\hbar$ is the reduced Planck's constant. This can then be substituted into the Schrödinger equation to give,
$-{\frac {\hbar ^{2}}{2m_{e}}}{\frac {d^{2}}{dx^{2}}}\psi (x)=E\psi (x)$
The above is simplified by multiplying both sides of the equation by $-2m_{e}$ and then dividing both sides by $\hbar ^{2}$.
$\hbar ^{2}{\frac {d^{2}}{dx^{2}}}\psi (x)=-2m_{e}E\psi (x)$
${\frac {d^{2}}{dx^{2}}}\psi (x)={\frac {-2m_{e}E}{\hbar ^{2}}}\psi (x)$

#### Finding a General Solution and Solving the Differential Equation

${\frac {d^{2}}{dx^{2}}}\psi (x)={\frac {-2m_{e}E}{\hbar ^{2}}}\psi (x)$ is a linear homogenous second order differential equation where $k={\frac {\sqrt {2m_{e}E}}{\hbar }}$, so,
$\psi (x)=A\sin(kx)+B\cos(kx)$
$\psi (x)=A\sin {\biggl{\frac {\sqrt {2m_{e}E}}{\hbar }}x{\biggr )}+B\cos {\biggl{\frac {\sqrt {2m_{e}E}}{\hbar }}x{\biggr )}$
Since the potential energy everywhere is zero, the electron can exist anywhere on the x-axis. The position of the particle exists across the interval $-\infty \leq x\leq \infty$, therefore the probability of finding the electron at any position $x$ is infinitely small and approaches zero.
${\frac {1}{\infty }}\approx 0$
Thus it can be assumed that $x=0$ for both terms of the general solution due to the absence of any boundary conditions. Putting $x=0$ into the general solution yields,
$\psi (x)=A\sin {\biggl{\frac {\sqrt {2m_{e}E}}{\hbar }}0{\biggr )}+B\cos {\biggl{\frac {\sqrt {2m_{e}E}}{\hbar }}0{\biggr )}$
$\psi (x)=0$
Therefore, the wavefunction collapses and is zero if the potential energy is zero everywhere along the x-axis, due to the probability of finding the electron at any position $x$ being infinitely small.

# Example 26

The square of the angular momentum of a hydrogen atom is measured to be $L^{2}$$=20\hbar ^{2}$. What are the possible values of the z-component of the orbital angular momentum, $L_{z}$, that could be measured for this atom?
Solution:
The eigenvalues of the square of the angular momentum operator (${\hat {L}}$$^2$) for a quantum mechanical system, such as an electron in a hydrogen atom, are given by:
${\hat {L}}^{2}Y_{l}^{m}(\theta ,\phi )=h^{2}l(l+1)Y_{l}^{m}(\theta ,\phi )$
where $Y_{l}^{m}(\theta ,\phi )$ are the spherical harmonics, which are eigenfunctions of ${\hat {L}}$$^2$, $l$ is the orbital quantum number, and $m$ is the magnetic quantum number. The given value for $L^{2}$ = 20 ħ2 , so we set up the equation:
$\hbar ^{2}l(l+1)=20\hbar ^{2}$
Dividing by $\hbar ^{2}$ and simplifying, we get:
$l(l+1)=20$
$l^{2}+l-20=0$
In this quadratic equation, $l$ can be factored to get:
$(l-4)(l+5)=0$
$l=4$
Since $l$ must be a non-negative integer. The magnetic quantum number $m$ can take on any integer value from $-l$ to $l$, thus for $l=4$, $m$ can be:
$m=-4,-3,-2,-1,0,1,2,3,4$
The z-component of the angular momentum, $L_{z}$, is quantized in units of $\hbar ^{2}$and given by:
$L_{z}=m\hbar$
$L_{z}=-4\hbar ,-3\hbar ,-2\hbar ,-1\hbar ,0\hbar ,1\hbar ,2\hbar ,3\hbar ,4\hbar$.
Therefore, the possible values of the z-component of the orbital angular momentum, $L_{z}$, that could be measured for the atom with a given $L^{2}$$=20\hbar ^{2}$ are $L_{z}=-4\hbar ,-3\hbar ,-2\hbar ,-1\hbar ,0\hbar ,1\hbar ,2\hbar ,3\hbar ,4\hbar$.

