\(\displaystyle \bold{(a)}\)
The photon will obtain the longest wave length when it has the least energy. In infinitely deep square well, this happens when the electron jumps from the first excited state level \(\displaystyle n = 2\) to the ground state level \(\displaystyle n = 1\).
We start with the photon energy formula.
\(\displaystyle E = hf = \frac{hc}{\lambda} = \frac{1.24 \times 10^3 \ \text{eV} \cdot \ \text{nm}}{\lambda}\)
In future Episodes, I will explain why \(\displaystyle hc = 1.24 \times 10^3 \ \text{eV} \cdot \ \text{nm}\)
We can use the same formula to find the change in energy between the level \(\displaystyle n = 2\) and the level \(\displaystyle n = 1\).
Then,
\(\displaystyle \Delta E = \frac{1.24 \times 10^3 \ \text{eV} \cdot \ \text{nm}}{\lambda}\)
From accumulated knowledge, we know that the energy levels formula for the electron in an infinite square well is:
\(\displaystyle E_n = \frac{n^2h^2}{8mL^2} = n^2E_1\) where \(\displaystyle E_1\) is the ground state energy.
Then,
\(\displaystyle \Delta E = \left(n^2_2 - n^2_1\right)E_1 =\frac{1.24 \times 10^3 \ \text{eV} \cdot \ \text{nm}}{\lambda}\)
This gives:
\(\displaystyle \lambda =\frac{1.24 \times 10^3 \ \text{eV} \cdot \ \text{nm}}{\left(n^2_2 - n^2_1\right)E_1} = \frac{1.24 \times 10^3 \ \text{eV} \cdot \ \text{nm}}{\left(2^2 - 1^2\right)8 \ \text{eV}} = 51.7 \ \text{nm}\)
