The inverse square law in electrostatic is `|F|=(e^(2))/((4pi epsi_(0))*r^(2))` for the force between an electrona nd a proton. The `((1)/(r))` 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 m force would be modified to `|F|=(e^(2))/((4pi epsi_(0))r^(2))[(1)/(r^(2))+(lambda)/(r)] exp(-lambdar)` where `lambda=(m_(p)c)/(h) and h=(h)/(2pi)`. Estimate the change in the ground state energy of a H-atom if `m_(p)` were `10^(-6)` times the mass of an electron

Explanation

As per the statemen,
`lambda=(m_(p)c)/(h)=(2pi m_(p)c)/(h)=(2pi (10^(-6)m_(e))c)/(h)`
`:. Lambda=(2xx3.14xx10^(-6)xx9.1xx10^(-31)xx3xx10^(8))/(6.625xx10^(-34))`
`:. Lambda=2.588xx10^(6)m^(-1)...(1)`
Note here `lambda=(m_(p)c)/(h)` is not wavelength but it is only constant. (Because its unit in `m^(-1))`.
Now Bohar radius `r_(B)=0.53Å=5.3xx10^(-11)m`
`:. lambda gt gt r_(B) ...(2)`
According to statement,
`|vecF|=(e^(2))/(4pi epsi_(0)) ((1)/(r^(2))+(lambda)/(r))e^(-lambdar) ....(3)`
Since `vecF` is a conservative force,
`|vecF|=(dU)/(dr)`
`:.dU=|vecF|dr`
`:.U= int|vecF|dr= int (e^(2))/(4pi epsi_(0))((1)/(r^(2))+(lambda)/(r))e^(-lambdar)dr ....(4)`
Suppose `x=(e^(-lambdar))/(r)=(1)/(r)xxe^(-lambdar)`
`:.(dx)/(dr)=(1)/(r) (e^(-lambdar)) (-lambda)+e^(lambda-r)(-(1)/(r^(2)))`
`:.dx=(-(lambda)/(r)e^(-lambdar)-(e^(-lambdar))/(r^(2))dr`
`:.dx=-((lambda^(-lambdar))/(r)+(e^(-lambdar))/(r^(2)))dr`
`rArr int (lambdae^(lambda-r))/(r)+(e^(-lambdar))/(r^(2))-dr=- intdx=-x=-(e^(-lambdar))/(r)....(5)`
From equation (4) and (5),
`U=(e^(2))/(4pi epsi_(0))(-(e^(-lambdar))/(r))`
`:.U-(e^(2))/(4pi epsi_(0))((e^(-lambdar))/(r))....(6)`
Now, according to Bohr.s first postulate, for n=1
`mvr=(h)/(2pi)`
`:.v=(h)/(2pi mr)....(7)`
Here
`F=(e^(2))/(4pi epsi_(0))((1)/(r^(2))+(lambda)/(r))e^(-lambdar)`
`:.(mv^(2))/(r)=(e^(2))/(4pi epsi_(0))((1)/(r^(2))+(lambda)/(r)) ( :. e^(-lambdar)~~1)`
`:. (m)/(r)xx(h^(2))/(4pi^(2)m^(2)r^(2))=(e^(2))/(4pi epsi_(0))((1)/(r^(2))+(lambda)/(r))` [ From equation (7)]
`:. (h^(2))/(4pi^(2)mr^(3))=(e^(2))/(4pi epsi_(0))((1)/(r^(2))+(lambda)/(r))`
`:.(h^(2))/(4pi^(2)m)=(e^(2))/(4pi epsi_(0))(r+lambdar^(2)) ...(8)`
Since no radiation is emitted for `r=r_(B)`, taking `lambda=0` from above equation `m_(p)=0`