# Example 27

Calculate the frequency of a photon needed to excite CO from the ground vibrational state to the first vibrational state (k=1860 N/m).
$v_{0}={\frac {1}{2\pi }}\cdot {\biggl{\frac {k}{\mu }}{\biggr )}^{\frac {1}{2}}\qquad where\ \mu ={\frac {m_{1}\cdot m_{2}}{m_{1}+m_{2}}}$
C= 12.011u, O =15.999u
Plug the values into the reduced mass equation:
$\mu ={\frac {12.011u\cdot 15.999u}{12.011u+15.999u}}$
$\mu =6.861u$
Convert the reduced mass from units of u to kg
$1u=1.66054\times 10^{-27}kg$
$\mu =1.139\times 10^{-26}kg$
$v_{0}={\frac {1}{2\pi }}\cdot {\biggl{\frac {1860N/m}{1.139\times 10^{-26}kg}}{\biggr )}^{\frac {1}{2}}\qquad$
$\therefore v_{0}=6.43\times 10^{13}s^{-1}$

# Example 28
## Question

For an electron in a 3D cube with a length of 6.00 Å, what is:

## Solution

(a) The mass of an electron must be known to determine the energy of the particle. To determine the energy state of an electron in a 3D box, the equation that relates the energy levels of a particle in a cube must be used with every variable in its SI units. The mass of an electron is known to be ${9.1093837{\text{×}}10^{-31}{\text{kg}}}$. Planck's constant is known to be ${6.62607015{\text{×}}10^{-34}{\text{m²·kg·s⁻¹}}}$. The length of the box is ${6.00{\text{x}}10^{-10}{\text{m}}}$ since one angstrom is known to be $10^{-10}{\text{m}}$. The energy of the electron in the cube can be calculated as follows,
    ${E}=\left({\frac {h^{2}}{8mL^{2}}}\right)\left({n_{\text{x}}^{2}+n_{\text{y}}^{2}+n_{\text{z}}^{2}}\right)$
    ${E}=\left({\frac {\left(6.62607015{\text{×}}10^{-34}{\text{m²·kg·s⁻¹}}\right)^{2}}{8{\text{×}}\left(9.1093837{\text{×}}10^{-31}{\text{kg}}\right){\text{×}}\left(6.00{\text{x}}10^{-10}{\text{m}}\right)^{2}}}\right)\left({{4^{2}}+{5^{2}}+{3^{2}}}\right)$
    ${E}=\left(1.68{\text{×}}10^{-19}{\text{J}}\right)\left(50\right)$
    ${E}=8.38{\text{×}}10^{-18}{\text{J}}$
(b) Degeneracy is defined as an energy state with an identical energy to that of a particle with a different set of quantum numbers. In this question, the energy of degenerate states must equal the following,
    ${E}=\left({\frac {h^{2}}{8mL^{2}}}\right)\left({{4^{2}}+{5^{2}}+{3^{2}}}\right)=\left({\frac {50h^{2}}{8mL^{2}}}\right)$
To determine the degeneracy of a system, the simplest method is to use a matrix to ensure a systematic process is followed. In this case, the sum of the square of all quantum numbers must equal 50. Therefore, the matrix is as follows,
    ${n_{\text{x}}}$ ${n_{\text{y}}}$ ${n_{\text{z}}}$ ${n_{\text{x}}^{2}+n_{\text{y}}^{2}+n_{\text{z}}^{2}}$ 4534² + 5² + 3² = 50 4354² + 3² + 5² = 50 5345² + 3² + 4² = 50 5435² + 4² + 3² = 50 3453² + 4² + 5² = 50 3543² + 5² + 4² = 50
This results in the identification of 6 different sets of quantum numbers, and thus, this energy state is said to be 6-fold degenerate.
(c) To determine the number of nodal planes, each axis must be considered individually. Mathematically, to calculate the number of nodal planes associated with the x-axis, the following equation is used,
    ${N_{\text{nodal planes x-axis}}=n_{\text{x}}-1}$
Therefore, in this example, the number of nodal planes in x-axis is,
    ${N_{\text{nodal planes x-axis}}=4-1=3}$
Similarly, for the y-axis and the z-axis, the number of nodal planes is,
    ${N_{\text{nodal planes y-axis}}=n_{\text{y}}-1=5-1=4}$
    ${N_{\text{nodal planes z-axis}}=n_{\text{z}}-1=3-1=2}$
Therefore, the total number of nodal planes in this system is 9. This was determined by summing the number of nodal planes associated with each individual axis.

