Problem

An electron moves with a speed of $v = 0.85c$. Find its total energy ($E$) and kinetic energy ($KE$) in electron volts.

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Data

• Speed of electron: $v = 0.85c$
• Rest mass of electron: $m_0 = 9.11 \times 10^{-31} \, \mathrm{kg}$
• Speed of light: $c = 3 \times 10^8 \, \mathrm{ms^{-1}}$

To Find:

1. Total energy ($E$) in $\mathrm{eV}$
2. Kinetic energy ($KE$) in $\mathrm{eV}$

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Prerequisite Concepts

1. Rest mass energy:

$$E_0 = m_0 c^2$$

2. Total energy of a relativistic particle:

$$E = \frac{E_0}{\sqrt{1 - \frac{v^2}{c^2}}}$$

3. Kinetic energy:

$$KE = E - E_0$$

4. Conversion from Joules to electron volts:

$$1 \, \mathrm{J} = 1.602 \times 10^{-19} \, \mathrm{eV}$$

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Solution

Step 1: Rest Mass Energy

Using the formula:

$$E_0 = m_0 c^2$$

Substitute the values:

$$E_0 = (9.11 \times 10^{-31}) \times (3 \times 10^8)^2$$ $$E_0 = 8.1918 \times 10^{-14} \, \mathrm{J}$$

Convert to $\mathrm{eV}$:

$$E_0 = \frac{8.1918 \times 10^{-14}}{1.602 \times 10^{-19}} \, \mathrm{eV}$$ $$E_0 = 0.511 \, \mathrm{MeV}$$

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Step 2: Total Energy

Using the relativistic energy formula:

$$E = \frac{E_0}{\sqrt{1 - \frac{v^2}{c^2}}}$$

Substitute the values:

$$E = \frac{0.511}{\sqrt{1 - \frac{(0.85c)^2}{c^2}}}$$ $$E = \frac{0.511}{\sqrt{1 - 0.7225}}$$ $$E = \frac{0.511}{\sqrt{0.2775}} = \frac{0.511}{0.527}$$ $$E = 0.970 \, \mathrm{MeV}$$

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Step 3: Kinetic Energy

Using the relationship:

$$KE = E - E_0$$

Substitute the values:

$$KE = 0.970 - 0.511$$ $$KE = 0.459 \, \mathrm{MeV}$$

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Answer

1. Total energy ($E$) of the electron:

$$E = 0.970 \, \mathrm{MeV}$$

2. Kinetic energy ($KE$) of the electron:

$$KE = 0.459 \, \mathrm{MeV}$$