Radioactive Decay — Question 4

Section 42-3 · Physics pages 7038–7039

Question 4 Integrated counts recover the activity at the start of a measurement A source has half-life 60.0 s. During a 60.0 s acquisition a detector records 3000 signal counts with constant efficiency .250 and negligible dead time or background. Each decay can contribute one count. source detector r Schematic; labels, not drawing lengths, specify the data. (a) Derive the relation between integrated counts and initial activity. (b) Find the initial activity and compare it with the measured mean activity inferred from counts per time.
Show solutionHide solution
Question 4 - Solution Integrated counts recover the activity at the start of a measurement source detector r (a) Method and calculation C = ε ∫T 0 A0e−λtdt = εA0(1 −e−λT )/λ. A0 = Cλ/[ε(1 −e−λT )] . (b) Interpretation and further calculation A0 = 3000(ln 2/60)/(.250 × .5) = 277.3 Bq . A = C/(εT) = 200 Bq, A/A0 = 1/(2 ln 2) = .7213 . Physical check Dividing counts by acquisition time returns an interval average. A rapidly decaying source requires the integral to recover its activity at a specified instant.

Question 4

Integrated counts recover the activity at the start of a measurement

A source has half-life 60.0 s. During a 60.0 s acquisition a detector records 3000 signal counts with constant efficiency .250 and negligible dead time or background. Each decay can contribute one count.

Diagram shown in the original SVG worksheet.

  1. Derive the relation between integrated counts and initial activity.

  2. Find the initial activity and compare it with the measured mean activity inferred from counts per time.

Question 4 - Solution

Integrated counts recover the activity at the start of a measurement

Diagram shown in the original SVG worksheet.

(a) Method and calculation

C=ε∫0T𝒜0e−λtdt=ε𝒜0(1−e−λT)/λ.C=\varepsilon\int_0^T\mathcal A_0e^{-\lambda t}dt=\varepsilon\mathcal A_0(1-e^{-\lambda T})/\lambda.𝒜0=Cλ/[ε(1−e−λT)].\boxed{\mathcal A_0=C\lambda/[\varepsilon(1-e^{-\lambda T})]}.

(b) Interpretation and further calculation

𝒜0=3000(ln⁡2/60)/(.250×.5)=277.3Bq.\boxed{\mathcal A_0=3000(\ln2/60)/(.250\times.5)=277.3\,\mathrm{Bq}}.𝒜¯=C/(εT)=200Bq,𝒜¯/𝒜0=1/(2ln⁡2)=.7213.\boxed{\overline{\mathcal A}=C/(\varepsilon T)=200\,\mathrm{Bq},\quad\overline{\mathcal A}/\mathcal A_0=1/(2\ln2)=.7213}.

Physical check

Dividing counts by acquisition time returns an interval average. A rapidly decaying source requires the integral to recover its activity at a specified instant.