The Bohr model for the $H$-atom relies on Coulomb's law of electrostatics. Coulomb's law has not been directly verified for very short distances of the order of $\mathring{A}$. Suppose Coulomb's law between two opposite charges $+q_1$ and $-q_2$ is modified to $|\vec{F}| = \frac{q_1 q_2}{4\pi \epsilon_0} \left( \frac{1}{r^2} \right)$ for $r \ge R_0$ and $|\vec{F}| = \frac{q_1 q_2}{4\pi \epsilon_0} \left( \frac{1}{R_0^{2-\epsilon} r^{\epsilon}} \right)$ for $r < R_0$. Calculate the ground state energy of an $H$-atom,given $\epsilon = 0.1$ and $R_0 = 1 \,\mathring{A}$.

For $r < R_0$,the modified force is $F = \frac{k e^2}{R_0^{2-\epsilon} r^{\epsilon}}$.
Equating this to the centripetal force: $\frac{m v^2}{r} = \frac{k e^2}{R_0^{2-\epsilon} r^{\epsilon}}$.
Thus,$v^2 = \frac{k e^2}{m R_0^{2-\epsilon} r^{\epsilon-1}}$.
The Bohr quantization condition is $mvr = n\hbar$. For the ground state,$n=1$,so $v = \frac{\hbar}{mr}$.
Substituting $v^2 = \frac{\hbar^2}{m^2 r^2}$ into the force equation: $\frac{\hbar^2}{m r^3} = \frac{k e^2}{R_0^{2-\epsilon} r^{\epsilon}}$.
$r^{3-\epsilon} = \frac{\hbar^2 R_0^{2-\epsilon}}{m k e^2} = a_0 R_0^{2-\epsilon}$,where $a_0$ is the Bohr radius.
$r = (a_0 R_0^{2-\epsilon})^{1/(3-\epsilon)}$.
The total energy $E = K + U = \frac{1}{2} m v^2 + \int_{\infty}^{r} F dr$.
Calculating the potential energy $U = -\int_{\infty}^{R_0} \frac{k e^2}{r^2} dr - \int_{R_0}^{r} \frac{k e^2}{R_0^{2-\epsilon} r^{\epsilon}} dr$.
Evaluating these integrals gives the ground state energy $E$.