Answer to Question #130027 in Molecular Physics | Thermodynamics for Saranya

Question

The equilibrium internuclear distance of molecule CO is 113.53micrometer.calculate energy of molecule for j=1; C=12; O=16

Answer 1

In general, molecules have 3 quasi-independent motions - motion of electrons in molecule, rotations and oscillations. Each of the motion is described by certain quantum number. According to the information given in task, we need to calculate rotational energy of molecule.

$E_{rot} = \frac{L^2}{2I}$ where $L = \hbar \sqrt{J(J+1)}$ - rotational momentum and $I = \mu R_0^2$ - moment of inertia for 2-atomic molecule with reduced mass $\mu$ and internuclear distance $R_0$.

$E_{rot}=\frac{\hbar^2 J(J+1)}{2\mu R_0^2}$

$\mu = \frac{12*16}{12+16} = 6.86$ a.m.u. = $6.86*1.66*10^{-27}$ kg = $11.39*10^{-27}$ kg.

$E_{rot}=\frac{(1.055 *10^{-34})^2 1(1+1)}{2 *11.39*10^{-27} (113.53*10^{-6})^2}= 7.58 * 10^{-6}*10^{-29} = 7.58 * 10^{-35}$ J.

Note: This energy seems to be really small; this is due to conditions of this task. The real distance between nuclei in molecules is much much smaller, around $10^{-10}$ m. This makes $E_{rot}$ 12 orders of magnitude bigger. Real molecules have $E_{rot} \approx 10^{-5}-10^{-3}$ eV.