Mean lifetime of a mixed sample

Problem

A sample contains two radioactive isotopes. Of isotope A there are initially $N_A(0) = 10^{6}$ atoms, with half-life $\tau_{1/2,A} = 1\ \text{day}$. Of isotope B there are $N_B(0) = 10^{7}$ atoms, with $\tau_{1/2,B} = 100\ \text{days}$. Calculate the total initial activity of the sample and the total activity after 50 days.

Solution

Decay constants. Recalling $\lambda = \dfrac{\ln 2}{\tau_{1/2}}$:

$\lambda_A = \frac{\ln 2}{1} = 0{,}693\ \text{day}^{-1}, \qquad \lambda_B = \frac{\ln 2}{100} = 0{,}00693\ \text{day}^{-1}$

Initial activities. The activity is $A = \lambda N$:

$A_{A}(0) = \lambda_A N_A(0) = 0{,}693 \cdot 10^{6} \approx 6{,}93 \cdot 10^{5}\ \text{day}^{-1}$

$A_{B}(0) = \lambda_B N_B(0) = 0{,}00693 \cdot 10^{7} \approx 6{,}93 \cdot 10^{4}\ \text{day}^{-1}$

Total initial activity:

$A_{\text{tot}}(0) = A_A(0) + A_B(0) \approx 6{,}93 \cdot 10^{5} + 6{,}93 \cdot 10^{4} \approx 7{,}6 \cdot 10^{5}\ \text{day}^{-1}$

After 50 days. Using $N(t) = N(0)\,2^{-t/\tau_{1/2}}$:

$N_A(50) = 10^{6} \cdot 2^{-50} \approx 0 \quad\Rightarrow\quad A_A(50) \approx 0$

Isotope A, with a 1-day half-life, has effectively vanished after 50 half-lives. For B:

$N_B(50) = 10^{7} \cdot 2^{-50/100} = 10^{7} \cdot 0{,}707 \approx 7{,}07 \cdot 10^{6}$

$A_B(50) = \lambda_B N_B(50) = 0{,}00693 \cdot 7{,}07 \cdot 10^{6} \approx 4{,}9 \cdot 10^{4}\ \text{day}^{-1}$

The total activity after 50 days is dominated by isotope B:

$\ev{A_{\text{tot}}(0) \approx 7{,}6 \cdot 10^{5}\ \text{d}^{-1}; \quad A_{\text{tot}}(50) \approx 4{,}9 \cdot 10^{4}\ \text{d}^{-1}}$

Connections

Topics: Nuclear physics Concepts: Radioactive decay