# Example 29
## Quantum Mechanical Harmonic Oscillator Wavefunction

A system undergoing harmonic motion around an equilibrium is known as a harmonic oscillator.
In quantum chemistry, the harmonic oscillator refers to a simplified model often used to describe how a diatomic molecule vibrates. This is because it behaves like two masses on a spring with a potential energy that depends on the displacement from the equilibrium, but the energy levels are quantized and equally spaced. The potential energy $V(x)={\frac {1}{2}}kx^{2}$ is non-zero and can theoretically range from $x=[-\infty {\text{ , }}\infty ]$.
The wavefunction for the quantum harmonic oscillator is given by the Hermite polynomials multiplied by the Gaussian function. The general form of the wavefunction is:
$\psi _{v}=N_{v}\cdot H_{v}\left({\sqrt {\frac {mw}{\hbar }}}x\right)\cdot \exp \left({\frac {-mw}{2\hbar }}x^{2}\right)$
With:
$v={\text{Quantum number}}$
$N_{v}={\text{Normalization factor}}$
$H_{v}=v{\text{-th Hermite polynomial}}$
The first four Hermite polynomials are:
${\begin{aligned}H_{0}(x)&=1,\\H_{1}(x)&=2x,\\H_{2}(x)&=4x^{2}-2,\\H_{3}(x)&=8x^{3}-12x.\\\end{aligned}}$
$m={\text{Mass of particle}}$
$w={\text{Angular frequency of the oscillator}}$
$\hbar ={\text{Reduced Planck's constant}}$
$x={\text{Position}}$

## Normalization of a Wavefunction

The probability of finding the particle in any state is given by the square of the wavefunction. Therefore normalizing a wavefunction in quantum mechanics means ensuring that the total probability of finding a particle in all possible positions is equal to 1. A normalized wavefunction is one that satisfies the normalization condition:
$\int _{-\infty }^{\infty }\mid \psi (x)\mid ^{2}dx=1$
Normalization is important because it ensures that the probability of finding the particle somewhere in space is 100%.

## Example

Show the derivation of the normalization factor of the v=1 state of the harmonic oscillator beginning from the unnormalized wavefunction.
$\psi _{v}=N_{v}\cdot H_{v}\left({\sqrt {\frac {mw}{\hbar }}}x\right)\cdot \exp \left({\frac {-mw}{2\hbar }}x^{2}\right)$
$N_{v}^{2}\int _{-\infty }^{\infty }H_{v}^{2}\left({\sqrt {\frac {mw}{\hbar }}}x\right)\cdot \exp \left({\frac {-mw}{\hbar }}x^{2}\right)dx=$
${\text{Let }}y=\left({\sqrt {\frac {mw}{\hbar }}}\right)x$
$N_{v}^{2}\int _{-\infty }^{\infty }H_{v}(y)H_{v}(y)\cdot \exp \left(-y^{2}\right){\sqrt {\frac {\hbar }{mw}}}dy=$
$N_{v}^{2}{\sqrt {\frac {\hbar }{mw}}}(-1)^{v}\int _{-\infty }^{\infty }H_{v}(y){\frac {d^{v}}{dy^{v}}}\exp \left(-y^{2}\right)dy=$
$N_{v}^{2}{\sqrt {\frac {\hbar }{mw}}}\int _{-\infty }^{\infty }\exp(-y^{2}){\frac {d^{v}}{dy^{v}}}H_{v}(y)dy=$
$N_{v}^{2}{\sqrt {\frac {\hbar }{mw}}}2^{v}v!{\sqrt {\pi }}=$
$N_{v}={\frac {1}{\sqrt {2^{v}v!}}}\left({\frac {mw}{\pi \hbar }}\right)^{1/4}$
Sub in v=1
$N_{1}={\frac {1}{\sqrt {2^{1}1!}}}\left({\frac {mw}{\pi \hbar }}\right)^{1/4}={\frac {1}{\sqrt {2}}}\left({\frac {mw}{\pi \hbar }}\right)^{1/4}$

## Check

Prove that $\psi _{1}$ is normalized (using notation from class)
$\psi _{1}=\left({\frac {4\alpha ^{3}}{\pi }}\right)^{1/4}x\exp \left({\frac {-\alpha x^{2}}{2}}\right)$, where $\alpha =\left({\sqrt {\frac {k\mu }{\hbar ^{2}}}}\right)$
$\int _{-\infty }^{\infty }\mid \psi _{1}\mid ^{2}dx=\left({\frac {4\alpha ^{3}}{\pi }}\right)^{1/2}\int _{-\infty }^{\infty }x^{2}\exp ^{-\alpha x^{2}}dx=\left({\frac {4\alpha ^{3}}{\pi }}\right)^{1/2}\left({\frac {1}{2\alpha }}\left({\frac {\pi }{\alpha }}\right)^{1/2}\right)=1$

