The momentum of a photon of energy `1 MeV` "in"` kg m//s` will be

Options

A. `0.33 xx 10^(6)`
B. `7 xx 10^(-24)`
C. `10^(-22)`
D. `5 xx 10^(-22)`

Explanation

To find the momentum of a photon with energy \(1 \, \text{MeV}\), we can use the relationship between energy and momentum for photons. The momentum \(p\) of a photon can be calculated using the formula:

\[ p = \frac{E}{c} \]

where:
- \(E\) is the energy of the photon,
- \(c\) is the speed of light in a vacuum, approximately \(3 \times 10^8 \, \text{m/s}\).

### Step-by-Step Solution:

1. **Convert Energy from MeV to Joules**:
The energy given is \(1 \, \text{MeV}\). To convert this to joules, we use the conversion factor:
\[ 1 \, \text{MeV} = 1.6 \times 10^{-13} \, \text{J} \]
Therefore,
\[ E = 1 \, \text{MeV} = 1.6 \times 10^{-13} \, \text{J} \]

2. **Use the Formula for Momentum**:
Now, we can substitute the energy \(E\) into the momentum formula:
\[ p = \frac{E}{c} \]
Substituting the values:
\[ p = \frac{1.6 \times 10^{-13} \, \text{J}}{3 \times 10^8 \, \text{m/s}} \]

3. **Calculate the Momentum**:
Performing the division:
\[ p = \frac{1.6 \times 10^{-13}}{3 \times 10^8} = \frac{1.6}{3} \times 10^{-21} \, \text{kg m/s} \]
\[ p \approx 0.533 \times 10^{-21} \, \text{kg m/s} \]
\[ p \approx 5.33 \times 10^{-22} \, \text{kg m/s} \]

4. **Final Result**:
Therefore, the momentum of the photon is approximately:
\[ p \approx 5.33 \times 10^{-22} \, \text{kg m/s} \]

### Conclusion:
The momentum of a photon with energy \(1 \, \text{MeV}\) is approximately \(5.33 \times 10^{-22} \, \text{kg m/s}\).

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