Radioactive Equilibrium

$\lambda_U \ll \lambda_{Th}$, so secular equilibrium applies: at equilibrium $A_{Th} = A_U$.

$$N_U = \frac{1.0}{238}\times 6.022\times 10^{23} = 2.53\times 10^{21}\ \text{atoms}$$

$$\lambda_U = \frac{\ln 2}{4.47\times 10^9\times 365.25\times 86400} = 4.91\times 10^{-18}\ \text{s}^{-1}$$

$$A_U = \lambda_U N_U = 1.24\times 10^4\ \text{Bq} = 12.4\ \text{kBq}$$

Hence $A_{Th} = 12.4$ kBq. The number of thorium atoms follows from $N = A/\lambda$:

$$N_{Th} = \frac{1.24\times 10^4}{\ln 2/(24.1\times 86400)} = 3.73\times 10^{10}\ \text{atoms}$$

$$m_{Th} = \frac{3.73\times 10^{10}}{6.022\times 10^{23}}\times 234 = 1.45\times 10^{-11}\ \text{g}$$

Answer: 12.4 kBq of ²³⁴Th — just 14.5 pg of it. That is the power of secular equilibrium: an immeasurably small mass carrying a very measurable activity.

$$\lambda_A = \frac{\ln 2}{66} = 1.050\times 10^{-2}\ \text{h}^{-1}, \qquad \lambda_B = \frac{\ln 2}{6} = 1.155\times 10^{-1}\ \text{h}^{-1}$$

(a) From (9):

$$t_{\max} = \frac{\ln(\lambda_B/\lambda_A)}{\lambda_B-\lambda_A} = \frac{\ln 11.0}{0.1050} = \frac{2.398}{0.1050} = 22.8\ \text{h}$$

(b) At $t_{\max}$ the activities are equal (D5), so evaluate the parent:

$$A_B(t_{\max}) = A_A(t_{\max}) = 4.0\times e^{-0.01050\times 22.8} = 4.0\times 0.787 = 3.15\ \text{mCi}$$

(Check with the full Bateman form (5): $4.0\times 1.10\times(0.787-0.0716) = 3.15$ mCi. ✓)

Answer: (a) 22.8 h, (b) 3.15 mCi. Note the equilibrium ratio $\lambda_B/(\lambda_B-\lambda_A) = 1.10$ — the technetium sits 10% above its parent, the signature of transient equilibrium.

$$N_t = \frac{1.0}{59}\times 6.022\times 10^{23} = 1.02\times 10^{22}\ \text{atoms}$$

(a) $$A_{\mathrm{sat}} = N_t\Phi\sigma = 1.02\times 10^{22}\times 1.0\times 10^{12}\times 37\times 10^{-24} = 3.78\times 10^{11}\ \text{Bq} = 10.2\ \text{mCi}$$

(b) $$\lambda = \frac{\ln 2}{5.27\times 365.25\times 86400} = 4.17\times 10^{-9}\ \text{s}^{-1}$$

$$A = A_{\mathrm{sat}}\left(1-e^{-\lambda t}\right) = 10.2\times\left(1-e^{-0.131}\right) = 10.2\times 0.123 = 1.26\ \text{mCi}$$

(c) $1-e^{-\lambda t} = 0.75 \Rightarrow e^{-\lambda t} = 1/4 \Rightarrow t = 2\ln 2/\lambda = 2T_{1/2} = 10.5$ y.

Answer: (a) 10.2 mCi, (b) 1.26 mCi, (c) 10.5 y. A 5-year-half-life product simply cannot be made to saturation in a reasonable time — the 1-year irradiation only reaches 12%.

Use the equal-λ result (10): $A_B(t) = A_{A0}\lambda t\,e^{-\lambda t}$. $dA_B/dt = 0$ gives $t_{\max} = 1/\lambda$:

$$t_{\max} = \frac{T_{1/2}}{\ln 2} = \frac{10.0}{0.693} = 14.4\ \text{min}$$

$$A_{B,\max} = 1000\times e^{-1} = 368\ \text{Bq}$$

Answer: 368 Bq at 14.4 min — the daughter can never exceed $1/e \approx 37$% of the initial parent activity when the decay constants are equal.

