Compton Scattering (DP IB Physics: HL): Revision Note

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

$E_i=h{f}_i=\frac{hc}{{\lambda }_i}\Rightarrow {\lambda }_i=\frac{hc}{E_i}$

• The energy of the incident photon, $E_i$ = 120 + 40 = 160 keV

${\lambda }_i=\frac{\left(6.63\times {10}^{-34}\right)\left(3.00\times {10}^8\right)}{\left(160\times {10}^3\right)\left(1.6\times {10}^{-19}\right)}$

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

(b) Change in photon wavelength

• The energy of the scattered photon, $E_f$ = 120 keV

${\lambda }_f=\frac{hc}{E_f}$

${\lambda }_f=\frac{\left(6.63\times {10}^{-34}\right)\left(3.00\times {10}^8\right)}{\left(120\times {10}^3\right)\left(1.6\times {10}^{-19}\right)}$

${\lambda }_f=1.04\times {10}^{-11}\mathrm{m}=0.0104\mathrm{nm}$ 

• Therefore, the change in wavelength is

$∆\lambda ={\lambda }_f-{\lambda }_i$

$∆\lambda =0.0104-0.00777=0.00263\mathrm{nm}=0.0026\mathrm{nm}\left(2\mathrm{s}.\mathrm{f}.\right)$ 

(c) Photon scattering angle

• The Compton formula is

$∆\lambda =\frac{h}{m_{e}c}\left(1-\cos \theta \right)\Rightarrow \cos \theta =\left(1-\frac{m_{e}c∆\lambda }{h}\right)$

• The scattering angle is therefore:

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

$\theta ={\cos }^{-1}\left(-0.0841\right)=94.8°=95°\left(2\mathrm{s}.\mathrm{f}.\right)$

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: 

$∆\lambda \propto \left(1-\cos \theta \right)$

• When θ = 0°, $\cos \theta =1$

So, $\left(1-\cos \theta \right)=0$ when θ = 0°

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

(b) Maximum change in photon wavelength

• When θ = 90°, $\cos \theta =0$

So, $\left(1-\cos \theta \right)=1$ when θ = 90°

• When θ = 180°, $\cos \theta =-1$

So, $\left(1-\cos \theta \right)=2$ when θ = 180°

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

In the unit conversions section of the data booklet, you are given the value $hc=1.99\times {10}^{-25}\mathrm{J}\mathrm{m}=1.24\times {10}^{-6}\mathrm{eV}\mathrm{m}$. 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 $hc=1.24\times {10}^{-6}\mathrm{eV}\mathrm{m}$ 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!