The inverse sequare law in electrostatic is `|F| = (e^(2))/((4piepsi_(0))r^(2))` for the force between an electron and a proton. The `(1/r)` dependence of `|F|` can be understood in quantum theo ry as being due to the fact that the particle of light (photon) is massless. If photons had a mass `m_(p)`, force would be modified to `|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.

Explanation

Here, we suppose mass of photon, `m_(p)=10^(-6)` electronic mass
i.e., `m_(p)=10^(-6)(0.5)MeV=5xx10^(-7)xx1.6xx10^(-13)J=0.8xx10^(-19)J`.
Now, `1/(lambda)=h/(m_(p)c)=(hc)/(m_(p)c^(2))=(10^(-34)xx3xx10^(8))/(0.8xx10^(-19))=4xx10^(-7)m` which is much larger than Bohr radius.
As `|vecF|=(e^(2))/(4pi in_(0))[1/(r^(2))+lambda/r] exp. (-lambdar).....(i) where lambda^(-1)=h/(m_(p)c)=4xx10^(-7)mgt gt r_(B)`
`:. lambdalt lt1/(r_(B)) or lambda_(B)lt lt1`
Now, `U(r)=(-e^(2))/(4pi in_(0)) (exp (-lambdar))/r......(ii)`
As `mvr=h. v=h/(mr). As exp (-lambdar)to1`, therefor, form (i) ` (mv^(2))/r=F=(e^(2))/(4pi in_(0))[1/(r^(2))+lambda/r]`
`:. m/r (h/(mr))^(2)=(e^(2))/(4pi in_(0))[1/(r^(2))+lambda/r]`
`(h^(2))/m=(e^(2))/(4pi in_(0)) r+lambdar^(2)........(iii)`
If `lambda=0, r_(B), then (h^(2))/m=(e^(2))/(4pi in_(0))r_(B)`.
As `lambda^(-1)gt gt r_(B)`, put `r=r_(B)+delta`
form (iii), `r_(B)=(r_(B)+delta)+lambda(r_(B)+delta)^(2)=r_(B)+delta+lambda(r_(B)^(2)+delta^(2)2delta r_(B))`
Neglecting `delta^(2)`, we get `0=lambda r_(B)^(2)+delta(1+2lambda r_(B)^(2))`
`delta=(-lambdar_(B)^(2))/(1+2lambdar_(B))=-lambdar_(B)^(2)(1-2lambda r_(B))`
`delta=-lambdar_(B)^(2) (As lambdar_(B)lt lt1)`
form (ii), `V(r)=(-e^(2))/(4pi in_(0)) (exp. (-lambda delta-lambdar_(B)))/(r_(B)+delta)=(-e^(2))/(4pi in_(0)) 1/(r_(B))[(1-delta/(r_(B))) (1-lambdar_(B))]=-27.2eV`
i.e., V(r) remai ns unchanged.
`K.E.=1/2mv^(2)=1/2m(h/(mr))^(2)=(h^(2))/(2m(r_(B)+delta)^(2)) = (h^(2))/(2mr_(B)^(2)) (1-(2delta)/(r_(B)))=(13.6eV)[1+2lambdar_(B)]`
Total energy `=(-e^(2))/(4pi in_(0)r_(B))+(h^(2))/(2mr_(B)^(2)) (1+2lambda r_(B))=-27.2+13.6(1+2lambdar_(B))eV.`
Change in ground state energy of hydrogen atom `=13.6xx2lambda_(B)eV=27.2lambdar_(B)eV`