Question

if the total energy of the first excited state of hydrogen atom is taken to be zero then the potential energy of the ground state of the hydrogen atom will be?( Ionisation energy of ground state H-atom is 13.6)

Answer 1

$$\begin{gathered} \mathrm{The}\mathrm{total}\mathrm{energy}\mathrm{of}\mathrm{hydrogen}\mathrm{atom}\mathrm{in}\mathrm{the}\mathrm{ground}\mathrm{state}=-13.6\mathrm{eV}(\mathrm{P}.\mathrm{E}=-27.2\mathrm{eV}+\mathrm{K}.\mathrm{E}=13.6\mathrm{eV}) \\ \mathrm{Total}\mathrm{energy}\mathrm{in}\mathrm{the}\mathrm{first}\mathrm{excited}\mathrm{state}=-3.4\mathrm{eV} \\ \mathrm{Now}\mathrm{the}\mathrm{energy}\mathrm{in}\mathrm{the}\mathrm{first}\mathrm{excited}\mathrm{state}(-3.4\mathrm{eV})\mathrm{is}\mathrm{taken}\mathrm{as}\mathrm{zero}.\mathrm{Then}\mathrm{the}\mathrm{whole}\mathrm{energy}\mathrm{spectrum}\mathrm{will}\mathrm{be}\mathrm{shoifted}\mathrm{by}3.4\mathrm{eV}. \\ \mathrm{Thus},\mathrm{potential}\mathrm{energy}\mathrm{in}\mathrm{ground}\mathrm{state}\mathrm{will}\mathrm{be}, \\ =-27.2+3.4\mathrm{eV}=-23.8\mathrm{eV} \end{gathered}$$