# The Compton Effect and Pressure of Light

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19.3The Compton Effect and Pressure of Light

In this Physics tutorial, you will learn:

• What is scattered (diffuse) radiation? When does it occur?
• What are the two components of scattered radiation?
• What is the Compton wavelength?
• How can we find the change in light wavelength during the Compton Effect?
• Does light exert any pressure on objects? How does this process occur?
• What are the factors affecting the light pressure?
• Is light a wave or a particle?

## Introduction

Where do we rely when supporting the idea that light is a wave? What about when light is considered as a particle?

Do you think atoms of matter can change the direction of light rays? Why?

Do you think light is able to exert pressure on objects? If yes, why we cannot feel it?

What is light really: a particle or a wave?

All these questions will get an answer after reading this tutorial, which makes a connection between classic and modern approach on light.

## What is the Compton Effect?

As we know from tutorial 12.7, the X-rays (otherwise known as Roentgen rays) represent a type of EM radiation of a much smaller wavelength than visible light. Their range varies from 0.1 nm to 0.02 nm, i.e. they are from 1000 to 20 000 times smaller than visible light. When this radiation falls on crystals, it interacts with the electrons of crystal atoms or molecules. As a result, the incident radiation scatters in different directions from the original. This scattered radiation is also known as "diffuse radiation". It is made up by two components: one has the same wavelength as the incident radiation and it is called the "coherent component" while the other has a greater wavelength than the incident radiation and is called "non-coherent component" of the diffused radiation.

The coherent component is obtained by the interaction of the incident radiation with the strongly bonded electrons to the atomic nucleus. These electrons are found in abundance in dielectric and semiconductor crystals but also in metals, in deep electronic layers. On the other hand, the non-coherent component of scattered radiation is obtained through the interaction of the incident X-radiation with the weakly bonded electrons such as the valence electrons of metal or graphite crystals.

The non-coherent scattering in graphite crystals was discovered by Arthur H. Compton in 1923. This is one of characteristic phenomena where EM radiation manifests its particle / quantum nature.

## Quantum Interpretation of Compton Scattering

The figure below shows the scheme of experiment made by Compton to explain the scattering effect when X-rays strike the surface of a crystal. The detector records the radiation scattered by θ degrees and also calculates the wavelength λ' of the scattered radiation. Compton found experimentally that the change in wavelength Δλ between the scattered radiation λ' and incident radiation λ depend only by the scattering angle θ. Mathematically, this dependence is written as

∆λ = λC ∙ (1 - cosθ)

The quantity λC (which obviously has the unit of wavelength) is known as the "Compton wavelength". It is a constant and its value is found experimentally (λC = 0.00243 nm = 2.43 × 10-12 m).

By analyzing the Compton formula, it is evident that the change in wavelength increases with the increase in the scattering angle. The figure below shows the data recorded by the detector for four different scattering angles (0°, 45°, 90° and 135°). The detector records in a given time the relative intensities of radiation and wavelengths for both components of diffused radiation - coherent and non-coherent. From the graphs, you can see that for EM waves moving in the original direction (θ = 0°) there is no Compton Effect (waves are not scattered). On the other hand, the effect is more visible for a 135° scattering angle than for 45° or 90° (coherent and non-coherent scattered wavelengths are more distant from each other, where coherent wavelengths move in the original direction of incident waves).

Compton's formula can be obtained theoretically only when considering the particle nature of X-radiation. For this purpose, Compton observed the interaction of photons produced by the X-radiation and the weakly bonded electrons. This interaction is a perfectly elastic collision in which both energy and momentum of the photon-electron system are conserved. Compton found theoretically that

∆λ = (h/m ∙ c) ∙ (1 - cosθ)

where m is the mass of electron (m = 9.1 × 10-31 kg), h is Planck's constant (h = 6.626 × 10-34 J · s) and c is the speed of light in vacuum (c = 3 × 108 m/s).

