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JEE Main Physics Study Material 2025 - Download PDF

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.

  • 69 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?

Photons absorbed in the matter are converted to heat. A source emitting n photon/sec of frequency ν is used to convert 1kg of ice at 0°C to water at 0°C. Then, the time T taken for the conversion

(a) decreases with increasing n, with ν fixed

(b) decreases with n fixed, ν increasing

(c) remains constant with n and ν changing such that nν = constant

(d) increases when the product nν increases

In a photoelectric effect experiment, the graph stopping potential V versus the reciprocal of wavelength obtained is shown in the figure. As the intensity of incident radiation is increased:

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The intensity of radiation from the source depends upon the distance 'd' of the point of observation from the source as

A photocell is illuminated by a small bright source placed 1 m away. When the same source of light is placed (1/2) m away, the number of electrons emitted by photocathode would

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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 =λ 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=hν=hcλ( in Joule )

where
where c= Speed of light, h= Plank's constant =6.6×1034 Jsec
ν= Frequency in Hz,λ= Wavelength of light.

E=12400(eV)λ(A0)

or we can write the energy of each photon as
Then ( n= the number of photons emitted per second) is given as

n= Power of source (W or Jsec) Energy of each photon (J)=PE=Phcλ=Pλhc(sec1)
 

 

 

 

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=EAt=PA( where Et=P= radiation power )
Its unit is W/m2 or Jm2sec
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=P4πr2 i.e I1r2
 

 

 

Photon Flux

The photon flux ϕϕ ) 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= Power of source (W or Jsec) Energy of each photon (J)=PE(sec1)


Similarly intensity I is given as

I=PA


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

ϕ= Intensity  Energy of each photon =IE=nAϕ=IE=Iλhc

- The photon flux ϕ ) due to a point isotropic source:

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

ϕ= number of photon per sec  sur face area of sphere of radius r=n4πr2
 

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=PE=Phν=Pλhc
where E= The energy of each photon
and Momentum of each photon is given as p=Ec=hλ
And we know the force is given as rate of change of momentum.
I.e For each photon F=dpdt
and for n photons per sec F=n(Δp)
For a black body, we get 100% absorption or a=1
i.e for this surface 100% of the photon will be absorbed

|Δp|=|0pi|=hλ


So Force is given as

F=n(Δp)=Pλhchλ=Pc

where P= Power

I=PAP=IA


So Force is given as

F=Pc=IAc

and radiation pressure is given as

 Pressure =FA=Ic
 

i.e For black body,

F=PC Pressure =Ic

- 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 pf=pi
So |Δp|=|pfpi|=|pipi|=2hλ

So Force is given as

F=n(Δp)=Pλhc2hλ=2Pc=2IAc

and radiation pressure is given as

 Pressure =FA=2Ic

- 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=aPc+2Prc=Pc(a+2r)=Pc((1r)+2r)=Pc(1+r)

and radiation pressure is given as

 Pressure =FA=PAc(1+r)=Ic(1+r)
 

Study it with Videos

Photons emitted by a source per second
Intensity of radiation
Photon Flux

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