Alpha Decay — Question 8

Section 42-4 · Physics pages 7066–7067

Question 8 A preformation probability multiplies the tunneling decay rate A model gives λ = SανP, where Sα is the alpha preformation probability, nu the barrier encounter frequency and P the penetration probability. Let ν = 1021 s−1, P = 10−25 and measured half-life T1/2 = 105 s. Sα νP λ Schematic; labels, not drawing lengths, specify the data. (a) Infer Sα and verify it is a probability. (b) If a second state has the same nu and P but half the preformation probability, find its half-life and relative activity for equal parent populations.
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Question 8 - Solution A preformation probability multiplies the tunneling decay rate Sα νP λ (a) Method and calculation λ = ln 2/T1/2 = 6.931 × 10−6 s−1. Sα = λ/(νP) = .06931 (0 < Sα < 1). (b) Interpretation and further calculation λ′ = λ/2 ⇒T1/2′ = 2.00 × 105 s . A = λN ⇒A ′/A = 1/2 (N′ = N). Physical check Penetration probability alone does not fix the lifetime. The nucleus must also contain the appropriate alpha-like configuration and supply an encounter rate.

Question 8

A preformation probability multiplies the tunneling decay rate

A model gives λ=SανP\lambda=S_\alpha\nu P, where SαS_\alpha is the alpha preformation probability, nu the barrier encounter frequency and P the penetration probability. Let ν=1021s−1\nu=10^{21}\,\mathrm{s^{-1}}, P=10−25P=10^{-25} and measured half-life T1/2=105sT_{1/2}=10^5\,\mathrm s.

Diagram shown in the original SVG worksheet.

  1. Infer SαS_\alpha and verify it is a probability.

  2. If a second state has the same nu and P but half the preformation probability, find its half-life and relative activity for equal parent populations.

Question 8 - Solution

A preformation probability multiplies the tunneling decay rate

Diagram shown in the original SVG worksheet.

(a) Method and calculation

λ=ln⁡2/T1/2=6.931×10−6s−1.\lambda=\ln2/T_{1/2}=6.931\times10^{-6}\,\mathrm{s^{-1}}.Sα=λ/(νP)=.06931(0<Sα<1).\boxed{S_\alpha=\lambda/(\nu P)=.06931}\quad(0<S_\alpha<1).

(b) Interpretation and further calculation

λ′=λ/2⇒T1/2′=2.00×105s.\lambda\prime=\lambda/2\Rightarrow\boxed{T_{1/2}\prime=2.00\times10^5\,\mathrm s}.𝒜=λN⇒𝒜′/𝒜=1/2(N′=N).\mathcal A=\lambda N\Rightarrow\boxed{\mathcal A\prime/\mathcal A=1/2}\quad(N\prime=N).

Physical check

Penetration probability alone does not fix the lifetime. The nucleus must also contain the appropriate alpha-like configuration and supply an encounter rate.