Radioactive Decay — Question 5

Section 42-3 · Physics pages 7040–7041

Question 5 Background subtraction increases the counting uncertainty A sample-plus-background run records G=2500 counts in 100 s. An independent background-only run records B=400 counts in the same duration. Both counts are Poisson distributed; source decay is negligible during measurement. G = 2500 − B = 400 independent counting runs Schematic; labels, not drawing lengths, specify the data. (a) Find the net count rate and its one-standard-deviation uncertainty. (b) If both acquisition times are increased by a factor of four at unchanged rates, find the new relative uncertainty.
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Question 5 - Solution Background subtraction increases the counting uncertainty G = 2500 − B = 400 independent counting runs (a) Method and calculation Rs = (G −B)/T = (2500 −400)/100 = 21.0 s−1 . σRs = √ G + B/T = √ 2900/100 = .5385 s−1 . (b) Interpretation and further calculation σRs/Rs = √ 2900/2100 = .02564 . T →4T : (σR/R)new = .02564/2 = .01282 = 1.282% . Physical check The background mean is subtracted, but its variance is added because it was measured independently. Subtracting uncertainties would underestimate the error.

Question 5

Background subtraction increases the counting uncertainty

A sample-plus-background run records G=2500 counts in 100 s. An independent background-only run records B=400 counts in the same duration. Both counts are Poisson distributed; source decay is negligible during measurement.

Diagram shown in the original SVG worksheet.

  1. Find the net count rate and its one-standard-deviation uncertainty.

  2. If both acquisition times are increased by a factor of four at unchanged rates, find the new relative uncertainty.

Question 5 - Solution

Background subtraction increases the counting uncertainty

Diagram shown in the original SVG worksheet.

(a) Method and calculation

Rs=(G−B)/T=(2500−400)/100=21.0s−1.\boxed{R_s=(G-B)/T=(2500-400)/100=21.0\,\mathrm{s^{-1}}}.σRs=G+B/T=2900/100=.5385s−1.\boxed{\sigma_{R_s}=\sqrt{G+B}/T=\sqrt{2900}/100=.5385\,\mathrm{s^{-1}}}.

(b) Interpretation and further calculation

σRs/Rs=2900/2100=.02564.\boxed{\sigma_{R_s}/R_s=\sqrt{2900}/2100=.02564}.T→4T:(σR/R)new=.02564/2=.01282=1.282%.T\to4T:\quad\boxed{(\sigma_R/R)_{\mathrm{new}}=.02564/2=.01282=1.282\%}.

Physical check

The background mean is subtracted, but its variance is added because it was measured independently. Subtracting uncertainties would underestimate the error.