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Found in: Page 1247

### Fundamentals Of Physics

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 $7.2×{10}^{4}\mathrm{m}/{\mathrm{s}}^{2}$.

See the step by step solution

## Step 1: The given data:

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

## Step 2: Understanding the concept of force of a Stern-Gerlach Experiment:

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, ${\mathrm{\mu }}_{\mathrm{B}}=9.27×{10}^{-24}\mathrm{J}/\mathrm{T}$

## Step 3: Calculation of the value of the acceleration of the silver atom:

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, $\mathrm{m}=1.8×{10}^{-25}\mathrm{kg}$)

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