How to score good marks in NEET 2025 - Tips for physics, chemistry & biology

Radiation Pressure MCQ - Practice Questions with Answers

Edited By admin | Updated on Sep 25, 2023 25:23 PM | #NEET

Quick Facts

Photons emitted by a source per second, Intensity of radiation, Photon Flux are considered the most difficult concepts.
Force exerted on a surface due to radiation are considered the most asked concepts.
37 Questions around this concept.

Solve by difficulty

EASY

If a source of power 4 kW produces 10²⁰ photons/second, the radiation belongs to a part of the spectrum called

$\gamma$ - rays

$X-rays$

ultraviolet rays

microwaves

View Solution

Concepts Covered - 4

Photons emitted by a source per second

Consider a point source of light emitting photons. And we want to find the number of Photons (n) emitted by this point source per second.

let the wavelength of light emitted by this = $\lambda$ and

the power of the source as P (in Watt or J/s)

As we know the energy of each photon is given by

$E=h \nu =\frac{h c}{\lambda} \ \ (in \ Joule)$

where

where c = Speed of light, h = Plank's constant = $6.6 \times 10^{-34} J-\sec$

$\nu$ = Frequency in Hz , $\lambda$ = Wavelength of light.

or we can write the energy of each photon as $E=\frac{12400 \ (e V)}{\lambda (A^{0})}$

Then ( n=the number of photons emitted per second) is given as

$n=\frac{Power \ of \ source \ (W \ or \ \frac{J}{sec} )}{Energy \ of \ each \ photon (J)}=\frac{P}{E}=\frac{P}{\frac{hc}{\lambda }}=\frac{P\lambda }{hc} \ (sec^{-1})$

Intensity of radiation

The intensity of light (I) : The intensity of any quantity is defined as that quantity per unit area.

So here, light energy (or radiation ) crossing per unit area normally per second is called intensity of light energy (or radiation ).

And the intensity I is given as

$I=\frac{E}{A t}=\frac{P}{A} \quad\left( where \ \ \frac{E}{t}=P=\text { radiation power }\right)$

Its unit is $W/m^2$ or $\frac{J}{m^2*sec}$

The intensity of light due to a point isotropic source:

Isotropic source means it emits radiation uniformly in all directions.

So The intensity I due to a point isotropic source at a distance r from it is given as

$I=\frac{P}{4 \pi r^{2}} \Rightarrow i.e \ \ I \propto \frac{1}{r^{2}}$

Photon Flux

The photon flux ( $\phi$ ) is defined as the number of photons incident on a normal surface per second per unit area.

As we know n ( the number of photons emitted per second) is given as

$n=\frac{Power \ of \ source \ (W \ or \ \frac{J}{sec} )}{Energy \ of \ each \ photon (J)}=\frac{P}{E} \ (sec^{-1})$

Similarly intensity I is given as

$I =\frac{P}{A}$

So The photon flux ( $\phi$ ) is given as ratio of Intensity (I) to Energy of each photon

$\phi =\frac{Intensity}{Energy \ of \ each \ photon}=\frac{I}{E}=\frac{n}{A}$

or $\phi =\frac{I}{E}=\frac{I\lambda }{hc}$

The photon flux ( $\phi$ ) due to a point isotropic source:

The photon flux ( $\phi$ ) due to a point isotropic source at a distance r from it is given as

$\phi = \frac{number \ of \ photon \ per \ sec }{surface \ area \ of \ sphere \ of \ radius \ r}=\frac{n}{4\pi r^2}$

Force exerted on a surface due to radiation

Radiation pressure/force- When photons fall on a surface they exert a pressure/force on the surface. The pressure/force experienced by the surface exposed to the radiation is known as Radiation pressure/force.

As we know

n=Number of emitted photons per sec is given as $n =\frac{P}{E}=\frac{P}{h \nu}=\frac{P \lambda}{h c}$