A diatomic gas molecule consists of two atoms of massseparated by a fixed distance drotating about an axis as indicated in given figure. Assuming that its angular momentum is quantized as in the Bohr model for the hydrogen atom, find

1. The possible angular velocities.
2. The possible quantized rotational energies.

The magnitude of the angular momentum of the electron i its orbit is restricted to the values,

$\mathit{L}\mathbf{=}\frac{\mathbf{n}\mathbf{h}}{\mathbf{2}\mathbf{\pi }}$…… (1)

Where, Lis angular momentum, h is plank’s constant, and n is quantum number.

We have given that, there are two atoms having mass m.

Distance between both atom is d.

We have to assume that angular momentum is quantized.

(a)

By equation (1) we have,

$L=\frac{nh}{2\mathrm{\pi }}$

Where, Lis angular momentum, his plank’s constant, and n is quantum number.

Also, we know that angular momentum is given by,

$L=l\omega$

Where, $\omega$is angular velocity and l is moment of inertia.

By the following figure, we can see that

$$\begin{gathered} l=m{r}^2+m{r}^2 \\ =2m{r}^2 \\ \mathrm{Also},r=\frac{d}{2} \end{gathered}$$

Now, by substituting value of L and l in equation (1) we get

$$\begin{gathered} l{\omega }_n=\frac{nh}{2\pi } \\ \left(2m{r}^2\right){\omega }_n=\frac{nh}{2\pi } \\ \left(2m\frac{d^2}{4}\right){\omega }_n=\frac{nh}{2\pi }\left(\because r=\frac{d}{2}\right) \\ {\omega }_n=\frac{nh}{\pi m{d}^2}.......\left(2\right) \end{gathered}$$

Hence, the possible angular velocities are ${\omega }_n=\frac{nh}{\pi m{d}^2}$.

(b)

By the following figure, we can see that

$$\begin{gathered} l=m{r}^2+m{r}^2 \\ l=2m{r}^2.......\left(3\right) \\ \mathrm{Also},r=\frac{d}{2} \end{gathered}$$

Now, rotational energies is given by,

$$\begin{gathered} E_n=\frac{1}{2}l{{\omega }_n}^2 \\ =\frac{1}{2}\left(2m{r}^2\right){\left(\frac{nh}{{\mathrm{\pi md}}^2}\right)}^2\left(byequation\left(2\right)and\left(3\right)\right) \\ =\left(m\frac{d^2}{4}\right)\left(\frac{n^2{h}^2}{{\mathrm{\pi }}^2{\mathrm{m}}^2{\mathrm{d}}^4}\right)\left(\because r=\frac{d}{2}\right) \\ =\left(\frac{n^2{h}^2}{4{\mathrm{\pi }}^2{\mathrm{md}}^2}\right) \end{gathered}$$

Hence, the possible quantized rotational energies is $E_n=\frac{n^2{h}^2}{4{\mathrm{\pi }}^2{\mathrm{md}}^2}$.