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 +q1, -q2 is modified to

F=q1q2(4πε0)1r2, rR0=q1q24πε01R02R0rε, rR0

Calculate in such a case, the ground state energy of a H-atom, if ε = 0.1, R0=1Å.

Hint: The electrostatic force of attraction between the positively charged nucleus and negatively charged electrons provides the necessary centripetal force.
Step 1: Find the radius of the orbit of the electron in the ground state.
Considering the case, when rR0=1Å
Let ε=2+δF=q1q24πε0R0δr2+δwhere, =q1q24πε0=(1.6×10-19)2×9×109                          23.04×10-29 N m2
The electrostatic force of attraction between the positively charged nucleus and negatively charged electrons (Coulombian force) provides the necessary centripetal force.
F=mv2r=R0δr2+δ or v2=R0δmr1+δmvr=r= mv= m[mR0δ]1/2r1/2+δ/2   [Applying Bohr's second postulates]Solving this for r, we getrn=[n2ħ 2mR0δ]11-δ where, rn is radius of nth orbit of electron.For n=1 and substituting the values of constant, we getr1=[ħ 2mR0δ]11-δ r1=[1.052×10-689.1×10-31×2.3×10-28×10+19]12.9 =8×10-11=0.08 nm               (<0.1 nm)
This is the radius of orbit of the electron in the ground state of the hydrogen atom.
Step 2: Find the kinetic energy and the potential energy of the electron in the ground state.
vn= mrn=nħ (mR0δn2ħ 2)11-δ For n=1, v1=ħ mr1=1.44×106 m/s [This is the speed of electron in ground state]KE=12mv12=9.43×10-19 J=5.9eV [This is the KE of electron in ground state]PE till R0=-R0 [This is the PE of electron in ground state at r=R0]
PE from R0 to r=+R0δR0rdrr2+δ=+R0δ-1-δ[1r1+δ]R0r [This is the PE of electron in ground state at R0 to r]=-R0δ1+δ[1r1+δ-1R01+δ]=-1+δ[R0δr1+δ-1R0]PE=-1+δ[R0δr1+δ-1R0+1+δR0]PE=--0.9[R0-1.9r-0.9-1.9R0]=2.30.9×10-18[(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 an electron in the ground state.