A beam of ligh incident on a surface has photons each of energy 1 mJ and intensity `25w//cm^(2)`. Find number of photons incident per second if surface area of `25cm^(2)`

Options

A. `6.3x10^(5)/s
B. `1.4xx10^(6)/s
C. `5.0xx10^(3)s`
D. `3.5xx10^(-6)Ns`

Answer

To solve the problem, we need to find the number of photons incident per second on a surface given the energy of each photon, the intensity of the light, and the surface area. Here's how we can approach the solution step-by-step:

### Step-by-Step Solution:

1. **Identify Given Values**:
- Energy of each photon, \( E = 1 \text{ mJ} = 1 \times 10^{-3} \text{ J} \)
- Intensity of the light, \( I = 25 \text{ W/cm}^2 \)
- Surface area, \( A = 25 \text{ cm}^2 \)

2. **Convert Units**:
- Since intensity is given in watts per square centimeter, we can keep it as is for our calculations. However, we will convert the area into square meters for consistency in SI units.
- \( A = 25 \text{ cm}^2 = 25 \times 10^{-4} \text{ m}^2 = 0.0025 \text{ m}^2 \)

3. **Calculate Power**:
- Power \( P \) can be calculated using the formula:
\[ P = I \times A \]
- Substitute the values:
\[ P = 25 \text{ W/cm}^2 \times 25 \text{ cm}^2 = 625 \text{ W} \]

4. **Calculate Number of Photons per Second**:
- The number of photons \( n \) incident per second can be calculated using the formula:
\[ n = \frac{P}{E} \]
- Substitute the values:
\[ n = \frac{625 \text{ W}}{1 \times 10^{-3} \text{ J}} = 625000 \text{ photons/s} \]
- This can also be expressed in scientific notation:
\[ n = 6.25 \times 10^5 \text{ photons/s} \]

5. **Final Answer**:
- The number of photons incident per second is \( 6.25 \times 10^5 \text{ photons/s} \).