Comparing the two formulae of Compton effect, it is clear that the Compton wavelength λC is

λC = h/m ∙ c

Let's find the value of Compton wavelength and compare it with the experimental value given earlier (2.43 × 10-12 m). Thus, substituting the known values, we obtain

λC = (6.626 × 10-34 J ∙ s)/(9.1 × 10-31 kg) ∙ (3 × 108 m/s)
= 2.4271 × 10-12 m
≈ 2.43 × 10-12 m

This value is equal to the experimental value found by Compton. Likewise, we can determine the value of Planck Constant using the value of Compton wave found experimentally and so on.

### Example 1

Calculate the wavelength of X-radiation scattered at 600 to the original direction if the incident wavelength is 0.02183 nm. (cos 600 = 0.500, sin 600 = 0.866)

### Solution 1

Clues:

θ = 30°
λ = 0.02183 nm = 2.183 × 10-11 m)
(λC = 2.43 × 10-12 m)
λ' = ?

Using the Compton Formula

∆λ = λC ∙ (1 - cosθ)

where Δλ = λ' - λ, we obtain for the scattered wavelength after substitutions

λ'- λ = λC ∙ (1 - cosθ)
λ' = λ + λC ∙ (1 - cosθ )
= (2.183×10-11 m) + (2.43 × 10-12 m) ∙ (1 - 0.500)
= (2.183×10-11 m) + (1.215 × 10-12 m)
= 3.398 × 10-12 m

## Pressure of Light

The interaction between light and matter is manifested in many ways such as through reflection, refraction, photoelectric effect, scattering etc. Among them there is the pressure effect of light as well (pressure exerted by light on the surface of objects).

The values of light pressure are very small; they vary between 10-6 Pa - 10-8 Pa. The phenomenon of light pressure was initially observed by Pyotr Lebedev at the early 1900s. He found that light pressure depends on the following factors:

1. coefficient of surface reflection r,
2. angle of incidence θ, and
3. intensity of incident beam I.

Combining these factors, Lebedev found the following formula for light pressure P:

P = I/c ∙ (1 + r) ∙ cos2 θ

We know that 0 ≤ r ≤ 1 where r = 0 for black bodies and r = 1 for perfect mirrors. Therefore, light pressure in perfect mirrors is twice the pressure produced by the same light beam on black bodies for the same incident angle.

In astronomic dimensions, the effect of light pressure is clearly observed in comets. When a comet approaches the Sun, its gaseous tail elongates and moves away from the Sun. This phenomenon occurs due to the high pressure of light emitted from the Sun.

Pressure produced by powerful lasers can reach values up to tens of atmospheres (millions of Pascals) as it is a beam with a concentrated energy on small surfaces. This is why laser beams are used in surgery and technology to cut things with high precision.

## Interpretation of Light Pressure

The phenomenon of light pressure can be explained correctly through both the wave and particle approach (classical and modern theory of light). This fact is a demonstration of the dual nature of light. Hence, we can say that light waves are made up by light particles (photons).

The particle interpretation of light pressure takes into consideration the collision of photons with the surface of objects. In every collision, small forces caused by the incident photons act on the object's surface, the resultant of which, gives the pressure of light on that surface.

The figure below shows a beam of parallel light waves incident to the surface of an object at an angle θ to the normal line. Let's denote by I the intensity of the incident light and by A the area of the zone when the light falls at the angle θ to the normal line. Light intensity represents the light energy incident on the unit area in every second. If ΔE is the light energy incident on the area A0 during a very short time interval Δτ, we obtain

∆E = I ∙ A0 ∙ ∆τ

On the other hand, based on the particle definition of light, it is clear that this energy ΔE is transported from the source to the surface by a number of photons ΔN, which are at any instant inside the imaginary cylinder of light where the A0 is the base of cylinder and c · Δτ its height. If we write by n the concentration of photons (i.e. the number of photons in the unit of volume), we have for the number ΔN of photons incident on the given surface during the time interval Δτ:

