Question

If I=0, then j=s,${\mathit{m}}_{\mathbf{j}}\mathbf{=}{\mathit{m}}_{\mathbf{s}}$, and the "good" states are the same $\left(\overline{)n{m}_s}\right)$for weak and strong fields. Determine${\mathit{E}}_{\mathbf{z}}^{\mathbf{1}}$(from Equation) and the fine structure energies (Equation 6.67), and write down the general result for the I=O Zeeman Effect - regardless of the strength of the field. Show that the strong field formula (Equation 6.82) reproduces this result, provided that we interpret the indeterminate term in square brackets as.

Short answer

The ${\mathrm{E}}_{\mathrm{Z}}^1=\frac{\mathrm{e}}{2\mathrm{m}}{\mathrm{B}}_{\mathrm{ext}}2{\mathrm{m}}_{\mathrm{s}}\mathrm{ħ}$and the fine structure energies is

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

Step-by-step solution

Definition of Zeeman Effect.

In the presence of a static magnetic field, the Zeeman Effect causes a spectral line to break into numerous components.

Step2: Structural isomers of carboxylic acids.

Use the equation 6.72,

$$\begin{gathered} E_z^1=\frac{e}{2m}{B}_{ext}.\left(L+2S\right) \\ =\frac{e}{2m}{\mathrm{B}}_{\mathrm{ext}}2{\mathrm{m}}_{\mathrm{s}}\mathrm{ħ} \end{gathered}$$

Fine structure energy (with j=1/2),

$$\begin{gathered} {\mathrm{E}}_{\mathrm{nj}}=-\frac{13.6eV}{n^2}\left[1+\frac{{\alpha }^2}{n^2}\left(n-\frac{3}{4}\right)\right] \\ {\mathrm{E}}_{\mathrm{tot}}=-\frac{13.6\mathrm{eV}}{{\mathrm{n}}^2}\left[1+\frac{{\mathrm{\alpha }}^2}{{\mathrm{n}}^2}\left(\mathrm{n}-\frac{3}{4}\right)\right]+2{\mathrm{m}}_{\mathrm{S}}{\mathrm{B}}_{\mathrm{ext}}\frac{\mathrm{eħ}}{2\mathrm{m}} \end{gathered}$$

Fine structure term is proportional to ${\alpha }^2$;

$$\begin{gathered} E_{fs}^1=-\frac{13.6eV}{n^4}{\alpha }^2\left(n-\frac{3}{4}\right) \\ =\frac{13.6\mathrm{eV}}{{\mathrm{n}}^3}{\mathrm{\alpha }}^2\left(\frac{3}{4\mathrm{n}}-1\right) \end{gathered}$$

{Same as equation } with the term in square brackets set equal to 1.

$E_{fs}^1=\frac{13.6eV}{n^3}{\alpha }^2\left\{\frac{3}{4n}-\left[\frac{I\left(I+1\right)-m_I{m}_s}{I\left(I+1/2\right)\left(I+1\right)}\right]\right\}$