Question 6 of 7: Counting Statistics of a Growing Daughter Activity

Question 6: Counting Statistics of a Growing Daughter Activity (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. Eight measured (time, count-rate) pairs plus the asymptotic count rate $C_\infty=3000$ $ ext{min}^{-1}$; the counts are Poisson-distributed (so $\sigma(C)=\sqrt{C}$); the variance-combination rule for a function $h(C)$.

Find. (a) $\sigma(C)$, $g(t)$ and $\sigma(g)$ at each finite $t$; (b) the physical meaning of $C_\infty$; (c) confirmation that $g(t)$ is linear in $t$, within the computed error bars; (d) the physical meaning of the slope; (e) the daughter's half-life.

Approach. Treat each tabulated count as Poisson so $\sigma(C)=\sqrt{C}$; propagate through $g(C)=-\ln(1-C/C_\infty)$ via $dg/dC=1/(C_\infty-C)$ to get $\sigma(g)$; then fit $g=a+bt$ by weighted least squares (weights $1/\sigma_g^2$) and read the daughter decay constant and half-life off the slope.

1. Part (a) — filling in the table. Counting is a Poisson process, so each tabulated count rate carries $\sigma(C)=\sqrt{C}$ (e.g. $\sigma(1000)=31.62$, $\sigma(2800)=52.92$). For $g=-\ln(1-C/C_\infty)$, the variance-combination rule with $h=g,\,x=C$ needs $dg/dC$: $$\frac{dg}{dC} = \frac{1}{C_\infty-C} \quad\Rightarrow\quad \sigma(g) = \left|\frac{dg}{dC}\right|\sigma(C) = \frac{\sqrt{C}}{C_\infty-C}$$ Both quantities are tabulated above for every finite $t$; note $\sigma(g)$ grows sharply as $C\to C_\infty$ (at $t=120$, $\sigma(g)=0.265$, an order of magnitude larger than at $t=0$), because the denominator $C_\infty-C$ shrinks as the daughter approaches full growth.
2. Part (b) — meaning of $C_\infty$. $C_\infty=3000$ $ ext{min}^{-1}$ is the count rate once the daughter has fully grown in and the system has reached its asymptotic (equilibrium) activity — the maximum count rate the detector will ever read from this sample, reached in the limit $t\to\infty$ once all transient growth is complete.
3. Part (c) — $g(t)$ is linear in $t$. A daughter growing in with decay constant $\lambda_D$ from some initial count-rate contribution follows $C(t)=C_\infty-(C_\infty-C_0)e^{-\lambda_D t}$, so $1-C(t)/C_\infty=\left(1-C_0/C_\infty\right)e^{-\lambda_D t}$ and $$g(t) = -\ln\!\left(1-\frac{C(t)}{C_\infty}\right) = \underbrace{-\ln\!\left(1-\frac{C_0}{C_\infty}\right)}_{a} + \lambda_D t$$ — exactly linear in $t$ with slope $\lambda_D$. A weighted least-squares fit ($w_i=1/\sigma_g^2(i)$, which correctly down-weights the highly uncertain large-$t$ points) through the eight tabulated points gives $$\boxed{g(t) \approx (0.410\pm0.011) + (0.01835\pm0.00072)\,t\quad(t\ \text{in hr})}$$ with goodness-of-fit $\chi^2/\text{dof}=1.70/6\approx0.28$ — a value of order unity (in fact comfortably below 1) confirms the straight-line model is consistent with the data within the statistical variability computed in part (a), which is exactly what the question asks to be shown. The fitted intercept ($0.410\pm0.011$) also agrees, within one standard deviation, with the directly-computed $g(0)=0.4055$ from the table — a useful independent consistency check on the fit.
4. Part (d) — meaning of the slope. From the derivation in part (c), the slope $dg/dt$ is the daughter's own decay constant, $\lambda_D$ — it is not an approximation or a derived proxy, but a direct physical rate constant read straight off the linearized data.
5. Part (e) — daughter half-life. Using the fitted slope $b=\lambda_D=0.01835\pm0.00072\ \text{hr}^{-1}$: $$T_{1/2} = \frac{\ln2}{\lambda_D}$$ $$\boxed{T_{1/2} = \frac{0.6931}{0.01835} \approx 37.8\pm1.5\ \text{hr}}$$