The ionisation energy of hydrogen atom is $$13.6\,eV.$$  Following Bohr’s theory, the energy corresponding to a transition between 3rd and 4th orbit is

Solution :
Total energy of electron for hydrogen like atom is given by
$$\eqalign{ & {E_n} = - \frac{{13.6{Z^2}}}{{{n^2}}} \cr & \therefore {E_3} = - \frac{{13.6}}{{{3^2}}}eV\,\,\left[ {Z = 1,n = 3} \right] \cr & = - 1.51\,eV \cr & {E_4} = - \frac{{13.6}}{{{4^2}}} = - 0.85\,eV \cr & \therefore {E_4} - {E_3} = 1.51 - 0.85 = 0.66\,eV \cr} $$