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 \[F=\frac{m{{v}^{2}}}{r}=\frac{{{e}^{2}}}{4\pi {{\varepsilon }_{0}}}\left( \frac{1}{{{r}^{2}}}+\frac{\beta }{{{r}^{3}}} \right)\] \[\text{and }mvr=\frac{nh}{2\pi }\Rightarrow v=\frac{nh}{2\pi mr}\] \[\therefore m{{\left( \frac{nh}{2\pi mr} \right)}^{2}}\times \frac{1}{r}=\frac{{{e}^{2}}}{4\pi {{\varepsilon }_{0}}}\left( \frac{1}{{{r}^{2}}}+\frac{\beta }{{{r}^{3}}} \right)\] or, \[\frac{{{a}_{0}}{{n}^{2}}}{{{r}^{3}}}=\frac{1}{{{r}^{2}}}+\frac{\beta }{{{r}^{3}}}\left( \therefore {{a}_{0}}=\frac{{{\varepsilon }_{0}}{{h}^{2}}}{m\pi {{e}^{2}}}\text{Given} \right)\] \[\therefore r={{a}_{0}}{{n}^{2}}-\beta \]You need to login to perform this action.
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