# Example 30
### Energy in the particle in a 1D box model

Energy in quantum mechanical systems is quantized, unlike in classically mechanical systems, where the energy can take any value. The energy of a quantum mechanical system may only hold specific discrete values, which are known as energy levels. Consider the model for the 1D particle in a box. In this model, the possible discrete energy levels are given by the relation,
    Energy levels of a particle in a 1D box ${\begin{aligned}&E_{n}={\frac {h^{2}n^{2}}{8mL^{2}}},\quad n=1,2,3,\ ...\\&\\&{\text{where}}\ h\ {\text{is Planck's constant}}\,(6.62607015\cdot 10^{-34}\,{\text{m}}^{2}{\text{kg/s}}),\\&n\ {\text{is the quantum number,}}\\&m\ {\text{is the mass of the particle, and}}\ L\ {\text{is the length of the box.}}\\\end{aligned}}$
The ground state of any quantum mechanical system is the state with the lowest possible energy (otherwise known as the zero-point energy of the system). Defined in the equation for the energy levels of a particle in a 1D box, the quantum number $n$ can hold any integer value equal to or greater than one and as such, the ground state of this model occurs when $n=1$. All energy levels above the ground state are referred to as excited states. There are an infinite number of these states and for the particle in a 1D box, are defined by the quantum number $n$, like the ground state. The first excited state is the energy level one above the ground state, in which $n=2$, the second excited state is the energy level one above the first excited state, where $n=3$, and so forth.

### Particle transitions between states

A particle in a 1D box can transition between energy levels by the means of three different processes, whose probability can be measured by Einstein coefficients. The transitional processes involve electromagnetic radiation in the form of photons. They are the following:
* Spontaneous emission: The spontaneous decay of a particle from a higher energy state to a lower energy state, emitting a photon of the transition energy.
* Stimulated emission: The decay of a particle from a higher energy state to a lower energy state stimulated by the presence of a photon of the same energy as the transition.
* Photon absorption: The promotion of a particle from a lower energy state to a higher energy state caused by the absorption of a photon of the transition energy.
The energy of a photon can be determined by the relationship between a photon's frequency of oscillation and energy, known as the Planck relation,
    Planck's relation ${\begin{aligned}&E=h\nu ,\\&\\&{\text{where}}\ h\ {\text{is Planck's constant,}}\\&E\ {\text{is the energy of the photon,}}\\&{\text{and}}\ \nu \ {\text{is the photon's frequency of oscillation in s}}^{-1}{\text{.}}\\\end{aligned}}$
In quantum chemistry, it is often more convenient to express frequency in terms of wavenumber (${\tilde {\nu }}$), which has units of ${\text{cm}}^{-1}$ compared to ${\text{s}}^{-1}$. There is a simple conversion as frequency and wavenumber are related to the speed of light by,
    ${\tilde {\nu }}={\frac {\nu }{c}}\,{\text{, where}}\ c\ {\text{is the speed of light in cm/s.}}$
As such, the Planck relation can be rewritten as,
    Planck's relation (wavenumber) $E=h{\tilde {\nu }}c$
This relation allows the calculation of the energy of a photon involved in one of the transitions between states, as the energy of the transition will be equal to the energy of the photon, whether it is the emitted photon in spontaneous emission, the stimulating photon in stimulated emission or the absorbed photon in the absorption transition.

## Example

What is the wavenumber (${\tilde {\nu }}_{photon}$) of electromagnetic radiation emitted when a proton in a one-dimensional box with length 472 nm decays from the fifth excited state to the fourth excited state?
SOLUTION: This transition is a spontaneous emission where the energy of the emitted photon can be determined by the difference between the energy levels of the transition,
    ${\begin{aligned}E_{photon}&=\Delta {E}\\&=E_{j}-E_{i},\\&{\text{where}}\ i\ {\text{and}}\ j\ {\text{are the final state and the excited state, respectively.}}\\\end{aligned}}$
As this transition occurs between the fourth and fifth excited states, $i$ and $j$ will be 5 and 6, respectively. From the energy levels of a particle in a 1D box,
    ${\begin{aligned}E_{6}-E_{5}&={\frac {h^{2}6^{2}}{8m_{p}L^{2}}}-{\frac {h^{2}5^{2}}{8m_{p}L^{2}}}\\&={\frac {11h^{2}}{8m_{p}L^{2}}},\\&{\text{where}}\ m_{p}\ {\text{is the mass of a proton in kg and}}\ L\ {\text{is the given length of the box in m,}}\ (1\ {\text{nm}}=1\cdot 10^{-9}{\text{m}}){\text{.}}\\\end{aligned}}$
Substituting these values into the above equation gives,
    ${\begin{aligned}E_{photon}&={\frac {11(6.62607015\cdot 10^{-34}\,{\text{m}}^{2}{\text{kg/s}})^{2}}{8(1.67262192\cdot 10^{-27}\,{\text{kg}})(472\cdot 10^{-9}\,{\text{m}})^{2}}}\\&=1.62\cdot 10^{-27}\,{\text{J}}\\\end{aligned}}$
Using the wavenumber variation of Planck's relation, the wavenumber of a photon of this exact energy can be determined and expressed in convenient units of ${\text{cm}}^{-1}$.
    ${\begin{aligned}{\tilde {\nu }}_{photon}&={\frac {E_{photon}}{hc}}{\text{, where}}\ h\ {\text{is Planck's constant and}}\ c\ {\text{is the speed of light expressed in cm/s.}}\\&={\frac {1.62\cdot 10^{-27}\,{\text{J}}}{(6.62607015\cdot 10^{-34}\,{\text{m}}^{2}{\text{kg/s}})(2.99792458\cdot 10^{10}\,{\text{cm/s}})}}\\&=8.16\cdot 10^{-5}\,{\text{cm}}^{-1}\\\end{aligned}}$

