What is the force exerted by a photon of intensity `1.4 k W m^(-2)` if it falls on a perfect absorber of radius 2 m ?

Options

A. `58.66xx10^(-6)N`
B. `10^8N`
C. `8.35xx10^(4)N`
D. `8.8xx10^(-8)N`

Answer

To find the force exerted by a photon of intensity \(1.4 \, \text{kW/m}^2\) on a perfect absorber of radius \(2 \, \text{m}\), we can follow these steps:

### Step 1: Calculate the Area of the Absorber
The area \(A\) of a circle is given by the formula:
\[ A = \pi r^2 \]
where \(r\) is the radius.

Given \(r = 2 \, \text{m}\):
\[ A = \pi (2)^2 = 4\pi \, \text{m}^2 \]

### Step 2: Convert Intensity to Standard Units
The intensity \(I\) is given as \(1.4 \, \text{kW/m}^2\). We need to convert this to watts:
\[ 1.4 \, \text{kW} = 1.4 \times 10^3 \, \text{W/m}^2 = 1400 \, \text{W/m}^2 \]

### Step 3: Use the Formula for Force
The force \(F\) exerted by the photons on the absorber can be calculated using the formula:
\[ F = \frac{I \cdot A}{c} \]
where:
- \(I\) is the intensity,
- \(A\) is the area,
- \(c\) is the speed of light (\(c \approx 3 \times 10^8 \, \text{m/s}\)).

### Step 4: Substitute the Values into the Formula
Substituting the values we have:
\[ F = \frac{(1400 \, \text{W/m}^2) \cdot (4\pi \, \text{m}^2)}{3 \times 10^8 \, \text{m/s}} \]

### Step 5: Calculate the Area
Calculating \(4\pi\):
\[ 4\pi \approx 4 \times 3.14 \approx 12.56 \, \text{m}^2 \]

### Step 6: Substitute and Calculate Force
Now substituting back into the force equation:
\[ F = \frac{(1400) \cdot (12.56)}{3 \times 10^8} \]
Calculating the numerator:
\[ 1400 \cdot 12.56 \approx 17584 \, \text{W} \]
Now substituting this into the force equation:
\[ F = \frac{17584}{3 \times 10^8} \approx 5.86 \times 10^{-5} \, \text{N} \]

### Final Answer
The force exerted by the photon on the perfect absorber is approximately:
\[ F \approx 5.86 \times 10^{-5} \, \text{N} \]