6.3 The compton effect (Page 2/12)

University physics volume 3 Page 4 / 12

We start with rearranging the terms in [link] and squaring it:

{[(E_{f} - {\tilde{E}}_{f}) + m_{0} c^{2}]}^{2} = E^{2} .

In the next step, we substitute [link] for $E^{2},$ simplify, and divide both sides by $c^{2}$ to obtain

{(E_{f} / c - {\tilde{E}}_{f} / c)}^{2} + 2 m_{0} c (E_{f} / c - {\tilde{E}}_{f} / c) = p^{2} .

Now we can use [link] to express this form of the energy equation in terms of momenta. The result is

{(p_{f} - {\tilde{p}}_{f})}^{2} + 2 m_{0} c (p_{f} - {\tilde{p}}_{f}) = p^{2} .

To eliminate $p^{2},$ we turn to the momentum equation [link] , rearrange its terms, and square it to obtain

{({\vec{p}}_{f} - {\vec{\tilde{p}}}_{f})}^{2} = p^{2} and {({\vec{p}}_{f} - {\vec{\tilde{p}}}_{f})}^{2} = p_{f}^{2} + {\tilde{p}}_{f}^{2} - 2 {\vec{p}}_{f} \cdot {\vec{\tilde{p}}}_{f} .

The product of the momentum vectors is given by [link] . When we substitute this result for $p^{2}$ in [link] , we obtain the energy equation that contains the scattering angle $θ :$

{(p_{f} - {\tilde{p}}_{f})}^{2} + 2 m_{0} c (p_{f} - {\tilde{p}}_{f}) = p_{f}^{2} + {\tilde{p}}_{f}^{2} - 2 p_{f} {\tilde{p}}_{f} cos θ .

With further algebra, this result can be simplified to

\frac{1}{{\tilde{p}}_{f}} - \frac{1}{p_{f}} = \frac{1}{m_{0} c} (1 - cos θ) .

Now recall [link] and write: $1 / {\tilde{p}}_{f} = λ^{'} / h$ and $1 / p_{f} = λ / h .$ When these relations are substituted into [link] , we obtain the relation for the Compton shift:

λ' - λ = \frac{h}{m_{0} c} (1 - cos θ) .

The factor $h / m_{0} c$ is called the Compton wavelength of the electron:

λ_{c} = \frac{h}{m_{0} c} = 0.00243 nm = 2.43 pm .

Denoting the shift as $Δ λ = λ' - λ,$ the concluding result can be rewritten as

Δ λ = λ_{c} (1 - cos θ) .

This formula for the Compton shift describes outstandingly well the experimental results shown in [link] . Scattering data measured for molybdenum, graphite, calcite, and many other target materials are in accord with this theoretical result. The nonshifted peak shown in [link] is due to photon collisions with tightly bound inner electrons in the target material. Photons that collide with the inner electrons of the target atoms in fact collide with the entire atom. In this extreme case, the rest mass in [link] must be changed to the rest mass of the atom. This type of shift is four orders of magnitude smaller than the shift caused by collisions with electrons and is so small that it can be neglected.

Compton scattering is an example of inelastic scattering , in which the scattered radiation has a longer wavelength than the wavelength of the incident radiation. In today’s usage, the term “Compton scattering” is used for the inelastic scattering of photons by free, charged particles. In Compton scattering, treating photons as particles with momenta that can be transferred to charged particles provides the theoretical background to explain the wavelength shifts measured in experiments; this is the evidence that radiation consists of photons.

Compton scattering

An incident 71-pm X-ray is incident on a calcite target. Find the wavelength of the X-ray scattered at a

30 °

angle. What is the largest shift that can be expected in this experiment?

Strategy

To find the wavelength of the scattered X-ray, first we must find the Compton shift for the given scattering angle,

θ = 30 ° .

We use [link] . Then we add this shift to the incident wavelength to obtain the scattered wavelength. The largest Compton shift occurs at the angle

θ

when

1 - cos θ

has the largest value, which is for the angle

θ = 180 ° .

Solution

The shift at

θ = 30 °

Δ λ = λ_{c} (1 - cos 30 °) = 0.134 λ_{c} = (0.134) (2.43) pm = 0.325 pm .

This gives the scattered wavelength:

λ' = λ + Δ λ = (71 + 0.325) pm = 71.325 pm .

The largest shift is

{(Δ λ)}_{max} = λ_{c} (1 - cos 180^{0}) = 2 (2.43 pm) = 4.86 pm .

Significance

The largest shift in wavelength is detected for the backscattered radiation; however, most of the photons from the incident beam pass through the target and only a small fraction of photons gets backscattered (typically, less than 5%). Therefore, these measurements require highly sensitive detectors.

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6.3 The compton effect (Page 2/12)

Compton scattering

Strategy

Solution

Significance

Summary

Conceptual questions

Problems

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