Parent-daughter decay series

Problem

A nuclide $A$ (mean lifetime $\tau_A = 1\ \text{hour}$) decays into nuclide $B$ (mean lifetime $\tau_B = 24\ \text{hours}$), which in turn decays into the stable nuclide $C$. At the initial instant $N_A(0) = 10^{12}$, $N_B(0) = 0$, $N_C(0) = 0$.

• (a) At what instant is the number of $B$ nuclei maximum?
• (b) What is the value of $N_B$ at that maximum?

Solution

Decay constants. From the mean lifetimes: $\lambda_A = 1/\tau_A = 1\ \text{h}^{-1}$ and $\lambda_B = 1/\tau_B = 1/24\ \text{h}^{-1}$.

Equation for $B$ and the Bateman solution. Nuclide $B$ is formed by the decay of $A$ and destroyed by decaying into $C$: $\dot N_B = \lambda_A N_A - \lambda_B N_B$. With $N_A(t) = N_{A0}e^{-\lambda_A t}$ and $N_B(0)=0$, integration (a first-order linear equation) gives the Bateman solution:

$N_B(t) = N_{A0}\,\frac{\lambda_A}{\lambda_A - \lambda_B}\left(e^{-\lambda_B t} - e^{-\lambda_A t}\right)$

(a) Time of the maximum. The maximum occurs where $\dot N_B = 0$, i.e. when the derivatives of the two exponentials balance:

$t^\ast = \frac{\ln(\lambda_A/\lambda_B)}{\lambda_A - \lambda_B} = \frac{\ln 24}{1 - 1/24} = \frac{3{,}178}{0{,}958} \approx 3{,}32\ \text{h}$

(b) Value of the maximum. Substituting $t^\ast$ into the solution. The prefactor is $\lambda_A/(\lambda_A-\lambda_B) = 1/0{,}958 \approx 1{,}043$; the exponentials are $e^{-\lambda_B t^\ast} = e^{-0{,}138} \approx 0{,}871$ and $e^{-\lambda_A t^\ast} = e^{-3{,}32} \approx 0{,}036$, with a difference $\approx 0{,}835$. Therefore:

$N_{B,\max} = 10^{12}\cdot 1{,}043\cdot 0{,}835 \approx 8{,}7\times 10^{11}$

Comment. Since $A$ decays much faster than $B$ ($\tau_A \ll \tau_B$), nearly all the $A$ nuclei transfer to $B$ before $B$ has time to decay: the maximum of $B$ is close to the initial value of $A$.

$\ev{\text{(a) } t^\ast\approx 3{,}32\ \text{h};\quad \text{(b) } N_{B,\max}\approx 8{,}7\times 10^{11}}$

Collegamenti

Argomenti: Fisica nucleare Concetti: Decadimento radioattivo Metodi: Separazione delle variabili