# Example 31

Write an example problem related to the conversion between wavelength, frequency, energy and wavenumber of electromagnetic radiation. Show each step in detail and explain the conversion.
Example: Carbon monoxide shows a sharp peak in its IR spectrum at 2143 cm$^{-1}$.
a) Calculate the frequency and wavelength of the molecule's vibration.
The question gives us the wavenumber which can be converted to frequency through the equation:
$v={\widehat {v}}*c$
where v is the frequency, ṽ is the wavenumber and c is the speed of light.
$v={\widehat {v}}*c$
$v=$ (2143cm$^{-1}$) * (2.99792458*10$^{10}$cm s$^{-1}$)
= 6.42455237494*10$^{13}$ s$^{-1}$
= 64.24 THz
To find the wavelength of the vibration we can use the equation:
${\hat {v}}$= ${1 \over \lambda }$
where $\lambda$ is the wavelength.
${\hat {v}}$= ${1 \over \lambda }$
This equation can be rearranged to solve for wavelength:
$\lambda$= ${1 \over {\hat {v}}}$
= ${1 \over 2143cm^{-1}}$
= 4.666356*10$^{-4}$ cm
This should be converted to nm since wavelength is usually written in nm.
λ = 4.666356*10$^{-4}$ cm * ${1nm \over (1*10^{-7}cm)}$
= 4.6*10$^3$nm
b) Calculate the energy of carbon monoxide at its ground state.
To find the energy of a molecule at its ground state we can use the equation:
$E=hv_{0}(v+\left({\frac {1}{2}}\right))$
where E is the energy, h is Planck's constant, $v_{0}$ is the frequency of the ground state, and $v$ is the quantum number of the energy level.
We can use this equation since carbon monoxide's vibrations act as a harmonic oscillator.
$E=hv_{0}(v+\left({\frac {1}{2}}\right))$
$E=h(6.424*10^{13}s^{-1})$$(0+\left({\frac {1}{2}}\right))$
$E=2.12847674*10^{-20}J$
c) Calculate the wavelength of a photon that excites an electron from the ground state up 2 levels.
In this case, the quantum number v is equal to 2. We can use the same equation used in b) to first solve the change in energy caused by the photon.
$E=hv_{0}(v+\left({\frac {1}{2}}\right))$
$E=h(6.424*10^{13}s^{-1})$$(2+\left({\frac {1}{2}}\right))$
$E=1.06423837*10^{-19}J$
Now that we have found the change in energy we can calculate the wavelength of the photon using the equation:
$E=\left({\frac {hc}{\lambda }}\right)$
Rearrange the equation to solve for λ:
$\lambda =\left({\frac {hc}{E}}\right)$
$\lambda =\left({\frac {hc}{1.06423837*10^{-19}J}}\right)$
$\lambda =0.0001866542cm$
Convert the wavelength to nm.
$\lambda =0.0001866542cm*\left({\frac {1nm}{1*10^{-7}cm}}\right)$
$\lambda =1.86*10^{-6}nm$
Therefore, a photon that is able to cause a transition from ground state to v = 2 must have a wavelength of 1.86 * 10$^{-6}$ nm.

