• # question_answer 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) $\frac{{{({{m}_{1}}+{{m}_{2}})}^{2}}{{n}^{2}}{{h}^{2}}}{2m_{1}^{2}m_{2}^{2}{{r}^{2}}}$        B) $\frac{{{n}^{2}}{{h}^{2}}}{2({{m}_{1}}+{{m}_{2}}){{r}^{2}}}$ C) $\frac{2{{n}^{2}}{{h}^{2}}}{({{m}_{1}}+{{m}_{2}}){{r}^{2}}}$                        D) $\frac{({{m}_{1}}+{{m}_{2}}){{n}^{2}}{{h}^{2}}}{2{{m}_{1}}{{m}_{2}}{{r}^{2}}}$

[d]  The energy of the system of two atoms of diatomic molecule, $E=\frac{1}{2}I{{\omega }^{2}}$ Where $I=$moment of inertia, $\omega =$Angular velocity$=\frac{L}{I},$ $L=$Angular momentum $I=\frac{1}{2}({{m}_{1}}{{r}_{1}}^{2}+{{m}^{2}}r_{2}^{2})$ Thus $E=\frac{1}{2}({{m}_{1}}{{r}_{1}}^{2}+{{m}_{2}}r_{2}^{2}){{\omega }^{2}}$ $E=\frac{1}{2}({{m}_{1}}{{r}_{1}}^{2}+{{m}_{2}}r_{2}^{2})\frac{{{L}^{2}}}{{{I}^{2}}}$ $L=n\frac{nh}{2n}$(According Bohr?s Hypothesis) $E=\frac{1}{2}({{m}_{1}}{{r}_{1}}^{2}+{{m}_{2}}{{r}_{2}}^{2})\frac{{{L}^{2}}}{{{({{m}_{1}}{{r}_{1}}^{2}+{{m}_{2}}r_{2}^{2})}^{2}}}$ $E=\frac{1}{2}\frac{{{L}^{2}}}{({{m}_{1}}{{r}_{1}}^{2}+{{m}_{2}}{{r}_{2}}^{2})}=\frac{{{n}^{2}}{{h}^{2}}}{8{{\pi }^{2}}{{({{m}_{1}}{{r}_{1}}^{2}+{{m}_{2}}r_{2}^{2})}^{2}}}$ $E=\frac{({{m}_{1}}+{{m}_{2}}){{n}^{2}}{{h}^{2}}}{8{{\pi }^{2}}{{r}^{2}},{{m}_{1}}{{m}_{2}}}\Rightarrow \frac{({{m}_{1}}+{{m}_{2}}){{n}^{2}}{{h}^{2}}}{2{{m}_{1}}{{m}_{2}}{{r}^{2}}}$ $\left[ \because {{r}_{1}}=\frac{{{m}_{2}}r}{{{m}_{1}}+{{m}_{2}}};{{r}_{2}}=\frac{{{m}_{2}}r}{{{m}_{1}}+{{m}_{2}}} \right]$