The momentum of a photon is `2 xx 10^(-16) gm - cm//sec`. Its energy is

Options

A. `0.61xx10^(-26)` erg
B. `2.0xx10^(-26)` erg
C. `6xx10^(-6) ` erg
D. `6xx10^(-8)` erg

Answer

To find the energy of a photon given its momentum, we can use the relationship between energy (E), momentum (p), and the speed of light (c). The formula we will use is:

\[ E = p \cdot c \]

### Step-by-Step Solution:

1. **Identify the given values:**
- Momentum \( p = 2 \times 10^{-16} \) gm-cm/s
- Speed of light \( c = 3 \times 10^{10} \) cm/s (since we are using the momentum in gm-cm/s, we will keep the speed of light in cm/s).

2. **Substitute the values into the energy equation:**
\[ E = p \cdot c \]
\[ E = (2 \times 10^{-16} \, \text{gm-cm/s}) \cdot (3 \times 10^{10} \, \text{cm/s}) \]

3. **Perform the multiplication:**
\[ E = 2 \times 3 \times 10^{-16} \times 10^{10} \]
\[ E = 6 \times 10^{-6} \, \text{gm-cm}^2/\text{s}^2 \]

4. **Convert the energy to appropriate units if necessary:**
- Since \( 1 \, \text{gm-cm}^2/\text{s}^2 = 10^{-7} \, \text{J} \), we can express the energy in Joules:
\[ E = 6 \times 10^{-6} \, \text{gm-cm}^2/\text{s}^2 = 6 \times 10^{-6} \times 10^{-7} \, \text{J} = 6 \times 10^{-13} \, \text{J} \]

5. **Final answer:**
The energy of the photon is \( 6 \times 10^{-6} \, \text{gm-cm}^2/\text{s}^2 \) or \( 6 \times 10^{-13} \, \text{J} \).