Radiation Pressure

Radiation pressure is the force exerted by electromagnetic radiation on a surface. This phenomenon occurs because photons, the particles of light, carry momentum and, when they strike a surface, they transfer that momentum, creating pressure. First theorized by James Clerk Maxwell and later confirmed experimentally, radiation pressure plays a crucial role in various scientific and technological applications. In everyday life, this effect is harnessed in technologies like solar sails used in space exploration, which rely on radiation pressure from the Sun to propel spacecraft. Additionally, radiation pressure is significant in understanding stellar dynamics and processes, such as the formation and behaviour of stars. In this article, we will discuss the concept of Radiation pressure, The intensity of Light, photon flux, radiation power and solved examples for better concept clarity.

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This Story also Contains

1. Radiation Pressure
2. The Intensity of Light
3. Radiation Pressure/Force
4. Solved Examples Based on Radiation Pressure
5. Summary

Radiation Pressure

Radiation pressure is the force exerted by electromagnetic radiation on a surface. This phenomenon occurs because photons, the particles of light, carry momentum and, when they strike a surface, they transfer that momentum, creating pressure. First theorized by James Clerk Maxwell and later confirmed experimentally, radiation pressure plays a crucial role in various scientific and technological applications. In everyday life, this effect is harnessed in technologies like solar sails used in space exploration, which rely on radiation pressure from the Sun to propel spacecraft.

Consider a point source of light-emitting photons. 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

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

where

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

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

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

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

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

The Intensity of Light

The intensity of light refers to the amount of energy the light wave carries per unit area per unit time. It is an important concept in physics and has wide-ranging applications in various fields, from photography to solar energy.

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

$I=\frac{E}{At}=\frac{P}{A}($ where $\frac{E}{t}=P=$ radiation power $)$

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

The Intensity of Light Due to a Point Isotropic Source

The intensity of light due to a point isotropic source, which emits light uniformly in all directions, is a crucial concept in understanding how light energy spreads out from a source. The intensity (I) of light from such a source decreases with the square of the distance (r) from the source.

An 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 $I\propto \frac{1}{r^2}$

The photon flux $(\varphi )$ 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

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

Similarly intensity I is given as

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

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

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

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

The Photon Flux $(\varphi )$ Due to a Point Isotropic Source

The photon flux due to a point isotropic source refers to the number of photons passing through a unit area per unit time. For a source emitting photons uniformly in all directions, the photon flux can be calculated using the power of the source and the energy of the individual photons.

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

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

Radiation Pressure/Force

Radiation pressure refers to the pressure exerted by electromagnetic radiation on a surface. This phenomenon occurs because light, although massless, carries momentum and when it interacts with a surface, it can transfer that momentum, exerting a 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 $n=\frac{P}{E}=\frac{P}{h\nu }=\frac{P\lambda }{hc}$

where E= The energy of each photon

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

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

I.e For each photon $F=\frac{dp}{dt}$

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

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

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

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

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

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 $=\frac{F}{A}=\frac{I}{c}$

i.e For black bodies,
$\begin{array}{rl}& F=\frac{P}{C}\\ & \text{Pressure}=\frac{I}{C}\end{array}$

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

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

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 $=\frac{F}{A}=\frac{P}{Ac}(1+r)=\frac{I}{c}(1+r)$

Solved Examples Based on Radiation Pressure

Example 1: If a source of power 4 kW produces 10²⁰ photons/second, the radiation belongs to a part of the spectrum called

1) $\gamma$-rays
2) $X-$ rays
3) ultraviolet rays
4) microwaves

Solution:

The energy of a photon

$E=h\nu =\frac{hc}{\lambda }$

wherein

$\begin{array}{rl}& h=\text{Plank's constant}\\ & \nu =\text{frequency of radiation}\\ & \lambda \to \text{wave length}\end{array}$

If the wavelength of the photon is & n is the number of photons emitted per second, then

$\begin{array}{rl}& P=n\frac{hc}{\lambda }\\ & 4\times {10}^3=\frac{{10}^{20}\times 6.62\times {10}^{-34}\times 3\times {10}^8}{\lambda }\\ & \lambda =\frac{19.8\times {10}^{-26}\times {10}^{20}}{4\times {10}^3}=4.96\times {10}^{-9}\\ & \lambda =49.6\text{AA}\end{array}$m

This wavelength represents X-rays

Hence, the answer is the option (2).

