The energy in MeV released due to transformation of 1 kg mass completely into energy, is (`c= 3 xx 10^8` m/s)

Options

A. `7.625xx10^9` MeV
B. `29xx10^5` MeV
C. `2.8xx10^(-28)` MeV
D. `5.625xx10^29` MeV

Answer

To solve the problem of finding the energy released when 1 kg of mass is completely transformed into energy, we will use Einstein's famous equation \(E = mc^2\).

### Step-by-Step Solution:

1. **Identify the mass (m)**:
Given that the mass is \(1 \text{ kg}\).

2. **Identify the speed of light (c)**:
Given that \(c = 3 \times 10^8 \text{ m/s}\).

3. **Apply Einstein's equation**:
We will substitute the values of mass and speed of light into the equation:
\[ E = mc^2 = 1 \text{ kg} \times (3 \times 10^8 \text{ m/s})^2 \]

4. **Calculate \(c^2\)**:
\[ c^2 = (3 \times 10^8)^2 = 9 \times 10^{16} \text{ m}^2/\text{s}^2 \]

5. **Calculate the energy (E)**:
\[ E = 1 \text{ kg} \times 9 \times 10^{16} \text{ m}^2/\text{s}^2 = 9 \times 10^{16} \text{ Joules} \]

6. **Convert Joules to electron volts**:
To convert Joules to electron volts, we use the conversion factor \(1 \text{ eV} = 1.6 \times 10^{-19} \text{ Joules}\):
\[ E (\text{in eV}) = \frac{9 \times 10^{16} \text{ Joules}}{1.6 \times 10^{-19} \text{ Joules/eV}} = 5.625 \times 10^{35} \text{ eV} \]

7. **Convert electron volts to mega electron volts**:
Since \(1 \text{ MeV} = 10^6 \text{ eV}\), we convert the energy to MeV:
\[ E (\text{in MeV}) = \frac{5.625 \times 10^{35} \text{ eV}}{10^6} = 5.625 \times 10^{29} \text{ MeV} \]

### Final Answer:
The energy released due to the transformation of 1 kg mass completely into energy is \(5.625 \times 10^{29} \text{ MeV}\).

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