GATE Nuclear and Radiochemistry — Decay Law, Q Values and Binding Energy
Nuclear chemistry is a small unit that gives back more than it takes. The mathematics is first-order kinetics, which you already know from chemical kinetics; the energetics is one conversion factor; and the qualitative part — decay modes, the band of stability, applications — is memorable because each rule has a physical reason. This guide sets out the formulas with their units, works four numericals in full, and lists the errors that most often turn a correct method into a wrong answer.
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The decay law — first-order kinetics with new names
A = λN A = A₀ e−λt τ (mean life) = 1/λ = t½ / ln 2
- N — number of radioactive nuclei remaining after time t.
- λ — decay constant, units of reciprocal time (s⁻¹, d⁻¹, y⁻¹). It is the probability per unit time that a given nucleus decays.
- t½ — half-life, the time for half the nuclei to decay. It does not depend on how much sample you have, or on temperature, pressure or chemical form.
- A — activity, the number of disintegrations per second. Units: 1 becquerel (Bq) = 1 disintegration per second; 1 curie (Ci) = 3.7 × 10¹⁰ Bq exactly by definition.
- τ — mean life, always longer than the half-life, by a factor 1/ln 2 ≈ 1.44.
Worked example 1 — fraction remaining, by two routes
³²P has t½ = 14.3 days. What fraction of a sample remains after 30.0 days?
Route 1 — via the decay constant.
λ = 0.6931 ÷ 14.3 = 0.04847 d⁻¹
λt = 0.04847 × 30.0 = 1.4541
N/N₀ = e−1.4541 = 0.234, i.e. about 23.4% remains
Route 2 — counting half-lives.
number of half-lives = 30.0 ÷ 14.3 = 2.098
N/N₀ = (½)2.098 = 2−2.098 = e−2.098 × 0.6931
= e−1.4541 = 0.234 ✓
The two routes agree, which is the check worth doing in the exam hall. A rough mental estimate also confirms it: two half-lives would leave 25%, and 30 days is slightly more than two half-lives, so slightly under 25% is exactly what we should expect.
Worked example 2 — activity of a weighed sample
Calculate the activity of 1.00 µg of pure ³²P (atomic mass ≈ 31.97 u, t½ = 14.3 days) in Bq and in Ci.
Step 1 — number of nuclei.
moles = 1.00 × 10⁻⁶ ÷ 31.97 = 3.128 × 10⁻⁸ mol
N = 3.128 × 10⁻⁸ × 6.022 × 10²³ = 1.883 × 10¹⁶ nuclei
Step 2 — decay constant in s⁻¹. The activity must be per second, so
convert the half-life first.
t½ = 14.3 × 24 × 3600 = 14.3 × 86,400 = 1.2355 × 10⁶ s
λ = 0.6931 ÷ 1.2355 × 10⁶ = 5.610 × 10⁻⁷ s⁻¹
Step 3 — activity.
A = λN = 5.610 × 10⁻⁷ × 1.883 × 10¹⁶ = 1.06 × 10¹⁰ Bq
Step 4 — convert to curies.
A = 1.06 × 10¹⁰ ÷ 3.7 × 10¹⁰ = 0.286 Ci
The trap in this question is Step 2. If you leave the half-life in days and report "Bq", the answer is wrong by a factor of 86,400. Fix the time unit before you touch the arithmetic.
Mass–energy: binding energy and Q values
1 u = 931.5 MeV (the conversion you will use every time)
The binding energy is the energy released when free nucleons come together to make a nucleus — equivalently, the energy needed to pull it apart. Dividing by the mass number gives the binding energy per nucleon, which is the real measure of stability. That curve rises steeply from hydrogen, peaks near mass number 56 (the iron–nickel region at roughly 8.8 MeV per nucleon), and falls slowly afterwards. Fusion of light nuclei and fission of heavy nuclei both release energy because both move the products towards that peak.
Worked example 3 — binding energy of the helium-4 nucleus
Given: mass of ¹H atom = 1.007825 u, neutron = 1.008665 u, ⁴He atom = 4.002603 u. Find the total binding energy and the binding energy per nucleon.
Step 1 — mass of the separate parts. ⁴He has 2 protons and 2 neutrons.
Using atomic masses throughout, the 2 electrons of the ⁴He atom are matched by the
2 electrons in the two ¹H atoms, so they cancel exactly.
2 × 1.007825 = 2.015650
2 × 1.008665 = 2.017330
total = 2.015650 + 2.017330 = 4.032980 u
Step 2 — mass defect.
Δm = 4.032980 − 4.002603 = 0.030377 u
Step 3 — convert to energy.
BE = 0.030377 × 931.5 = 28.30 MeV
Step 4 — per nucleon.
28.30 ÷ 4 = 7.07 MeV per nucleon
This is why the alpha particle is so stable and why alpha emission, rather than the emission of separate nucleons, is a favoured decay route for heavy nuclei.
Worked example 4 — Q value of an alpha decay
²³⁸U → ²³⁴Th + ⁴He. Atomic masses: ²³⁸U = 238.050788 u, ²³⁴Th = 234.043601 u, ⁴He = 4.002603 u. Find Q.
Step 1 — total mass of products.
234.043601 + 4.002603 = 238.046204 u
Step 2 — mass difference.
Δm = 238.050788 − 238.046204 = 0.004584 u
Step 3 — energy released.
Q = 0.004584 × 931.5 = 4.27 MeV
Two points that matter. First, Q is positive, so the decay is energetically allowed — that is what a Q value tells you. Second, the alpha particle does not carry all 4.27 MeV: momentum must be conserved, so the recoiling ²³⁴Th nucleus takes a share. The alpha's kinetic energy is Q × (A − 4)/A = 4.27 × 234/238 = 4.20 MeV. Questions that ask for "the energy of the alpha particle" rather than "the energy released" are testing exactly this distinction.
