Question

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.

Short answer

1. The possible angular velocities are${\omega }_n=\frac{nh}{\pi m{d}^2}$.
2. The possible quantized rotational energies are $E_n=\frac{n^2{h}^2}{4{\pi }^{2}m{d}^2}$.

Step-by-step solution

The Bohr model for Hydrogen.

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.

Identification of the given data.

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.

Finding the possible angular velocities.

(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 ${\omega }_n=\frac{nh}{\pi m{d}^2}$.

Finding the possible quantized rotational energies.

(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 $E_n=\frac{n^2{h}^2}{4{\mathrm{\pi }}^2{\mathrm{md}}^2}$.