Question 6.3: Interpreting the scattering of X-rays by a graphite sheet as......
Interpreting the scattering of X-rays by a graphite sheet as caused by the collision of a photon with an electron, A. H. Compton showed in 1923 that
\lambda ^{\prime } = \lambda+ \frac{2h}{mc} \sin^{2}\frac{\theta}{2} .(6.134)
The energy and momentum transfer from the X-ray to the electron is known as the Compton effect. Derive Compton’s formula, where λ is the wavelength of the incoming X-ray, \lambda ^{\prime } is the wavelength of the outgoing X-ray, m is the mass of the electron and θ is the angle between the outgoing and the incoming X-rays.
In Fig. 6.6 a photon of frequency ν collides with an electron at rest and is deflected by an angle θ while its frequency ν diminishes to ν^{\prime }, whereas the electron’s recoil angle is θ^{\prime }. Let P and P^{\prime } be the initial and final four-momenta of the photon, with Q and Q^{\prime } being those of the electron. Conservation of four-momentum requires
P + Q = P^{\prime } + Q^{\prime }. (6.135)
Inasmuch as we are not interested in the recoiling electron, it turns out to be convenient to eliminate the undesirable four- vector Q^{\prime } by isolating it in (6.135) and squaring the resulting equation:
Q^{\prime }²= (P + Q − P^{\prime })² = P² + Q² + P^{\prime }² + 2Q · P − 2P · P^{\prime }− 2Q · P^{\prime }. (6.136)
Since Q² = Q^{\prime }² = m²c² and P² = P^{\prime }² = 0, we are left with
P · P^{\prime } = Q · P − Q · P^{\prime }. (6.137)
Calculating the right-hand side of this equation in the laboratory frame, in which Q = (mc, 0, 0, 0), we find Q · P = mhν and Q · P^{\prime } = mhν^{\prime }, while
P · P^{\prime } =\frac{hν}{c}\frac{hν^{\prime }}{c} \left(1-\hat{n} \cdot \hat{n}^{\prime }\right)= \frac{h^{2}vν^{\prime }}{c^{2}} \left(1- \cos \theta\right). (6.138)
Inserting these results into (6.137), we finally get
\frac{hvν^{\prime }}{c^{2}}\left(1- \cos \theta\right)=m\left(v-ν^{\prime }\right). (6.139)
It is easy to check that in terms of the semi-angle θ/2 and of the wavelength λ = c/ν this last equation becomes identical to Compton’s formula (6.134).
