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

Considering the case, when `r le R_(0) = 1 Å`
Let `epsi = 2 + delta`
`F = (q_(1)q_(2))/(4piepsi_(0)).(R_(0)^(delta))/(r^(2+delta))`
where, `(q_(1)q_(2))/(4pi_(0)epsi_(0)) = (1.6xx10^(-19))^(2) xx 9 xx 10^(9)`
`= 23.04 xx 10^(-29) Nm^(2)`
The electrostatic force of attraction between positively charged nucleus and negatively charged electrons (Columbian force) provides necessary centripetal force.
`= (mv^(2))/(r)` or `v^(2) = (^^R^(0))/(mr^(1+delta))"........."(i)`
`mvr = nh. r = (nh)/(mv) = (nh)/(m)[(m)/(^^R_(0)^(delta))]^(1//2) r^(1//2+delta//2)`
[Applying Bohr's second postulates]
solving this for r,we get `r_(n) = [(n^(2)h^(2))/(m^^R_(0)^(delta))]^((1)/(1-delta))`
where, `r_(n)` is radius of nth orbit of electron,
For `n = 1` and substituting the values of constant, we get
`r_(1) = [(h^(2))/(m^^R_(0)^(delta))]^((1)/(1-delta))`
`r_(1) = [(1.05^(2)xx10^(-68))/(9.1 xx 10^(-31) xx 2.3 xx 10^(-28) xx 10^(+19))]^((1)/(2.9))`
`= 8 xx10^(-11)`
`= 0.08 nm , (lt 0.1 nm)`
This is the radius of orbit of electron in ground state of hydrogen atom.
`v_(n) = (nh)/(mr_(n)) = nh((m^^R_(0)^(delta))/(n^(2)h^(2)))^((1)/(1-delta))`
For `n = 1, v_(1) - (h)/(mr_(1)) = 1.44 xx 10^(6) m//s`
[This is the speed of electron in ground state]
`PE` till `R_(0) = - (^^)/(R_(0))` , [This is the PE of electron in ground state at ` r = R_(0)`]
PE from `R_(0)` to `r = + ^^ R_(0)^(delta)int_(R_(0))^(delta)(dr)/(r^(2+delta)) = + (^^R_(0)^(delta))/(-1-delta)[(1)/(r^(1+delta))]_(R_(0))^(r)`
[This is PE of electron in ground state at `R_(0)` to `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))]`
`PE=-(^^)/(1+delta)[(R_(0)^(delta))/r^(1+delta)-(1)/(R_(0))+(1+delta)/R_(0)]`
`PE = - (^^)/(-0.9)[(R_(0)^(-1.9))/(r^(-0.9))-(1.9)/(R_(0))]`
`= (2.3)/(0.9) xx 10^(-18) [(0.8)^(0.9) - 1.9]J = - 17.3 eV`
Total energy is `(-17.3 + 5.9) = - 11.4 eV`
This is the required TE of electron in ground state.