A diatomic molecule is made of two masses $m_1$ and $m_2$ which are separated by a distance $r$. If we…

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)
  1. $\frac{\left(m_1+m_2\right)^2 n^2 h^2}{2 m_1^2 m_2^2 r^2}$
  2. $\frac{n^2 h^2}{2\left(m_1+m_2\right) r^2}$
  3. $\frac{2 n^2 h^2}{\left(m_1+m_2\right) r^2}$
  4. $\frac{\left(m_1+m_2\right) n^2 h^2}{2 m_1 m_2 r^2}$

Solution

$r_1=\frac{m_2 r}{m_1+m_2} ; r_2=\frac{m_1 r}{m_1+m_2}$ $\left(l_1+l_2\right) \omega=\frac{n h}{2 \pi}=nh$ $K . E=\frac{1}{2}\left(l_1+l_2\right) \omega^2=\frac{n^2 h^2\left(m_1+m_2\right)}{2 m_1 m_2 r^2}$

Asked in: JEE Main 2012 (Offline)

Practice more Structure of Atoms and Nuclei questions on Aicharya