Holooly Plus Logo

Question 17.5.2: The internuclear distance (bond length) of carbon monoxide m......

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.

Step-by-Step
The 'Blue Check Mark' means that this solution was answered by an expert.
Learn more on how do we answer questions.
\begin{array}{l}r=1.13 \mathrm{~A}^{\circ}=1.13 \times 10^{-10} \mathrm{~m} \\  \\\mu=\frac{m_{1} m_{2}}{m_{1}+m_{2}}=\frac{1.99 \times 10^{-26} \mathrm{~kg} \times 2.66 \times 10^{-26} \mathrm{~kg}}{1.99 \times 10^{-26} \mathrm{~kg}+2.66 \times 10^{-26} \mathrm{~kg}}=1.14 \times 10^{-26} \mathrm{~kg}\end{array}

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}

Related Answered Questions

Question: 17.4.4

Verified Answer:

• A band at 3080 \mathrm{cm}^{–1} d...
Question: 17.4.3

Verified Answer:

CH_3COOH spectrum will have the f...