if an electron and a positron annihilitate , them the energy relasesed is

Options

A. `3.2xx10^(-13)J`
B. `1.6xx10^(-13)J`
C. `4.8xx10^(-13)J`
D. `6.4x10^(-13)J`

Explanation

To find the energy released when an electron and a positron annihilate, we can use the principles of mass-energy equivalence as described by Einstein's equation, \(E = mc^2\).

### Step-by-Step Solution:

1. **Understand the Annihilation Process**:
- An electron (with mass \(m_e\)) and a positron (which has the same mass as the electron) annihilate each other. This process converts their mass into energy.

2. **Identify the Mass of the Electron and Positron**:
- The mass of an electron (and positron) is approximately \(m_e = 9.1 \times 10^{-31} \, \text{kg}\).

3. **Calculate the Total Mass Involved**:
- Since both the electron and positron are involved in the annihilation, the total mass \(m\) is:
\[ m = m_e + m_e = 2m_e = 2 \times 9.1 \times 10^{-31} \, \text{kg} = 1.82 \times 10^{-30} \, \text{kg} \]

4. **Use Einstein's Equation**:
- According to Einstein's mass-energy equivalence principle, the energy \(E\) released during the annihilation can be calculated using:
\[ E = mc^2 \]
- Where \(c\) is the speed of light, approximately \(c = 3 \times 10^8 \, \text{m/s}\).

5. **Substitute the Values**:
- Plugging in the values:
\[ E = (1.82 \times 10^{-30} \, \text{kg}) \times (3 \times 10^8 \, \text{m/s})^2 \]
- Calculate \(c^2\):
\[ c^2 = (3 \times 10^8)^2 = 9 \times 10^{16} \, \text{m}^2/\text{s}^2 \]
- Now substitute \(c^2\) into the energy equation:
\[ E = 1.82 \times 10^{-30} \times 9 \times 10^{16} \]

6. **Final Calculation**:
- Calculate the energy:
\[ E = 1.638 \times 10^{-13} \, \text{J} \]
- Rounding this gives approximately:
\[ E \approx 1.6 \times 10^{-13} \, \text{J} \]

### Conclusion:
The energy released when an electron and a positron annihilate is approximately \(1.6 \times 10^{-13} \, \text{J}\).