Compton Scattering Slide

PHY-1109 Physics

Compton Scattering Slide Compton Effect (Compact) Compton Effect (Detailed)

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The Compton Effect

According to the quantum theory of light, photons behave like particles except for their lack of rest mass. How far can this analogy be carried? For instance, can we consider a collision between a photon and an electron as if both were billiard balls of rest mass. How far can this analogy be carried? For instance, can we consider a collision between a photon and an electron as if both were billiard balls?

An x-ray photon strikes an electron (assumed to be initially at rest in the laboratory coordinate system) and is scattered away from its original direction of motion while the electron receives an impulse and begins to move.

We can think of the photon as losing an amount of energy in the collision that is the same as the kinetic energy $T$ gained by the electron, although actually separate photons are involved.

The scattering of a photon by an electron is called the Compton effect. Energy and momentum are conserved in such an event, and the scattered photon has less energy than the incident photon.

Image/Diagram Note:

• Description: Diagram showing an Incident photon ($E=hv$, $p=hv/c$) striking a Target electron ($E=m_0{c}^2$, $p=0$). A scattered photon ($E=h{v}^{\prime }$, $p=h{v}^{\prime }/c$) is shown scattered at angle $\varphi$. A Scattered electron ($E=\sqrt{m_0^2{c}^4+p^2{c}^2}$, $p=p$) is shown recoiling at angle $\theta$.
• Vector Diagram Note: Vector diagram of the momenta and their components of the incident and scattered photons and the scattered electron. (The vector diagram shows the momentum conservation, with $p$ having components $p\cos \theta$ and $p\sin \theta$; the scattered photon momentum $h{v}^{\prime }/c$ having components $(h{v}^{\prime }/c)\cos \varphi$ and $(h{v}^{\prime }/c)\sin \varphi$; and the incident photon momentum $hv/c$ being purely horizontal).

If the initial photon has the frequency $v$ associated with it, the scattered photon has the lower frequency $v^{\prime }$, where:

Loss in photon energy = Gain in electron energy

$hv-h{v}^{\prime }\equiv T\text{ (4)}$

However, we know that:

$E=\sqrt{m_0^2{c}^4+p^2{c}^2}$

The momentum of a massless particle is related to its energy by the formula:

$E=pc$

Since $E=hv$:

For a photon, its momentum is:

$p=\frac{E}{c}=\frac{hv}{c}\text{ (5)}$

Momentum, unlike energy, is a vector quantity that incorporates direction as well as magnitude, and in the collision momentum must be conserved in each of two mutually perpendicular directions. When more than two bodies participate in a collision, momentum must be conserved in each of three mutually perpendicular directions.

The directions we choose here are that of the original photon and one perpendicular to it in the plane containing the electron and the scattered photon.

The initial photon momentum is $hv/c$, the scattered photon momentum is $h{v}^{\prime }/c$, and the initial and final electron momenta are respectively $0$ and $p$. In the original photon direction:

Initial momentum = Final momentum

$\frac{hv}{c}+0=\frac{h{v}^{\prime }}{c}\cos \varphi +p\cos \theta \text{ (6)}$

And perpendicular to this direction:

Initial momentum = Final momentum

$0=\frac{h{v}^{\prime }}{c}\sin \varphi -p\sin \theta \text{ (7)}$

The angle $\varphi$ is that between the directions of the initial and scattered photons, and $\theta$ is that between the directions of the initial photon and the recoil electron.

From Eqns. (4), (6), and (7) we can find a formula that relates the wavelength difference between initial and scattered photons with the angle between their directions, both of which are readily measurable quantities (unlike the energy and momentum of the recoil electron).

The first step is to multiply Eqns. (6) and (7) by $c$ and rewrite them as:

$pc\cos \theta =hv-h{v}^{\prime }\cos \varphi$

$pc\sin \theta =h{v}^{\prime }\sin \varphi$

By squaring each of these equations and adding the new ones together, the angle $\theta$ is eliminated, leaving:

$p^2{c}^2=(hv{)}^2-2(hv)(h{v}^{\prime })\cos \varphi +(h{v}^{\prime }{)}^2\text{ (8)}$

We equate the two expressions for the total energy of a particle:

$E=T+m_0{c}^2$

$E=\sqrt{m_0^2{c}^4+p^2{c}^2}$

We have:

$(T+m_0{c}^2{)}^2=m_0^2{c}^4+p^2{c}^2$

$p^2{c}^2=T^2+2m_0{c}^{2}T$

Since:

$T=hv-h{v}^{\prime }$

Then:

