CSIR-NET Nuclear and Radiochemistry — Decay, Q Values and Counting

By Aniket Bhardwaj · 14 September 2026 · CSIR-NET Chemistry

Nuclear and radiochemistry is a small, self-contained corner of the CSIR-NET Chemical Sciences syllabus, and that is exactly why it is worth learning properly. The physics does not branch endlessly the way organic mechanisms do. There are perhaps six equations, a handful of decay modes and one set of unit conversions. Once those are secure, the questions become arithmetic you can finish in a minute. This article covers the decay law, activity and specific activity, Q values, binding energy, radioactive equilibrium and counting statistics, with every number actually computed.

The decay law and the quantities that follow from it

Radioactive decay is first order in the number of undecayed nuclei. Nothing else about the nucleus matters — not temperature, not pressure, not chemical form.

N = N₀ e^−λt · λ = ln 2 / t_½ · A = λN · mean life τ = 1/λ = t_½ / ln 2

N — number of undecayed nuclei; N₀ its value at t = 0.
λ — decay constant, units s⁻¹ (or y⁻¹, h⁻¹). It is the probability that one nucleus decays per unit time.
A — activity, decays per second. 1 becquerel (Bq) = 1 decay s⁻¹; 1 curie (Ci) = 3.7 × 10¹⁰ Bq, by definition.
t_½ — half-life. Note λ = ln 2 / t_½, not 1/t_½. That single slip is the most common arithmetic error in this topic.

Because A = λN, activity and number of nuclei decay with the same exponential, so A = A₀e^−λt as well. You may work in either currency, but never mix the two inside one calculation.

Worked example 1 — specific activity of cobalt-60

Cobalt-60 has t_½ = 5.27 years. Find the activity of 1.00 mg of pure ⁶⁰Co, in Bq and in Ci.

Step 1 — decay constant.
λ = ln 2 / t_½ = 0.6931 / 5.27 y = 0.13152 y⁻¹.
One year = 365.25 × 86 400 s = 3.15576 × 10⁷ s.
λ = 0.13152 ÷ (3.15576 × 10⁷) = 4.1676 × 10⁻⁹ s⁻¹.

Step 2 — number of nuclei.
n = 1.00 × 10⁻³ g ÷ 59.93 g mol⁻¹ = 1.6686 × 10⁻⁵ mol.
N = 1.6686 × 10⁻⁵ × 6.022 × 10²³ = 1.0048 × 10¹⁹ nuclei.

Step 3 — activity.
A = λN = 4.1676 × 10⁻⁹ × 1.0048 × 10¹⁹ = 4.19 × 10¹⁰ Bq.
In curies: (4.188 × 10¹⁰) ÷ (3.7 × 10¹⁰) = 1.13 Ci.

Cross-check. That works out at roughly 1100 Ci per gram, which is the order of magnitude quoted for high-specific-activity ⁶⁰Co sources — a good sign the arithmetic is sound.

Decay modes and the displacement rules

Every question that asks "what is the product?" is answered by conserving mass number A and charge Z. Learn the table below and you never have to reason it out again.

Mode	Emitted	Change in Z	Change in A	Note
α	⁴He nucleus	−2	−4	Common above Z ≈ 83
β⁻	electron + antineutrino	+1	0	Neutron-rich nuclei
β⁺	positron + neutrino	−1	0	Needs Q > 2m_ec² = 1.022 MeV
Electron capture	X-rays, neutrino	−1	0	Competes with β⁺; no energy threshold
γ / internal conversion	photon / orbital electron	0	0	De-excitation only

Nuclear reactions are written compactly as ¹⁴N(α, p)¹⁷O, read as "target(incoming, outgoing)product". Balance A and Z on both sides and the missing particle is forced — there is never any ambiguity.

Worked example 2 — Q value of an alpha decay

For ²¹⁰Po → ²⁰⁶Pb + α, using atomic masses ²¹⁰Po = 209.982874 u, ²⁰⁶Pb = 205.974465 u and ⁴He = 4.002603 u, find Q and the kinetic energy of the emitted alpha particle.

Step 1 — mass difference.
Products: 205.974465 + 4.002603 = 209.977068 u.
Δm = 209.982874 − 209.977068 = 0.005806 u.

Step 2 — energy. 1 u ≡ 931.494 MeV/c².
Q = 0.005806 × 931.494 = 5.408 MeV.

Step 3 — how the energy is shared. Momentum conservation gives the alpha the fraction (A − 4)/A of Q:
T_α = 5.408 × 206/210 = 5.408 × 0.98095 = 5.305 MeV.
The recoiling ²⁰⁶Pb nucleus carries the remaining 0.103 MeV.

Why atomic masses may be used here. Po has 84 electrons; Pb (82) plus He (2) also totals 84, so the electron masses cancel exactly. For β⁺ decay they do not cancel, and you must subtract 2m_ec² = 1.022 MeV.

Binding energy — the curve that explains fission and fusion

BE = [Z · m(¹H) + N · m_n − m(atom)] × 931.494 MeV

Worked example 3 — binding energy per nucleon of ⁵⁶Fe (m = 55.934936 u, Z = 26, N = 30; m(¹H) = 1.007825 u, m_n = 1.008665 u).