Why the electrons cancelled here: ²³⁸U has 92 electrons in its atomic mass; the products have 90 (Th) + 2 (He) = 92. Using atomic masses on both sides is therefore exact for alpha decay.
Decay modes and when each occurs
The controlling idea is the neutron-to-proton ratio. Light stable nuclei sit close to n/p = 1; heavier stable nuclei need progressively more neutrons to dilute proton–proton repulsion, so the band of stability curves upward, reaching about n/p = 1.5 for the heaviest stable nuclides. A nucleus decays in whichever way moves it towards that band.
| Mode | Emitted | Change in Z, A | Occurs when |
|---|---|---|---|
| Alpha (α) | ⁴He nucleus | Z − 2, A − 4 | Heavy nuclei, generally Z > 83 |
| Beta minus (β⁻) | electron + antineutrino | Z + 1, A unchanged | Too many neutrons: a neutron becomes a proton |
| Beta plus (β⁺) | positron + neutrino | Z − 1, A unchanged | Too few neutrons; needs Q > 1.022 MeV |
| Electron capture (EC) | neutrino; X-rays follow | Z − 1, A unchanged | Too few neutrons; competes with β⁺ and works at lower Q |
| Gamma (γ) | photon | No change | An excited daughter nucleus relaxes |
| Spontaneous fission | two large fragments + neutrons | Large change | Very heavy nuclei |
The Q-value bookkeeping differs between the beta modes, and this is examined. Using atomic masses: for β⁻, Q = [M(parent) − M(daughter)] × 931.5 MeV, because the emitted electron is already counted in the daughter atom's electrons. For β⁺, you must subtract two electron masses: Q = [M(parent) − M(daughter) − 2me] × 931.5 MeV, and since 2mec² = 1.022 MeV, positron emission is impossible unless the atomic mass difference exceeds that. Electron capture has no such threshold, which is why proton-rich nuclei with small mass differences decay by EC rather than β⁺.
Nuclear reactions and cross-sections
A nuclear reaction is written in the compact form X(a, b)Y, meaning target X is struck by projectile a, emits b and becomes Y. For example, ¹⁴N(α, p)¹⁷O. Both mass number and charge must balance on the two sides — this is the fastest way to identify a missing particle in an exam question.
The probability of a reaction is expressed as a cross-section, with the unit barn: 1 barn = 10⁻²⁴ cm² = 10⁻²⁸ m². A large cross-section means the reaction happens readily at that projectile energy.
Radioanalytical methods worth knowing
- Neutron activation analysis (NAA): the sample is irradiated with neutrons, stable nuclides become radioactive, and the characteristic gamma energies identify the elements while the intensities quantify them. It is non-destructive and extremely sensitive for many trace elements.
- Isotope dilution analysis: a known amount of a labelled compound is added to the sample and the mixture is equilibrated; the drop in specific activity after isolating a pure portion gives the original amount. Its great advantage is that you do not need quantitative recovery — only a pure fraction.
- Radiometric titration and radiotracers: following an element through a process by its radiation, used to establish reaction pathways and exchange rates.
- The Szilard–Chalmers effect: after neutron capture, the recoil breaks the chemical bond holding the atom, so the radioactive product ends up in a different chemical form and can be separated from the bulk target. This is how carrier-free (high specific activity) preparations are obtained.
- Counting statistics: radioactive decay is random, so a total of N counts carries a standard deviation of √N. Recording 10,000 counts gives σ = 100, a relative precision of 1%; to halve that uncertainty you must count four times as long. This simple result is a favourite short question.
Common mistakes that cost marks
- Mixing time units. λ and t must be in the same unit; activity in Bq demands λ in s⁻¹.
- Confusing λ with t½. They are inversely related through ln 2, not equal. Likewise mean life ≠ half-life.
- Reporting the Q value as the alpha particle's kinetic energy. The recoiling daughter takes a share, given by the momentum-conservation split.
- Mixing nuclear and atomic masses in one calculation. Pick atomic masses and stay with them; then check whether the electrons cancel for the decay mode you are dealing with.
- Forgetting the 1.022 MeV threshold for β⁺ emission.
- Using 931.5 MeV/u as if it were MeV/kg, or forgetting the conversion entirely and quoting an answer in u.
- Assuming half-life changes with temperature or chemical form. To the accuracy of any exam question, it does not — nuclear decay is not a chemical process.
- Treating a curie as an SI unit. The SI unit is the becquerel; 1 Ci = 3.7 × 10¹⁰ Bq.
Preparation map
| Sub-topic | Question style | Priority |
|---|---|---|
| Decay law and half-life | Fraction remaining, time elapsed, activity | High — quick, certain marks |
| Activity units | Bq ↔ Ci; activity of a weighed sample | High |
| Binding energy | Δm → MeV, per nucleon | High |
| Q values | Allowed or not; energy shared with recoil | High |
| Decay modes and n/p ratio | Predict the mode; balance a reaction | Medium — conceptual |
| Fission and fusion | Why both release energy | Medium |
| Radioanalytical methods | Match a technique to its purpose | Medium |
Half-life problems are pure practice. Once you have set up N = N₀e−λt a dozen times, the method becomes automatic and you stop making unit slips. The free Half-Life calculator works between the decay constant, the half-life, the elapsed time and the fraction remaining, so you can check a whole set of practice questions in minutes and catch a wrong answer while you still remember what you did.
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