The necessary centripetal force for the rotation of electron is supplied by the electrostatic force between the electron and nucleus. $\frac{mv^2}{r} = \left(\frac{1}{4\pi\varepsilon_0}\right)\left(\frac{e^2}{r^2}\right) [\text{putting } Z=1]$ Or, $mv^2 = \frac{e^2}{4\pi\varepsilon_0 r} \dots (i)$ From Bohr's theory, $mvr = \frac{nh}{2\pi}$$\therefore v = \frac{nh}{2\pi m r}$ Putting in equation (i) $m\left(\frac{nh}{2\pi m r}\right)^2 = \frac{e^2}{4\pi\varepsilon_0 r}$ Or, $r = \frac{\varepsilon_0 n^2 h^2}{\pi m e^2}$ In general, $r_n = \frac{\varepsilon_0 n^2 h^2}{\pi m e^2}$ $\therefore r_n \propto n^2$ (b) $E_n \propto -\frac{1}{n^2}$ Energy is minimum at $n=1$ As $n$ increases, the enrgy becomes less negative, which means energy increases.