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(C) In a hydrogen atom,the total energy $E$,kinetic energy $K$,and potential energy $U$ are related by $U = 2E$ and $K = -E$.For the first excited sta

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$-6.8$

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(C) In a hydrogen atom,the total energy $E$,kinetic energy $K$,and potential energy $U$ are related by $U = 2E$ and $K = -E$.For the first excited state $(n=2)$,the total energy is $E_2 = -13.6 / 2^2 = -3.4 \ eV$.The potential energy in the first excited state is $U_2 = 2E_2 = 2(-3.4) = -6.8 \ eV$.If we shift the potential energy scale such that the new potential energy $U'_2 = 0$,we have added $+6.8 \ eV$ to the potential energy.Since the total energy $E = K + U$,and kinetic energy $K$ remains unchanged by a shift in the potential energy reference point,the new total energy $E'$ is $E' = E + \Delta U$.For the ground state $(n=1)$,the original total energy is $E_1 = -13.6 \ eV$.The shift applied is $\Delta U = +6.8 \ eV$.Therefore,the new total energy in the ground state is $E'_1 = -13.6 + 6.8 = -6.8 \ eV$.

The total energy of a hydrogen atom in its ground state is $-13.6 \ eV$. If the potential energy in the first excited state is taken as zero,then the total energy in the ground state will be ..... $eV$.

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