Compton scattering

Simon Greaves

Research Institute of Electrical Communication, Tohoku University, Sendai

QM3

1 / 14

Compton scattering I

Scattering of (low energy) X-rays first observed by Barkla in 1909.

The scattered X-rays had the same wavelength as the incident

X-rays.

Thomson’s theory (1900): Incident X-rays cause electrons to

oscillate at the same frequency. Electrons also emit

electromagnetic radiation at this frequency.

Compton studied higher energy X-rays and found a wavelength

shift for some scattered X-rays.

Compton scattering occurs from free, or weakly bound electrons,

i.e. electron binding energy ≪ X-ray energy. It is independent of

the type of material used.

2 / 14

Compton scattering II

First measured by Arthur Compton in 1922. A photon loses part of

its energy to an electron during a collision. The lower photon

energy results in a lower frequency through the E = hf relationship.

The change in frequency is dependent upon the angle of incidence.

h/mc = 2.426×10−12 m for an electron.

3 / 14

Compton scattering III

As the scattering angle increases,

two distinct peaks appear.

The Compton effect suggests

that light has a particulate

nature.

Compton won the Nobel prize

in 1927 for the discovery of this

effect.

The shift in wavelength

becomes greater as the

scattering angle increases.

4 / 14

X-rays I

X-rays are generated by accelerating electrons emitted from a

cathode.

X-rays are emitted when the accelerated electrons collide with the

target material, e.g. Cu, W, Mo etc.

5 / 14

X-rays II

The incident electron ejects an electron from a target atom.

An electron in the target atom fills the vacancy, losing energy as it

does so.

The energy lost is emitted as an X-ray photon.

6 / 14

X-rays III

X-rays have fixed wavelengths that depend on the target material.

7 / 14

Relativistic equations I

Relativistic energy

E =mc2

(

1 −v2

c2

)1

2

(as v → 0 , E → mc2)

Relativistic momentum

p =mv

(

1 −v2

c2

)1

2

(m is the electron mass)

Somv

p=

mc2

Eand v =

pc2

E

8 / 14

Relativistic equations II

Substitute into E to get

E

(

1 −p2c4

E2c2

)

1

2

= mc2

So

E2− p2c2 = m2c4 and E2 = m2c4 + p2c2

This is the energy of a moving particle with mass m and momentum

p.

For a stationary particle (p = 0) E = mc2.

For a photon (m = 0) E = cp.

9 / 14

Conservation of momentum I

The incident photon has energy E0 and momentum p0. The

scattered photon has energy E1 and momentum p1 and is scattered

through an angle θ.

The recoiling electron has momentum p2. It recoils at an angle φ.

From conservation of momentum we have

p0 = p1 cos θ + p2 cosφ along the direction of incidence and

p1 sin θ + p2 sinφ = 0 perpendicular to the direction of incidence.

10 / 14

Conservation of momentum II

Square and collect p2 terms to get

p2

2 sin2 φ+ p2

2 cos2 φ = p2

1 sin2 θ + (p0 − p1 cos θ)

2

Then

p2

2= p2

1sin2 θ + p2

0− 2p0p1 cos θ + p2

1cos2 θ

So

p2

2 = p2

0 + p2

1 − 2p0p1 cos θ

or

p2

2= (p0 − p1)

2 + 2p0p1(1 − cos θ)

11 / 14

Conservation of energy I

Next, we deal with the energy. Assuming the electron is at rest

before the collision (E = mc2) we have

E0 + mc2 = E1 + (m2c4 + p2

2c2)1/2

Therefore

E0 − E1 = (m2c4 + p2

2c2)1/2

− mc2 = c(p0 − p1)

where c(p0 − p1) is the change in photon energy (E = cp).

So

(m2c4 + p2

2c2)1/2 = c(p0 − p1) + mc2

or

m2c4 + p2

2c2 = m2c4 + c2(p0 − p1)

2 + 2mc3(p0 − p1)

12 / 14

Conservation of energy II

Divide by c2 to get

p2

2= (p0 − p1)

2 + 2mc(p0 − p1)

We can combine the expressions for p2 obtained from the

conservations of energy and momentum to give

mc(p0 − p1) = p0p1(1 − cos θ)

Now, for the photon

E = cp = hf = hc/λ so p = h/λ

Then

p0 − p1 =h

λ0

−h

λ1

=h(λ1 − λ0)

λ0λ1

and p0p1 =h2

λ0λ1

13 / 14

Final result

We havemch(λ0 − λ1)

λ0λ1

=h2(1 − cos θ)

λ0λ1

Finally

(λ0 − λ1) = ∆λ =h(1 − cos θ)

mc

The quantity h/mc is known as the Compton wavelength of the

electron. It is independent of the material and incident photon

wavelength.

∆λ increases for larger θ. This experiment also demonstrates the

particulate nature of light.

14 / 14