`(h^(2))/(4pi^(2)m)=(e^(2))/(4pi epsi_(0))(r_(B)) .....(9)`
Comparing equation (8) and (9),
`r_(B)=r+lambda r^(2).....(10)`
Suppose `r=r_(B)+dr ....(11)` (where dr is diffrentially small)
Now,
`(e^(-lambdar))/(r)=(e^(-lambda(r_(B)+dr)))/(r_(B)+dr)`
`=(1)/(r_(B)+dr){1-lambda (r_(B)+dr)}`
`=(1-lambdar_(B)-lambdadr)/(r_(B)+dr)`
`~~(1-lambda r_(B))/(r_(B)(1+(dr)/(r_(B)))) ( :. lambda dr` is negligible)
`:.(e^(-lambdar))/(r)=((1-lambdar_(B))/(r_(B)))(1+(dr)/(r_(B)))^(-1)`
`=((1-lambdar_(B))/(r_(B)))(1-(dr)/(r_(B)))`
`=(1-lambdar_(B))((1)/(r_(B))-(dr)/(r_(B)^(2)))`
`=(1-0)((1)/(r_(B))-0)`
`( :. lambdar_(B) lt lt tl lt 1 and (dr)/(r_(B)^(2))` is negligible)
`=(1)/(r_(B)) ....(12)`
From equation (6) and (12),
`U=-(e^(2))/(4pi epsi_(0))((1)/(r_(B)))`
`=-(ke^(2))/(r_(B)) ( :. k=(1)/(4 pi epsi_(0)))`
`=-((9xx10^(9))(1.6xx10^(-19))^(2))/((0.53xx10^(-10))`
`=-4.374xx10^(10-38+10)J`
`=-4.374xx10^(-18)J`
`:.U=-(4.374xx10^(-18))/(1.6xx10^(-19))eV`
`=-2.717xx10^(1)eV`
`:.U=-27.17eV`
In the present, case, kinetic energy of electron
`K=(1)/(2)mv^(2)=(1)/(2)m((h^(2))/(4pi^(2)m^(2)r^(2)))` [ from equ. (7)]
`:.K= (h^(2))/(8 pi^(2)mr^(2))`
`=(h^(2))/(8pi^(2)m(r_(B)+dr)^(2))`
`=(h^(2))/(8pi^(2)m{r_(B)(1+(dr)/(r_(B)))}^(2))`
`=(h^(2))/(8pi^(2)mr_(B)^(2))(1+(dr)/(r_(B)))^(-2)`
`:.K=(h^(2))/(8 pi^(2)mr_(B)^(2))(1-(2dr)/(r_(B)))....(14)`
(According to binomial theorem)
From equation `(10),r_(B)=r+lambdar^(2)`
But according to equation (11), `r=r_(B)+dr` and so
`r_(B)=r_(B)+dr+lambda(r_(B)+dr)^(2)`
`:.0=dr+lambdar_(B)^(2)+2lambdar_(B)dr+lambda dr^(2)` `:. 0~~ dr+lambdar_(B)^(2)`
`( :.2lambdar_(B)dr and lambda dr^(2)` are negligible here)
`:.dr =-lambda_(B)^(2)....(15)`
From equation (14) and (15) `K=(h^(2))/(8pi^(2)mr_(B)^(2)){1-(2)/(r_(B))(-lambda_(B)^(2))}`
`:.K=(h^(2))/(8pi^(2)mr_(B)^(2))(1+2lambdar_(B))`
`:.K=13.6(1+2lambdar_(B))....(16)`
`( :. (h^(2))/(8pi^(2)mr_(B)^(2))=13.6eV)`
Adding equation (13) and (16), total energy in final condition,
`E_(f)=13.6(1+2lambdar_(B))-27.2eV`
but `E_(i)=-13.6eV`
Required change of total energy,
`DeltaE=E_(f)-E_(i)`
`=13.6(1+2lambdar_(B))-27.2-(-13.6)`
`=13.6(1+2lambdar_(B))-13.6`
`=13.6+27.2lambdar_(B)-13.6`
`:.DeltaE=27.2lambdar_(B)eV`