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)

$\frac{\left(m_{1}+m_{2}\right)n^{2}h^{2}}{2m_{1}m_{2}r^{2}}$

Answer 2

$m_{1}r_{1}=m_{2}r_{2}$ $r_{1}+r_{2}=r$ $\therefore r_{1}=\frac{m_{2}r}{m_{1}+m_{2}}$ $r_{2}=\frac{m_{2}r}{m_{1}+m_{2}}$ $\therefore \varepsilon=\frac{1}{2}I\omega^{2}$ $=\frac{1}{2}\left(m_{1}r^{2}_{1}+m_{2}r_{2}^{2}\right).\omega^{2} ............\left(i\right)$ $mvr=\frac{h}{2\pi}=I\omega$ $\omega=\frac{nh}{2\pi I} \,\varepsilon = \frac{1}{2}I. \frac{n^{2}h^{2}}{4\pi^{2}I^{2}}$ $=\frac{n^{2}h^{2}}{8\pi ^{2}} \frac{1}{\left(m_{1}r^{2}_{1}+m_{2}r^{2}_{2}\right)}$ $=\frac{n^{2}h^{2}}{8\pi ^{2}} \frac{1}{m_{1} \frac{m^{2}_{2}r^{2}_{0}}{\left(m_{1}+m_{2}\right)^{2}} }+m_{2} \frac{m^{2}_{1}r^{2}}{\left(m_{1}+m_{2}\right)^{2}}$ $=\frac{n^{2}h^{2}}{8\pi^{2}r^{2}} \frac{\left(m_{1}+m_{2}\right)^{2}}{m_{1}m_{2}\left(m_{1}+m_{2}\right)}=\frac{\left(m_{1}m_{2}\right)n^{2}h^{2}}{8\pi^{2}r^{2}m_{1}m_{2}}$