question_answer

In the Bohr’s model of hydrogen- like atom the force between the nucleus and the electron is modified as \[\operatorname{F}=\frac{{{e}^{2}}}{4\pi {{\in }_{0}}}\left( \frac{1}{{{\operatorname{r}}^{2}}}+\frac{\beta }{{{\operatorname{r}}^{3}}} \right),\] Where \[\beta \]is a constant. For this atom, the radius of the \[{{\operatorname{n}}^{\operatorname{th}}}\] orbit in terms of the Bohr radius \[\left( {{a}_{0}}\frac{{{\in }_{0}}{{\operatorname{h}}^{2}}}{\operatorname{m}\pi {{e}^{2}}} \right)\] is   [JEE ONLINE 23-04-2013]

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 :

[c] As

and

or,

or