The inverse square law in electrostatics is F= $\frac{{\mathit{e}}^2}{4\mathit{\pi }{\mathit{\epsilon }}_0{\mathit{r}}^2}$ for the force between an electron and proton . the (1/r) dependence of F can be understood in quantum theory as being due to fact the particles of light photon is massless. If photon had a mass m_p. force would be modified to F= $\frac{{\mathit{e}}^2}{4\mathit{\pi }{\mathit{\epsilon }}_0{\mathit{r}}^2}\left[\frac{1}{{\mathit{r}}^2}+\frac{\mathit{\lambda }}{\mathit{r}}\right]\mathbf{e}\mathbf{x}\mathbf{p}?(-\mathit{\lambda }\mathit{r})$  where $\mathit{\lambda }=\frac{{\mathit{m}}_{\mathit{p}}\mathit{c}\mathit{}}{\mathit{h}}$  and n= h/2 $\mathit{\pi }.$ estimate the change in the ground state energy of H atom if m_p were 10^-6 times the mass of electron.

This is a Long Answer Type Questions as classified in NCERT Exemplar

Explanation- m_p= 10^-6

M_pc²=10^{-6 $*electronmass*c^2$}

 = 0.8 $*{10}^{-19}J$

Wavelength associated with both of them is same

$\frac{h}{m_{p}c}=\frac{hc}{m_p{c}^2}=\frac{{10}^{-34}*3*{10}^8}{0.8*{10}^{-19}}=4*{10}^{-7}m$

U(r)=- $\frac{e^2}{4\pi {\epsilon }_o}\frac{\mathrm{e}\mathrm{x}\mathrm{p}?(-\lambda t)}{r}$

Mvr=h

V=h/mr

Mv²/r=( $\frac{e^2}{4\pi {\epsilon }_o}$ ) ( $\frac{1}{r^2}+\frac{\lambda }{r}$ )

$\frac{h^2}{m{r}^3}=\left(\frac{e^2}{4\pi {\epsilon }_o}\right)\left(\frac{1}{r^2}-\frac{\lambda }{r}\right)$

$\frac{h^2}{m}=\frac{e^2}{4\pi {\epsilon }_o}{r}_B$

$kinecticwillbedoubleofgroundstateso$

13.6 $*2=27.2eV$