Question

Using ${\mathit{f}}^{\mathbf{2}}\mathbf{=}{\mathit{L}}^{\mathbf{2}}\mathbf{+}{\mathit{S}}^{\mathbf{2}}\mathbf{+}\mathbf{2}\mathit{L}\mathbf{-}\mathit{S}$to eliminate L - S. as wellas $\mathit{L}\mathbf{=}\sqrt{\mathbf{l}\mathbf{(}\mathbf{l}\mathbf{+}\mathbf{1}\mathbf{)}\mathbf{h}\mathbf{,}\mathbf{S}}\mathbf{=}\sqrt{\mathbf{s}\mathbf{(}\mathbf{s}\mathbf{+}\mathbf{1}\mathbf{)}}\mathbf{}\mathbf{and}\mathbf{}\sqrt{\mathbf{j}\mathbf{(}\mathbf{j}\mathbf{+}\mathbf{1}\mathbf{)}\mathbf{h}}$, obtain equation (8- 32 )from the equation that precedes it.

Short answer

answer is${\mu }_{j_{avg}}=\frac{e\sout{h}}{2m}\left(\frac{3j\left(j+1\right)-l(l+1)+s(s+1)}{2\sqrt{j(j+1)}}\right)$

Step-by-step solution

Concept of the Average, Square and Magnitude of the Angular Momentum.

The average total magnetic dipole moment${\mathbf{\mu }}_{{\mathbf{j}}_{\mathbf{avg}}}$is given by

${\mathit{\mu }}_{{\mathbf{j}}_{\mathbf{a}\mathbf{v}\mathbf{g}}}\mathbf{=}\frac{\mathbf{e}}{\mathbf{2}\mathbf{m}}\frac{{\mathbf{L}}^{\mathbf{2}}\mathbf{+}\mathbf{2}{\mathbf{S}}^{\mathbf{2}}\mathbf{+}\mathbf{3}\overrightarrow{\mathbf{L}}\mathbf{-}\overrightarrow{\mathbf{S}}}{\mathbf{J}}$ …(1)

The square of the total angular momentum${\mathbf{J}}^{\mathbf{2}}$is given by

${\mathit{J}}^{\mathbf{2}}\mathbf{=}{\mathit{L}}^{\mathbf{2}}\mathbf{+}{\mathit{S}}^{\mathbf{2}}\mathbf{+}\mathbf{2}\overrightarrow{\mathbf{L}}\mathbf{-}\overrightarrow{\mathbf{S}}$ …(2)

The magnitudes of the angular momenta J,Land Sare given by

$\mathit{J}\mathbf{=}\sqrt{\mathbf{j}\mathbf{(}\mathbf{j}\mathbf{+}\mathbf{1}\mathbf{)}}\sout{\mathbf{h}}$ …(3)

$\mathit{L}\mathbf{=}\sqrt{\mathbf{l}\mathbf{(}\mathbf{l}\mathbf{+}\mathbf{1}\mathbf{)}}\sout{\mathbf{h}}$ …(4)

$\mathit{S}\mathbf{=}\sqrt{\mathit{s}\mathbf{(}\mathit{s}\mathbf{+}\mathbf{1}\mathbf{)}}\sout{\mathbf{h}}$ …(5)

Determine the equation

From Eq. (2) we can write

$i-\overrightarrow{S}=\frac{1}{2}\left(J^2-L^2-S^2\right)$ …(6)

By plugging in Eq. (3) into Eq. (1) we get

$$\begin{gathered} {\mu }_{j_{avg}}=\frac{e}{2m}\left(\frac{L^2+2S^2+{\displaystyle \frac{3}{2}}{J}^2-{\displaystyle \frac{3}{2}}{L}^2-{\displaystyle \frac{3}{2}}{S}^2}{J}\right) \\ =\frac{e}{2m}\left(\frac{3J^2-L^2+S^2}{2J}\right) \end{gathered}$$ …(7)

By plugging in Eqs. (3), (4) and (5) into Eq. (7) we get

${\mu }_{j_{vg}}=\frac{e\sout{h}}{2m}\left(\frac{2j(j+1)-l(l+1)+s(s+1)}{2\sqrt{j(j+1)}}\right)$