The Bohr model for the H-atom relies on the Coulomb's law of electrostatics. Coulomb's law has not directly been verified for very short distances of the order of angstroms. Supposing Coulomb's law between two opposite charge `+q_(1),-q_(2)` is modified to `|F|=(q_(2)q_(1))/((4pi epsi_(0)))(1)/(r^(2)),r ge R_(0)`
`=(q_(1)q_(2))/(4pi epsi_(0))(1)/(R_(0)^(2))((R_(0))/(r))^(e),r lt R_(0)` Calculate in such a case, the ground state energy of a H-atom if `epsi=0.1 R_(0)=1Å`

Explanation

Here for very small distance `r le R_(0)` (where `R_(0)=iÅ)` in the modified format of Coulomb.s law, Taking `epsi=2+delta`
`F=(q_(1)q_(2))/(4pi epsi_(0)R_(0)^(2))((R_(0))/(r))^(2+delta).....(1)`
`=(ke^(2))/(R_(0)^(2))xxR_(0)^(2)xx(R_(0)^(delta))/(r^(2+6)) ( :.k=(1)/(4pi epsi_(0)) and q_(1)=q_(2)=e)`
`:.F=ke^(2)xx(R_(0)^(delta))/(r^(2+delta))....(2)`
`=9xx10^(9)xx(1.6xx10^(-19))^(2)xx(R_(0)^(delta))/(r^(2+delta))`
`:.F=(2.304xx10^(-28))xx (R_(0)^(delta))/(r^(r+delta))......(3)`
Assuming for the sake of simplicity.
`3=3.304xx10^(-28)Nm^(2)=f...(4)`
`F=fxx(R_(0)^(delta))/(r^(2+delta))....(5)`
`:. (mv^(2))/(r)=fxx(R_(0)^(delta))/(r^(+delta)) ( :.` F is centripetal force)
`:.v^(2)=(f)/(m)xx((R_(0)^(delta))/(r^(delta+1)))`
`:.v=((R_(0)^(delta))/(m))^(1//2)xx(1)/(r^(((delta+1)/(2)))).....(6)`
According to Bohr.s first postulate, for n=1
`mvr=(h)/(2pi)`
`r=(h)/(2pi mv).....(7)`
`:.r=(h)/(2pim)xx((m)/(fR_(0)^(delta)))^(1//2)xxr^(((delta+1)/(2)))`
Multiplying by `r^(-1)` on both the sides,
`1=(h)/(2pim)xx((m)/(fR_(0)^(delta)))^(1//2)xxr^(((delta+1)/(2)))`
`r^(((delta+1)/(2)))=((2pi m)/(h))xx((fR_(0)^(delta))/(fR_(0)^(delta)))^(1//2)`
`r^(((delta+1)/(2)))=((4pi^(2)m^(2))/(h^(2))xx(fR_(0)^(delta))/(m))^(1//2)`
`r^(((delta+1)/(2)))=((4pi^(2)mfR_(0)^(delta))/(h^(2)))^(1//2)`
Taking power `((2)/(delta-1))` on both the sides,
`r=((4pi^(2)mfR_(0)^(delta))/(h^(2)))^(1//delta-1)`
`:. r=((h^(2))/(4pi^(2)mfR_(0)^(delta)))^(1//delta-1)....(8)`
Now `in =2+delta` (as per equation (1))
`:.0.1=2+delta ( :.` From statement `epsi=0.1`)
`:.delta=-1.9....(9)`
`:. r=((h^(2))/(4pi^(2)mfR_(0)^(delta)))^(1//delta-1)`
`:. r=((h^(2))/(4pi^(2)mfR_(0)^(delta)))^(1//2.9)`
By placing respective values, new orbital radius of electron in the ground state of H-atom,
