The Photon, the Quantum of Light — Question 5

Section 38-1 · Physics pages 6260–6261

Question 5 Total power and photon rate reveal an unknown two-color mixture A beam contains photons of energies 2.00 eV and 3.00 eV only. Its total power is 4.00 mW and total photon rate is 1016 s−1. Use 1 eV = 1.60 × 10−19 J. E (eV) ˙N 2 3 Schematic; labels, not drawing lengths, specify the data. (a) Find the fraction of photons in each energy group. (b) Find the optical power carried by each group and check both conservation constraints.
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Question 5 - Solution Total power and photon rate reveal an unknown two-color mixture E (eV) ˙N 2 3 (a) Method and calculation E = P/ ˙N = 2.50 eV, 2f + 3(1 −f) = 2.50. f = 1/2 , ˙N2 = ˙N3 = 5.00 × 1015 s−1. (b) Interpretation and further calculation P2 = 1.60 mW , P3 = 2.40 mW . P2 + P3 = 4.00 mW, ˙N2 + ˙N3 = 1016 s−1. Physical check The inferred mean energy lies between the two component energies; otherwise the proposed two-color model would be impossible.

Question 5

Total power and photon rate reveal an unknown two-color mixture

A beam contains photons of energies 2.00 eV and 3.00 eV only. Its total power is 4.00mW4.00\,\mathrm{mW} and total photon rate is 1016s−110^{16}\,\mathrm{s^{-1}}. Use 1eV=1.60×10−19J1\,\mathrm{eV}=1.60\times10^{-19}\,\mathrm J.

Diagram shown in the original SVG worksheet.

  1. Find the fraction of photons in each energy group.

  2. Find the optical power carried by each group and check both conservation constraints.

Question 5 - Solution

Total power and photon rate reveal an unknown two-color mixture

Diagram shown in the original SVG worksheet.

(a) Method and calculation

E¯=P/Ṅ=2.50eV,2f+3(1−f)=2.50.\overline E=P/\dot N=2.50\,\mathrm{eV},\quad2f+3(1-f)=2.50.f=1/2,Ṅ2=Ṅ3=5.00×1015s−1.\boxed{f=1/2},\qquad\dot N_2=\dot N_3=5.00\times10^{15}\,\mathrm{s^{-1}}.

(b) Interpretation and further calculation

P2=1.60mW,P3=2.40mW.\boxed{P_2=1.60\,\mathrm{mW}},\qquad\boxed{P_3=2.40\,\mathrm{mW}}.P2+P3=4.00mW,Ṅ2+Ṅ3=1016s−1.P_2+P_3=4.00\,\mathrm{mW},\quad\dot N_2+\dot N_3=10^{16}\,\mathrm{s^{-1}}.

Physical check

The inferred mean energy lies between the two component energies; otherwise the proposed two-color model would be impossible.