The diagram shows the lowest four energy leve | vedclass

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(B) The energy levels are $E_1 = -12\, eV$,$E_2 = -7\, eV$,$E_3 = -3\, eV$,and $E_4 = -1\, eV$.The electron starts at $E_4 = -1\, eV$ and ends at $E_1

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$4$

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(B) The energy levels are $E_1 = -12\, eV$,$E_2 = -7\, eV$,$E_3 = -3\, eV$,and $E_4 = -1\, eV$.The electron starts at $E_4 = -1\, eV$ and ends at $E_1 = -12\, eV$. The total energy released is $\Delta E = (-1\, eV) - (-12\, eV) = 11\, eV$.Since the electron emits exactly two photons,the sum of the energies of the two photons must be $11\, eV$ $(h
u_1 + h
u_2 = 11\, eV)$.Possible transitions from $E_4 = -1\, eV$ to $E_1 = -12\, eV$ in two steps:$1$. $E_4 	o E_3 	o E_1$: Energies are $(-1 - (-3)) = 2\, eV$ and $(-3 - (-12)) = 9\, eV$.$2$. $E_4 	o E_2 	o E_1$: Energies are $(-1 - (-7)) = 6\, eV$ and $(-7 - (-12)) = 5\, eV$.Possible photon energies are $2\, eV, 9\, eV, 6\, eV, 5\, eV$.Comparing these with the options,$4\, eV$ is not among the possible photon energies.

The diagram shows the lowest four energy levels for an electron in a hypothetical atom. The electron is excited to the $-1\, eV$ level and transitions to the lowest energy state by emitting exactly two photons. Which of the following energies could not belong to either of the photons (in $, eV$)?

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