Nuclear Fission: The Basic Process — Question 7

Section 43-1 · Physics pages 7164–7165

Question 7 A fission-product yield links a burst to later activity A short burst contains 2.00 × 1010 fissions. One radioactive product has independent yield .0600 nucleus per fission and half-life 2.00 h. Ignore parent feeding and loss of product atoms. fission product daughter emission Schematic; labels, not drawing lengths, specify the data. (a) Calculate its initial inventory and initial activity in becquerels. (b) Find its activity after 6 h and the number of its decays in that interval.
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Question 7 - Solution A fission-product yield links a burst to later activity fission product daughter emission (a) Method and calculation N0 = .0600(2.00 × 1010) = 1.20 × 109 , λ = ln 2/7200 s−1. A0 = λN0 = 1.15525 × 105 Bq . (b) Interpretation and further calculation A(6 h) = A0 2−3 = 1.44406 × 104 Bq . Ndec = N0(1 −2−3) = 1.05 × 109 . Physical check Independent yield counts directly formed nuclei. A cumulative yield would require careful treatment of precursors to avoid double counting.

Question 7

A fission-product yield links a burst to later activity

A short burst contains 2.00×10102.00\times10^{10} fissions. One radioactive product has independent yield .0600 nucleus per fission and half-life 2.00 h. Ignore parent feeding and loss of product atoms.

Diagram shown in the original SVG worksheet.

  1. Calculate its initial inventory and initial activity in becquerels.

  2. Find its activity after 6 h and the number of its decays in that interval.

Question 7 - Solution

A fission-product yield links a burst to later activity

Diagram shown in the original SVG worksheet.

(a) Method and calculation

N0=.0600(2.00×1010)=1.20×109,λ=ln⁡2/7200s−1.\boxed{N_0=.0600(2.00\times10^{10})=1.20\times10^9},\quad\lambda=\ln2/7200\,\mathrm{s^{-1}}.A0=λN0=1.15525×105Bq.\boxed{A_0=\lambda N_0=1.15525\times10^5\,\mathrm{Bq}}.

(b) Interpretation and further calculation

A(6h)=A02−3=1.44406×104Bq.\boxed{A(6\,\mathrm h)=A_0\,2^{-3}=1.44406\times10^4\,\mathrm{Bq}}.Ndec=N0(1−2−3)=1.05×109.\boxed{N_{\mathrm{dec}}=N_0(1-2^{-3})=1.05\times10^9}.

Physical check

Independent yield counts directly formed nuclei. A cumulative yield would require careful treatment of precursors to avoid double counting.