Question

Light with an energy flux of $9W/c{m}^2$ falls on a non-reflecting surface at normal incidence. If the surface has an area of 20 $c{m}^2$. The total momentum delivered for complete absorption in one hour is:
$$\begin{gathered} \text{A}\text{. 2}\text{.16}\times \text{1}{\text{0}}^{-4}kgm{s}^{-1} \\ \text{B}\text{. 1}\text{.16}\times \text{1}{\text{0}}^{-3}kgm{s}^{-1} \\ \text{C}\text{. 2}\text{.16}\times \text{1}{\text{0}}^{-3}kgm{s}^{-1} \\ \text{D}\text{. 3}\text{.16}\times \text{1}{\text{0}}^{-4}kgm{s}^{-1} \end{gathered}$$

Answer 1

Hint: The energy of a photon is equal to the product of momentum and speed of light. Energy flux or intensity of incident light is the energy delivered to the surface per unit area and per unit time.

Formula used:
The energy-flux relation is given as:
$$I={\displaystyle \frac{E}{At}}$$
where I represents the intensity or energy flux of the light incident on a surface, E is the energy transferred to the surface having surface A and t represents the time duration of exposure to the light.
Energy of a photon is given as:
$E=pc$
where p is the momentum of the photon which travels with velocity c. The value of c is $3\times {10}^{8}m/s$

Complete step-by-step answer:
We are given that light having energy flux of $9W/c{m}^2$ is incident at right angle (normally) to a non-reflecting surface which has surface area equal to $20c{m}^2$.
$$\begin{gathered} I=9W/c{m}^2 \\ A=20c{m}^2 \end{gathered}$$
The light is incident for a duration of one hour on the surface.
$t=1hr=3600s$
We know that energy flux is equal to the energy delivered to a surface per unit area and per unit time. Therefore we have the following relation for energy flux.
$$I={\displaystyle \frac{E}{At}}$$
Substituting all the known values, we get
$$\begin{gathered} 9={\displaystyle \frac{E}{20\times 3600}} \\ \Rightarrow E=9\times 20\times 3600 \\ \Rightarrow E=648000W \end{gathered}$$
Now using this value of energy we can calculate the momentum transferred to the surface as follows:
$$\begin{gathered} p={\displaystyle \frac{E}{c}} \\ ={\displaystyle \frac{64800}{3\times {10}^8}} \\ =2.16\times {10}^{-4}kgm/s \end{gathered}$$
Hence, the correct answer is option A.

Note: In the expression of energy of photons, we have total momentum of the incident light and obtained value of momentum is valid because we are given that in 1 hour all the energy or momentum of the light is absorbed.