Question

Consider the (eight) $\mathbf{n}\mathbf{=}\mathbf{2}\mathbf{}\mathbf{states}\mathbf{,}\mathbf{}\mathbf{|}\mathbf{2}{\mathbf{lm}}_{\mathbf{l}}{\mathbf{m}}_{\mathbf{s}}\mathbf{⟩}\mathbf{.}$Find the energy of each state, under strong-field Zeeman splitting. Express each answer as the sum of three terms: the Bohr energy, the fine-structure (proportional to${\mathbf{a}}^{\mathbf{2}}$), and the Zeeman contribution (proportional to${\mathbf{\mu }}_{\mathbf{B}}{\mathbf{B}}_{\mathbf{ext}}\mathbf{.}$). If you ignore fine structure altogether, how many distinct levels are there, and what are their degeneracies?

Short answer

Only five of the eight distinct energies remain when fine structure is ignored: There are three degenerate energies.

Step-by-step solution

Definition of Zeeman splitting.

When radiation (such as light) originates in a magnetic field, Zeeman splitting occurs, which is the splitting of a single spectral line into two or more lines of different frequencies.

Step2: Derivation.

For $\mathrm{n}=2$, there are eight different states $|2{\mathrm{lm}}_{\mathrm{l}}{\mathrm{m}}_{\mathrm{s}}⟩$:

$$\begin{gathered} \mathrm{i})\mathrm{l}=0\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}{\mathrm{m}}_{\mathrm{l}}=0\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}{\mathrm{m}}_{\mathrm{s}}=-\frac{1}{2}, \\ \mathrm{ii})\mathrm{l}=0\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}{\mathrm{m}}_{\mathrm{l}}=0\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}{\mathrm{m}}_{\mathrm{s}}=\frac{1}{2}, \\ \mathrm{iii})\mathrm{l}=1\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}{\mathrm{m}}_{\mathrm{l}}=-1\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}{\mathrm{m}}_{\mathrm{s}}=-\frac{1}{2}, \\ \mathrm{iv})\mathrm{l}=1\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}{\mathrm{m}}_{\mathrm{l}}=-1\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}{\mathrm{m}}_{\mathrm{s}}=\frac{1}{2}, \\ \mathrm{v})\mathrm{l}=1\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}{\mathrm{m}}_{\mathrm{l}}=0\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}{\mathrm{m}}_{\mathrm{s}}=-\frac{1}{2}, \\ \mathrm{vi})\mathrm{l}=1\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}{\mathrm{m}}_{\mathrm{l}}=0\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}{\mathrm{m}}_{\mathrm{s}}=\frac{1}{2}, \\ \mathrm{vii})\mathrm{l}=1\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}{\mathrm{m}}_{\mathrm{l}}=1\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}{\mathrm{m}}_{\mathrm{s}}=-\frac{1}{2}, \\ \mathrm{viii})\mathrm{l}=1\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}{\mathrm{m}}_{\mathrm{l}}=1\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}{\mathrm{m}}_{\mathrm{s}}=1/2 \\ \mathrm{i})\mathrm{l}=0\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}{\mathrm{m}}_{\mathrm{l}}=0\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}{\mathrm{m}}_{\mathrm{s}}=-\frac{1}{2}, \\ \mathrm{ii})\mathrm{l}=0\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}{\mathrm{m}}_{\mathrm{l}}=0\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}{\mathrm{m}}_{\mathrm{s}}=\frac{1}{2}, \\ \mathrm{iii})\mathrm{l}=1\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}{\mathrm{m}}_{\mathrm{l}}=-1\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}{\mathrm{m}}_{\mathrm{s}}=-\frac{1}{2}, \\ \mathrm{iv})\mathrm{l}=1\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}{\mathrm{m}}_{\mathrm{l}}=-1\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}{\mathrm{m}}_{\mathrm{s}}=\frac{1}{2}, \\ \mathrm{v})\mathrm{l}=1\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}{\mathrm{m}}_{\mathrm{l}}=0\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}{\mathrm{m}}_{\mathrm{s}}=-\frac{1}{2}, \\ \mathrm{vi})\mathrm{l}=1\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}{\mathrm{m}}_{\mathrm{l}}=0\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}{\mathrm{m}}_{\mathrm{s}}=\frac{1}{2}, \\ \mathrm{vii})\mathrm{l}=1\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}{\mathrm{m}}_{\mathrm{l}}=1\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}{\mathrm{m}}_{\mathrm{s}}=-\frac{1}{2}, \\ \mathrm{viii})\mathrm{l}=1\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}{\mathrm{m}}_{\mathrm{l}}=1\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}{\mathrm{m}}_{\mathrm{s}}=\frac{1}{2} \end{gathered}$$

Total energy is equal to:

