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(D) In a standard hydrogen atom,the total energy of the $n$-th state is given by $E_n = -\frac{13.6}{n^2} \, eV$.For the ground state $(n=1)$,$E_1 = -

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(D) In a standard hydrogen atom,the total energy of the $n$-th state is given by $E_n = -\frac{13.6}{n^2} \, eV$.For the ground state $(n=1)$,$E_1 = -13.6 \, eV$.For the first excited state $(n=2)$,$E_2 = -\frac{13.6}{2^2} = -3.4 \, eV$.The potential energy $(P.E.)$ is related to the total energy $(E)$ by the relation $P.E. = 2E$.Thus,for the ground state,$P.E. = 2 	imes (-13.6) = -27.2 \, eV$.If we shift the energy scale such that the potential energy in the ground state is assumed to be $0 \, eV$,we must add $27.2 \, eV$ to the energy of all states.Under this new scale,the energy of the first excited state becomes $E'_2 = E_2 + 27.2 = -3.4 + 27.2 = 23.8 \, eV$.This matches the given condition. Therefore,the potential energy of the hydrogen atom in the ground state is assumed to be $0 \, eV$.

If the energy in the first excited state in a hydrogen atom is $23.8 \, eV$,then the potential energy of a hydrogen atom in the ground state can be assumed to be ....... $eV$.

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