Question

A small object at rest, absorbs a light pulse of power $20 \,mW$ and duration $300\, ns$. Assuming speed of light as $3 \times 10^8 \,m / s$, the momentum of the object becomes equal to :

• A. $2 \times 10^{-17} kg \,m / s$
• B. $1 \times 10^{-17} kg \,m / s$
• C. $3 \times 10^{-17} kg\, m / s$
• D. $0.5 \times 10^{-17} kg \,m / s$

Answer 1

The Correct Option is A

To solve this problem, we need to determine the momentum gained by the object after absorbing a light pulse. The momentum of a light pulse can be related to its energy through the formula:

\(p = \frac{E}{c}\)

where \(p\) is the momentum, \(E\) is the energy of the light pulse, and \(c\) is the speed of light.

Let's calculate the energy of the light pulse. Given that the power \(P\) is 20 mW and the duration \(t\) is 300 ns:

1. Convert the power to watts and the time to seconds for consistency in units:

• \(P = 20 \, \text{mW} = 20 \times 10^{-3} \, \text{W} = 0.02 \, \text{W}\)
• \(t = 300 \, \text{ns} = 300 \times 10^{-9} \, \text{s} = 3 \times 10^{-7} \, \text{s}\)

2. Calculate the energy \(E\) of the pulse using the formula \(E = P \times t\):

• \(E = 0.02 \, \text{W} \times 3 \times 10^{-7} \, \text{s} = 6 \times 10^{-9} \, \text{J}\)

3. Using the momentum formula \(p = \frac{E}{c}\) and substituting the values:

• \(c = 3 \times 10^{8} \, \text{m/s}\)
• \(p = \frac{6 \times 10^{-9} \, \text{J}}{3 \times 10^{8} \, \text{m/s}}\)
• \(p = 2 \times 10^{-17} \, \text{kg} \, \text{m/s}\)

Therefore, the momentum of the object becomes equal to \(2 \times 10^{-17} \, \text{kg} \, \text{m/s}\), which matches the correct answer.