∆N = n ∙ A0 ∙ c ∙ ∆τ

For simplicity, we will consider only the case of monochromatic light, i.e. when all incident photons have the same frequency. Obviously, the energy of each of them is

Eph = h ∙ f

while the total energy ΔE transported by (all) ΔN photons is

ΔE = (h ∙ f) ∙ ∆N
= (h ∙ f) ∙ n ∙ A0 ∙ c ∙ ∆τ

Comparing the two expressions found for the energy of incident light (photons), we obtain the following relationship between the intensity of light beam and energy of incident photons:

I ∙ A0 ∙ ∆τ = (h ∙ f) ∙ n ∙ A0 ∙ c ∙ ∆τ

or

I = (h ∙ f) ∙ n ∙ c

The quantity (h ∙ f) ∙ n represents the volume density of photons energy and it is denoted by w. Hence, we obtain for the pressure of light

P = I/c ∙ (1 + r) ∙ cos2 θ
= w ∙ (1 + r) ∙ cos2 θ

### Example 2

A monochromatic light beam of intensity 8 × 10-2 W/m2 and wavelength equal to 500 nm is incident at 37° to the normal line on a metal surface of reflection coefficient r = 0.9. Calculate:

1. Pressure of the light beam
2. The number of incident photons per unit of volume
3. The number of photons incident on a 0.2 m2 section of the metal surface during 2 seconds

### Solution 2

Clues:

I = 8 × 10-2 W/m2
λ = 500 nm = 5 × 10-7 m
θ = 37° (cos 37° = 0.8)
r = 0.9
(c = 3 × 108 m/s)
(h = 6.626 × 10-34 J · s)

1. P = ?
Using the equation of light pressure
P = I/c ∙ (1 + r) ∙ cos2 θ
we obtain after substitutions:
P = (8 × 10-2 W/m2 )/(3 × 108 m/s) ∙ (1 + 0.9) ∙ (0.8)2
= 3.243 × 10-10 Pa
2. n = ?
The concentration of photons (the number of photons per unit volume) giving that I = w · c, is
n = w/h ∙ f
= I/c ∙ h ∙ f
= I ∙ λ/h ∙ c2
= (8 × 10-2 W/m2 ) ∙ (5 × 10-7 m)/(6.626 × 10-34 J ∙ s) ∙ (3 × 108 m/s)2
= 6.7 × 108 photons/m3
3. ΔN = ?
The number of photons incident on the given surface (A0 = 0.2 m2) during the time interval Δτ = 2 s, is
∆N = n ∙ A0 ∙ c ∙ ∆τ
= (6.7 × 108 photons/m3 ) ∙ (0.2 m2 ) ∙ (3 × 108 m/s) ∙ (2 s)
= 8.04 × 108 photons

## Wave Interpretation of Light

The wave interpretation of light is based on the action exerted by the electric component E (do not confuse it with energy, this is the electric field caused by light as an EM wave) and the magnetic one B of light waves on matter electrons. Electronic currents are induced under the effect of electric field and the magnetic Lorentz Force acts on them. This action determines the pressure effect of light. According to this theory, even if the existence of photons is not considered, the light pressure is calculated by the two equations

P = I/c ∙ (1 + r) ∙ cos2 θ

and

P = w ∙ (1 + r) ∙ cos2 θ

where the quantity w bears the meaning of volume density of light wave.

### Is Light a Particle of a Wave?

The debates on this topic originate more than 300 years ago, when Newton and Huygens were still alive and have continues until about one century ago. Nowadays, the idea of dual nature (wave-particle) of light is widely accepted in the scientific circles. As stated earlier, we call "photons" the particles of light. However, we use the wave approach to describe the way in which photons propagate in space. We will explain more in detail this dual nature not only of light but also of matter in the following articles.

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