Question 6 of 7: Radioactive In-Growth — Count-Rate Table, Linearity, and Daughter Half-Life

Question 6: Radioactive In-Growth — Count-Rate Table, Linearity, and Daughter Half-Life (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. Count-rate table above, $C_\infty=3000$ counts/min; each $C$ value is the count recorded in a fixed one-minute counting interval.

Find. (a) the completed table ($\sigma(C)$, $g(t)$, $\sigma(\lambda t)$ for every row); (b) the physical meaning of $C_\infty$; (c) a demonstration that $g(t)$ is linear in $t$; (d) the physical meaning of the slope; (e) the half-life of the daughter.

Approach. Treat each one-minute count as Poisson-distributed ($\sigma(C)=\sqrt{C}$), propagate that uncertainty through $g(t)=-\ln(1-C/C_\infty)$ via the given variance rule, then fit $g(t)=a+bt$ by weighted least squares; the growth law $C_\infty-C(t)=(C_\infty-C_0)e^{-\lambda t}$ predicts exactly this linear form with slope $b=\lambda$, the decay constant of the in-growing daughter.

1. Part (a) — completing the table. With $\sigma(C)=\sqrt{C}$ (Poisson counting statistics on the one-minute count) and, by the given variance rule, $$\sigma(g) = \left|\frac{\partial g}{\partial C}\right|\sigma(C) = \frac{\sigma(C)}{C_\infty-C}$$ (since $\partial g/\partial C=\partial[-\ln(1-C/C_\infty)]/\partial C=1/(C_\infty-C)$), and identifying the table's final column with the same propagated uncertainty on the linear model $g(t)=\lambda t+\text{const.}$ (so $\sigma(\lambda t)=\sigma(g)$), the completed table is:

   Completed table

   $t$ (hr)	$C$	$\sigma(C)=\sqrt{C}$	$g(t)=-\ln(1-C/C_\infty)$	$\sigma(\lambda t)=\sigma(C)/(C_\infty-C)$
   0	1000	31.6	0.4055	0.0158
   4	1150	33.9	0.4834	0.0183
   8	1300	36.1	0.5680	0.0212
   16	1500	38.7	0.6931	0.0258
   32	1900	43.6	1.0033	0.0396
   48	2200	46.9	1.3218	0.0586
   80	2500	50.0	1.7918	0.1000
   120	2800	52.9	2.7081	0.2646
   $\infty$	3000	—	undefined (log singularity)	—

2. Part (b) — physical meaning of $C_\infty=3000$. $C_\infty$ is the saturation (asymptotic) count rate the daughter's activity approaches once its own production and decay have reached a steady state — physically, once essentially all of the initially present parent activity available to feed the daughter has been converted, the daughter's count rate stops rising and plateaus at $C_\infty$. It represents full secular in-growth of the daughter under the (implicit) assumption that the parent supply driving the growth is effectively constant over the daughter's own growth timescale.
3. Part (c) — linearity of $g(t)$. If the daughter count rate grows from an initial value $C_0$ toward $C_\infty$ purely exponentially with the daughter's own decay constant $\lambda$, $$C_\infty - C(t) = (C_\infty-C_0)\,e^{-\lambda t}$$ then $$g(t) = -\ln\!\left(\frac{C_\infty-C(t)}{C_\infty}\right) = \ln\!\left(\frac{C_\infty}{C_\infty-C_0}\right) + \lambda t$$ which is exactly linear in $t$, with intercept $g(0)=\ln[C_\infty/(C_\infty-C_0)]$ and slope $\lambda$. A weighted least-squares fit of the completed table (weights $1/\sigma(g)^2$) gives $$\boxed{g(t) \approx 0.410 + 0.01835\,t\ \ (t\ \text{in hr})}$$ and every one of the eight tabulated points agrees with this line to well within its own $\sigma(g)$ (largest deviation $0.86\sigma$, at $t=80$ hr) — confirming $g(t)$ is linear in $t$ within the statistical variability of the counting data, exactly as the growth-law derivation predicts.
4. Part (d) — significance of the slope. Since $g(t)=\lambda t+\text{const.}$, the slope $dg/dt$ is the daughter's own radioactive decay constant $\lambda$ — the same $\lambda$ that governs how fast the daughter itself would decay away once formed. It is a purely intrinsic nuclear property of the daughter, independent of the parent's activity or of how much daughter has already built up at any given time.
5. Part (e) — daughter half-life. From the fitted slope $\lambda=0.01835\pm0.00072\ \text{hr}^{-1}$: $$T_{1/2} = \frac{\ln 2}{\lambda}$$ $$\boxed{T_{1/2} \approx 37.8\ \text{hr}\ \ (\pm 1.5\ \text{hr, from the fit uncertainty})}$$