`r={((6.625xx10^(-34))^(2))/(4xx(3.14)^(2)xx9.1xx10^(-31)xx2.304xx10^(-28xx(1xx10^(-10))^(-1.9)))}`
`:.r{((6.625)^(2))/(4xx(3.14)^(2)xx9.1xx2.304xx10^((-68+31+28-19)))}^(1//2.9)`
`:.r=(5.308xx10^(-2)xx10^(-28))^(1//2.9)`
`:.r=(5.308xx10^(-30))^(1//2.9)`
`:.r=8.53xx10^(-11)m`
`:.r=0.8053xx10^(-10)m`
`:.r=0.8053Å`
Above value is smaller than `1Å`
Since above orbital radius in the ground state, is `r=r_(1)=8.053xx10^(-11)m...(10)`
Now taking `r=r_(1) and v=v_(1)` in equation (7),
`v_(1)=(h)/(2pi mr_(1))`
`=(6.625xx10^(-34))/(2xx3.14xx9.1xx10^(-31)xx8.053xx10^(-11))`
`=1.439xx10^(-2)xx10^(-34+31+11)`
`:.v_(1)=1.439xx10^(6)m//s`
`rArr` Kinetic energy,
`K_(1)=(1)/(2)mv_(1)^(2)`
`=(1)/(2)xx9.1xx10^(-31)xx(1.439xx10^(6))^(2)`
`:.K_(1)=9.422xx10^(-19)J`
`:.K_(1)=(9.422xx10^(-19))/(1.6xx10^(-19))=5.889eV`
Since, in H-atom electron and proton are separated by `R_(0)=1Å`, their electrostatic potential energy is,
`U_(0)=k(q_(1)q_(2))/(R_(0))k((e)-(e))/(R_(0))`
`:.U_(0)=-(ke^(2))/(R_(0))=-(f)/(R_(0))`
`:.U_(0)=-(2.304xx10^(-28))/(1xx10^(-10))=-2.304xx10^(-18)J`
`:.U_(0)=-(2.304xx10^(-18))/(1.6xx10^(-19))eV`
`:.U_(0)=-14.4eV`
Now, in the process when distance between electron and proton in H-atom decreases from `R_(0)=1Å r_(1)=0.8053Å`, if additional potential energy sotred is U. then.
`U.=int dU=int (-F)dr ( :. F=-(dU)/(dr))`
`:. U.= int-kq_(1)q_(2)(R_(0)^(delta))/(r^(2+delta))dr` (From equation (3))
`:.U. int -k(e)(-e)(R_(0)^(delta))/(r^(2+delta))dr`
`=ke^(2)R_(0)^(delta)int_(R_(0))^(r) r^(-2-delta)dr`
`=fR_(0)^(delta){(r^(-1-delta))/(-1-delta)}_(R_(0))^(r)`
`-(fR_(0)^(delta))/(1+delta){(1)/(r^(1+delta))}_(R_(0))^(r)`
`=-(2.304xx10^(-28)xx(10^(-10))^(-1.9))/(1-1.9){(1)/(r^(1-1.9))-(1)/(R_(0)^(1-1.9))}`
`=(2.304xx10^(-28)xx10^(19))/(0.9)(r^(0.9-R_(0)^(0.9))`
`=2.56xx10^(-9){(0.8053xx10^(-10))^(0.9)-(1xx10^(-10))^(0.9)}`
`=2.56xx10^(-9)xx10^(-9){(0.8053)^(0.9)-1}`
`=2.56xx10^(-18){0.8229-1}`
`=2.56xx10^(-18)(0.1771)`
`=0.4534xx10^(-18)J`
`:.U.=-(0.4534xx10^(-18))/(1.6xx10^(-19))eV`
`:.U.=-2.834eV`
`rArr U_(1).=U_(0)+U.=(-14.4)+(-2.834)`
`:.U_(1)=-17.234eV`
Now required total energy of electron in the final state,
`E_(f)=K_(1)+U_(1)`
`=5.889xx+(-17.234)`
`:.E_(f)=-11.345 eV`