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)

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

Answer 1

Answer: (D)

$m_1{r}_1=m_2{r}_2$
$r_1+r_2=r$
$\therefore r_1=\frac{m_{2}r}{m_1+m_2}$
$r_2=\frac{m_{2}r}{m_1+m_2}$
$\therefore \epsilon =\frac{1}{2}I{\omega }^2$
$=\frac{1}{2}\left(m_1{r}_1^2+m_2{r}_2^2\right).{\omega }^2............\left(i\right)$
$mvr=\frac{h}{2\pi }=I\omega$
$\omega =\frac{nh}{2\pi I}<br/>\epsilon =\frac{1}{2}I.\frac{n^2{h}^2}{4{\pi }^2{I}^2}$
$=\frac{n^2{h}^2}{8{\pi }^2}\frac{1}{\left(m_1{r}_1^2+m_2{r}_2^2\right)}$
$=\frac{n^2{h}^2}{8{\pi }^2}\frac{1}{m_1\frac{m_2^2{r}_0^2}{{\left(m_1+m_2\right)}^2}}+m_2\frac{m_1^2{r}^2}{{\left(m_1+m_2\right)}^2}$
$=\frac{n^2{h}^2}{8{\pi }^2{r}^2}\frac{{\left(m_1+m_2\right)}^2}{m_1{m}_2\left(m_1+m_2\right)}=\frac{\left(m_1{m}_2\right)n^2{h}^2}{8{\pi }^2{r}^2{m}_1{m}_2}$