Question

A pulse of light in a duration of $50\text{ns}$ is absorbed completely by a small object initially at rest. Power of the pulse is $300\text{mW}$ and the speed of light $3\times {10}^8\text{m/s}$. The change in momentum of the object is

• A. $5\times {10}^{-16}\text{kg m/s}$
• B. $5\times {10}^{-17}\text{kg m/s}$
• C. $5\times {10}^{-18}\text{kg m/s}$
• D. $5\times {10}^{-20}\text{kg m/s}$

Answer 1

The correct option is B $5\times {10}^{-17}\text{kg m/s}$

Given:
$t=50\times {10}^{-9}\text{s},P=300\times {10}^{-3}\text{W},c=3\times {10}^8\text{m/s}$

As, $\text{Total Incident energy}E=Pt$

$\lambda =\frac{h}{p}$ and $E=\frac{hc}{\lambda }$

$\therefore$ From above equations, the momentum of the pulse of light

$p=\frac{E}{c}=\frac{Pt}{c}$

As, the pulse of light is completely absorbed by the object. Therefore, the change in momentum of the object is equal to the momentum of the pulse of light.


$\therefore p=\frac{300\times {10}^{-3}\times 50\times {10}^{-9}}{3\times {10}^8}=5\times {10}^{-17}\text{kgm/s}$

Hence, option $\left(\text{B}\right)$ is correct.