# Example 32
## The Wavefunction of a Particle in a One-Dimensional Box

Despite the wavefunction having very little physical interpretation, the solution to the wavefunction of a particle is beneficial in order to determine the probability of such particle existing in a given area. When deriving the equation for the wavefunction of a particle in a one-dimensional box, we first begin by defining the Hamiltonian operators (i.e., kinetic and potential energy) for the given system, and substituting these values into the Schrodinger equation.
In quantum mechanics, the Schrodinger equation is given as,
    The Schrodinger Equation ${\hat {H}}\psi =E\psi$ $\left({\hat {T}}+{\hat {V}}\right)\psi =E\psi$
where ${\hat {H}}$ is the Hamiltonian operator, ${\hat {T}}$ is the kinetic energy operator, ${\hat {V}}$ is the potential energy operator, $\psi$ is the wavefuntion, and $E$ is the total energy of the system. The first equation is a simplification of the second equation, since ${\hat {H}}={\hat {T}}+{\hat {V}}$. Mathematically, the wavefunction $\psi$ is an eigenfunction of the Hamiltonian operator ${\hat {H}}$ with an eigenvalue of $E$.

### Defining the Hamiltonian

For the case of a particle in a one-dimensional box, its potential energy is equal to zero within the constraints of such box (i.e., $0\leq x\leq L$), but the potential energy of the particle is infinite outside of these constraints:
    Potential Energy Range of Particle in 1D Box $V(x)={\begin{cases}\infty ,&x<0\\0,&0\leq x\leq L\\\infty ,&x>L\end{cases}}$
Additionally, the kinetic energy operator in the $x$ dimension is given as ${\hat {T}}=-{\frac {{\hslash }^{2}}{2m}}{\frac {{d}^{2}}{{dx}^{2}}}$. Therefore, since $V(x)=0$ inside of the box, the values defined for ${\hat {T}}$ and ${\hat {V}}$ can be substituted into the Schrodinger equation, which simplifies to ${\frac {{d}^{2}\psi (x)}{{dx}^{2}}}=-{\frac {2E\cdot m}{{\hslash }^{2}}}\cdot \psi (x)$.

### Boundary Conditions of the Wavefunction

The boundary conditions must then be identified for a particle in a one-dimensional box. The constraints of the box have been previously defined as the position of the particle, $x$, being contained within a box of length $L$. This means that the particle cannot exist when $x=L$ or when $x=0$. The boundary conditions for the value of the wavefunction must therefore be $\psi (0)=0$ and $\psi (L)=0$.

### General Solution

This is a second-order linear homogenous equation with respect to $\psi (x)$, and has a general solution of
$\psi (x)=A\cdot \sin(k\cdot x)+B\cdot \cos(k\cdot x)$, where $k={\frac {\sqrt {2\cdot m\cdot E}}{\hslash }}$
Thus, the solution for the given second-order linear homogenous equation is
$\psi (x)=A\cdot \sin \left({\frac {\sqrt {2\cdot m\cdot E}}{\hslash }}\cdot x\right)+B\cdot \cos \left({\frac {\sqrt {2\cdot m\cdot E}}{\hslash }}\cdot x\right)$
To determine the specific solution for this wavefunction, the constants $A$ and $B$ must be found.

### Determining the Unknown Variables

By substituting the first boundary condition that $\psi (0)=0$ into the general solution of the wavefunction, the value of $B$ may be found.
$\psi (0)=A\cdot \sin \left({\frac {\sqrt {2\cdot m\cdot E}}{\hslash }}\cdot 0\right)+B\cdot \cos \left({\frac {\sqrt {2\cdot m\cdot E}}{\hslash }}\cdot 0\right)$
$\psi (0)=A\cdot \sin(0)+B\cdot \cos(0)=0$
$A\cdot 0+B\cdot 1=0$
$\therefore B=0$
We now have $\psi (x)=A\cdot \sin \left({\frac {\sqrt {2\cdot m\cdot E}}{\hslash }}\cdot x\right)$. The second boundary condition, $\psi (a)=0$, may be applied.
$\psi (L)=A\cdot \sin \left({\frac {\sqrt {2\cdot m\cdot E}}{\hslash }}\cdot L\right)=0$
The value of A cannot be equal to zero, otherwise the value of the wavefunction would be equal to zero at any position of $x$. So, $\sin \left({\frac {\sqrt {2\cdot m\cdot E}}{\hslash }}\cdot L\right)=0$.
The sine function is periodic, therefore, there are multiple values of $x$ at which $\sin(x)=0$. The values of x in which this is true are integer multiples of $\pi$. Thus, ${\frac {\sqrt {2\cdot m\cdot E}}{\hslash }}\cdot L=n\pi$. This can be further simplified by noting that the sine function is symmetric, therefore, the negative value of some $n$ will have the same wavefunction as its positive value.
Additionally, substituting $n=0$ into our above equation would result in a value of 0 for the wavefunction at every position.
$\psi (x)=A\cdot \sin(0\cdot \pi \cdot x)=A\cdot \sin(0)=0$ Therefore, $n\geq 1$.
We can then plug ${\frac {\sqrt {2\cdot m\cdot E}}{\hslash }}={\frac {n\pi }{L}}$ back into our wavefunction equation, which gives $\psi (x)=A\cdot \sin \left({\frac {n\pi x}{L}}\right)$
The second property of a wavefunction is that it is normalized, meaning that the integral of the square of the wavefunction (i.e., the probability distribution) over all space must be equal to 1. We can now apply this to the above function to determine the constant value of $A$.
$\int _{0}^{L}(\psi (x))^{2}dx=1$
$\int _{0}^{L}A^{2}\sin ^{2}\left({\frac {n\pi x}{L}}\right)dx=1$
$A^{2}\int _{0}^{L}\sin ^{2}\left({\frac {n\pi x}{L}}\right)dx=1$
The tabulated solution to this integral gives a constant value for $A$ as
$A={\sqrt {\frac {2}{L}}}$
Therefore, the finalized specific solution of the wavefunction for a particle in a one-dimensional box is as follows.
$\psi (x)={\sqrt {\frac {2}{L}}}\sin \left({\frac {n\pi x}{L}}\right)$, where $n=1,2,3...$

