How Lasers Work — Question 7

Section 40-11 · Physics pages 6784–6785

Question 7 An emission cross section converts threshold gain into an inversion density An equal-degeneracy transition has gain g = σ∆n, where σ = 3.00 × 10−23 m2 and ∆n is inversion per volume. Required gain is gth = .300 m−1. The active volume is Va = 2.00 × 10−8 m3, with a nearly empty lower level and upper lifetime τ = 200 µs. gain medium output L pump Schematic; labels, not drawing lengths, specify the data. (a) Find threshold inversion density and number of excited atoms. (b) Find the absorbed pump power needed to replace spontaneous decay at threshold if each useful excitation requires one pump photon of energy 2.50 eV. Use 1 eV = 1.602 × 10−19 J and unit pump quantum efficiency.
Show solutionHide solution
Question 7 - Solution An emission cross section converts threshold gain into an inversion density gain medium output L pump (a) Method and calculation ∆nth = gth/σ = 1.00 × 1022 m−3 . Nth = ∆nthVa = 2.00 × 1014 . (b) Interpretation and further calculation Rth = Nth/τ = 1.00 × 1018 s−1. Pabs,th = Rth(2.50)(1.602 × 10−19) = .4005 W . Physical check This is an ideal absorbed threshold power. Pump delivery losses, imperfect excitation efficiency, spatial overlap, and additional decay paths would increase an actual incident threshold.

Question 7

An emission cross section converts threshold gain into an inversion density

An equal-degeneracy transition has gain g=σΔng=\sigma\Delta n, where σ=3.00×10−23m2\sigma=3.00\times10^{-23}\,\mathrm{m^2} and Δn\Delta n is inversion per volume. Required gain is gth=.300m−1g_{\mathrm{th}}=.300\,\mathrm{m^{-1}}. The active volume is Va=2.00×10−8m3V_a=2.00\times10^{-8}\,\mathrm{m^3}, with a nearly empty lower level and upper lifetime τ=200μs\tau=200\,\mu\mathrm s.

Diagram shown in the original SVG worksheet.

  1. Find threshold inversion density and number of excited atoms.

  2. Find the absorbed pump power needed to replace spontaneous decay at threshold if each useful excitation requires one pump photon of energy 2.50eV2.50\,\mathrm{eV}. Use 1eV=1.602×10−19J1\,\mathrm{eV}=1.602\times10^{-19}\,\mathrm J and unit pump quantum efficiency.

Question 7 - Solution

An emission cross section converts threshold gain into an inversion density

Diagram shown in the original SVG worksheet.

(a) Method and calculation

Δnth=gth/σ=1.00×1022m−3.\boxed{\Delta n_{\mathrm{th}}=g_{\mathrm{th}}/\sigma=1.00\times10^{22}\,\mathrm{m^{-3}}}.Nth=ΔnthVa=2.00×1014.\boxed{N_{\mathrm{th}}=\Delta n_{\mathrm{th}}V_a=2.00\times10^{14}}.

(b) Interpretation and further calculation

Rth=Nth/τ=1.00×1018s−1.R_{\mathrm{th}}=N_{\mathrm{th}}/\tau=1.00\times10^{18}\,\mathrm{s^{-1}}.Pabs,th=Rth(2.50)(1.602×10−19)=.4005W.\boxed{P_{\mathrm{abs,th}}=R_{\mathrm{th}}(2.50)(1.602\times10^{-19})=.4005\,\mathrm W}.

Physical check

This is an ideal absorbed threshold power. Pump delivery losses, imperfect excitation efficiency, spatial overlap, and additional decay paths would increase an actual incident threshold.