Alpha Decay — Question 10

Question 10

Recoil of an excited daughter Doppler-shifts its gamma ray

An alpha decay at rest leaves an excited daughter of rest energy $Mc^2=180000\,\mathrm{MeV}$ with recoil kinetic energy $T_D=.100\,\mathrm{MeV}$. It emits a photon of rest-frame energy $E_0=1.00\,\mathrm{MeV}$ before slowing. Use $E_\gamma\simeq E_0(1+\beta\cos\theta)$ to first order in beta.

1. Find the recoil speed as a fraction of c.

2. Find the forward and backward photon energies and their separation.

Question 10 - Solution

Recoil of an excited daughter Doppler-shifts its gamma ray

(a) Method and calculation

$$T_D\simeq\tfrac12Mc^2\beta^2\Rightarrow\boxed{\beta=\sqrt{2T_D/(Mc^2)}}.$$$$\boxed{\beta=\sqrt{.200/180000}=1.054\times10^{-3}}.$$

(b) Interpretation and further calculation

$$\boxed{E_+=1.001054\,\mathrm{MeV},\quad E_-=.998946\,\mathrm{MeV}}.$$$$\boxed{E_+-E_-=2\beta E_0=.002108\,\mathrm{MeV}=2.108\,\mathrm{keV}}.$$

Physical check

The angle is measured relative to daughter motion. Slowing before emission reduces this Doppler spread; the supplied E0 already denotes the photon energy in the daughter rest frame.