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Q13P

Expert-verifiedFound in: Page 1247

Book edition
10th Edition

Author(s)
David Halliday

Pages
1328 pages

ISBN
9781118230718

**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?**

The acceleration of the silver atom as it passes through the deflecting magnet in the experiment is $7.2\times {10}^{4}\mathrm{m}/{\mathrm{s}}^{2}$.

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,

$\mathrm{F}={\mathrm{\mu}}_{\mathrm{B}}=\frac{\mathrm{dB}}{\mathrm{dz}}\phantom{\rule{0ex}{0ex}}\mathrm{ma}={\mathrm{\mu}}_{\mathrm{B}}=\frac{\mathrm{dB}}{\mathrm{dz}}$

….. (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, $\mathrm{m}=1.8\times {10}^{-25}\mathrm{kg}$)

$\mathrm{a}=\frac{{\mathrm{\mu}}_{\mathrm{s}}\frac{\mathrm{dB}}{\mathrm{dz}}}{\mathrm{m}}\phantom{\rule{0ex}{0ex}}=\frac{\left(9.27\times {10}^{-24}\mathrm{J}/\mathrm{T}\right)\times 1.4\times {10}^{3}\mathrm{T}/\mathrm{m}}{1.8\times {10}^{-25}\mathrm{kg}}\phantom{\rule{0ex}{0ex}}=7.2\times {10}^{4}\mathrm{m}/{\mathrm{s}}^{2}$

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

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