IPhO 1983 · Theory problem 4

Compton scattering

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A photon of wavelength \(\lambda_i\) is scattered by a moving, free electron. As a result the electron stops and the resulting photon of wavelength \(\lambda_0\) scattered at an angle \(\theta=60^\circ\) with respect to the direction of the incident photon, is again scattered by a second free electron at rest. In this second scattering process a photon with wavelength of \(\lambda_f=1{,}25\times10^{-10}\,m\) emerges at an angle \(\theta=60^\circ\) with respect to the direction of the photon of wavelength \(\lambda_0\). Find the de Broglie wavelength for the first electron before the interaction. The following constants are known:

\(h=6{,}6\times10^{-34}\,J\cdot s\) - Planck’s constant
\(m=9{,}1\times10^{-31}\,kg\) - mass oh the electron
\(c=3{,}0\times10^8\,m/s\) - speed of light in vacuum

The purpose of the problem is to calculate the values of the speed, momentum and wavelength of the first electron.

To characterize the photons the following notation are used:

Table 4.1

initial photon photon – after the first scattering final photon
momentum \(\vec p_i\) \(\vec p_0\) \(\vec p_f\)
energy \(E_i\) \(E_0\) \(E_f\)
wavelength \(\lambda_i\) \(\lambda_i\) \(\lambda_f\)

To characterize the electrons one uses

Table 4.2

first electron before collision first electron after collision second electron before collision Second electron after collision
momentum \(\vec p_{1e}\) \(0\) \(0\) \(\vec p_{2e}\)
energy \(E_{1e}\) \(E_{0e}\) \(E_{0e}\) \(E_{2e}\)
speed \(\vec v_{1e}\) \(0\) \(0\) \(\vec v_{2e}\)

Official English text of the problem.

Answers

  • Problem IV (7 points) — The first electron’s initial speed, momentum magnitude, and de Broglie wavelength are \[ \boxed{v_{1e}\approx5.8\times10^6\ \mathrm{m\,s^{-1}},} \] \[ \boxed{p_{1e}\approx5.3\times10^{-24}\ \mathrm{kg\,m\,s^{-1}},} \] \[ \boxed{\lambda_{\mathrm{dB}}\approx1.24\times10^{-10}\ \mathrm m.} \] The intermediate photon wavelength is \(\lambda_0\approx1.238\times10^{-10}\,\mathrm m\), and \(\lambda_i=\lambda_f=1.25\times10^{-10}\,\mathrm m\).

Solution

Problem IV. There are no numbered subparts in the supplied statement.

We apply conservation of energy and momentum to each collision. For a photon, \[ E=\frac{hc}{\lambda},\qquad p=\frac{E}{c}=\frac h\lambda, \] while an electron satisfies \[ E_e^2=m^2c^4+p_e^2c^2. \] We initially use exact relativistic relations.

The second scattering determines \(\lambda_0\).

The second electron is initially at rest, so \[ E_0+mc^2=E_f+E_{2e}, \qquad \vec p_0=\vec p_f+\vec p_{2e}. \] The angle between the incoming and outgoing photons is \(\theta\), giving \[ p_{2e}^2c^2=E_0^2+E_f^2-2E_0E_f\cos\theta. \] Substitution into the electron’s energy–momentum relation yields \[ (mc^2+E_0-E_f)^2 =m^2c^4+E_0^2+E_f^2-2E_0E_f\cos\theta. \] After cancellation, \[ mc^2(E_0-E_f)=E_0E_f(1-\cos\theta). \] Dividing by \(E_0E_f\) and substituting \(E=hc/\lambda\), we obtain the Compton shift: \[ \lambda_f-\lambda_0=\frac{h}{mc}(1-\cos\theta). \]

With the given constants, \[ \frac{h}{mc} =\frac{6.6\times10^{-34}} {(9.1\times10^{-31})(3.0\times10^8)} =2.41758\times10^{-12}\ \mathrm m. \] Since \(\cos60^\circ=\tfrac12\), \[ \Delta\lambda=\frac{h}{2mc} =1.20879\times10^{-12}\ \mathrm m, \] and therefore \[ \lambda_0 =1.25\times10^{-10}-1.20879\times10^{-12} =1.2379121\times10^{-10}\ \mathrm m. \]

The first scattering determines \(\lambda_i\).

The first electron comes to rest after the collision. Thus \[ E_{1e}+E_i=mc^2+E_0, \qquad \vec p_{1e}+\vec p_i=\vec p_0. \] Consequently, \[ E_{1e}=mc^2+E_0-E_i \] and \[ p_{1e}^2c^2=E_0^2+E_i^2-2E_0E_i\cos\theta. \] Using the electron’s energy–momentum relation again, \[ (mc^2+E_0-E_i)^2 =m^2c^4+E_0^2+E_i^2-2E_0E_i\cos\theta, \] which simplifies to \[ mc^2(E_0-E_i)=E_0E_i(1-\cos\theta). \] Hence \[ \lambda_i-\lambda_0=\frac{h}{mc}(1-\cos\theta). \]

The photon gains energy in this collision, so its wavelength decreases. Because both scattering angles are equal, the decrease in the first collision equals the increase in the second: \[ \boxed{\lambda_i=\lambda_f=1.25\times10^{-10}\ \mathrm m.} \]

Initial electron speed.

All the first electron’s initial kinetic energy is transferred to the photon: \[ K_{1e}=E_0-E_i =hc\left(\frac1{\lambda_0}-\frac1{\lambda_i}\right). \] Numerically, \[ K_{1e}=1.54674\times10^{-17}\ \mathrm J. \] Writing \(E_{1e}=\gamma mc^2\), we have \[ \gamma=1+\frac{K_{1e}}{mc^2} =1.0001888569. \] Therefore \[ v_{1e} =c\sqrt{1-\frac1{\gamma^2}} =5.82963\times10^6\ \mathrm{m\,s^{-1}}, \] so \[ \boxed{v_{1e}\approx5.8\times10^6\ \mathrm{m\,s^{-1}}.} \]

Since \(v_{1e}/c\approx0.0194\), the nonrelativistic approximation \(K_{1e}\approx mv_{1e}^2/2\) would also be adequate at the precision of the supplied data.

Initial electron momentum.

Momentum conservation gives its direction and magnitude through \[ \vec p_{1e}=\vec p_0-\vec p_i, \] \[ p_{1e} =h\sqrt{\frac1{\lambda_0^2} +\frac1{\lambda_i^2} -\frac{2\cos60^\circ}{\lambda_0\lambda_i}}. \] Substituting the wavelengths, \[ p_{1e}=5.30597\times10^{-24}\ \mathrm{kg\,m\,s^{-1}}, \] in agreement with \(p_{1e}=\gamma mv_{1e}\). Thus \[ \boxed{p_{1e}\approx5.3\times10^{-24}\ \mathrm{kg\,m\,s^{-1}}.} \]

Initial electron de Broglie wavelength.

The de Broglie relation is \[ \lambda_{\mathrm{dB}}=\frac{h}{p_{1e}}. \] Equivalently, using \(\cos60^\circ=\tfrac12\), \[ \lambda_{\mathrm{dB}} =\frac{\lambda_0\lambda_i} {\sqrt{\lambda_0^2+\lambda_i^2-\lambda_0\lambda_i}}. \] Therefore \[ \lambda_{\mathrm{dB}} =\frac{6.6\times10^{-34}}{5.30597\times10^{-24}} =1.24388\times10^{-10}\ \mathrm m, \] or \[ \boxed{\lambda_{\mathrm{dB}}\approx1.24\times10^{-10}\ \mathrm m.} \]

The solution was checked against the official solution.