Courses for Kids
Free study material
Offline Centres
Store Icon

What is the force exerted by a photon of intensity ${{1}}{{.4 kW}}{{{m}}^{{{ - 2}}}}$ if it falls on perfect absorber of radius ${{2 m}}$?
$A) {{2}}{{.35 \times 1}}{{{0}}^{{{ - 4}}}}{{ N}} \\
B) {{1}}{{{0}}^{{8}}}{{ N}} \\
C) {{8}}{{.35 \times 1}}{{{0}}^{{4}}}{{ N}} \\
D) {{8}}{{.8 \times 1}}{{{0}}^{{{ - 8}}}}{{ N}}$

Last updated date: 16th May 2024
Total views: 43.8k
Views today: 1.43k
43.8k+ views
Hint: For finding out the force exerted by any particle, recall the formulas that include intensity or radius ( or area ) or both. As there is no such direct relation. Now, try to find out other formulas where force can be easily determined. For sure there will be some indirect relation which on further simplification lands on the required quantity.

Complete step by step solution:
Given: Intensity of photon, ${{I = 1}}{{.4 kW}}{{{m}}^{{{ - 2}}}} = 1.4{{ \times 1}}{{{0}}^{{3}}}{{ W}}{{{m}}^{{{ - 2}}}}$
Radius of photon, ${{r = 2 m}}$
As there is no direct relation of finding out force exerted by the photon with given quantities. So, we need to first calculate the pressure exerted by the photon. Once the value of exerted pressure is known then we will proceed further using formula, ${{force = pressure \times area}}$. Since the value of radius is given in the question, we can easily determine the value of the area of the photon.
Pressure exerted by a photon, ${{P = }}\dfrac{{{{\text{Intensity} (I)}}}}{{{{\text{velocity of light} (c)}}}}$
We know that velocity of light in vacuum, ${{c = 3 \times 1}}{{{0}}^{{8}}}{{ m/s}}$
Substituting the quantities in above formula, we get
$\Rightarrow {{P = }}\dfrac{{{{1}}{{.4 \times 1}}{{{0}}^{{3}}}}}{{{{3 \times 1}}{{{0}}^{{8}}}}}\dfrac{{{N}}}{{{{{m}}^{{2}}}}}$
Now, force exerted by a photon is given by the formula
$\Rightarrow {{F = P \times A}}$
Where ${{P = }}$pressure and ${{A = }}$area
Area, ${{A = 4\pi }}{{{r}}^{{2}}}{{ = 4 \times }}\dfrac{{{{22}}}}{{{7}}}{{ \times }}{{{2}}^{{2}}}{{ }}{{{m}}^{{2}}}$
Now substituting the value of area and pressure in above formula, we get
$\Rightarrow {{F = }}\dfrac{{{{1}}{{.4 \times 1}}{{{0}}^{{3}}}}}{{{{3 \times 1}}{{{0}}^{{8}}}}}{{ \times 4 \times }}\dfrac{{{{22}}}}{{{7}}}{{ \times }}{{{2}}^{{2}}}{{ = 2}}{{.35 \times 1}}{{{0}}^{{{ - 4}}}}{{ N}}$
Therefore, force exerted by a photon is ${{2}}{{.35 \times 1}}{{{0}}^{{{ - 4}}}}{{ N}}$.

Thus, option (A) is the correct choice.

Note: A photon is a massless and stable elementary particle that doesn’t have any charge. Since the photons are a massless quantity so, the photons always moves with the velocity of light in vacuum i.e. ${{3 \times 1}}{{{0}}^{{8}}}{{ m/s}}$. Keep in mind that before substituting the required quantities in formula make sure that all the quantities are in their SI unit.