A small object at rest,absorbs a light pulles of power `20mW` and duration `300ns`.Assuming speed of light is `3times10^(8)m/s`,the momentum of the object become equal to:

Options

A. `2times10^(-17)kgm/s`
B. `1times10^(-17)kgm/s`
C. `3times10^(-17)kgm/s`
D. `0.5times10^(-17)kgm/s`

Answer

To solve the problem, we need to find the momentum of an object that absorbs a light pulse. The steps to solve this problem are as follows:

### Step 1: Understand the relationship between power, energy, and time.
Power (P) is defined as the rate of energy transfer over time. The relationship can be expressed as:
\[ P = \frac{E}{t} \]
where \( E \) is the energy absorbed and \( t \) is the duration.

### Step 2: Calculate the energy absorbed by the object.
Given:
- Power \( P = 20 \, \text{mW} = 20 \times 10^{-3} \, \text{W} \)
- Duration \( t = 300 \, \text{ns} = 300 \times 10^{-9} \, \text{s} \)

Using the formula for power:
\[ E = P \times t \]
Substituting the values:
\[ E = (20 \times 10^{-3} \, \text{W}) \times (300 \times 10^{-9} \, \text{s}) \]
\[ E = 20 \times 300 \times 10^{-3} \times 10^{-9} \]
\[ E = 6000 \times 10^{-12} \]
\[ E = 6 \times 10^{-9} \, \text{J} \]

### Step 3: Relate energy to momentum.
For massless particles like photons, the relationship between energy (E) and momentum (p) is given by:
\[ E = p \cdot c \]
where \( c \) is the speed of light.

### Step 4: Solve for momentum.
Rearranging the equation to find momentum:
\[ p = \frac{E}{c} \]
Substituting the values:
- \( E = 6 \times 10^{-9} \, \text{J} \)
- \( c = 3 \times 10^{8} \, \text{m/s} \)

Calculating momentum:
\[ p = \frac{6 \times 10^{-9}}{3 \times 10^{8}} \]
\[ p = 2 \times 10^{-17} \, \text{kg m/s} \]

### Final Answer:
The momentum of the object becomes equal to \( 2 \times 10^{-17} \, \text{kg m/s} \).

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