Alpha Decay — Question 3

Question 3

Consistent atomic masses give the alpha Q value directly

Supplied neutral-atom masses are $M_P=210.000000u$, $M_D=205.993000u$, and $M_{\mathrm{He}}=4.002603u$. Use $uc^2=931.5\,\mathrm{MeV}$. Electronic binding differences may be neglected.

1. Calculate Q and determine whether decay is energetically allowed.

2. Using daughter-to-alpha mass ratio 206:4, predict the emitted alpha kinetic energy.

Question 3 - Solution

Consistent atomic masses give the alpha Q value directly

(a) Method and calculation

$$Q=(M_P-M_D-M_{\mathrm{He}})c^2.$$$$\boxed{Q=.004397(931.5)=4.09581\,\mathrm{MeV}>0}.$$

(b) Interpretation and further calculation

$$T_\alpha=Q\frac{M_D}{M_D+m_\alpha}\simeq Q\frac{206}{210}.$$$$\boxed{T_\alpha=4.01779\,\mathrm{MeV}}.$$

Physical check

Neutral parent electrons equal the total electrons in a neutral daughter plus neutral helium, so rest masses cancel consistently. Positive Q is necessary but does not determine the decay rate.