The rest energy of an electron is 0.511 MeV. The electron is accelerated from rest to a velocity 0.5 c. The change in its energy will be

Options

A. 0.105 MeV
B. 0.079 MeV
C. 0.051 MeV
D. 0.026 MeV

Explanation

To solve the problem of finding the change in energy of an electron accelerated from rest to a velocity of 0.5c, we will use the concepts of relativistic energy.

### Step-by-Step Solution:

1. **Identify the Rest Energy**:
The rest energy (E₀) of the electron is given as 0.511 MeV.

2. **Calculate the Lorentz Factor (γ)**:
The Lorentz factor (γ) is given by the formula:
\[ \gamma = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}} \]
Here, \( v = 0.5c \). Therefore,
\[ \frac{v^2}{c^2} = (0.5)^2 = 0.25 \]
Now, substituting this into the Lorentz factor formula:
\[ \gamma = \frac{1}{\sqrt{1 - 0.25}} = \frac{1}{\sqrt{0.75}} = \frac{1}{\sqrt{\frac{3}{4}}} = \frac{2}{\sqrt{3}} \approx 1.155 \]

3. **Calculate the Total Energy (E)**:
The total energy (E) of the electron when it is moving at velocity 0.5c is given by:
\[ E = \gamma m_0 c^2 \]
Where \( m_0 \) is the rest mass of the electron. Since we know the rest energy \( E₀ = m_0 c^2 = 0.511 \text{ MeV} \), we can write:
\[ E = \gamma E_0 = 1.155 \times 0.511 \text{ MeV} \]
Now calculating this:
\[ E \approx 0.590 \text{ MeV} \]

4. **Calculate the Change in Energy (ΔE)**:
The change in energy (ΔE) is the difference between the total energy at velocity 0.5c and the rest energy:
\[ \Delta E = E - E_0 = 0.590 \text{ MeV} - 0.511 \text{ MeV} \approx 0.079 \text{ MeV} \]

### Final Answer:
The change in energy of the electron when accelerated from rest to a velocity of 0.5c is approximately **0.079 MeV**.