Question 4 of 7: Inverse Compton Scattering — Generating 10 MeV Photons

Question 4: Inverse Compton Scattering — Generating 10 MeV Photons (10 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. $\lambda_0=2h/m_ec$; incident photon $\lambda_i=10\ \mu\text{m}$ (CO$_2$ laser); target final photon energy $E_f=10$ MeV; $m_ec^2=0.511$ MeV.

Find. (a) dimensional proof; (b) numeric comparison $\lambda_0$ vs. $\lambda_i$; (c) the condition for the approximation (2); (d) electron speed $v$; (e) total electron energy; (f) electron kinetic energy; (g) accelerating voltage.

Approach. Part (a)–(c) are dimensional/limiting-case analysis of the given formula. For (d)–(g), use the ultra-relativistic limit of equation (2) ($\lambda_f/\lambda_i=(1-\beta)/(1+\beta)\to1/4\gamma^2$ as $\beta\to1$) together with $E_f/E_i=\lambda_i/\lambda_f$ to solve for the Lorentz factor $\gamma$, then read off $v$, total energy, kinetic energy, and the equivalent accelerating voltage.

1. Part (a) — dimensions of $\lambda_0=2h/m_ec$. $[h]=\text{J}\cdot\text{s} =\text{kg}\,\text{m}^2\text{s}^{-1}$, $[m_e]=\text{kg}$, $[c]=\text{m}\,\text{s}^{-1}$, so $$[\lambda_0] = \frac{\text{kg}\,\text{m}^2\text{s}^{-1}}{\text{kg}\cdot\text{m}\,\text{s}^{-1}} = \text{m}$$ $$\boxed{[\lambda_0]=\text{m} \ \ \text{(a length, consistent with a wavelength)}}$$
2. Part (b) — numeric value of $\lambda_0$ vs. the CO$_2$ laser line. $$\lambda_0 = \frac{2h}{m_ec} = \frac{2(6.626\times10^{-34})}{(9.109\times10^{-31})(2.998\times10^{8})}$$ $$\boxed{\lambda_0 \approx 4.85\times10^{-12}\ \text{m} = 4.85\ \text{pm}}$$ Compared with the CO$_2$ laser's $\lambda_i=10\ \mu\text{m}=1\times10^{4}$ pm, $$\frac{\lambda_i}{\lambda_0} = \frac{10\ \mu\text{m}}{4.85\ \text{pm}} \approx 2.1\times10^{6}$$ i.e. $\lambda_0$ is about two million times smaller than the incident laser wavelength — utterly negligible by comparison.
3. Part (c) — condition for the simplified equation (2). In equation (1), the $\lambda_0$ term is added directly to $\lambda$ (both multiplied by the same $(1-\beta)/(1+\beta)$ factor), so it can be dropped whenever $$\boxed{\lambda \gg \lambda_0 = 2h/m_ec}$$ i.e. whenever the incident photon's wavelength is much larger than the (fixed, $\sim$5 pm) Compton-like length scale $\lambda_0$. Part (b) shows this holds by six orders of magnitude for a CO$_2$-laser photon, which is exactly why the approximation (2) is safe to use for the rest of this question.
4. Part (d) — electron velocity $v$ to generate 10 MeV photons. Using equation (2) and $E\propto1/\lambda$ (so $E_f/E_i=\lambda_i/\lambda_f$): $$\frac{E_f}{E_i} = \frac{\lambda_i}{\lambda_f} = \frac{1+\beta}{1-\beta}, \qquad \beta=v/c$$ The incident photon energy is $E_i=hc/\lambda_i=1240\ \text{eV}\cdot\text{nm}/10{,}000\ \text{nm} \approx0.124$ eV. For an ultra-relativistic electron ($\beta\to1$), $1-\beta\approx1/2\gamma^2$ and $1+\beta\approx2$, so $(1+\beta)/(1-\beta)\approx4\gamma^2$: $$4\gamma^2 \approx \frac{E_f}{E_i} = \frac{10\times10^{6}\ \text{eV}}{0.124\ \text{eV}} \approx 8.07\times10^{7}$$ $$\gamma \approx \sqrt{\frac{8.07\times10^7}{4}} \approx 4490$$ $$\beta = \sqrt{1-1/\gamma^2} \approx 0.99999998,\qquad \boxed{v = \beta c \approx 2.99792\times10^{8}\ \text{m/s}}$$ (the electron falls short of $c$ by only about 7.4 m/s — an extreme but physically ordinary regime, the same one used in real Compton-backscatter gamma-ray sources).
5. Part (e) — total electron energy at $v$. $$E_{\text{tot}} = \gamma m_ec^2 = 4490\times0.511\ \text{MeV}$$ $$\boxed{E_{\text{tot}} \approx 2295\ \text{MeV} \approx 2.29\ \text{GeV}}$$
6. Part (f) — kinetic energy at $v$. $$KE = (\gamma-1)m_ec^2 = E_{\text{tot}} - m_ec^2 = 2295 - 0.511$$ $$\boxed{KE \approx 2294\ \text{MeV}}$$ (the 0.511 MeV rest-mass energy is a negligible correction at this $\gamma$).
7. Part (g) — accelerating voltage. An electron accelerated through potential difference $V$ gains kinetic energy $KE=eV$, so numerically $V$ in volts equals $KE$ in eV: $$\boxed{V = \frac{KE}{e} \approx 2.29\times10^{9}\ \text{V} = 2.29\ \text{GV}}$$ (in practice such an electron beam is built up over many accelerating stages, e.g. a linac or synchrotron, rather than a single 2.3 GV potential — the voltage quoted is the energy equivalent, not a literal single-gap design value).