Radioactive Decay — Question 8

Question 8

A radioactive daughter first accumulates and then decays

A parent P decays entirely to radioactive daughter D, which decays to a stable product. Initially $N_P=N_0$ and $N_D=0$. Their constants satisfy $\lambda_D=2\lambda_P=2\lambda$.

1. Solve $\dot N_D=\lambda N_0e^{-\lambda t}-2\lambda N_D$.

2. Find the time and size of the maximum daughter population and compare parent and daughter activities then.

Question 8 - Solution

A radioactive daughter first accumulates and then decays

(a) Method and calculation

$$\frac{d}{dt}(N_De^{2\lambda t})=\lambda N_0e^{\lambda t}\Rightarrow\boxed{N_D=N_0(e^{-\lambda t}-e^{-2\lambda t})}.$$$$\dot N_D=0\Rightarrow e^{-\lambda t}=2e^{-2\lambda t}\Rightarrow\boxed{t_{\max}=\ln2/\lambda}.$$

(b) Interpretation and further calculation

$$\boxed{N_{D,\max}=N_0(1/2-1/4)=N_0/4}.$$$$\boxed{\mathcal A_D=2\lambda(N_0/4)=\lambda N_0/2=\mathcal A_P}\quad\text{at }t_{\max}.$$

Physical check

The daughter population peaks when its production and destruction rates are equal. Equal activities do not imply equal populations if decay constants differ.