26 × 1.007825 = 26.20345 u
30 × 1.008665 = 30.25995 u
Sum of the separated parts = 56.46340 u
Δm = 56.46340 − 55.934936 = 0.528464 u
BE = 0.528464 × 931.494 = 492.26 MeV
BE/A = 492.26 ÷ 56 = 8.79 MeV per nucleon — the well-known maximum region of the binding-energy curve. Nuclei lighter than this release energy by fusing; heavier ones release energy by splitting. That single number explains both fusion and fission in one line.

Parent–daughter relations and secular equilibrium

When a long-lived parent feeds a short-lived daughter (λ₁ ≪ λ₂), the daughter builds up until it decays exactly as fast as it is formed. That is secular equilibrium, and it gives the most useful shortcut in the topic.

At secular equilibrium: A₁ = A₂, i.e. λ₁N₁ = λ₂N₂, so N₂/N₁ = λ₁/λ₂ = t_½(daughter) / t_½(parent)

Worked example 4. ²²⁶Ra (t_½ = 1600 y) decays to ²²²Rn (t_½ = 3.82 d). What is the atom ratio at equilibrium?

Convert to the same unit: 1600 y × 365.25 = 584 400 d.
N(Rn)/N(Ra) = 3.82 ÷ 584 400 = 6.54 × 10⁻⁶.

So a gram of radium holds only a vanishingly small number of radon atoms at any instant, yet the two have equal activity. Candidates who confuse "amount" with "activity" get this backwards every time.

When the half-lives are merely comparable (λ₁ < λ₂, but not far smaller) you get transient equilibrium, in which the activity ratio settles at A₂/A₁ = λ₂/(λ₂ − λ₁). If the parent is shorter-lived than the daughter, no equilibrium is reached at all — the parent simply disappears.

Counting statistics — the one piece of statistics this topic expects

Radioactive counting follows Poisson statistics, so the standard deviation of a total count N is just √N. Everything else follows from that.

σ = √N · relative error = σ/N = 1/√N

Worked example 5. A sample gives 2500 counts. What is the percentage error, and how much longer must you count to halve it?

σ = √2500 = 50 counts, so the relative error = 50 ÷ 2500 = 0.020 = 2.0%.
For 1.0% we need 1/√N = 0.010, so N = 10 000 counts — four times as many, and therefore four times the counting time. Precision improves only as the square root of effort, which is why long counts are unavoidable for weak sources.

Applied radiochemistry worth a line in your notes

Tracers. A radioisotope is chemically identical to its stable partner, so it follows the same pathway and can still be detected at vanishingly small concentration.
Szilard–Chalmers reaction. Recoil following neutron capture breaks the chemical bond holding the atom, letting the newly formed radionuclide be separated from the bulk target — giving a product of much higher specific activity.
Generators. The ⁹⁹Mo/^99mTc pair is the classic transient-equilibrium system used in nuclear medicine; the short-lived daughter is eluted repeatedly from the longer-lived parent.
Neutron activation analysis identifies elements from the characteristic γ energies of the activated nuclides. It is an elemental technique, not a molecular one — a distinction questions like to test.
Magic numbers 2, 8, 20, 28, 50, 82 and 126 mark closed nuclear shells and unusual stability — the nuclear analogue of noble-gas configurations.

Mistakes that cost marks

Writing λ = 1/t_½. It is ln 2 / t_½. The 0.693 is missed in a surprising number of scripts.
Mixing time units. If λ is in y⁻¹, t must be in years. Convert once, at the start, and write the unit beside every number.
Nuclear versus atomic masses. Tables list atomic masses, which include the electrons. For α and β⁻ decay they cancel; for β⁺ you must subtract 1.022 MeV; for electron capture you must not.
Confusing amount with activity. At secular equilibrium the activities are equal, not the numbers of atoms.
Treating half-life as chemistry. Decay constants are unaffected by temperature, pressure or oxidation state.
Forgetting the recoil share. The alpha particle carries (A−4)/A of Q, not the whole of it.
Averaging counts wrongly. σ = √N applies to the total count, not to a count rate; convert to rate only after you have the error on the total.

Where this appears in the paper

Sub-topic	Typical question form
Decay law	Fraction remaining after n half-lives; time to fall to a given activity
Activity and units	Bq ↔ Ci conversion; specific activity of a stated mass
Q values	Energy released from a mass table; is a given decay mode energetically allowed?
Binding energy	BE/A comparison; why fusion of light nuclei releases energy
Equilibria	Secular vs transient; atom or activity ratio for a parent–daughter pair
Counting	Percentage error from a count total; counting time for a target precision
Applications	Tracer design, Szilard–Chalmers, activation analysis, generators

Treat that table as a topic map, not as a prediction of the paper. The number of questions and the distribution of marks change between sessions, so always read the current official notification and syllabus rather than any secondary source, including this one.

Check every decay calculation in seconds. The Half-Life Calculator takes the half-life, the initial amount and the elapsed time and returns the decay constant, the amount remaining and the fraction decayed — so you can verify a full page of decay arithmetic before you trust it.

Open the Half-Life Calculator →

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