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) \[\frac{2{{n}^{2}}{{h}^{2}}}{\left( {{m}_{1}}+{{m}_{2}} \right){{r}^{2}}}\]
D) \[\frac{\left( {{m}_{1}}+{{m}_{2}} \right){{n}^{2}}{{h}^{2}}}{2{{m}_{1}}{{m}_{2}}{{r}^{2}}}\]
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 \[I=\frac{1}{2}\left( {{m}_{1}}r_{1}^{2}+{{m}_{2}}r_{2}^{2} \right)\] Thus, \[E=\frac{1}{2}\left( {{m}_{1}}r_{1}^{2}+{{m}_{2}}r_{2}^{2} \right){{\omega }^{2}}\] ?.(i) \[E=\frac{1}{2}\left( {{m}_{1}}r_{1}^{2}+{{m}_{2}}r_{2}^{2} \right)\frac{{{L}^{2}}}{{{I}^{2}}}\] ?(ii) \[L=n\,\hbar \] (According Bohr's Hypothesis) \[E=\frac{1}{2}\left( {{m}_{1}}r_{1}^{2}+{{m}_{2}}r_{2}^{2} \right)\frac{{{L}^{2}}}{{{\left( {{m}_{1}}r_{1}^{2}+{{m}_{2}}r_{2}^{2} \right)}^{2}}}\] \[E=\frac{1}{2}\frac{{{L}^{2}}}{\left( {{m}_{1}}r_{1}^{2}+{{m}_{2}}r_{2}^{2} \right)}=\frac{{{n}^{2}}{{h}^{2}}}{8{{\pi }^{2}}{{r}^{2}}{{m}_{1}}m{{ & }_{2}}}\]\[E=\frac{1}{2}\frac{{{L}^{2}}}{\left( {{m}_{1}}r_{1}^{2}+{{m}_{2}}r_{2}^{2} \right)}=\frac{{{n}^{2}}{{h}^{2}}}{8{{\pi }^{2}}{{r}^{2}}{{m}_{1}}m{{ & }_{2}}}\] \[E=\frac{\left( {{m}_{1}}+{{m}_{2}} \right){{n}^{2}}{{h}^{2}}}{8{{\pi }^{2}}{{r}^{2}}m{{ & }_{1}}{{m}_{2}}}\]
You need to login to perform this action.
You will be redirected in
3 sec