$\lambda_{Bi} > \lambda_{Po}$: no equilibrium — the polonium peaks and then decays alone. Using (9):

$$\lambda_{Bi} = \frac{\ln 2}{5.0} = 0.1386\ \text{d}^{-1}, \qquad \lambda_{Po} = \frac{\ln 2}{138} = 5.02\times 10^{-3}\ \text{d}^{-1}$$

$$t_{\max} = \frac{\ln(0.00502/0.1386)}{0.00502-0.1386} = \frac{-3.317}{-0.1336} = 24.8\ \text{d}$$

At $t_{\max}$, $A_{Po} = A_{Bi}(t_{\max})$ (D5):

$$A_{Po,\max} = 1.0\times e^{-0.1386\times 24.8} = 3.19\times 10^{-2}\ \text{MBq} = 31.9\ \text{kBq}$$

Answer: 24.8 d; 31.9 kBq. After the bismuth has vanished, the polonium simply decays with its own 138-day half-life.

After equilibrium is reached, $t$ is large compared with the daughter's half-life, so $e^{-\lambda_B t}$ is negligible in the Bateman activity (5):

$$A_B(t) \approx A_{A0}\frac{\lambda_B}{\lambda_B-\lambda_A}\,e^{-\lambda_A t}, \qquad A_A(t-x) = A_{A0}\,e^{-\lambda_A(t-x)}$$

Setting $A_A(t-x) = A_B(t)$ and cancelling $A_{A0}e^{-\lambda_A t}$:

$$e^{\lambda_A x} = \frac{\lambda_B}{\lambda_B-\lambda_A} \;\;\Longrightarrow\;\; x = \frac{1}{\lambda_A}\ln\!\left(\frac{\lambda_B}{\lambda_B-\lambda_A}\right)$$

Now let $\lambda_A/\lambda_B \to 0$. Write $\frac{\lambda_B}{\lambda_B-\lambda_A} = \frac{1}{1-\lambda_A/\lambda_B} \approx 1 + \frac{\lambda_A}{\lambda_B}$ and use $\ln(1+\epsilon) \approx \epsilon$:

$$x \approx \frac{1}{\lambda_A}\cdot\frac{\lambda_A}{\lambda_B} = \frac{1}{\lambda_B} \qquad\blacksquare$$

Physical meaning: in the near-secular limit the daughter lags the parent by one daughter mean life — the B-atoms you count at time $t$ were, on average, born one mean life $1/\lambda_B$ earlier.

Principle: neutrons convert a fraction of the target nuclei into a radioactive product; the induced activity is proportional to the mass of the element present. A standard of known mass irradiated simultaneously with the sample lets flux, cross section and irradiation history cancel (comparator method, (16)).

$$\lambda = \frac{\ln 2}{2.25} = 0.308\ \text{min}^{-1}$$

At end-of-irradiation the activities are in the ratio of the Al masses, $m/10.5$; each then decays for its own cooling time:

$$\frac{1.37\times 10^3}{5.37\times 10^3} = \frac{m}{10.5}\,e^{-\lambda(10.0-5.0)}$$

$$m = 10.5\times\frac{1.37}{5.37}\times e^{0.308\times 5.0} = 10.5\times 0.2551\times 4.667 = 12.5\ \mu\text{g}$$

$$\text{Al content} = \frac{12.5\ \mu\text{g}}{0.240\ \text{g}} = 52.1\ \mu\text{g/g}$$

Answer: 52.1 μg/g (≈ 52 ppm). Note the sample was counted later, so its raw count rate must be corrected up by $e^{\lambda\Delta t}$ — the classic slip is to correct in the wrong direction.

$$\lambda_{Mo} = \frac{\ln 2}{66} = 1.050\times 10^{-2}\ \text{h}^{-1}, \qquad \lambda_{Tc} = \frac{\ln 2}{6} = 1.155\times 10^{-1}\ \text{h}^{-1}$$

From (9):

$$t_{\max} = \frac{\ln(11.0)}{0.1050} = 22.8\ \text{h}$$

At $t_{\max}$, $A_{Tc} = A_{Mo}(t_{\max})$ (D5):

$$A_{Tc}(t_{\max}) = 4.0\times e^{-0.01050\times 22.8} = 4.0\times 0.787 = 3.15\ \text{mCi}$$

Answer: 22.8 h; 3.15 mCi. (Same working as Example E2, §4.)

