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 find the energy levels of the system and the ground-state energy of 3 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 next energy levels from $ E_2 $ to $E_6$ are given by $$E=n^2E_1$$
Thus,
$$
E_2 = (2^2) E_1 = 4 \times 1.51 \, \text{eV} =\color{red}{\bf 6.04} \, \text{eV}
$$
$$
E_3 = (3^2) E_1 = 9 \times 1.51 \, \text{eV} = \color{red}{\bf 13.6} \, \text{eV}
$$
$$
E_4 = (4^2) E_1 = 16 \times 1.51 \, \text{eV} = \color{red}{\bf 24.2} \, \text{eV}
$$
$$
E_5= (5^2) E_1 = 25\times 1.51 \, \text{eV} = \color{red}{\bf 37.8} \, \text{eV}
$$
$$
E_6= (6^2) E_1 = 36\times 1.51 \, \text{eV} = \color{red}{\bf 54.4} \, \text{eV}
$$
According to the selection rule $ \Delta n = \text{odd} $, the allowed transitions are from $ n = 6 $ to lower energy levels with odd quantum numbers. Therefore, the possible transitions are: $ 6 \to 5 $, and $ 6 \to 3 $
The Pauli Exclusion Principle prohibits transitions into the $ n = 1 $ state because it is already occupied by two electrons (one spin-up, one spin-down).
See the graph below.
$$\color{blue}{\bf [b]}$$
The energy of the emitted photon is equal to the energy difference between the initial and final energy levels of the electron and it is given by
$$
E_{\text{photon}}=\Delta E = \frac{hc}{\lambda}
$$
Solve for $ \lambda $:
$$
\lambda = \frac{hc}{\Delta E}
$$
For $ 6 \to 5 $:
$$
\lambda_{6 \to 5} = \frac{hc}{ (E_6-E_5)e}
$$
where $e$ for converting from eV to J
Plug the known;
$$
\lambda_{6 \to 5} = \frac{(6.63 \times 10^{-34})(3\times 10^8)}{ (54.4-37.8)(1.6\times 10^{-19})}=\color{red}{\bf 75.0}\;\rm nm
$$
For $ 6 \to 3 $
$$
\lambda_{6 \to 3} = \frac{hc}{ (E_6-E_3)e}
$$
$$
\lambda_{6 \to 3} = \frac{(6.63 \times 10^{-34})(3\times 10^8)}{ (54.4-13.6)(1.6\times 10^{-19})}=\color{red}{\bf 30.5}\;\rm nm
$$