$p^2{c}^2=(hv-h{v}^{\prime }{)}^2+2m_0{c}^2(hv-h{v}^{\prime })$

Substituting this value of $p^2{c}^2$ in Eq. (8), we finally obtain:

$2m_0{c}^2(hv-h{v}^{\prime })=2(hv)(h{v}^{\prime })(1-\cos \varphi )\text{ (9)}$

This relationship is simpler when expressed in terms of wavelength. Dividing Eqn. (9) by $2h^2{c}^2$:

$\frac{m_{0}c}{h}\left(\frac{v}{c}-\frac{v^{\prime }}{c}\right)=\frac{v{v}^{\prime }}{c^2}(1-\cos \varphi )$

Since $v/c=1/\lambda$ and $v^{\prime }/c=1/{\lambda }^{\prime }$:

$\frac{m_{0}c}{h}\left(\frac{1}{\lambda }-\frac{1}{{\lambda }^{\prime }}\right)=\frac{1}{\lambda {\lambda }^{\prime }}(1-\cos \varphi )$

${\lambda }^{\prime }-\lambda =\frac{h}{m_{0}c}(1-\cos \varphi )\text{ (10)}$

Equation (10) was derived by Arthur H. Compton in the early 1920s, and the phenomenon it describes, which he was the first to observe, is known as the Compton effect. It constitutes very strong evidence in support of the quantum theory of radiation.

Eqn. (10) gives the change in wavelength expected for a photon that is scattered through the angle $\varphi$ by a particle of rest mass $m_0$. This change is independent of the wavelength $\lambda$ of the incident photon.

The quantity:

${\lambda }_c=\frac{h}{m_{0}c}\text{ (11)}$

is called the Compton wavelength of the scattering particle.

For an electron ${\lambda }_c=0.02426\text{ A˚}(2.426\times 10^{-12}\text{ m})$.

In terms of ${\lambda }_c$, Eq. (10) becomes:

${\lambda }^{\prime }-\lambda ={\lambda }_c(1-\cos \varphi )\text{ (12)}$

The Compton wavelength gives the scale of the wavelength change of the incident photon.

From Eqn. (12) we note that the greatest wavelength change possible corresponds to $180^{\circ }$, when the wavelength change will be twice the Compton wavelength ${\lambda }_c$.

Because ${\lambda }_c=2.426\text{ pm}$ for an electron, and even less for other particles owing to their larger rest masses.

The maximum wavelength change in the Compton effect is $4.852\text{ pm}$.

Changes of this magnitude or less are readily observable only in x-rays: the shift in wavelength for visible light is less than $0.01$ percent of the initial wavelength, whereas for x-rays of $0.1\text{ nm}$ it is several percent.

The Compton effect is the chief means by which x-rays lose energy when they pass through matter.

Image/Diagram Note: Experimental demonstration of the Compton effect.

• Description: Figure shows a “Source of monochromatic x-rays,” passing through “Collimators” and striking a target (unlabeled). An “Unscattered x-ray” beam continues straight, while a “Scattered x-ray” beam travels toward an “X-ray spectrometer” at an angle $\varphi$. The path of the spectrometer is shown by a dashed arc.

The experimental demonstration of the Compton effect is straightforward. A beam of x-rays of a single, known wavelength is directed at a target, and the wavelengths of the scattered x-rays are determined at various angles.

The results exhibit the wavelength shift predicted by Eqn. (10), but at each angle the scattered x-rays also include many that have the initial wavelength.

This is not hard to understand. In deriving Eqn. (10) it was assumed that the scattering particle is able to move freely, which is reasonable since many of the electrons in matter are only loosely bound to their parent atoms.

Image/Diagram Note: Experimental confirmation of Compton scattering.

• Description: Four graphs showing “Relative intensity” versus “Wavelength” for scattering angles $\varphi =0^{\circ },45^{\circ },90^{\circ },$ and $135^{\circ }$. The graph for $0^{\circ }$ shows a single peak. The graphs for $45^{\circ },90^{\circ },$ and $135^{\circ }$ each show two peaks (one at the original wavelength and one shifted wavelength, $\Delta \lambda$).
• Caption: The greater the scattering angle, the greater the wavelength change, in accord with Eqn. (10).

Other electrons, however, are very tightly bound and when struck by a photon, the entire atom recoils instead of the single electron.

In this event the value of $m_0$ to use in Eq. (10) is that of the entire atom, which is tens of thousands of times greater than that of an electron, and the resulting Compton shift is accordingly so small as to be undetectable.

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Now that you have the complete source content, are you ready to dive into understanding the key concepts, like how Compton scattering confirms the particle nature of light? Or perhaps we can focus on deriving one of the core equations?