$\mathrm{E}=-\frac{13.6\mathrm{eV}}{{\mathrm{n}}^2}+{\mathrm{\mu }}_{\mathrm{B}}{\mathrm{B}}_{\mathrm{ext}}({\mathrm{m}}_{\mathrm{l}}+2{\mathrm{m}}_{\mathrm{s}})+\frac{13.6\mathrm{eV}}{{\mathrm{n}}^3}{\mathrm{\alpha }}^2\left[\frac{3}{4\mathrm{n}}-\frac{\mathrm{l}(\mathrm{l}+1)-{\mathrm{m}}_{\mathrm{l}}{\mathrm{m}}_{\mathrm{s}}}{\mathrm{l}\left(\mathrm{l}+\frac{1}{2}\right)(\mathrm{l}+1)}\right]$

role="math" localid="1656064856692" $$\begin{gathered} \mathrm{i})\mathrm{E}=-\frac{13.6\mathrm{eV}}{{\mathrm{n}}^2}-{\mathrm{\mu }}_{\mathrm{B}}{\mathrm{B}}_{\mathrm{ext}}+\frac{13.6\mathrm{eV}}{{\mathrm{n}}^3}{\mathrm{\alpha }}^2\left[\frac{3}{4\mathrm{n}}-1\right]. \\ \mathrm{ii})\mathrm{E}=-\frac{13.6\mathrm{eV}}{{\mathrm{n}}^2}+{\mathrm{\mu }}_{\mathrm{B}}{\mathrm{B}}_{\mathrm{ext}}+\frac{13.6\mathrm{eV}}{{\mathrm{n}}^3}{\mathrm{\alpha }}^2\left[\frac{3}{4\mathrm{n}}-1\right]. \\ \mathrm{iii})\mathrm{E}=-\frac{13.6\mathrm{eV}}{{\mathrm{n}}^2}-2{\mathrm{\mu }}_{\mathrm{B}}{\mathrm{B}}_{\mathrm{ext}}+\frac{13.6\mathrm{eV}}{{\mathrm{n}}^3}{\mathrm{\alpha }}^2\left[\frac{3}{4\mathrm{n}}-\frac{3}{2}\right]. \\ \mathrm{iv})\mathrm{E}=-\frac{13.6\mathrm{eV}}{{\mathrm{n}}^2}+\frac{13.6\mathrm{eV}}{{\mathrm{n}}^3}{\mathrm{\alpha }}^2\left[\frac{3}{4\mathrm{n}}-\frac{5}{6}\right]. \\ \mathrm{v})\mathrm{E}=-\frac{13.6\mathrm{eV}}{{\mathrm{n}}^2}-{\mathrm{\mu }}_{\mathrm{B}}{\mathrm{B}}_{\mathrm{ext}}+\frac{13.6\mathrm{eV}}{{\mathrm{n}}^3}{\mathrm{\alpha }}^2\left[\frac{3}{4\mathrm{n}}-\frac{2}{3}\right]. \\ \mathrm{vi})\mathrm{E}=-\frac{13.6\mathrm{eV}}{{\mathrm{n}}^2}+{\mathrm{\mu }}_{\mathrm{B}}{\mathrm{B}}_{\mathrm{ext}}+\frac{13.6\mathrm{eV}}{{\mathrm{n}}^3}{\mathrm{\alpha }}^2\left[\frac{3}{4\mathrm{n}}-\frac{2}{3}\right]. \\ \mathrm{vii})\mathrm{E}=-\frac{13.6\mathrm{eV}}{{\mathrm{n}}^2}+2{\mathrm{\mu }}_{\mathrm{B}}{\mathrm{B}}_{\mathrm{ext}}+\frac{13.6\mathrm{eV}}{{\mathrm{n}}^3}{\mathrm{\alpha }}^2\left[\frac{3}{4\mathrm{n}}-\frac{5}{6}\right]. \\ \mathrm{viii})\mathrm{E}=-\frac{13.6\mathrm{eV}}{{\mathrm{n}}^2}+\frac{13.6\mathrm{eV}}{{\mathrm{n}}^3}{\mathrm{\alpha }}^2\left[\frac{3}{4\mathrm{n}}-\frac{1}{2}\right]. \end{gathered}$$

If fine structure is ignored,$\mathrm{a}=0$.

Therefore, the following energies:

$\mathrm{E}=-\frac{13.6\mathrm{eV}}{{\mathrm{n}}^2}-{\mathrm{\mu }}_{\mathrm{B}}{\mathrm{B}}_{\mathrm{ext}}$Degree of degeneracy: 2

$\mathrm{E}=-\frac{13.6\mathrm{eV}}{{\mathrm{n}}^2}+{\mathrm{\mu }}_{\mathrm{B}}{\mathrm{B}}_{\mathrm{ext}}$Degree of degeneracy: 2

$\mathrm{E}=-\frac{13.6\mathrm{eV}}{{\mathrm{n}}^2}$Degree of degeneracy: 2

$\mathrm{E}=-\frac{13.6\mathrm{eV}}{{\mathrm{n}}^2}-2{\mathrm{\mu }}_{\mathrm{B}}{\mathrm{B}}_{\mathrm{ext}}$Degree of degeneracy: 1

$\mathrm{E}=-\frac{13.6\mathrm{eV}}{{\mathrm{n}}^2}+2{\mathrm{\mu }}_{\mathrm{B}}{\mathrm{B}}_{\mathrm{ext}}$Degree of degeneracy: 1

Now only there are five distinct energies instead of eight.