Answer
$44\Big(\frac{h^2}{8mL^2}\Big)$
Work Step by Step
The quantized energies for an electron trapped in a one-dimensional infinite potential well of width $L$
$E_{n}=\frac{n^2h^2}{8mL^2},$
where $m$ is the electron mass and $n$ is a quantum number.
According to the current problem, there are seven electrons confined to the one dimensional infinite potential well. As the electrons do not electrically interact with one another, we can determine the ground state configuration of the system by placing the seven electrons in the well one by one by maintaining the Pauli exclusion principle; that is, no two electrons can have the same set of values for their quantum numbers $n$ and $m_s$.
$(i)$ First, the electrons will go into the lowest energy level $E_{1}$. According to the Pauli exclusion principle, two electrons can accommodate in the energy state $E_{1}$. Thus the energy level $E_{1}$ will be fully occupied by two electrons with quantum numbers $(n,m_s)$ of $(1,-\frac{1}{2})$ and $(1,\frac{1}{2})$.
$(ii)$ Thereafter, the electrons will go into next higher energy level. Thus, the energy level $E_{2}$ will be fully occupied by two electrons with quantum numbers $(n_,m_s)$ of $(2,-\frac{1}{2})$ and $(2,\frac{1}{2})$.
$(iii)$ Then, the next higher energy level $E_{3}$ will be fully occupied by two electrons with quantum numbers $(n,m_s)$ of $(3,-\frac{1}{2})$ and $(3,\frac{1}{2})$.
$(iv)$ At last, the remaining seventh electron will go into the next higher level, which is the $E_{4}$ level. Thus, the energy level $E_{4}$ will be partially occupied by one electron with quantum numbers $(n,m_s)$ of $(4,-\frac{1}{2})$ or $(4,\frac{1}{2})$.
Now, the total energy $E_{gr}$ is the sum of the energies of the individual electrons in the system’s ground-state configuration. Therefore,
$E_{gr}=2E_{1}+2E_{2}+2E_{3}+E_{4}$
or, $E_{gr}=2\times\Big(\frac{h^2}{8mL^2}\Big)+2\times4\Big(\frac{h^2}{8mL^2}\Big)+2\times9\Big(\frac{h^2}{8mL^2}\Big)+16\Big(\frac{h^2}{8mL^2}\Big)$
or, $E_{gr}=44\Big(\frac{h^2}{8mL^2}\Big)$
$\therefore$ The energy of the ground state of the system is $44\Big(\frac{h^2}{8mL^2}\Big)$
