Past exam of the short options paper of the physics Prelims of the University of Oxford / 2019 / S3 / 2

Compton scattering transfers energy and momentum from a photon to an electron. For an initially stationary, approximately free electron, a photon scattered through a nonzero angle generally has lower frequency and longer wavelength; the recoiling electron carries the difference. The effect demonstrates the photon momentum $p_{\gamma }=h\nu /c=h/\lambda$, beyond a purely elastic classical scattering description.

The middle panel of the figure in solution shows the incident photon along $+x$, the outgoing photon at angle $\theta$ above that direction, and the recoiling electron at deflection angle $\varphi$ below it. Each arrow labels its particle and momentum; the initial electron is at rest at the interaction point. The scattered photon and electron have opposite transverse momentum components, while their forward components sum to the incident photon momentum.

Take a stationary initial electron of mass $m_e$, a photon incident along $x$, and a nonrelativistic recoil velocity of magnitude $v$. Choose $\theta$ above $x$ for the scattered photon and $\varphi$ below $x$ for the electron. Conservation of momentum givesThe electron's unchanged rest energy cancels in conservation of energy, leavingEquivalently use $h\nu /c=h/\lambda$ and $h\nu =hc/\lambda$. The nonrelativistic limit requires $v/c\ll 1$, and treating the electron as free requires its binding energy to be negligible on the scale of the collision.

For complete photon backscatter $\theta =\pi$, the electron recoils forward and momentum conservation givesCombining this with conservation of energy and writing $\Delta \lambda ={\lambda }^{\prime }-\lambda$ givesFor $\Delta \lambda \ll \lambda ,{\lambda }^{\prime }$, the final ratio is approximately four, soWithin the stated nonrelativistic equations the unsimplified relation is implicit and has higher-order recoil corrections; the displayed final approximation is all that the small-shift assumption supports. Exact relativistic Compton recoil kinematics gives $\Delta \lambda =(h/m_{e}c)(1-\cos \theta )$, and therefore exactly $2{\lambda }_C$ at backscatter. Agreement at leading order does not make the nonrelativistic derivation exact at arbitrary photon energy.

The ordinary electron Compton wavelength isA photon of that wavelength has energy equal to the electron rest energy, $hc/{\lambda }_C=m_e{c}^2$. It therefore sets the characteristic scale at which photon recoil cannot generally be treated nonrelativistically, and it sets the magnitude of the Compton wavelength shift. The reduced Compton wavelength, ${\bar{\lambda }}_C=\hslash /(m_{e}c)={\lambda }_C/(2\pi )$, is a related but different convention. Localization on this reduced scale needs momenta of order $m_{e}c$, so a nonrelativistic single-particle description becomes inadequate. Neither wavelength is a measured geometrical radius of the electron.

Retain the backscattering geometry from part (c). With $\lambda =10\mathrm{nm}$, ${\lambda }^{\prime }=\lambda +2{\lambda }_C$ differs from $\lambda$ by only $4.85\times 10^{-4}$ of its value. Momentum conservation givesThe Lorentz-factor correction is of order $v^2/c^2\simeq 2.4\times 10^{-7}$, so the electron recoil is safely nonrelativistic. Its kinetic energy is only about $0.060\mathrm{eV}$, compared with incoming photon energy about $124\mathrm{eV}$. This validates the free-electron recoil approximation's relativistic aspect; actual binding or thermal motion in matter can require separate corrections. No larger recoil occurs at another scattering angle for the same initial photon.