Thermonuclear Fusion: The Basic Process — Question 10

Section 43-5 · Physics pages 7250–7251

Question 10 Detector efficiencies must be removed before inferring fusion branches A hypothetical D-D ensemble has two branches: one emits a neutron, the other a proton. Background-subtracted counts are 1200 neutrons and 1800 protons, with efficiencies .200 and .300 respectively. Assume one counted-type particle per event, negligible dead time and Poisson count uncertainties. D–D events n, ϵn p, ϵp Schematic; labels, not drawing lengths, specify the data. (a) Infer total branch events and branching fractions. (b) Propagate the count uncertainties to the corrected neutron-to-proton event ratio, treating efficiencies as exact.
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Question 10 - Solution Detector efficiencies must be removed before inferring fusion branches D–D events n, ϵn p, ϵp (a) Method and calculation Nn = Cn/ϵn = 1200/.200 = 6000, Np = 1800/.300 = 6000 . Ntotal = 12000, bn = bp = .500 . (b) Interpretation and further calculation r = Cn/ϵn Cp/ϵp = 1, σr r = √ 1 Cn + 1 Cp . r = 1.000 ± .0373 . Physical check Raw count fractions are biased when efficiencies differ. Background subtraction would add uncertainty if its error were not negligible as stipulated.

Question 10

Detector efficiencies must be removed before inferring fusion branches

A hypothetical D-D ensemble has two branches: one emits a neutron, the other a proton. Background-subtracted counts are 1200 neutrons and 1800 protons, with efficiencies .200 and .300 respectively. Assume one counted-type particle per event, negligible dead time and Poisson count uncertainties.

Diagram shown in the original SVG worksheet.

  1. Infer total branch events and branching fractions.

  2. Propagate the count uncertainties to the corrected neutron-to-proton event ratio, treating efficiencies as exact.

Question 10 - Solution

Detector efficiencies must be removed before inferring fusion branches

Diagram shown in the original SVG worksheet.

(a) Method and calculation

Nn=Cn/ϵn=1200/.200=6000,Np=1800/.300=6000.\boxed{N_n=C_n/\epsilon_n=1200/.200=6000,\quad N_p=1800/.300=6000}.Ntotal=12000,bn=bp=.500.\boxed{N_{\mathrm{total}}=12000,\quad b_n=b_p=.500}.

(b) Interpretation and further calculation

r=Cn/ϵnCp/ϵp=1,σrr=1Cn+1Cp.r=\frac{C_n/\epsilon_n}{C_p/\epsilon_p}=1,\quad\frac{\sigma_r}{r}=\sqrt{\frac1{C_n}+\frac1{C_p}}.r=1.000±.0373.\boxed{r=1.000\pm.0373}.

Physical check

Raw count fractions are biased when efficiencies differ. Background subtraction would add uncertainty if its error were not negligible as stipulated.