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(D) For an electron in a one-dimensional box of length $L$,the energy levels are given by $E_{n} = n^{2} \left( \frac{h^{2}}{8 m_{e} L^{2}} \right) =

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the total energy of the second excited state is $U_{0} + 8 \alpha$

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(D) For an electron in a one-dimensional box of length $L$,the energy levels are given by $E_{n} = n^{2} \left( \frac{h^{2}}{8 m_{e} L^{2}} \right) = n^{2} \alpha$,where $n = 1, 2, 3, \dots$.According to the Pauli exclusion principle,each energy level can hold two electrons (one spin-up,one spin-down).For $4$ electrons,the ground state configuration fills the $n=1$ level ($2$ electrons) and the $n=2$ level ($2$ electrons).The ground state energy $U_{0}$ is the sum of the energies of these $4$ electrons: $U_{0} = 2(E_{1}) + 2(E_{2}) = 2(1^{2} \alpha) + 2(2^{2} \alpha) = 2 \alpha + 8 \alpha = 10 \alpha$.The first excited state occurs when one electron is promoted from $n=2$ to $n=3$. The new total energy is $U_{0} - E_{2} + E_{3} = 10 \alpha - 4 \alpha + 9 \alpha = 15 \alpha = U_{0} + 5 \alpha$.The second excited state occurs when one electron is promoted from $n=1$ to $n=3$. The new total energy is $U_{0} - E_{1} + E_{3} = 10 \alpha - 1 \alpha + 9 \alpha = 18 \alpha = U_{0} + 8 \alpha$.Thus,option $(d)$ is correct.

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