Saturation activity. During irradiation the product is made at the constant rate $N_t\Phi\sigma$ and decays with $\lambda$: $dN^*/dt = N_t\Phi\sigma - \lambda N^*$. With $N^*(0)=0$, the integrating-factor solution is (12), and multiplying by $\lambda$ gives the stated $A = N_t\Phi\sigma(1-e^{-\lambda t})$. As $t\to\infty$, $e^{-\lambda t}\to 0$ and production exactly balances decay:

$$A_{\mathrm{sat}} = N_t\Phi\sigma$$

Half-saturation. Put $A = A_{\mathrm{sat}}/2$: $1-e^{-\lambda t} = 1/2$, so $e^{-\lambda t} = 1/2$, $\lambda t = \ln 2$, and $t = \ln 2/\lambda = T_{1/2}$. ∎

This exact two-parter was repeated in 2022 as parts (a) and (b) — the single most-asked derivation in the chapter.

Identical to the 2021 question above — see its solution. Working from $dN^*/dt = N_t\Phi\sigma - \lambda N^*$: (a) $A_{\mathrm{sat}} = N_t\Phi\sigma$; (b) half-saturation gives $e^{-\lambda t} = 1/2$, i.e. $t = T_{1/2}$. When the same derivation is asked three years running, it is not a suggestion — it is a syllabus.

Note: the printed scheme mislabels the (n,γ) product as ²⁰⁹Bi; it is ²¹⁰Bi, and the question itself asks for the ²¹⁰Bi disintegration rate — solved accordingly.

$$N_t = \frac{1.0}{209}\times 6.022\times 10^{23} = 2.88\times 10^{21}\ \text{atoms}$$

$$R = N_t\Phi\sigma = 2.88\times 10^{21}\times 1.0\times 10^{12}\times 0.2\times 10^{-24} = 5.76\times 10^{8}\ \text{s}^{-1}$$

$$\lambda = \frac{\ln 2}{5.0\times 86400} = 1.60\times 10^{-6}\ \text{s}^{-1}, \qquad t = 3.0\ \text{h} = 1.08\times 10^4\ \text{s}$$

$$A = R\left(1-e^{-\lambda t}\right) = 5.76\times 10^{8}\times\left(1-e^{-0.0173}\right) = 5.76\times 10^{8}\times 0.01718 = 9.90\times 10^{6}\ \text{Bq}$$

Answer: 9.90 × 10⁶ disintegrations s⁻¹ (≈ 0.27 μCi). The 3 h irradiation is far short of the 5-day half-life, so only ~1.7% of the saturation activity (5.76 × 10⁸ Bq) is reached — for $\lambda t \ll 1$ you may use $A \approx R\lambda t$ directly.

Repeat of the 2020 (MCHEM-0102) question: $t_{\max} = \ln(11.0)/0.1050 = 22.8$ h, and $A_{Tc}(t_{\max}) = 4.0\,e^{-0.01050\times 22.8} = 3.15$ mCi. Answer: 22.8 h; 3.15 mCi.

NAA (neutron activation) is superb for medium and heavy elements but blind to several light elements: H, He, Li, Be have tiny thermal-neutron capture cross sections, and C, N, O give products (e.g. short-lived β⁺ emitters) that are hard to count by γ-spectrometry. CPAA (charged-particle activation) uses protons, deuterons or α-particles, whose Coulomb-barrier reactions — e.g. ¹²C(d,n)¹³N, ¹⁶O(p,α)¹³N, ¹⁴N(p,α)¹¹C — determine exactly those light elements with high sensitivity.

Second, the probes sample different depths: neutrons penetrate the bulk, while charged particles stop within micrometres, so CPAA is surface-sensitive and NAA is a bulk technique. Between them they cover the whole periodic table and the whole sample. Hence complementary.

The sample is irradiated with neutrons; (n,γ) capture converts part of each element into a radioactive isotope whose activity is proportional to the mass of that element. Measuring the characteristic radiation of the product — compared against a standard irradiated simultaneously (comparator method) — gives the elemental concentration, non-destructively and down to ppm–ppb levels. (Fuller statement in §2.5.)

