A) \[{{\operatorname{r}}_{\operatorname{n}}}={{\operatorname{a}}_{0}}\operatorname{n}-\beta \]
B) \[{{\operatorname{r}}_{\operatorname{n}}}={{\operatorname{a}}_{0}}{{\operatorname{n}}^{2}}+\beta \]
C) \[{{\operatorname{r}}_{\operatorname{n}}}={{\operatorname{a}}_{0}}{{\operatorname{n}}^{2}}-\beta \]
D) \[{{\operatorname{r}}_{\operatorname{n}}}={{\operatorname{a}}_{0}}\operatorname{n}+\beta \]
Correct Answer: C
Solution :
As\[F=\frac{m{{v}^{2}}}{r}=\frac{{{e}^{2}}}{4\pi {{\in }_{0}}}\left( \frac{1}{{{r}^{2}}}+\frac{B}{{{r}^{3}}} \right)\] and\[mvr=\frac{nh}{2\pi }\Rightarrow v=\frac{nh}{2\pi mr}\] \[\therefore \]\[m{{\left( \frac{nh}{2\pi mr} \right)}^{2}}\times \frac{1}{r}=\frac{{{e}^{2}}}{4\pi {{\in }_{0}}}\left( \frac{1}{{{r}^{2}}}+\frac{B}{{{r}^{3}}} \right)\] or,\[\frac{1}{{{r}^{2}}}+\frac{B}{{{r}^{3}}}=\frac{m{{n}^{2}}{{h}^{2}}4\pi {{\in }_{0}}}{4{{\pi }^{2}}{{m}^{2}}{{e}^{2}}{{r}^{3}}}\] or \[\frac{{{a}_{0}}{{n}^{2}}}{{{r}^{3}}}=\frac{1}{{{r}^{2}}}+\frac{B}{{{r}^{3}}}\] \[\left( \because {{a}_{0}}=\frac{{{\in }_{0}}{{h}^{2}}}{m\pi {{e}^{2}}}\text{Given} \right)\]\[\therefore \]\[r={{a}_{0}}{{n}^{2}}-B\]
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