# Quantum Chemistry/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.