Example 2: A 2 mW laser operates at a wavelength of 500 nm. The number of photons that will be emitted per second is $n\times {10}^{15}$. What is the value of n? [Given Plank's constant $h=6.6\times {10}^{-34}\mathrm{Js}$, speed of light $c=3.0\times {10}^8\mathrm{m}/\mathrm{s}$ ]

1) 5

2) 1.5

3) 2

4) 1

Solution:

$\begin{array}{rl}& P=\frac{n\cdot hc}{\lambda }\\ & n=\frac{p\lambda }{hc}\\ & n=\frac{2\times {10}^{-3}\times 500\times {10}^{-9}}{6.6\times {10}^{-34}\times 3\times {10}^8}\\ & n=5\times {10}^{15}\end{array}$

Hence, the answer is the Option (1).

Example 3: A beam of light has two wavelengths 4972 Å and 6216 Å with a total intensity of $3.6\times {10}^{-3}{\mathrm{Wm}}^{-2}$ equally distributed among the two wavelengths. The beam falls normally on an area of 1 cm² of a clean metallic surface of work function 2.3 eV. Assume that there is no loss of light by reflection and that each capable photon ejects one electron. The number of photoelectrons liberated in 2s is approximately :

1) $6\times {10}^{11}$
2) $9\times {10}^{11}$
3) $11\times {10}^{11}$
4) $15\times {10}^{11}$

Solution:

$\begin{array}{rl}{\lambda }_1& =4972A^{\circ },{\lambda }_2=6216A^{\circ }\\ I=3.6& \times {10}^{-3}w/m^2\end{array}$

Intensity with each wavelength $=1.8\times {10}^{-3}\mathrm{W}/{\mathrm{m}}^2$

the energy of a photon is given by

$\begin{array}{rl}& E=\frac{hc}{\lambda }=\frac{6.62\times {10}^{-34}\times 3\times {10}^8}{\lambda }\\ & E=12.4\times {10}^3/\lambda \\ & E_1=12.4\times {10}^3/{\lambda }_1=2.493\mathrm{eV}=3.98\times {10}^{-19}\mathrm{J}=2.48\mathrm{eV}\\ & E_2=12.4\times {10}^3/{\lambda }_2=3.189\times {10}^{-19}\mathrm{J}=(1.99\mathrm{eV})\end{array}$

And work function is

$\varphi =2.3\mathrm{eV}$

Since only $E_1>\varphi$

So only photons corresponding to ${\lambda }_1$ will be able to eject photoelectrons

So $N/\sec =\frac{P}{E}=\frac{IA}{E}$

here $A=1{\mathrm{cm}}^2$

$N/\sec =\frac{1.8\times {10}^{-3}}{3.984\times {10}^{-19}\times {10}^4}=0.45\times {10}^{12}$

In 2 sec N will be

$\mathrm{N}=9\times {10}^{11}$

Hence, the answer is the option (2).

Example 4: A metal plate of the area $1\times {10}^{-4}{\mathrm{m}}^2$ is illuminated by a radiation of intensity 16 mW/ m². The work function of the metal is 5eV. The energy of the incident photons is 10 eV and only 10% of it produces photoelectrons. The number of emitted photoelectrons per second and their maximum energy, respectively, will be: $[1\mathrm{eV}=1.6\times {10}^{-19}\mathrm{J}]$

1) ${10}^{10}$ and 5 eV
2) ${10}^{12}$ and 5 eV
3) ${10}^{11}$ and 5 eV
4) ${10}^{14}$ and 10 eV

Solution:

$\begin{array}{rl}K{E}_{max}& =E-\varphi \\ & =10\mathrm{eV}-5\mathrm{eV}=5\mathrm{eV}\\ \frac{n}{t}=\frac{IA}{E}& =\frac{16\times {10}^{-3}\times {10}^{-4}}{10\times 1.6\times {10}^{-18}}={10}^{11}\end{array}$

Hence, the answer is the option (3).

Example 5: 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:

1) The straight line shifts to the right

2) The slope of the straight line gets more steep

3) The straight line shifts to the left

4) The graph does not change

Solution:

$\begin{array}{rl}& {\mathrm{eV}}_{\mathrm{s}}=\mathrm{hv}-\mathrm{w}\\ & V_s=\frac{hv}{e}-\frac{w}{e}\end{array}$

Frequency and work function are constant therefore graph does not change.

Summary

Radiation pressure is the force exerted by light on a surface due to the momentum of photons. This pressure is harnessed in technologies like solar sails for spacecraft propulsion and plays a critical role in astrophysics and climate science. Key concepts include the intensity of light, photon flux, and the resulting radiation force on different surfaces. Practical applications and solved examples demonstrate how radiation pressure and related principles are applied in real-world scenarios, from identifying types of radiation to calculating the number of emitted photoelectrons and their energy.