## Example

The wavefunction of an electrion moving along a platinum wire with length $a$ is given by $\psi (x)=B(x^{3}-3x^{2}+2x)$, where $x$ is the position of the electron. Outside the boundaries of this wire, the value of the wavefunction is zero ($\psi (x)=0$). Determine the equation of the wavefunction of this electron by calculating $B$, the normalization constant, in terms of $a$. Is this normalized wavefunction a valid wavefunction? Explain why or why not.
As previously mentioned, one of the properties of a wavefunction is that it must be normalized. When normalizing the wavefunction, a normalization constant appears. Therefore, to determine the normalization constant $B$ for this wavefunction, we must integrate over all space and set it equal to 1.
$\int _{0}^{L}(\psi (x))^{2}dx=1$
In this case, the length of the wire is an arbitrary constant $a$. Thus, $L=a$. Additionally, the equation for the wavefunction is $B(x^{3}-3x^{2}+2x)$, which can be substituted in for $\psi (x)$.
$\int _{0}^{a}(B(x^{3}-3x^{2}+2x))^{2}dx=1$
$B^{2}\int _{0}^{a}x^{6}+9x^{4}+4x^{2}dx=1$
$B^{2}\left[{\frac {x^{7}}{7}}+{\frac {9x^{5}}{5}}+{\frac {4x^{3}}{3}}\right]_{0}^{a}=1$
$B^{2}\left({\frac {a^{7}}{7}}+{\frac {9a^{5}}{5}}+{\frac {4a^{3}}{3}}\right)=1$
$B^{2}={\frac {1}{\frac {15a^{7}+329a^{3}}{105}}}$
$\therefore B={\sqrt {\frac {105}{15a^{7}+329a^{3}}}}$
To ensure that this is a valid wavefunction, one of two properties have been satisfied: the function is normalized. We must now ensure that the boundary conditions are valid as well (i.e., $\psi (0)=0$ and $\psi (a)=0$).
$\psi (0)={\sqrt {\frac {105}{15a^{7}+329a^{3}}}}\cdot 0=0$
The first boundary condition is valid.
$\psi (a)={\sqrt {\frac {105}{15a^{7}+329a^{3}}}}(a^{3}-3a^{2}+2a)\neq 0$
The value of the wavefunction at $x=a$ will never be equal to zero, since the length of the wire is always a nonzero value. Thus, because this wavefunction disobeys the second property of $\psi (0)=0$ and $\psi (a)=0$, it is not considered a valid wavefunction.

# Example 33
## 1. Background Information[edit | edit source]
#### 1.1 Wave Function[edit | edit source]

Wave is a verb meaning "to move back and forth". Oscillations happen when a build-up of an opposing force, such water waves, springs, or pendulums, counteracts displacements in either direction. The reason it's called a "wave" function is because it satisfies a wave equation, known as the Schrödinger equation. This equation describes how the wave function evolves over time, much like how classical wave equations describe the propagation of waves in classical physics. So, the term "wave function" reflects both the wave-like behavior of quantum entities and the mathematical form of the function, which satisfies a wave equation. The Schrödinger equation is also 2nd order; for free electron, it has the same oscillatory “wave” solution and is given by;
${\frac {\ d^{2}\Psi (x)}{dx^{2}}}=-{\frac {\ 2mE}{\hbar ^{2}}}*\Psi (x)$

