Determine the speed at which the kinetic energy of an electron is equal to twice its rest energy.

Options

A. 0.45c
B. 0.87c
C. 0.63c
D. 0.94c

Answer

To determine the speed at which the kinetic energy of an electron is equal to twice its rest energy, we can follow these steps:

### Step 1: Write the expression for kinetic energy (KE)
The relativistic kinetic energy (KE) of a particle is given by the formula:
\[ KE = \gamma m_e c^2 - m_e c^2 \]
where \(\gamma = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}}\) is the Lorentz factor, \(m_e\) is the rest mass of the electron, and \(c\) is the speed of light.

### Step 2: Set the kinetic energy equal to twice the rest energy
The rest energy (\(E_0\)) of the electron is given by:
\[ E_0 = m_e c^2 \]
We want to find the speed at which the kinetic energy is twice the rest energy:
\[ KE = 2E_0 = 2m_e c^2 \]

### Step 3: Substitute the expression for KE into the equation
Substituting the expression for kinetic energy into the equation gives:
\[ \gamma m_e c^2 - m_e c^2 = 2m_e c^2 \]
This simplifies to:
\[ \gamma m_e c^2 = 3m_e c^2 \]

### Step 4: Divide both sides by \(m_e c^2\)
Dividing both sides by \(m_e c^2\) yields:
\[ \gamma = 3 \]

### Step 5: Substitute the expression for \(\gamma\)
Now, substituting the expression for \(\gamma\):
\[ \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}} = 3 \]

### Step 6: Solve for \(v^2\)
Squaring both sides gives:
\[ 1 - \frac{v^2}{c^2} = \frac{1}{9} \]
Rearranging this gives:
\[ \frac{v^2}{c^2} = 1 - \frac{1}{9} = \frac{8}{9} \]

### Step 7: Solve for \(v\)
Taking the square root of both sides:
\[ v = c \sqrt{\frac{8}{9}} = \frac{c \sqrt{8}}{3} = \frac{2\sqrt{2}}{3}c \approx 0.943c \]

### Conclusion
The speed at which the kinetic energy of an electron is equal to twice its rest energy is approximately \(0.943c\).