Compton Scattering (HL) | HL IB Physics Revision Notes 2025

Compton Scattering

Compton scattering can be observed when a high energy photon (typically X-ray or gamma) interacts with an orbital electron
- This phenomenon is further evidence of the particle nature of light

Compton scattering of an X-ray photon with an orbital electron

The Compton Effect is defined as:

The interaction of a high-energy photon with an orbital electron which causes an increase in the wavelength of the photon and the ejection of the electron

During the collision, the photon transfers some of its energy to the orbital electron
Because of this transfer of energy:
- The photon is deflected from its initial path
- The photon's wavelength increases, as its energy decreases
- The electron involved is ejected from the atom
The electron and photon are deflected in different directions due to conservation of momentum

The Compton Formula

The Compton scattering formula is given by

$∆ λ = \frac{h}{m_{e} c} (1 - \cos θ)$

Where:
- $∆ λ$ = change in the wavelength of the photon $(λ_{f} - λ_{i})$ (m)
- $h$ = Planck constant
- $m_{e}$ = mass of an electron (kg)
- $c$ = speed of light (m s⁻¹)
- $θ$ = scattering angle of the photon (°)

The constant $\frac{h}{m_{e} c}$ is known as the Compton wavelength

The equation tells us:
- The reduced wavelength of the photon depends on the scattering angle
- The greater the scattering angle, the longer the wavelength

This equation assumes that the electron is initially at rest before the interaction

Worked example

An X-ray photon collides with a stationary orbital electron. The scattered photon has an energy of 120 keV and the recoiling electron has an energy of 40 keV.

Determine

(a)

the wavelength of the incident X-ray photon.

(b)

the change in wavelength of the photon.

(c)

the scattering angle of the photon.

Answer:

(a) Initial photon wavelength

Photon energy and wavelength are related by

$E_{i} = h f_{i} = \frac{h c}{λ_{i}} \Rightarrow λ_{i} = \frac{h c}{E_{i}}$

The energy of the incident photon, $E_{i}$ = 120 + 40 = 160 keV

$λ_{i} = \frac{(6.63 \times 10^{- 34}) (3.00 \times 10^{8})}{(160 \times 10^{3}) (1.6 \times 10^{- 19})}$

$λ_{i} = 7.77 \times 10^{- 12} m = 0.0078 nm$ (2 s.f.)

(b) Change in photon wavelength

The energy of the scattered photon, $E_{f}$ = 120 keV

$λ_{f} = \frac{h c}{E_{f}}$

$λ_{f} = \frac{(6.63 \times 10^{- 34}) (3.00 \times 10^{8})}{(120 \times 10^{3}) (1.6 \times 10^{- 19})}$

$λ_{f} = 1.04 \times 10^{- 11} m = 0.0104 nm$

Therefore, the change in wavelength is

$∆ λ = λ_{f} - λ_{i}$

$∆ λ = 0.0104 - 0.00777 = 0.00263 nm = 0.0026 nm (2 s . f .)$

(c) Photon scattering angle

The Compton formula is

$∆ λ = \frac{h}{m_{e} c} (1 - \cos θ) \Rightarrow \cos θ = (1 - \frac{m_{e} c ∆ λ}{h})$

The scattering angle is therefore:

$\cos θ = 1 - \frac{(9.11 \times 10^{- 31}) \times (3.00 \times 10^{8}) \times (0.00263 \times 10^{- 9})}{6.63 \times 10^{- 34}} = - 0.0841$

$θ = \cos^{- 1} (- 0.0841) = 94.8 ° = 95 ° (2 s . f .)$

Worked example

Deduce the scattering angle at which

(a)

no change in photon wavelength is observed

(b)

the largest change in photon wavelength is observed

Answer:

(a) No change in photon wavelength

From the Compton formula:

$∆ λ \propto (1 - \cos θ)$

When θ = 0°, $\cos θ = 1$

So, $(1 - \cos θ) = 0$ when θ = 0°

Therefore, when θ = 0°, the change in photon wavelength will be zero

(b) Maximum change in photon wavelength

When θ = 90°, $\cos θ = 0$

So, $(1 - \cos θ) = 1$ when θ = 90°

When θ = 180°, $\cos θ = - 1$

So, $(1 - \cos θ) = 2$ when θ = 180°

Therefore, when θ = 180°, the change in photon wavelength will be twice the Compton wavelength of the electron

Examiner Tip

In the unit conversions section of the data booklet, you are given the value $h c = 1.99 \times 10^{- 25} J m = 1.24 \times 10^{- 6} eV m$ . You can use this value to save time typing into your calculator.

You may get a slightly different answer due to the slight differences in rounding. For example, in the worked example above, if you used $h c = 1.24 \times 10^{- 6} eV m$ throughout, you would get an answer of 99° instead of 95° in part (c). This would be fine in an exam situation as examiners will allow for the discrepancy - just as long as your working is clear!

Compton Scattering (HL) (HL IB Physics)

Revision Note