You are studying the absorption of electromagnetic radiation by electrons in a crystal structure. The situation is well described by an electron in a cubical box of side length L. The electron is initially in the ground state. (a) You observe that the longest-wavelength photon that is absorbed has a wavelength in

air of λ = 624 nm. What is L? (b) You find that λ = 234 nm is also absorbed when the initial state is still the ground state. What is the value of n² for the final state in the transition for which this wavelength is absorbed, where

${\mathbf{n}}^{\mathbf{2}}\mathbf{=}{\mathbf{n}}_{\mathbf{X}}^{\mathbf{2}}\mathbf{+}{\mathbf{n}}_{\mathbf{Y}}^{\mathbf{2}}\mathbf{+}{\mathbf{n}}_{\mathbf{Z}}^{\mathbf{2}}$

What is the degeneracy of this energy level (including the degeneracy due to electron spin)?

Longest wavelength is observed for the photon that has the smallest energy transition.

For the cubical box, the required transition is between the ground state (1,1,1) and the first excited state (2,1,1). The photon energy is,

$∆\mathrm{E}=\frac{\mathrm{hc}}{\mathrm{\lambda }}$ ...(i)

The energy of the particle in a cubical box,

${\mathrm{E}}_{{\mathrm{n}}_{\mathrm{X}},{\mathrm{n}}_{\mathrm{Y}},{\mathrm{n}}_{\mathrm{Z}}}=\frac{\left({\mathrm{n}}_{\mathrm{X}}^2+{\mathrm{n}}_{\mathrm{Y}}^2+{\mathrm{n}}_{\mathrm{Z}}^2\right){\mathrm{\pi }}^2{\sout{\mathrm{h}}}^2}{2{\mathrm{mL}}^2}$

For the energy in transition substitute the n values in the above equation,

$∆\mathrm{E}=\left[\left(2^2+1^2+1^2\right)-\left(1^2+1^2+1^2\right)\right]\frac{{\mathrm{\pi }}^2{\sout{\mathrm{h}}}^2}{2{\mathrm{mL}}^2}$

This energy difference is equal to energy difference in equation (i),

$$\begin{gathered} \frac{\mathrm{hc}}{624\mathrm{nm}}=\frac{3{\mathrm{\pi }}^2{\sout{\mathrm{h}}}^2}{2{\mathrm{mL}}^2} \\ =\frac{3{\mathrm{h}}^2}{8{\mathrm{mL}}^2} \end{gathered}$$

Solving for L,

$$\begin{gathered} \mathrm{L}=\sqrt{\frac{3\mathrm{h}\left(624\mathrm{nm}\right)}{8\mathrm{mc}}} \\ =\sqrt{\frac{3\left(6.626\times {10}^{-34}\mathrm{J}·\mathrm{s}\right)\left(624\mathrm{m}\right)}{8\left(9.11\times {10}^{-31}\mathrm{kg}\right)\left(3.00\times {10}^8\mathrm{m}/\mathrm{s}\right)}} \\ =7.53\times {10}^{-10}\mathrm{m} \\ =0.753\mathrm{nm} \end{gathered}$$

For the final state, the principal quantum number is n,

From part (a) the wavelength is denoted as λ₁ and from part (b), the wavelength is denoted as λ₂.

So, for part (a),

$\frac{\mathrm{hc}}{{\mathrm{\lambda }}_1}=\frac{3{\mathrm{\pi }}^2{\sout{\mathrm{h}}}^2}{2\mathrm{mL}}$

So, for part (b),

$\frac{\mathrm{hc}}{{\mathrm{\lambda }}_2}=\left({\mathrm{n}}^2-3\right)\frac{{\mathrm{\pi }}^2{\sout{\mathrm{h}}}^2}{2{\mathrm{mL}}^2}$

Divide the above two equations,

$$\begin{gathered} \frac{{\displaystyle \frac{\mathrm{hc}}{{\mathrm{\lambda }}_2}}}{{\displaystyle \frac{\mathrm{hc}}{{\mathrm{\lambda }}_1}}}=\frac{{\mathrm{\lambda }}_1}{{\mathrm{\lambda }}_2} \\ =\frac{624}{234} \\ =\frac{{\mathrm{n}}^2-3}{3} \end{gathered}$$

Solve for n,

$$\begin{gathered} n^2=3+3\left(\frac{624}{234}\right) \\ =3+8 \\ =11 \end{gathered}$$

The three possibilities for n = 11 is,

$\left(1,1,3\right),\left(1,3,1\right),\mathrm{and}\left(3,1,1\right)$

Therefore the degeneracy is 6 including the spin degeneracy.