In the Bohr's model of hydrogen-like atom, the force between the nucleus and the electron is modifie

Question

In the Bohr's model of hydrogen-like atom, the force between the nucleus and the electron is modified as $F=\frac{e^2}{4 \pi \varepsilon_0}\left(\frac{1}{r^2}+\frac{\beta}{r^3}ight)$, where $\beta$ is a constant. For this atom, the radius of the $n^{	ext {th }}$ orbit in terms of the Bohr radius $\left(a_0=\frac{\varepsilon_0 h^2}{m \pi e^2}ight)$ is

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(c) : $F=\frac{m v^2}{r}=\frac{e^2}{4 \pi \varepsilon_0}\left(\frac{1}{r^2}+\frac{\beta}{r^3}ight)$ Also, $m v r=\frac{n h}{2 \pi} \Rightarrow v=\frac{n h}{2 \pi m r}$ Put this value in equation (i), \ $ m\left(\frac{n h}{2 \pi m r}ight)^2 	imes \frac{1}{r}=\frac{e^2}{4 \pi \varepsilon_0}\left(\frac{1}{r^2}+\frac{\beta}{r^3}ight) ; \frac{1}{r^2}+\frac{\beta}{r^3}=\frac{m n^2 h^2 4 \pi \varepsilon_0}{4 \pi^2 m^2 e^2 r^3} $ \ Given $a_0=\frac{\varepsilon_0 h^2}{m \pi e^2}$; So, $\frac{a_0 n^2}{r^3}=\frac{1}{r^2}+\frac{\beta}{r^3} ; a_0 n^2=r+\beta$; \ For $n^{	ext {th }}$ atom, $r_n=a_0 n^2-\beta$