While working in a magnetics lab, you conduct an experiment in which a hydrogen atom in the n= 1 state is in a magnetic field of magnitude B. A photon of wavelength l (in air) is absorbed in a transition from the m_s= - 1/2 to the m_s= + 1/2 state. The wavelengths l as a function of Bare given in the table.

(a) Graph the data in the table as photon frequency fversus B, where f= c/λ. Find the slope of the straight line that gives the best fit to the data. (b) Use your results of part (a) to calculate $\mathbf{\left|}{\mathbf{\mu }}_{\mathbf{z}}\mathbf{\right|}$, the magnitude of the spin magnetic moment. (c) Let $\mathit{\gamma }\mathbf{=}\frac{\left|{\mu }_z\right|}{\left|S_z\right|}$denote the gyromagnetic ratio for electron spin. Use your result of part (b) to calculate $\mathit{\gamma }$. What is the value of $\mathit{\gamma }\mathbf{/}\mathbf{(}\mathit{e}\mathbf{/}\mathbf{2}\mathbf{}\mathit{m}\mathbf{(}\mathbf{\left)}\mathbf{\right)}$given by your experimental data?

First, the table of frequency and B has to be determined from the given wavelength values in the table given in question via the expression,

$\mathrm{f}=\frac{\mathrm{c}}{\mathrm{\lambda }}$

For

$$\begin{gathered} \lambda =0.0214\mathrm{m} \\ \mathrm{f}=\frac{2.998\times {10}^8\mathrm{m}/\mathrm{s}}{0.0214\mathrm{m}} \\ =1.40\times {10}^{10}\mathrm{Hz} \end{gathered}$$

Similar calculations for all other wavelength values will produce the following table as a result,

The graph can now be plotted as

The slope that best fits the straight line is $2.84\times {10}^{10}\mathrm{Hz}/\mathrm{T}$

The energy difference between the two levels is the energy of the photon, E = hf

$$\begin{gathered} ∆\mathrm{E}=2{\mathrm{\mu }}_{\mathrm{z}}\mathrm{B} \\ =\mathrm{hf} \end{gathered}$$

Solve for f,

$\mathrm{f}=\left(\frac{2{\mathrm{\mu }}_{\mathrm{z}}}{\mathrm{h}}\right)\mathrm{B}$

From part (a), the f versus B graph is a straight line and so from the above relation, $\left(\frac{2{\mathrm{\mu }}_{\mathrm{z}}}{\mathrm{h}}\right)$is the slope.

Solve for the spin magnetic moment,

$$\begin{gathered} {\mu }_z=\frac{1}{2}h(slope) \\ =\frac{1}{2}\left(6.626\times {10}^{-34}\mathrm{J}\cdot \mathrm{s}\right)\left(2.84\times {10}^{10}\mathrm{Hz}/\mathrm{T}\right) \\ =9.41\times {10}^{-24}\mathrm{J}/\mathrm{T} \end{gathered}$$

Thus, the value of the spin magnetic moment is $9.41\times {10}^{-24}\hspace{0.33em}\mathrm{J}/\mathrm{T}$.

From given question,

$$\begin{gathered} \gamma =\frac{\left|{\mu }_z\right|}{\left|S_z\right|} \\ =\frac{\left|{\mu }_z\right|}{ħ/2} \\ =\frac{2\left(9.41\times {10}^{-24}\mathrm{J}/T\right)}{1.055\times {10}^{-34}\mathrm{J}\cdot \mathrm{s}} \\ =1.784\times {10}^{11}\mathrm{Hz}/\mathrm{T} \end{gathered}$$

Also,

$$\begin{gathered} \frac{\gamma }{\mathrm{e}/2\mathrm{m}}=\frac{2\gamma \mathrm{m}}{\mathrm{e}} \\ =\frac{2\left(1.784\times {10}^{11}\mathrm{Hz}/\mathrm{T}\right)\left(9.11\times {10}^{-31}\mathrm{kg}\right)}{1.602\times {10}^{-19}\mathrm{C}} \\ =2.03 \end{gathered}$$