JEE Chemistry Nuclear Chemistry Notes Set 03

Radioactive Decay Law

Radioactive decay is a statistical, first order process: every nucleus has the same fixed chance of decaying in a given time, and it has no memory of its past.
\( \frac{dN}{dt}=-\lambda N,\quad N=N_0e^{-\lambda t} \), where \( \lambda \) is the decay constant (unit time⁻¹).

Half-Life, Mean Life and Activity

Half-life \( t_{1/2}=\frac{0.693}{\lambda} \) and mean life \( \tau=\frac1\lambda=1.44\,t_{1/2} \).
After n half-lives the amount left is \( N_0\left(\frac12\right)^n \), with n = total time / half-life.
Activity \( A=\lambda N=A_0e^{-\lambda t} \). SI unit: becquerel (1 Bq = 1 disintegration per second). 1 curie = 3.7 × 10¹⁰ Bq, 1 rutherford = 10⁶ Bq.
If the nuclide is also being produced, then dN/dt equals the rate of production minus the activity; the two are equal only for a pure decaying sample.

Radioactive Dating

Carbon dating compares the present ¹⁴C/¹²C ratio of dead matter with that of living matter: \( t=\frac{2.303\,t_{1/2}}{0.693}\log\frac{N_0}{N} \).
Rock dating uses the Pb : U ratio: the original uranium is [U] + [Pb], so \( t=\frac{2.303}{\lambda}\log\left(1+\frac{[Pb]}{[U]}\right) \).

Nuclear Stability and Modes of Decay

Stable nuclei lie in a belt in the n–p plot; n/p is about 1 for light nuclei and rises to about 1.5 for heavy ones.
n/p too high: β⁻ emission (n → p + e⁻), which raises Z by 1 and leaves A unchanged.
n/p too low: positron emission, K-electron capture (both raise n/p) or α emission for heavy nuclei (A falls by 4, Z by 2).
γ emission removes excess energy and changes neither A nor Z.

Nuclear Fission and Fusion

Fission: a heavy nucleus splits into two lighter nuclei with the release of neutrons and large energy, for example ¹₀n + ²³⁵₉₂U → ¹⁴⁴₅₄Xe + ⁹⁰₃₈Sr + 2 ¹₀n. A chain reaction needs at least the critical mass; control rods (boron or cadmium) absorb neutrons.
Fusion: light nuclei combine into a heavier nucleus and release energy because of the mass defect. It needs temperatures of 10⁶ to 10⁷ K (thermonuclear reactions) and powers the sun (proton–proton chain: 4¹₁H → ⁴₂He + 2 positrons + 24.7 MeV).

Illustration 1. A radioactive sample has a half-life of 20 minutes. What fraction remains after 1 hour?
Answer:
1 hour = 3 half-lives, so the fraction left is \( \left(\frac12\right)^3=\frac18 \).

Illustration 2. Estimate the age of a wooden relic whose ¹⁴C activity is 25 % of that of fresh wood (t_1/2 = 5730 years).
Answer:
25 % = (1/2)², so two half-lives have elapsed.
Age = 2 × 5730 = 11460 years.

Illustration 3. Find the number of α and β⁻ particles emitted in the change ²³²₉₀Th → ²⁰⁸₈₂Pb.
Answer:
α particles = (232 − 208)/4 = 6, which lowers Z by 12 to 78.
To reach Z = 82, 4 β⁻ particles are needed.