Alpha Decay — Question 4

Question 4

Separated alpha lines reveal excited daughter states

A parent at rest has two alpha-decay branches to the ground state and to one excited state of the same daughter. Their alpha kinetic energies are 5.00 MeV and 4.20 MeV. Use daughter-to-alpha mass ratio 196:4, equal for both branches to this accuracy.

1. Infer the daughter excitation energy including recoil.

2. If the excited state emits one photon, estimate its photon energy neglecting photon recoil, and explain why it differs from the alpha-line separation.

Question 4 - Solution

Separated alpha lines reveal excited daughter states

(a) Method and calculation

$$Q_j=T_{\alpha,j}(1+m_\alpha/M_D).$$$$\boxed{E_*=(5.00-4.20)(1+4/196)=.81633\,\mathrm{MeV}}.$$

(b) Interpretation and further calculation

$$\boxed{E_\gamma\simeq E_*=.81633\,\mathrm{MeV}}.$$$$\boxed{E_*-.800=.01633\,\mathrm{MeV}}\quad\text{is the change in daughter recoil energy}.$$

Physical check

Excitation reduces both outgoing kinetic energies. A precision photon calculation would also include gamma recoil and possible Doppler shifts.