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(C) The energy of an electron in the $n^{th}$ orbit of a hydrogen atom is given by $E_n = -13.6 / n^2 \, eV$.For the ground state $(n_1 = 1)$,the ener

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$4$

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(C) The energy of an electron in the $n^{th}$ orbit of a hydrogen atom is given by $E_n = -13.6 / n^2 \, eV$.For the ground state $(n_1 = 1)$,the energy is $E_1 = -13.6 \, eV$.When the atom absorbs a photon of energy $12.75 \, eV$,the new energy $E_n$ becomes:$E_n = E_1 + 12.75 \, eV = -13.6 + 12.75 = -0.85 \, eV$.Using the formula $E_n = -13.6 / n^2$,we have:$-0.85 = -13.6 / n^2$$n^2 = 13.6 / 0.85 = 16$$n = 4$.Therefore,the principal quantum number of the excited state is $4$.

When a photon of energy $12.75 \, eV$ is completely absorbed by a hydrogen atom in its ground state,the principal quantum number of the excited state will be:

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