The inverse sequare law in electrostaticis |F| = (e^(2))/((4piepsi_(0))r^(2))for the force between an electron and a proton. The (1/r) dependenceof |F| can be understood in quantum theo ry as being due to the factthat the particle of light (photon) is massless. Ifphotonshad a mass m_(p), force would bemodifiedto |F| = (e^(2))/((4piepsi_(0))pi^(2)) [ (1)/(r^(2)) + (lambda)/(r)] .exp(-lambdar) where lambda = (m_(p)c)/(h) and h = (h)/(2pi). Estimate the change in the gound state energy of a H-atom if m_(p)were 10^(-6) times the mass of the electron.

`18.6lambdar_(B)`
`-27.2`
`27.2lambdar_B`
`-lambdar_b`

Solution :As `lambda=(m_p c)/ħ=(m_pc^2)/(ħc)=((10^(-6)m_e)c^2)/(ħc)`
`=(10^(-6)[0.51](1.6xx10^(-13)J))/((1.5xx10^(-34) Js)(3xx10^8 ms^(-1))`
`=0.26xx10^7 m^(-1) " " [because m_e c^2=0.51 MeV]`
`r_B` (Bohr's radius ) =0.51Å=`0.51xx10^(-10)m`
or `lambdar_B=(0.26xx10^7 m^(-1))(0.51xx10^(-10))=0.14xx10^(-3) " lt lt " 1`
Further , as |F|=`(e^2/(4piepsilon_0))[1/R^2+lambda/r]e^(-lambdar)`...(i)
and `|F|=(dU)/(dr), U_r=int|F|dr=(e^2/(4piepsilon_0))int ((lambdae^(-lambdar))/r+e^(-lambdar)/r^2) dr`
If `z=e^(-lambdar)/r=1/r(e^(-lambdar)),(dz)/(dr)=[1/r(e^(-lambdar))(-lambda)+(e^(-lambdar))(-1/r^2)]`
or `dz=-[(lambdae^(-lambdar))/r+e^(-lambdar)/r^2]dr`
Thus `int((lambdae^(-lambdar))/r+e^(-lambdar)/r^2)drimplies - intdz=- z=-e^(-lambdar)/r`
`=-(e^2/(4piepsilon_0))(e^(-lambdar)/r)`...(ii)
We know that, `mvr=ħimplies v=ħ/(MR)`, and `(mv^2)/r=F=(e^2/(4piepsilon_0))(1/r^2+lambda/r)`
[PUTTING `e^(-lambdar~~` 1 in eqn.(i)]
Thus , `(m/r)(ħ^2/(m^2r^2))=(e^2/(4piepsilon_0))(1/r^2+lambda/r)`
or `ħ^2/m=(e^2/(4piepsilon_0))(r^2+lambda/r)`
When `lambda=0` , `r=r_B` and `ħ^2/m=(e^2/(4piepsilon_0))r_B` ...(iv)From EQNS. (iii) and (iv) , `r_B=r+lambdar^2`
Let `r=r_B=delta` so that form (iii)
`r_B=(r_B+delta)+lambda(r_B^2+delta^2+2deltar_B)`
or `0=lambdar_B^2+delta(1+2lambdar_B)` (neglecting `delta^2`)
or `delta=-(lambdar_B^2)/((1+2lambdar_B))=(-lambdar_B^2)(1+2lambdar_B)^(-1)`

`=(-lambdar_B^2)(1-2lambdar_B)=-lambdar_B^2 " " (because lambdar_B lt lt 1)`
From eqn. (ii) `U_r=-(e^2/(4piepsilon_0))(e^(-lambda(r_B+delta)))/((r_B+delta))`
`=-(e^2/(4piepsilon_0)1/r_B)(1-delta/r_B)(1-lambdar_B)~~ - e^2/(4piepsilon_0r_B)=-27.2 eV`
`[because e^(-lambda(r_B+delta))~~1-lambda(r_B+delta)=1-lambdar_B-lambdadelta~~1-deltar_B]`
and `1/((r_B+delta))=1/(r_B(1+delta//r_B))=1/r_B(1+delta/r_B)^(-1) =1/r_B(1-delta/r_B)`
Further, KE of the electron, `K=1/2mv^2=1/2m(ħ^2/(m^2r^2))`
`=ħ^2/(2mr^2)=ħ^2/(2m(r_B+delta)^2)=ħ^2/(2mr_B^2+(1+delta//r_B)^2)`
`=(ħ^2/(2mr_B^2))(1+delta/r_B)^(-2)=(ħ^2/(2mr_B^2))(1-(2delta)/r_B)`
`=(13.6)(1+2lambdar_B)eV`
(as `ħ^2/(2mr_B^2)=13.6 eV and delta=-lambdar_B^2`)
Total ENERGY of H-atom in the ground state =final energy - initial energy
`=(-.^13.6+27.2lambdar_B)eV-(-13.6 eV)=(27.2 lambdar_B)eV`