A) \[{{r}_{n}}={{a}_{0}}n-\beta \]
B) \[{{r}_{n}}={{a}_{0}}{{n}^{2}}+\beta \]
C) \[{{r}_{n}}={{a}_{0}}{{n}^{2}}-\beta \]
D) \[{{r}_{n}}={{a}_{0}}n+\beta \]
Correct Answer: C
Solution :
[c] As \[F=\frac{m{{v}^{2}}}{r}=\frac{{{e}^{2}}}{4\pi {{\in }_{0}}}\left( \frac{1}{{{r}^{2}}}+\frac{\beta }{{{r}^{3}}} \right)\] and \[mvr=\frac{nh}{2\pi }\]\[\Rightarrow \]\[v=\frac{nh}{2\pi mr}\] \[\therefore \] \[m{{\left( \frac{nh}{2\pi mr} \right)}^{2}}\times \frac{1}{r}=\frac{{{e}^{2}}}{4\pi {{\in }_{0}}}\left( \frac{1}{{{r}^{2}}}+\frac{\beta }{{{r}^{3}}} \right)\] or, \[\frac{1}{{{r}^{2}}}+\frac{\beta }{{{r}^{3}}}=\frac{m{{n}^{2}}{{h}^{2}}4\pi {{\in }_{0}}}{4{{\pi }^{2}}{{m}^{2}}{{e}^{2}}{{r}^{3}}}\] or, \[\frac{{{a}_{0}}{{n}^{2}}}{{{r}^{3}}}=\frac{1}{{{r}^{2}}}+\frac{\beta }{{{r}^{3}}}\] \[\left( \because {{a}_{0}}=\frac{{{\in }_{0}}{{h}^{2}}}{m\pi {{e}^{2}}}\,\,\text{Given} \right)\] For \[{{n}^{th}}\]atom \[\therefore \] \[{{r}_{n}}={{a}_{0}}{{n}^{2}}-\beta \]
You need to login to perform this action.
You will be redirected in
3 sec
