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Compton Scattering (DP IB Physics)

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Katie M

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Katie M

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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

6-11-2-compton-scattering_ocr-al-physics

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

increment lambda space equals space fraction numerator h over denominator m subscript e c end fraction open parentheses 1 space minus space cos space theta close parentheses

  • Where:

    • increment lambda = change in the wavelength of the photon open parentheses lambda subscript f space minus space lambda subscript i close parentheses (m)

    • h = Planck constant

    • m subscript e = mass of an electron (kg)

    • c = speed of light (m s−1)

    • theta = scattering angle of the photon (°)

  • The constant fraction numerator h over denominator m subscript e c end fraction 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 subscript i space equals space h f subscript i space equals space fraction numerator h c over denominator lambda subscript i end fraction space space space space space rightwards double arrow space space space space space space lambda subscript i space equals space fraction numerator h c over denominator E subscript i end fraction

  • The energy of the incident photon, E subscript i = 120 + 40 = 160 keV

lambda subscript i space equals space fraction numerator open parentheses 6.63 cross times 10 to the power of negative 34 end exponent close parentheses open parentheses 3.00 cross times 10 to the power of 8 close parentheses over denominator open parentheses 160 cross times 10 cubed close parentheses open parentheses 1.6 cross times 10 to the power of negative 19 end exponent close parentheses end fraction

lambda subscript i space equals space 7.77 cross times 10 to the power of negative 12 end exponent space straight m space equals space 0.0078 space nm (2 s.f.)

(b) Change in photon wavelength

  • The energy of the scattered photon, E subscript f = 120 keV

lambda subscript f space equals fraction numerator space h c over denominator E subscript f end fraction

lambda subscript f space equals space fraction numerator open parentheses 6.63 cross times 10 to the power of negative 34 end exponent close parentheses open parentheses 3.00 cross times 10 to the power of 8 close parentheses over denominator open parentheses 120 cross times 10 cubed close parentheses open parentheses 1.6 cross times 10 to the power of negative 19 end exponent close parentheses end fraction

lambda subscript f space equals space 1.04 cross times 10 to the power of negative 11 end exponent space straight m space equals space 0.0104 space nm 

  • Therefore, the change in wavelength is

increment lambda space equals space lambda subscript f space minus space lambda subscript i

increment lambda space equals space 0.0104 space minus space 0.00777 space equals space 0.00263 space nm space equals space 0.0026 space nm space open parentheses 2 space straight s. straight f. close parentheses 

(c) Photon scattering angle

  • The Compton formula is

increment lambda space equals space fraction numerator h over denominator m subscript e c end fraction open parentheses 1 space minus space cos space theta close parentheses space space space space space rightwards double arrow space space space space space cos space theta space equals space open parentheses 1 space minus space fraction numerator m subscript e c increment lambda over denominator h end fraction close parentheses

  • The scattering angle is therefore:

cos space theta space equals space 1 space minus space fraction numerator open parentheses 9.11 cross times 10 to the power of negative 31 end exponent close parentheses cross times open parentheses 3.00 cross times 10 to the power of 8 close parentheses cross times open parentheses 0.00263 cross times 10 to the power of negative 9 end exponent close parentheses over denominator 6.63 cross times 10 to the power of negative 34 end exponent end fraction space equals space minus 0.0841

theta space equals space cos to the power of negative 1 end exponent open parentheses negative 0.0841 close parentheses space equals space 94.8 degree space equals space 95 degree space open parentheses 2 space straight s. straight f. close parentheses

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: 

increment lambda space proportional to space open parentheses 1 space minus space cos space theta close parentheses

  • When θ = 0°, cos space theta space equals space 1

So, open parentheses 1 minus space cos space theta close parentheses space equals space 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 space theta space equals space 0

So, open parentheses 1 minus space cos space theta close parentheses space equals space 1 when θ = 90°

  • When θ = 180°, cos space theta space equals space minus 1

So, open parentheses 1 minus space cos space theta close parentheses space equals space 2 when θ = 180°

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

Examiner Tips and Tricks

In the unit conversions section of the data booklet, you are given the value h c space equals space 1.99 cross times 10 to the power of negative 25 end exponent space straight J space straight m space equals space 1.24 space cross times space 10 to the power of negative 6 end exponent space eV space straight 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 space equals space 1.24 space cross times space 10 to the power of negative 6 end exponent space eV space straight 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!

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    Katie M

    Author: Katie M

    Expertise: Physics

    Katie has always been passionate about the sciences, and completed a degree in Astrophysics at Sheffield University. She decided that she wanted to inspire other young people, so moved to Bristol to complete a PGCE in Secondary Science. She particularly loves creating fun and absorbing materials to help students achieve their exam potential.