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

For `m_(p) = 10^(-6)` times, the mass of an electron , the energy associated with it is given by
`m_(p)c^(2) = 10^(-6) xx "electron mass" xx c^(2)`
`= 10^(-6) xx 0.5 MeV`
`= 10^(-6) xx 0.5 xx 1.6 xx 10^(-19)`
`= 0.8 xx 10^(-13) J`
The wavelength associated with is given by
`(h)/(m_(p)c) = (hc)/(m_(p)c^(2)) = (10^(-34) xx 3 xx 10^(8))/(0.8 xx 10^(-19))`
`~~ 4 xx 10^(-7) m gt gt` Bohr radius
`|F| = (e^(2))/(4piepsi_(0))[(1)/(r^(2)) + (lambda)/(r)]"exp"(-lambdar)`
where `lambda^(-1) = (h)/(m_(p)c) ~~ 4 xx 10^(-7)m gt gt r_(B)`
`:. lambda lt lt 1/(r_(B))` i.e., `lambdar_(B) lt lt 1`
`U(r) = -(e)/(4piepsi_(0)). ("exp"(-lambdar))/(r)`
`mvr = h :. v = (h)/(mr)`
Also, `(mv^(2))/(r) = ~~ ((e^(2))/(4piepsi_(0)))[(1)/(r^(2)) + (lambda)/(r)]`
`:. (h^(2))/(mr^(3)) = ((e^(2))/(4piepsi_(0)))[(1)/(r^(2))+(lambda)/(r)]`
`:. (h^(2))/(m) = ((e^(2))/(4piepsi_(0)))[r+pir^(2)]`
If `lambda = 0 , r = r_(B) = h/m.(4piepsi_(0))/(e^(2))`
`(h^(2))/(m) = (e^(2))/(4piepsi_(0)).r_(B)`
Since, `lambda^(-1) gt gt r_(B)`, put `r = r_(B) + delta`
`:. r_(B) = r_(B) + delta + lambda(r_(B)^(2) + delta^(2) + 2deltar_(B)),` neglect `delta^(2)`
or `0 = lambdar_(B)^(2) + delta(1 + 2lambdar_(B))`
`delta = (-lambdar_(B)^(2))/(1+2lambdar_(B)) ~~ lambdar_(B)^(2) (1+2lambdar_(B)) = - lambdar_(B)^(2)`
Since, `lambdar_(B) lt lt 1`
`:. V(r) = -(e^(2))/(4piepsi_(0)) . (exp(-lambdadelta - lambdar_(B)))/(r_(B)+delta)`
`:. V(r)= - (e^(2))/(4piepsi_(0))1/(r_(B))[(1-(delta)/(r_(B))).(1-lambdar_(B))]`
`~=(-27.2 eV)` remains unchanged
`KE = -(1)/(3) mv^(2) = 1/2m.(h^(2))/(mr^(2)) = (h^(2))/(2(r_(B)+delta)^(2)) = (h^(2))/(2r_(B)^(2)) (1-(2delta)/(r_(B)))`
`= (13.6eV) [1+2lambdar_(B)]`
Total energy `= - (e^(2))/(4piepsi_(0)r_(B))+(h^(2))/(2r_(B)^(2)) [1+2lambdar_(B)]`
`= - 27.2 + 13.6 [1+2lambdar_(B)] eV`
Change in energy ` = 13.6 xx 2lambdar_(B)eV = 27.2 lambdar_(B) eV`