Question

Light is incident normally on a completely absorbing surface with energy flux of $25Wc{m^{ - 2}}$. If the surface has an area of $25c{m^2}$, the momentum transferred to the surface in 40 min time duration would be
(A) $5.0 \times {10^{ - 3}}Ns$
(B) $3.5 \times {10^{ - 6}}Ns$
(C) $1.4 \times {10^{ - 6}}Ns$
(D) $6.3 \times {10^{ - 4}}Ns$

Answer 1

Energy of radiation can be equal to the product of its momentum and the speed of light. Use the known energy flux to determine the energy. (you can use its unit as guide)
Formula used: In this solution we will be using the following formulae;
$E = pc$, where $E$ is the energy of a photon (or radiation), $p$ is the momentum of the photon and $c$ is the speed of light in vacuum.
$P = \dfrac{E}{t}$ where $P$ is equal to power (or radiant power in this case), and $t$ is time.
$\Phi = \dfrac{P}{A}$ where $\Phi$ is energy flux and $A$ is the area of the surface.

Complete Step-by-Step Solution:
In this case radiation is fully absorbed by a surface hence all energy of the radiation falling on it may be converted to motion of some sort such as motion of electrons or atoms in general. The momentum of the photons would be transferred to the electrons or atoms.
The momentum associated with a photon of a particular energy can be related as follows
$E = pc$, where $E$ is the energy of a photon (or radiation), $p$ is the momentum of the photon, and $c$ is the speed of light in vacuum.
To calculate the energy, we recall that
$P = \dfrac{E}{t}$ where $P$ is equal to power (or radiant power in this case), and $t$ is time., and
$\Phi = \dfrac{P}{A}$ where $\Phi$ is energy flux and $A$ is the area of the surface.
Hence,
$P = \Phi A$
Inserting into equation $P = \dfrac{E}{t}$ and rearranging we get,
$E = \Phi At$
Hence,
$E = 25\left( {25} \right)40 \times 60 = 1500000J$ (since 1 minute is equal to 60 seconds)
Hence,
$1500000 = p\left( {3 \times {{10}^8}} \right)$
$\Rightarrow p = \dfrac{{1500000}}{{3 \times {{10}^8}}} = 0.005Ns$ or $5 \times {10^{ - 3}}Ns$

Hence, the correct option is A.

Note: For clarity, observe that we did not have to convert the energy flux and area to their SI unit. This is simply because, the energy flux is given in $W/c{m^2}$ and area is given in $c{m^2}$, hence if they are multiplied, the cm cancels out (along with any conversion factor) hence leaving the SI unit of power, the Watt.