Alpha Decay — Question 2

Question 2

Two-body recoil partitions the alpha-decay energy

A parent initially at rest decays to an alpha particle of mass $m_\alpha=4u$ and daughter of mass $M_D=196u$. The available kinetic energy is Q=5.00 MeV. Use nonrelativistic recoil kinematics.

1. Derive the kinetic-energy shares from momentum conservation.

2. Evaluate both energies and find the ratio of alpha to daughter speeds.

Question 2 - Solution

Two-body recoil partitions the alpha-decay energy

(a) Method and calculation

$$\mathbf p_D=-\mathbf p_\alpha,\quad T_D/T_\alpha=m_\alpha/M_D.$$$$\boxed{T_\alpha=Q\frac{M_D}{M_D+m_\alpha},\quad T_D=Q\frac{m_\alpha}{M_D+m_\alpha}}.$$

(b) Interpretation and further calculation

$$\boxed{T_\alpha=4.900\,\mathrm{MeV},\quad T_D=.1000\,\mathrm{MeV}}.$$$$m_\alpha v_\alpha=M_Dv_D\Rightarrow\boxed{v_\alpha/v_D=196/4=49}.$$

Physical check

Equal and opposite momenta do not imply equal kinetic energies. The lighter alpha particle receives most of Q.