Beta Decay — Question 7

Section 42-5 · Physics pages 7084–7085

Question 7 The same phase-space model predicts count fractions and endpoint sensitivity Use the model dΓ/dT = C0T 2(Q −T)2, with the same constant C0 for two otherwise identical transitions. Coulomb and electron-mass effects are neglected. T dΓ/dT Q/4 Schematic; labels, not drawing lengths, specify the data. (a) Find how total decay rate scales with Q and the half-life ratio if Q doubles. (b) For a normalized spectrum, calculate the fraction of electrons with T < Q/4.
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Question 7 - Solution The same phase-space model predicts count fractions and endpoint sensitivity T dΓ/dT Q/4 (a) Method and calculation Γ = C0 ∫Q 0 T 2(Q −T)2dT = C0Q5/30 . Q →2Q : Γ′/Γ = 32, T1/2′/T1/2 = 1/32 . (b) Interpretation and further calculation F(x) = 30 ∫x 0 u2(1 −u)2du = 10x3 −15x4 + 6x5. F(1/4) = .103515625 ≃10.35% . Physical check The fifth-power rule here assumes unchanged nuclear matrix element and the stated limiting spectrum. Nuclear selection rules can outweigh a simple Q comparison.

Question 7

The same phase-space model predicts count fractions and endpoint sensitivity

Use the model dΓ/dT=C0T2(Q−T)2d\Gamma/dT=C_0T^2(Q-T)^2, with the same constant C0C_0 for two otherwise identical transitions. Coulomb and electron-mass effects are neglected.

Diagram shown in the original SVG worksheet.

  1. Find how total decay rate scales with Q and the half-life ratio if Q doubles.

  2. For a normalized spectrum, calculate the fraction of electrons with T<Q/4T<Q/4.

Question 7 - Solution

The same phase-space model predicts count fractions and endpoint sensitivity

Diagram shown in the original SVG worksheet.

(a) Method and calculation

Γ=C0∫0QT2(Q−T)2dT=C0Q5/30.\Gamma=C_0\int_0^QT^2(Q-T)^2dT=\boxed{C_0Q^5/30}.Q→2Q:Γ′/Γ=32,T1/2′/T1/2=1/32.Q\to2Q:\quad\boxed{\Gamma\prime/\Gamma=32,\quad T_{1/2}\prime/T_{1/2}=1/32}.

(b) Interpretation and further calculation

F(x)=30∫0xu2(1−u)2du=10x3−15x4+6x5.F(x)=30\int_0^xu^2(1-u)^2du=10x^3-15x^4+6x^5.F(1/4)=.103515625≃10.35%.\boxed{F(1/4)=.103515625\simeq10.35\%}.

Physical check

The fifth-power rule here assumes unchanged nuclear matrix element and the stated limiting spectrum. Nuclear selection rules can outweigh a simple Q comparison.