# Quantum Chemistry/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,
$f={\frac {3.01\times 10^{-33}{\text{ J}}}{6.6261\times 10^{-34}{\text{ J}}{\text{ s}}}}$
$f=4.54{\text{s}}^{-1}=4.54{\text{Hz}}$