The inverse square law in electrostatics is $|\vec F| = \frac{{{e^2}}}{{4\pi { \in _0}{r^2}}}$ for the force between an electron and a proton. The $\frac{1}{r^2}$ dependence of $|\vec F|$ can be understood in quantum theory as being due to the fact that the particle of light (photon) is massless. If photons had a mass $m_p$,the force would be modified to $|\vec F| = \frac{{{e^2}}}{{4\pi { \in _0}}}\left( {\frac{1}{{{r^2}}} + \frac{\lambda }{r}} \right)\left( {{e^{ - \lambda r}}} \right)$ where $\lambda = \frac{{{m_p}c}}{\hbar }$ and $\hbar = \frac{h}{{2\pi }}$. Estimate the change in the ground state energy of a $H$-atom if $m_p$ were $10^{-6}$ times the mass of an electron.

(A) Given $\lambda = \frac{m_p c}{\hbar} = \frac{2 \pi m_p c}{h}$. With $m_p = 10^{-6} m_e$,$\lambda = \frac{2 \pi (10^{-6} \times 9.1 \times 10^{-31} \text{ kg}) (3 \times 10^8 \text{ m/s})}{6.63 \times 10^{-34} \text{ J s}} \approx 2.58 \times 10^6 \text{ m}^{-1}$.
Since the Bohr radius $r_B \approx 5.3 \times 10^{-11} \text{ m}$,we have $\lambda r_B \approx 1.37 \times 10^{-4} \ll 1$.
Using the modified potential $U(r) = -\frac{e^2}{4\pi \epsilon_0} \frac{e^{-\lambda r}}{r}$,we expand the exponential for small $\lambda r$: $U(r) \approx -\frac{e^2}{4\pi \epsilon_0} \frac{1-\lambda r}{r} = -\frac{e^2}{4\pi \epsilon_0} (\frac{1}{r} - \lambda)$.
The change in potential energy is $\Delta U = U_{modified} - U_{coulomb} = \frac{e^2}{4\pi \epsilon_0} \lambda$.
Substituting $\frac{e^2}{4\pi \epsilon_0} = k e^2 \approx 1.44 \text{ eV nm} = 1.44 \times 10^{-9} \text{ eV m}$,the energy shift $\Delta E \approx \Delta U = (1.44 \times 10^{-9} \text{ eV m}) (2.58 \times 10^6 \text{ m}^{-1}) \approx 3.7 \times 10^{-3} \text{ eV}$.