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Physics / Chapter 38: Photons and Matter Waves / Section 38-3

Photons Have Momentum — Question 1

Section 38-3 · Physics pages 6292–6293

Question 1 Absorbing one photon gives a measurable impulse even when recoil energy is tiny A stationary small absorber of mass M = 10−15 kg absorbs one photon of wavelength 663 nm. Use h = 6.63 × 10−34 J s and c = 3.00 × 108 m/s. Treat recoil nonrelativistically and neglect the tiny mass correction. photons detector M Schematic; labels, not drawing lengths, specify the data. (a) Find the impulse and recoil speed. (b) Find recoil kinetic energy and its fraction of the photon energy.
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Question 1 - Solution Absorbing one photon gives a measurable impulse even when recoil energy is tiny photons detector M (a) Method and calculation pγ = h/λ = 1.00 × 10−27 kg m/s , J = pγ. v = pγ/M = 1.00 × 10−12 m/s . (b) Interpretation and further calculation KR = p2 γ/(2M) = 5.00 × 10−40 J . Eγ = pγc = 3.00 × 10−19 J, KR/Eγ = 1.66667 × 10−21 . Physical check Most of the absorbed energy becomes internal energy rather than translational recoil. The calculated speed justifies the nonrelativistic approximation.

Question 1

Absorbing one photon gives a measurable impulse even when recoil energy is tiny

A stationary small absorber of mass M=10−15kgM=10^{-15}\,\mathrm{kg} absorbs one photon of wavelength 663 nm. Use h=6.63×10−34Jsh=6.63\times10^{-34}\,\mathrm{J\,s} and c=3.00×108m/sc=3.00\times10^8\,\mathrm{m/s}. Treat recoil nonrelativistically and neglect the tiny mass correction.

Diagram shown in the original SVG worksheet.

  1. Find the impulse and recoil speed.

  2. Find recoil kinetic energy and its fraction of the photon energy.

Question 1 - Solution

Absorbing one photon gives a measurable impulse even when recoil energy is tiny

Diagram shown in the original SVG worksheet.

(a) Method and calculation

pγ=h/λ=1.00×10−27kgm/s,J=pγ.\boxed{p_\gamma=h/\lambda=1.00\times10^{-27}\,\mathrm{kg\,m/s}},\quad J=p_\gamma.v=pγ/M=1.00×10−12m/s.\boxed{v=p_\gamma/M=1.00\times10^{-12}\,\mathrm{m/s}}.

(b) Interpretation and further calculation

KR=pγ2/(2M)=5.00×10−40J.\boxed{K_R=p_\gamma^2/(2M)=5.00\times10^{-40}\,\mathrm J}.Eγ=pγc=3.00×10−19J,KR/Eγ=1.66667×10−21.E_\gamma=p_\gamma c=3.00\times10^{-19}\,\mathrm J,\quad\boxed{K_R/E_\gamma=1.66667\times10^{-21}}.

Physical check

Most of the absorbed energy becomes internal energy rather than translational recoil. The calculated speed justifies the nonrelativistic approximation.

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