In the Bohr's model of hydrogen-like atom the force between the...

As $F=\frac{m^2}{r}=\frac{e^2}{4 \pi \epsilon_0}\left(\frac{1}{r^2}+\frac{B}{r^3}\right)$ and $\mathrm{mvr}=\frac{\mathrm{nh}}{2 \pi} \Rightarrow \mathrm{v}=\frac{\mathrm{nh}}{2 \pi \mathrm{mr}}$ $\therefore \mathrm{m}\left(\frac{\mathrm{nh}}{2 \pi \mathrm{mr}}\right)^2 \times \frac{1}{\mathrm{r}}=\frac{\mathrm{e}^2}{4 \pi \epsilon_0}\left(\frac{1}{\mathrm{r}^2}+\frac{\mathrm{B}}{\mathrm{r}^3}\right)$ or, $\frac{1}{\mathrm{r}^2}+\frac{\mathrm{B}}{\mathrm{r}^3}=\frac{\mathrm{mn}^2 \mathrm{~h}^2 4 \pi \epsilon_0}{4 \pi^2 \mathrm{~m}^2 \mathrm{e}^2 \mathrm{r}^3}$ \begin{aligned} & \text{ or, } \frac{\mathrm{a}_0 \mathrm{n}^2}{\mathrm{r}^3}=\frac{1}{\mathrm{r}^2}+\frac{\mathrm{B}}{\mathrm{r}^3} \\ & \left(\because \mathrm{a}_0=\frac{\epsilon_0 \mathrm{~h}^2}{\mathrm{~m} \pi \mathrm{e}^2} \text{ Given }\right) \\ & \therefore \mathrm{r}=\mathrm{a}_0 \mathrm{n}^2-\mathrm{B} \end{aligned}