The inverse square law in electrostatic is F=e2(4πε0)r2 for the force between an electron and a proton. The (1r) dependence of 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 mp, the force would be modified to F=e2(4πε0)r[1r2+λr]·exp(-λr) where λ=mpc ħ and  ħ=h2π. Estimate the change in the ground state energy of a H-atom if mp were 10-6 times the mass of an electron.

Hint: The electrostatic force provides the required centripetal force.
Step 1: Find the wavelength.
For mp=10-6 times the mass of an electron, the energy associated with it is given by-
mpc2=10-6×electron mass×c210-6×0.5 MeV10-6×0.5×1.6×10-230.8×10-19 J
The wavelength associated with it is given by:
 ħmpc= ħcmpc2=10-34×3×1080.8×10-194×10-7 m>>Bohr radius
Step 2: Find the potential energy.
F=e24πε0[1r2+λr]exp(-λr)where, λ-1= ħmpc4×10-7m>>rBλ<<1rB i.e., λrB<<1
U(r)=-e24πε0·exp(-λr)rmvr= ħ  v= ħmrAlso mv2r=(e24πε0)[1r2+λr] ħ2mr3=(e24πε0)[1r2+λr] ħ2m=(e24πε0)[r+πr2]
if λ=0; r=rB= ħm·4πε0e2 ħ2m=e24πε0·r
Since, λ-1>>r, put r=rB+δrB=rB+δ+λ(rB2+δ2+2δrB); neglect δ2or 0=λrB2+δ(1+2λrB)δ=-λrB21+2λrBλrB2(1-2λrB)=-λrB2
Since, λrB<<1V(r)=-e24πε0·exp(-λδ-λrB)rB+δV(r)=-e24πε01rB[(1-δrB)·(1-λrB)](-27.2eV) remains unchangedStep 3: Find the kinetic energy.KE=-12mv2=12m· ħ2m2r2= ħ22mrB+δ2= ħ22mrB21+δrB-2= ħ22mrB21-2δrB
Total energy=-e24πε0rB+ħ22mrB2[1+2λrB]=-27.2+13.6[1+2λrB]eV
Change in energy=1.36×2λrBeV=27.2λrBeV