Exercises — Applications — radiocarbon dating, medical (Tc-99m, I-131), RTG (Pu-238 in spacecraft)

WHAT the question tests: matching the job to the property of the radiation.

• (a) C-14, ${\beta }^{-}$, $t_{1/2}=5730$ yr — its half-life matches the age range of organic things.
• (b) Tc-99m, $\gamma$ (140 keV), $t_{1/2}=6$ h — a photon escapes the body so a camera sees it; low dose.
• (c) I-131, ${\beta }^{-}$ (+ some $\gamma$), $t_{1/2}=8$ d — short-range $\beta$ dumps energy locally to kill cells; the thyroid concentrates iodine on its own.
• (d) Pu-238, $\alpha$, $t_{1/2}=87.7$ yr — dense heat, decades of steady power, no sunlight needed.

Rule: "See it $=\gamma$; Slay it $=\beta /\alpha$."

False. The mass number is $99$ either way. The m stands for metastable — an excited nuclear energy state that later drops down and releases the extra energy as a $140\text{ keV}$ $\gamma$ photon: ${}^{99m}\text{Tc}\to {}^{99}\text{Tc}+\gamma$ Same nucleus, same $99$ nucleons — only the internal energy changed.

WHY $\lambda$ first: every age/fraction formula needs $\lambda$, and it is fixed by the half-life alone. $\lambda =\frac{\ln 2}{t_{1/2}}=\frac{0.6931}{5730}=1.21\times 10^{-4}{\text{yr}}^{-1}$

WHAT connects heat to decay: heat output $\propto$ decays per second $\propto N$, so heat follows the same $e^{-\lambda t}$ curve. After exactly one half-life, $N$ (and heat) is halved: $P=200\times \frac{1}{2}=100\text{W}.$

WHY count half-lives: $24\text{ h}$ is a whole number of half-lives, so no logs needed. $n=\frac{24}{6}=4\text{half-lives},\text{fraction}={\left(\frac{1}{2}\right)}^4=\frac{1}{16}=\mathrm{0.0625.}$ So about $6.25\%$ remains — the dose self-limits quickly.

Step 1 — get $\lambda$. $\lambda =\frac{\ln 2}{5730}=1.21\times 10^{-4}{\text{yr}}^{-1}$. Why: converts a ratio into a time. Step 2 — form the depletion ratio. $\frac{A_0}{A}=\frac{15.3}{9.2}=1.663$. Why: the age formula needs how much the activity has dropped. Step 3 — invert the exponential with a log. $t=\frac{1}{\lambda }\ln \frac{A_0}{A}=\frac{\ln (1.663)}{1.21\times 10^{-4}}=\frac{0.5088}{1.21\times 10^{-4}}\approx 4200\text{yr}.$ Why the log: time lives inside the exponent of $A_0{e}^{-\lambda t}$; the natural log is the exact operation that pulls it out.

Picture quality: a gamma camera detects photons that escape the body. The ${\beta }^{-}$ from I-131 is stopped within millimetres of tissue, so it never reaches the camera — only the secondary $\gamma$ does. Tc-99m emits a pure, clean $140\text{ keV}$ $\gamma$ tuned to the camera, giving a sharper image. Dose: every ${\beta }^{-}$ dumps all its energy inside the patient (short range) — pure damage, no diagnostic benefit. Plus I-131's $8$-day half-life keeps irradiating for over a week, versus Tc-99m fading in hours. Conclusion: the very property that makes I-131 a good scalpel (energy deposited locally) makes it a bad camera.

(a) WHAT "how many half-lives" means: solve $(1/2{)}^n=0.30$. Take logs of both sides: $n\ln (1/2)=\ln 0.30\Rightarrow n=\frac{\ln 0.30}{\ln 0.5}=\frac{-1.204}{-0.6931}=1.737$ half-lives. (b) $t=n\cdot t_{1/2}=1.737\times 5730\approx 9950\text{yr}.$ Cross-check via $\lambda$: $t=\frac{1}{\lambda }\ln \frac{1}{0.30}=\frac{1.204}{1.21\times 10^{-4}}\approx 9950\text{yr}.$ ✓ Both routes agree — because "count half-lives" and "use $\lambda$" are the same equation wearing different clothes.

WHY this is a "solve for $t$" problem: we know the final fraction and want the time. $e^{-\lambda t}=0.60,\lambda =\frac{\ln 2}{87.7}=7.90\times 10^{-3}{\text{yr}}^{-1}.$ Take logs to free $t$ from the exponent: $t=\frac{-\ln 0.60}{\lambda }=\frac{0.5108}{7.90\times 10^{-3}}\approx 64.6\text{yr}.$ So the probe stays above the $60\%$ threshold for about 65 years — which is why Voyager (launched 1977) still returns data decades on.

(a) WHY use $\lambda$ here: $120\text{ h}$ is not a whole number of $66$-h half-lives, so count via the exponential. $\lambda =\frac{\ln 2}{66}=0.01050{\text{h}}^{-1},A=40e^{-\lambda \cdot 120}\text{GBq}.$ Exponent: $\lambda t=0.01050\times 120=1.260$, so $A=40e^{-1.260}=40\times 0.2837=11.35\text{GBq}.$ (b) Fraction $=e^{-1.260}=0.284$, i.e. about $28\%$ of launch strength. (c) After only $5$ days the parent Mo-99 has already dropped to under a third of its yield, so the daily "milk" of Tc-99m keeps shrinking. By ~2 weeks there is too little to make useful doses — hence a weekly replacement cycle. (Also links to Nuclear Reactions and Transmutation: the whole supply chain rests on Mo-99 $\to$ Tc-99m decay.)

(a) With $\lambda =1.21\times 10^{-4}{\text{yr}}^{-1}$:

• X: $t_X=\frac{\ln 2}{\lambda }=\frac{0.6931}{1.21\times 10^{-4}}\approx 5730\text{yr}$ (exactly one half-life — makes sense, activity halved).
• Y: $t_Y=\frac{\ln 4}{\lambda }=\frac{1.386}{1.21\times 10^{-4}}\approx 11460\text{yr}$ (two half-lives). (b) The proof: since $A_0/A=4=2^2=(A_0/A{)}_X^2$, and $t=\frac{1}{\lambda }\ln (A_0/A)$, $t_Y=\frac{1}{\lambda }\ln (2^2)=\frac{2}{\lambda }\ln 2=2t_X.$ WHY it works: the log turns a squaring of the ratio into a doubling of the time. Whenever one sample's remaining fraction is the square of another's, its age is exactly double. That is the log's defining behaviour — $\ln (x^2)=2\ln x$ — showing up as physics.

Fraction left over whole half-lives ::: $(1/2{)}^n$ Fraction over any time ::: $e^{-\lambda t}$ with $\lambda =\ln 2/t_{1/2}$ Time from a ratio ::: $t=\frac{1}{\lambda }\ln (A_0/A)$ Why a log appears at all ::: time sits inside an exponent; $\ln$ is the exact operation that brings it down