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