The Bohr model for the H-atom relies on the Coulomb's law of electrostatics . Coulomb's law has not directly been varified for very short distances of the order of angstroms. Suppos-ing Coulomb's law between two oppsite charge `+q_(1),-q_(2)` is modified to `|vec(F)|=(q_(1)q_(2))/((4piepsilon_(0))r^(2))1/r^(2),rgeR_(0)`
` =(q_(1)q_(2))/((4piepsilon_(0))r^(2))1/R_(0)^(2)(R_(0)/r)^(epsilon), rleR_(0)`
Calculate in such a case , the ground state enenergy of H-atom , if `epsilon= 0.1,R_(0)=1Å`

Explanation

We are given `|vecF|=(q_(1)q_(2))/((4piepsilon_(0))) 1/(r^(2)) , rgeR_(0) |vecE|=(q_(1)q_(2))/((4piepsilon_(0))) 1/(R_(0)^(2)) ((R_(0))/(r^(epsilon)))^(epsilon) , rleR_(0)`
`in=0.1, R_(0)=1Å`.
Let `in=2+delta`.
`:. F=(q_(1)q_(2))/((4piepsilon_(0))) (R_(0)^(delta))/(r^(2+delta))=^^(R_(0)^(delta))/(r^(2+delta))` for `rleR_(0)`
Where `^^=(q_(1)q_(2))/(4piepsilon_(0))=9xx10^(9) (1.6xx10^(-19))^(2)=23.04xx10^(-29)`
So, `F=(mv^(2))/r=^^ (R_(0)^(delta))/(r^(2+delta)) or v^(2)=(^^R_(0)^(delta))/(mr^(1+delta)).......(i)`
(i) As `mvr=nh, r=(nh)/(mv)` Using (i), `r=(nh)/m[m/(^^(R_(0)^(delta))]^(1//2) r^(1/2+delta/2)`
Solving this for r, we get `r_(n)=[(n^(2)h^(2))/(m^^R_(0)^(delta))]^(1/(1-delta))`
for `n=1, r_(1)=[(n^(2)h^(2))/(m^^R_(0)^(delta))]^(1/(1-delta))=[(1.05^(2)xx10^(-68))/((9.1xx10^(-31))(23.04xx10^(-29))10^(19))]^(1/2.9)=8xx10^(-11)m=0.08nmlt0.1nm`
(ii) form `v_(n)=(nh)/(mr_(n))=nh((^^R_(0)^(delta))/(n^(2)h^(2)))^(1/(1-delta))` for `n=1, v_(1)=h/(mr_(1))=1.44xx10^(6)m//s`
(iii) `K.E. =1/2mv_(1)^(2)=9.43xx10^(-19)J=5.9eV` P.E. `(till R_(0))=-(^^)/(R_(0))`
P.E. form `R_(0)` to `r=^^R_(0)^(delta)int_(R_(0))^(r) (dr)/(r^(2+delta)) =(^^R_(0)^(delta))/(-1-delta) [1/(r^(1+delta))]_(R_(0))^(r) =-(^^R_(0)^(delta))/(1+delta) [1/(r^(1+delta))-1/(R_(0)^(1+delta))]=-(^^)/(1+delta)[R_(0)^(delta)/(r^(1+delta))-1/(R_(0))]`
P.E. `=-(^^)/(1+delta)[(R_(0)^(delta)/(r^(1+delta))-1/(R_(0))+(1+delta)/(R_(0))]=(-^^)/(-0.9) [(R_(0)^(-1.9))/(r^(-0.9))-1.9/(R_(0))]=(2.3xx10^(-28))/0.9 [(0.8)^(0.9)-1.9]"joule"=-17.3eV`
`:.` Total energy =`(-17.3+5.9)eV=-11.4eV`