Answer
See the detailed answer below.
Work Step by Step
$$\color{blue}{\bf [a]}$$
We are given a one-dimensional rigid box with infinite potential walls, and its length is $ L = 0.50 \, \text{nm} $. The goal is to calculate the energy levels of the system and the ground-state energy of five electrons in this box.
The energy levels for a particle in a one-dimensional box are given by:
$$
E_n = \frac{n^2 h^2}{8mL^2}
$$
Where $ n $ is the quantum number ($ n = 1, 2, 3, \ldots $),
We need to find the lowest energy level, $ E_1 $, when $ n = 1 $:
$$
E_1 = \frac{h^2}{8mL^2}
$$
Substituting the known:
$$
E_1 = \frac{(6.63 \times 10^{-34})^2 }{8(9.11 \times 10^{-31} )(0.50 \times 10^{-9} )^2} =\bf 2.41 \times 10^{-19} \,\rm \text{J}
$$
$$
E_1 =\color{red}{\bf 1.51} \, \text{eV}
$$
Therefore, the second energy level $ E_2 $ and the third energy level $ E_3 $ are given by $$E=n^2E_1$$
Thus,
$$
E_2 = (2^2) E_1 = 4 \times 1.51 \, \text{eV} =\color{red}{\bf 6.04} \, \text{eV}
$$
$$
E_3 = (3^2) E_1 = 9 \times 1.51 \, \text{eV} = \color{red}{\bf 13.6} \, \text{eV}
$$
The Pauli Exclusion Principle allows only two electrons to occupy each energy level; one with spin up and one with spin down.
We have five electrons, so:
Two electrons will occupy the $ n = 1 $ level (energy $ 1.51 \, \text{eV} $),
Two electrons will occupy the $ n = 2 $ level (energy $ 6.04 \, \text{eV} $),
The fifth electron will occupy the $ n = 3 $ level (energy $ 13.6 \, \text{eV} $).
The energy level diagram below shows the distribution of electrons across the different energy levels.
$$\color{blue}{\bf [b]}$$
The total ground-state energy $ E $ is the sum of the energies of all the electrons:
$$
E = 2E_1 + 2E_2 + E_3
$$
Substitute the values:
$$
E = 2(1.51 ) + 2(6.04 ) + 13.6 =\color{red}{\bf 28.7} \, \text{eV}
$$
