The internuclear distance (bond length) of carbon monoxide molecule is 1.13 Å. Calculate the energy (in joules and eV) of this molecule in the first excited rotational level. Also calculate the angular velocity of the molecule. Given atomic masses of ^{12}C=1.99×10^{-26} kg; ^{16}O = 2.66× 10^{-26} kg.
moment of inertia I=\mu r^{2}\\
\quad \quad \qquad I=1.14 \times 10^{-26} \mathrm{~kg} \times\left(1.13 \times 10^{-10} \mathrm{~m}\right)^{2}=1.46 \times 10^{-46} \mathrm{~kg} \mathrm{~m}^{2}Energy of a rigid diatomic molecule is given by
E_{J}=\frac{h^{2}}{8 \pi^{2} I} J(J+1) \text { Joule }For the first excited level, J=1; hence
\begin{array}{l}E_{1}=\frac{\left(6.626 \times 10^{-34}\right)^{2} \times 1(1+1)}{8 \times 3.14 \times 3.14 \times 1.46 \times 10^{-46} \mathrm{~kg} \mathrm{~m}^{2}} \\ \\ \quad=7.61 \times 10^{-23} \text { Joule }\\\end{array}\\since 1 eV = 1.602 × 10^{-19} J
E_1=\frac{7.61×10^{-23}}{1.602×10^{-19} J eV^{-1}}\\\quad =4.76 \times 10^{-4} \mathrm{eV}\\ \text { angular velocity } \omega=\sqrt{\frac{2 E_{1}}{I}}=\sqrt{\frac{2 \times 7.61 \times 10^{-23} \mathrm{~J}}{1.46 \times 10^{-46} \mathrm{~kg} \mathrm{~m}^{2}}}=3.32 \times 10^{11} \mathrm{~radians}^{-1}