A photon of wavelength strikes an electron of mass that is free and at rest

PHYSICS · Multi-part · Fermi University Physics · Relativity & Gravitation · Quantum Mechanics · Origins of quantum theory

Quantum Mechanics: From the Photon to the Hydrogen Atom, Ch Light in lumps: blackbody radiation, the photoelectric effect and Compton scattering

Problem

A photon of wavelength λ\lambda strikes an electron of mass mem_e that is free and at rest. The photon is scattered through an angle θ\theta and leaves with wavelength λ′\lambda'; the electron recoils. Energy and momentum are conserved, the electron's energy obeys E2=p2c2+me2c4E^2 = p^2c^2 + m_e^2c^4, and a photon of wavelength λ\lambda has energy hc/λhc/\lambda and momentum h/λh/\lambda. ((a)) Show that λ′−λ=hmec(1−cos⁡θ)\lambda' - \lambda = \dfrac{h}{m_ec}(1 - \cos\theta). ((b)) X-rays of wavelength 62.062.0 pm are scattered through 60.0∘60.0^\circ by a nearly free electron at rest. Find the scattered wavelength, in pm. ((c)) Find the kinetic energy carried off by the recoiling electron, in eV. ((d)) Take the incoming photon along the xx axis and the scattered photon 60.0∘60.0^\circ above it. Using the xx and yy components of momentum conservation, find the angle ϕ\phi below the xx axis at which the electron recoils. Use h=6.626×10−34h = 6.626\times10^{-34} J s, me=9.109×10−31m_e = 9.109\times10^{-31} kg, c=3.00×108 m s−1c = 3.00\times10^{8}\ \mathrm{m\,s^{-1}} and hc=1240hc = 1240 eV nm =1.240×106= 1.240\times10^{6} eV pm.

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