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(B) The electron starts at $-1\,eV$ and ends at $-12\,eV$. The total energy released is $\Delta E = (-1\,eV) - (-12\,eV) = 11\,eV$.Since the electron

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(B) The electron starts at $-1\,eV$ and ends at $-12\,eV$. The total energy released is $\Delta E = (-1\,eV) - (-12\,eV) = 11\,eV$.Since the electron transitions by emitting exactly two photons, the sum of the energies of the two photons must be $11\,eV$ $(E_1 + E_2 = 11\,eV)$.Possible intermediate states are $-3\,eV$ and $-7\,eV$.Case $1$: Transition via $-3\,eV$ level:Photon $1$: $(-1\,eV) - (-3\,eV) = 2\,eV$Photon $2$: $(-3\,eV) - (-12\,eV) = 9\,eV$Case $2$: Transition via $-7\,eV$ level:Photon $1$: $(-1\,eV) - (-7\,eV) = 6\,eV$Photon $2$: $(-7\,eV) - (-12\,eV) = 5\,eV$Possible photon energies are $2\,eV, 9\,eV, 6\,eV, 5\,eV$.Comparing with the options, $4\,eV$ is not among the possible photon energies.

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