(i) RNAA — Radiochemical Neutron Activation Analysis (chemical separation after irradiation, before counting). (ii) FNAA — Fast Neutron Activation Analysis (fast, e.g. 14 MeV, neutrons instead of thermal). (iii) CPAA — Charged Particle Activation Analysis (protons/deuterons/α-particles instead of neutrons).

In short: secular is the limiting case of transient as $\lambda_A/\lambda_B \to 0$, where the fixed ratio $\lambda_B/(\lambda_B-\lambda_A)$ collapses to 1.

Same proof as the 2020 question with $\lambda_A \to \lambda_1$, $\lambda_B \to \lambda_2$: from the equilibrium Bateman form, $e^{\lambda_1 x} = \lambda_2/(\lambda_2-\lambda_1)$, so $x = (1/\lambda_1)\ln[\lambda_2/(\lambda_2-\lambda_1)] \to 1/\lambda_2$ as $\lambda_1/\lambda_2 \to 0$. ∎

From D6: $N_A = N_{A0}e^{-\lambda t}$ and $N_B = N_{A0}\lambda t\,e^{-\lambda t}$. For C, $dN_C/dt = \lambda N_B - \lambda_1 N_C$; multiply by the integrating factor $e^{\lambda_1 t}$:

$$\frac{d}{dt}\left(N_C e^{\lambda_1 t}\right) = \lambda^2 N_{A0}\,t\,e^{(\lambda_1-\lambda)t}$$

With $\alpha = \lambda_1-\lambda$, $\int_0^t \tau e^{\alpha\tau}d\tau = \big(e^{\alpha t}(\alpha t-1)+1\big)/\alpha^2$, giving

$$N_C(t) = \frac{\lambda^2 N_{A0}}{\alpha^2}\left[e^{-\lambda_1 t} + e^{-\lambda t}(\alpha t-1)\right]$$

Multiplying by $\lambda_1$ and writing $A_0 = \lambda N_{A0}$:

$$A_C(t) = A_0\,\frac{\lambda_1\lambda^2}{(\lambda_1-\lambda)^2}\,\left[e^{-\lambda_1 t} + e^{-\lambda t}\big((\lambda_1-\lambda)t-1\big)\right]\ \blacksquare$$

Sanity check: at $t=0$ the bracket is $1-1=0$, so $A_C(0)=0$ as required. This is the boxed result (11) of §3.

$\lambda_A \ll \lambda_B$ (parent half-life enormously longer than the daughter's). Then $A_B \to A_A$: daughter activity equals parent activity.

Greater: $A_B/A_A = \lambda_B/(\lambda_B-\lambda_A) > 1$. For ⁹⁹Mo/⁹⁹ᵐTc it is 1.10.

$A_B(t) = A_{A0}\,\dfrac{\lambda_B}{\lambda_B-\lambda_A}\,(e^{-\lambda_A t}-e^{-\lambda_B t})$.

At $t_{\max} = \ln(\lambda_B/\lambda_A)/(\lambda_B-\lambda_A)$. For ⁹⁹Mo/⁹⁹ᵐTc this is ≈ 22.8 h — roughly daily milking.

$A_{\mathrm{sat}} = N_t\Phi\sigma$ — the activity at which production balances decay. Strictly reached only at infinite irradiation; 99% is reached after ≈ 6.64 product half-lives.

A standard irradiated simultaneously with the sample cancels $N_t$, $\Phi$, $\sigma$ and the irradiation history — no absolute flux or cross-section values are needed, only the activity ratio with decay corrections.

Secular: ²³⁸U → ²³⁴Th. Transient: ⁹⁹Mo → ⁹⁹ᵐTc. No equilibrium: ²¹⁰Bi → ²¹⁰Po.

The parent's: both activities vary as $e^{-\lambda_A t}$. Plotting $\ln A_B$ vs $t$ gives a line of slope $-\lambda_A$ — this is how the ⁹⁹ᵐTc eluted from a generator is seen to decay with the 66 h half-life of its ⁹⁹Mo parent.