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) \[\frac{({{m}_{1}}+{{m}_{2}}){{n}^{2}}{{h}^{2}}}{2{{m}_{1}}{{m}_{2}}{{r}^{2}}}\]
Correct Answer: D
Solution :
[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) |
| \[E=\frac{1}{2}({{m}_{1}}{{r}_{1}}^{2}+{{m}_{2}}{{r}_{2}}^{2})\frac{{{L}^{2}}}{{{({{m}_{1}}{{r}_{1}}^{2}+{{m}_{2}}r_{2}^{2})}^{2}}}\] |
| \[E=\frac{1}{2}\frac{{{L}^{2}}}{({{m}_{1}}{{r}_{1}}^{2}+{{m}_{2}}{{r}_{2}}^{2})}=\frac{{{n}^{2}}{{h}^{2}}}{8{{\pi }^{2}}{{({{m}_{1}}{{r}_{1}}^{2}+{{m}_{2}}r_{2}^{2})}^{2}}}\] |
| \[E=\frac{({{m}_{1}}+{{m}_{2}}){{n}^{2}}{{h}^{2}}}{8{{\pi }^{2}}{{r}^{2}},{{m}_{1}}{{m}_{2}}}\Rightarrow \frac{({{m}_{1}}+{{m}_{2}}){{n}^{2}}{{h}^{2}}}{2{{m}_{1}}{{m}_{2}}{{r}^{2}}}\] |
| \[\left[ \because {{r}_{1}}=\frac{{{m}_{2}}r}{{{m}_{1}}+{{m}_{2}}};{{r}_{2}}=\frac{{{m}_{2}}r}{{{m}_{1}}+{{m}_{2}}} \right]\] |
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