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# Radiation Pressure - Practice Questions & MCQ

Edited By admin | Updated on Sep 18, 2023 18:34 AM | #JEE Main

## Quick Facts

• Photons emitted by a source per second, Intensity of radiation, Photon Flux are considered the most difficult concepts.

• 49 Questions around this concept.

## Solve by difficulty

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

A point source of light emits photons of energy 3.5  eV each. If the power of the source is 2.0 W, then the number of photons emitted per second is:

A light bulb emits photons of various wavelengths ranging from 400 nm to 700 nm. If the power of the light bulb is 100 W, what is the total number of photons emitted per second?

A 100 W light bulb emits $5.0 \times 10^{18}$ photons per second. Determine emitted photons' wavelength. $\left(\right. Take h=6.6 \times 10^{-34} \mathrm{Js}$ and $\left.c=3.0 \times 10^8 \mathrm{~m} / \mathrm{s}\right)$

A laser diode emits light of wavelength $\lambda=780 \mathrm{~nm}$ with a power of P = 20 mW. How many photons are emitted per second from the laser diode?

A radiation of energy $E$ falls normally on a perfectly reflecting surface. The momentum transferred to the surface is :

## 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

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

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 intensityis 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 intensitydue 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  intensityis 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 $n =\frac{P}{E}=\frac{P}{h \nu}=\frac{P \lambda}{h c}$

where E= The energy of each photon

and Momentum of each photon is given as $p= \frac{E}{c}= \frac{h}{\lambda}$

And we know the force is given as rate of change of momentum.

I.e For each photon $F=\frac{dp}{dt}$

and for n photons per sec $F=n(\Delta p)$

For a black body, we get 100 % absorption or a=1

i.e for this surface 100% of the photon will be absorbed

so $| \Delta p|=|0-p_i|=\frac{h}{\lambda }$

So Force is given as $F=n(\Delta p)=\frac{P\lambda }{hc}*\frac{h}{\lambda }=\frac{P}{c}$

where P=Power

As $I=\frac{P}{A}\Rightarrow P=IA$

So Force is given as $F=\frac{P}{c}=\frac{IA}{c}$

and radiation pressure is given as $Pressure=\frac{F}{A}=\frac{I}{c}$

$\begin{array}{c}{\text { i.e For black body, }} \\ {\qquad \begin{array}{l}{F=\frac{P}{C}} \\ {Pressure=\frac{I}{c}}\end{array}}\end{array}$

• For perfectly reflecting surface (i.e mirror)

i.e r=1

i.e  for this surface 100% of the photon will be reflected

i.e $p_f=-p_i$

So $| \Delta p|=|p_f-p_i|=|-p_i-p_i|=\frac{2h}{\lambda }$

So Force is given as $F=n(\Delta p)=\frac{P\lambda }{hc}*\frac{2h}{\lambda }=\frac{2P}{c}=\frac{2IA}{c}$

and radiation pressure is given as $Pressure=\frac{F}{A}=\frac{2I}{c}$

• For neither perfectly reflecting nor perfectly absorbing body

i.e body having Absorption coefficient=a and reflection coefficient=r

and we have $a+r=1$

So Force is given as $F=\frac{aP}{c}+\frac{2Pr}{c}=\frac{P}{c}(a+2r)=\frac{P}{c}((1-r)+2r)=\frac{P}{c}(1+r)$

and radiation pressure is given as $Pressure=\frac{F}{A}=\frac{P}{Ac}(1+r)=\frac{I}{c}(1+r)$

## Study it with Videos

Photons emitted by a source per second