What is the acceleration of a silver atom as it passes through the deflecting magnet in the Stern–Gerlach experiment of Fig. 40-8 if the magnetic field gradient is 1.4 T/mm?

Gradient of the magnetic field, $\frac{\mathrm{dB}}{\mathrm{dz}}=1.4\mathrm{T}/\mathrm{mm}$

Using the formula of force from second Newton's law in the force value of the Stern-Gerlach Experiment, we can get the required acceleration by substituting the obtained data from the experiment.

Formulas:

The force is due to Newton’s second law of motion,

F = ma

….. (1)

The force obtained due to gradient change in a magnetic field,

$$\begin{gathered} \mathrm{F}={\mathrm{\mu }}_{\mathrm{B}}=\frac{\mathrm{dB}}{\mathrm{dz}} \\ \mathrm{ma}={\mathrm{\mu }}_{\mathrm{B}}=\frac{\mathrm{dB}}{\mathrm{dz}} \end{gathered}$$

….. (2)

Where, the Bohr magneton,${\mathrm{\mu }}_{\mathrm{B}}=9.27\times {10}^{-24}\mathrm{J}/\mathrm{T}$

Using the value of force from equation (2) in equation (1), the value of the acceleration of the silver atoms as follows: (Using the data given in the Sample problem, $\mathrm{m}=1.8\times {10}^{-25}\mathrm{kg}$)

$$\begin{gathered} \mathrm{a}=\frac{{\mathrm{\mu }}_{\mathrm{s}}{\displaystyle \frac{\mathrm{dB}}{\mathrm{dz}}}}{\mathrm{m}} \\ =\frac{\left(9.27\times {10}^{-24}\mathrm{J}/\mathrm{T}\right)\times 1.4\times {10}^3\mathrm{T}/\mathrm{m}}{1.8\times {10}^{-25}\mathrm{kg}} \\ =7.2\times {10}^4\mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$

Hence, the value of the acceleration of the silver atom is$7.2\times {10}^4\mathrm{m}/{\mathrm{s}}^2$ .