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

Calculate in such a case, the ground state energy of a H-atom, if $\mathrm{\epsilon }$ = 0.1, ${R}_{0}=1\mathrm{Å}.$

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 $r\le {R}_{0}=1\mathrm{Å}$
The electrostatic force of attraction between the positively charged nucleus and negatively charged electrons (Coulombian force) provides the necessary centripetal force.
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.
Total energy is (-17.3 + 5.9)=-11.4 eV
This is the required TE of an electron in the ground state.