# Quantum Chemistry/Example 12

Write a question and its solution that shows the specific selection rule for a quantum rigid rotor.

Consider an N2 molecule with a bond length of 1.09Å.

(a) Calculate the energy at the quantum number 3 using the specific selection rule for a rigid rotor.

Solution: This bond length is given (1.09Å), and the reduced mass (μ) and the inertia (I) can be calculated to determine the energy at quantum number (J) 3. The bond length must be changed to the SI unit of m.

$\mu ={\frac {\mu _{N}\mu _{N}}{\mu _{N}+\mu _{N}}}$ $={\frac {14.0067u\cdot 14.0067u}{14.0067u+14.0067u}}\cdot ({\frac {1.6605\times 10^{-}27kg}{1u}})$ $=1.1629\times 10^{-26}kg$ Now, calculating the inertia (I) value:

$I=\mu \cdot (r_{e})^{2}$ $=(1.1629\times 10^{-26}kg)\cdot (1.09\times 10^{-10}m)^{2}$ $=1.3816\times 10^{-46}kg\cdot m^{2}$ Using the inertia (I) the energy can be calculated with it's relation to the quantum number of rotational spectroscopy:

$E=2(J+1)\cdot ({\frac {\hbar ^{2}}{2I}})$ $=2(3+1)\cdot ({\frac {(1.05457\times 10^{-34}J\cdot s)^{2}}{2(1.3816\times 10^{-46}kg\cdot m^{2})}})$ $=8\cdot (4.0247\times 10^{-23}J)$ $=3.2197\times 10^{-23}J$ (b) Calculate the quantum number if the translational energy is 2.415 × 10-22J. In reference to part (a), does this value adhere to the specific selection rule? Why or why not?

$J={\frac {E\cdot I}{\hbar ^{2}}}-1$ $={\frac {(2.415\times 10^{-22}J)\cdot (1.3816\times 10^{-46}kg\cdot m^{2})}{(1.05457\times 10^{-34}J\cdot s)^{2}}}-1$ $=2$ This transitional change does adhere to the specific selection rule for a quantum rigid rotor, because the transitional change is within the $\Delta J=\pm 1$ .

(c) Calculate the quantum number if the translational energy is 4.830 × 10-22J. In reference to part (a), does this value adhere to the specific selection rule? Why or why not?

$J={\frac {E\cdot I}{\hbar ^{2}}}-1$ $={\frac {(4.830\times 10^{-22}J)\cdot (1.3816\times 10^{-46}kg\cdot m^{2})}{(1.05457\times 10^{-34}J\cdot s)^{2}}}-1$ $=5$ This transitional change does not adhere to the specific selection rule for a quantum rigid rotor, because the transitional change is not within the $\Delta J=\pm 1$ . Compared to part (a), it has a change of $\Delta J=\pm 2$ .