#### 1.2 Free Electron[edit | edit source]

An electron is said to be free if it can move along the x-axis in any direction without any force acting on it (i.e., 𝒱 𝑥 = 0, 𝑥 = −∞, ∞ ). It is not bound by an external force, or equivalently not in a region where its potential energy varies. In quantum mechanics, it indicates that the particle is in a region of uniform potential, which is often set to zero in the area of interest because any point in space can have an arbitrary potential set to zero.
The wave function of a free electron is typically represented as a plane wave, which is a solution to the Schrödinger equation. The Schrödinger equation is a fundamental equation in quantum mechanics that describes how the quantum state of a physical system changes with time. The wave function of a free electron can be written as:
$\psi (x,t)=A*e^{i(kx-wt))$
where:
- is the wave function, which depends on position x and time t,
- is the amplitude of the wave,
- k is the wave number, which is related to the momentum of the electron,
- is the angular frequency, which is related to the energy of the electron,
- is the imaginary unit.

## Example
#### 2.1 Derive the wave function of a free electron[edit | edit source]

Time-independent Schrödinger wave equation is given by;
${\widehat {H}}\Psi =E\psi$
$({\hat {\mathrm {T} }}+{\widehat {V}})\psi =E\psi$
Next define the Hamiltonian,
${\widehat {H}}={\widehat {T}}+{\widehat {V}}$
Substitute the potential energy operator
${\widehat {V}}=V(x)=0$ (free electron)
and the kinetic energy operator
${\widehat {T}}=$ ${\frac {-\hbar ^{2}}{2m}}{\frac {\ d^{2}}{dx^{2}}}$
$({\hat {\mathrm {T} }}+{\widehat {V}})\psi =E*\psi$
To obtain the Schrödinger equation for a free particle, rearrange the differential equation; isolate 2nd derivative on LHS
$-{\frac {\hbar ^{2}}{2m}}{\frac {\ d^{2}\Psi (x)}{dx^{2}}}=E\Psi (x)$
and make the substitution
$k^{2}={\frac {2mE}{\hbar ^{2}}}$
Since is the kinetic energy,
$\mathrm {E} ={\frac {p^{2}}{2m}}$
p and the wave vector k are related, p = $\hbar$k
we also could recognize that ${\frac {2mE}{\hbar ^{2}}}$ is just $k^{2}$as shown in this equation
${\frac {2mE}{\hbar ^{2}}}=({\frac {2m}{\hbar ^{2}}})({\frac {p^{2}}{2m}})$
$=({\frac {2m}{\hbar ^{2}}})({\frac {\hbar ^{2}k^{2}}{2m}})$
$=k^{2}$
after rearranging and substitution of result from equation, the result is
$({\frac {d^{2}}{dx^{2}}}+k)\psi (x)=$ $0$
This linear second-order differential equation can be solved by separating into two factors, and each is set equal to 0.$({\frac {d^{2}}{dx^{2}}}+k^{2})$$\psi (x)$$=$ $({\frac {d}{dx}}+ik)$ $({\frac {d}{dx}}-ik)$$\psi (x)$$=$$0$Equation is true if either
$({\frac {d}{dx}}+ik)$$\psi (x)$$=$$0$
$({\frac {d}{dx}}-ik)$$\psi (x)$$=$$0$
By concurrently rearranging and denoting the two equations and their solutions with a + and a - sign, the result is
${\frac {d\psi _{\pm }(x)}{\psi _{\pm }(x)}}$ $=$ $\pm$ $ik$ $dx$
which translates to
$ln$ ${\psi _{\pm }(x)}$ $=$ $\pm$ $ikx$ $+$${C_{\pm }(x)}$
and finally
${\psi _{\pm }(x)}$$=$ $A_{\pm }$$e^{i}$$^{k}$$^{x}$
Any function that satisfies the requirements of wave functions and is a solution to this differential equation is a valid wave function for this system.

## Reference[edit | edit source]

* Libretexts. (2022, April 21). 5.1: The Free Particle. Chemistry LibreTexts. https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Book_Quantum_States_of_Atoms_and_Molecules_(Zielinksi_et_al)/_Translational_States/_The_Free_Particle
* Anderson, J.M. Introduction to Quantum Chemistry, 1969, W.A. Benjamin, Inc, pg.81-91.
* ↑ Feller, S. E.; Blaich, C. F. Error Estimates for Fitted Parameters: Application to HCl/DCl Vibrational/Rotational Spectroscopy, J. Chem. Educ., 2001, 78 (3), 409. DOI: 10.1021/ed078p409