Problem

An electron, with a mass of $9.11 \times 10^{-31} \, \mathrm{kg}$, moves with a speed of $0.75c$. Find its relativistic momentum and compare this value with the momentum calculated from the classical expression.

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Data

• Rest mass of electron: $m_0 = 9.11 \times 10^{-31} \, \mathrm{kg}$
• Speed of electron: $v = 0.75c$
• Speed of light: $c = 3 \times 10^8 \, \mathrm{ms}^{-1}$

To Find:

1. Relativistic momentum ($p_{\text{rel}}$)
2. Classical momentum ($p_{\text{class}}$)

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Prerequisite Concepts

1. Relativistic momentum:

$$p_{\text{rel}} = \frac{m_0 v}{\sqrt{1 - \frac{v^2}{c^2}}}$$

where $m_0$ is the rest mass, $v$ is the speed of the electron, and $c$ is the speed of light.

2. Classical momentum:

$$p_{\text{class}} = m_0 v$$

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Solution

Step 1: Relativistic Momentum

Using the formula for relativistic momentum:

$$p_{\text{rel}} = \frac{m_0 v}{\sqrt{1 - \frac{v^2}{c^2}}}$$

Substitute the given values:

$$p_{\text{rel}} = \frac{(9.11 \times 10^{-31}) \times (0.75 \times 3 \times 10^8)}{\sqrt{1 - \frac{(0.75c)^2}{c^2}}}$$

Simplify the denominator:

$$\sqrt{1 - \frac{(0.75)^2 c^2}{c^2}} = \sqrt{1 - 0.5625} = \sqrt{0.4375} = 0.6614$$

Now calculate:

$$p_{\text{rel}} = \frac{(9.11 \times 10^{-31}) \times (0.75 \times 3 \times 10^8)}{0.6614}$$ $$p_{\text{rel}} = \frac{2.05 \times 10^{-22}}{0.6614} = 3.10 \times 10^{-22} \, \mathrm{kg \, ms^{-1}}$$

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Step 2: Classical Momentum

Using the classical formula:

$$p_{\text{class}} = m_0 v$$

Substitute the values:

$$p_{\text{class}} = (9.11 \times 10^{-31}) \times (0.75 \times 3 \times 10^8)$$ $$p_{\text{class}} = 2.05 \times 10^{-22} \, \mathrm{kg \, ms^{-1}}$$

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Answer

1. Relativistic Momentum:

$$p_{\text{rel}} = 3.10 \times 10^{-22} \, \mathrm{kg \, ms^{-1}}$$

2. Classical Momentum:

$$p_{\text{class}} = 2.05 \times 10^{-22} \, \mathrm{kg \, ms^{-1}}$$

Comparison:

The relativistic momentum is approximately $60\%$ greater than the classical momentum, highlighting the importance of relativistic effects at high speeds.