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)

Question

Dr. Abubacker · Accepted Answer

(d) [image:13_image29.emf]
$m_{1}r_{1} = m_{2}r_{2}$
$r_{1} + r_{2} = r$
$	herefore r_{1} = \frac{m_{2}r}{m_{1} + m_{2}},r_{2} = \frac{m_{1}r}{m_{1} + m_{2}}$
$	herefore\varepsilon = \frac{1}{2}I\omega^{2}$
$= \frac{1}{2}\left( m_{1}r_{1}^{2} + m_{2}r_{2}^{2} ight).\omega^{2}$ …(i)
$mvr = \frac{nh}{2\pi} = I\omega$
$\Rightarrow \omega = \frac{nh}{2\pi l}$
$	herefore\varepsilon = \frac{1}{2}I.\frac{n^{2}h^{2}}{4\pi^{2}I^{2}}$
$= \frac{n^{2}h^{2}}{8\pi^{2}}.\frac{1}{\left( m_{1}r_{1}^{2} + m_{2}r_{2}^{2} ight)}$
$= \frac{n^{2}h^{2}}{8\pi^{2}}\frac{1}{m_{1}\frac{m_{2}^{2}r^{2}}{\left( m_{1} + m^{2} ight)^{2}} + m_{2}\frac{m_{1}^{2}r^{2}}{\left( m_{1} + m_{2} ight)^{2}}}$
$= \frac{n^{2}h^{2}}{8\pi^{2}r^{2}}\frac{\left( m_{1} + m_{2} ight)^{2}}{m_{1}m_{2}(m_{1} + m_{2})} = \frac{\left( m_{1} + m_{2} ight)n^{2}h^{2}}{8\pi^{2}r^{2}m_{1}m_{2}}$
$= \frac{\left( m_{1} + m_{2} ight)n^{2}\hslash^{2}}{2m_{1}m_{2}r^{2}}$

Q. 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)

Explanation

Reviewed by Dr. Abubacker

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