Question

A diatomic molecule is made of two masses $m_1$ and $m_2$ which are separated by a distance r. If we calculate its rotational energy by applying Bohr's rule of angular momentum quantization, its energy will be given by: (n is an integer) $(h=\frac{h}{2\pi })$

• A. $\frac{(m_1+m_2{)}^2{n}^2{h}^2}{2m_1^2{m}_2^2{r}^2}$
• B. $\frac{n^2{h}^2}{2(m_1+m_2)r^2}$
• C. $\frac{2n^2{h}^2}{2(m_1+m_2)r^2}$
• D. $\frac{(m_1+m_2)n^2{h}^2}{2m_1{m}_2{r}^2}$

Answer 1

The correct option is D $\frac{(m_1+m_2)n^2{h}^2}{2m_1{m}_2{r}^2}$

According to the problem:
$\frac{\left(m_1+m_2\right)n^2{h}^2}{2m_1{m}_2{r}^2}$

Rotational Kinetic energy of the two body system rotating about their centre o of mass is

$RKE=\frac{1}{2}\mu w^2{r}^2$

where $\mu =\frac{m_1{m}_2}{m_1+m_2}$=reduced mass

and angular momentum, $L=\mu w{r}^2=\frac{nh}{2\pi }$

$w=\frac{2n}{2\pi \mu r^2}$

$RKE=\frac{1}{2}\mu w^2{r}^2=\frac{1}{2}\mu .\left(\frac{nh}{2\pi \mu r^2}\right)r^2$

$\frac{n^2{h}^2}{8{\pi }^2\mu r^2}=\frac{n^2{h}^2}{2\mu r^2}n=\frac{h}{2\pi }$

$\frac{\left(m_1+m_2\right)n^2{h}^2}{2m_1{m}_2{r}^2}$