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(B) The energy of an electron in the ground state of a hydrogen atom is $E_1 = -13.6 \, eV$. The ionization energy of the excited state is given as $+

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$12.75$

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(B) The energy of an electron in the ground state of a hydrogen atom is $E_1 = -13.6 \, eV$. The ionization energy of the excited state is given as $+0.85 \, eV$,which means the energy of the electron in that excited state is $E_n = -0.85 \, eV$. When the electron returns from the excited state to the ground state,the energy of the emitted photon is given by the difference in energies: $\Delta E = E_n - E_1 = -0.85 \, eV - (-13.6 \, eV) = 12.75 \, eV$.

The ionization energy of an electron in an excited state of a hydrogen atom is $+0.85 \, eV$. What will be the energy of the emitted photon in $eV$ when it returns to the ground state?

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