Question

A photon is completely reflected by a surface. If $4\times {10}^{12}$ such photons, having a wavelength of $0.8\text{nm}$ are bouncing back from the surface, the change in momentum of the surface is,

• A. $3.32\times {10}^{-9}\text{Ns}$
• B. $3.3\times {10}^{-6}\text{Ns}$
• C. $3.32\times {10}^{15}\text{Ns}$
• D. $6.64\times {10}^{-12}\text{Ns}$

Answer 1

The correct option is D $6.64\times {10}^{-12}\text{Ns}$

$n=4\times {10}^{12}$
$\lambda =0.8\text{nm}$
The change in momentum due to collision with one photon $=2p$

$\therefore \Delta p=2\times \frac{nh}{\lambda }=\frac{2\times 4\times {10}^{12}\times 6.63\times {10}^{-34}}{0.8\times {10}^{-9}}$

$=6.63\times {10}^{-12}\text{Ns}$

Hence, $\left(D\right)$ is the correct answer.