Radioactivity: Alpha, Beta, Gamma & the Decay Law

Becquerel's Lucky Accident (1896)

A. H. Becquerel discovered radioactivity purely by accident. Studying fluorescence of compounds under visible light, he illuminated pieces of uranium-potassium sulphate, wrapped them in black paper, and separated the package from a photographic plate by a piece of silver. After several hours, the developed plate showed blackening — something emitted by the compound had penetrated both the black paper and the silver.

Later experiments showed radioactivity is a nuclear phenomenon: an unstable nucleus undergoes decay — radioactive decay.

The three decays of nature

Decay	What is emitted	Character
Alpha ($\alpha$)	a helium nucleus $^{4}_{2}$He	charge +2e, mass ~4 u
Beta ($\beta$)	electrons ($\beta^-$) or positrons ($\beta^+$)	positron = electron's antiparticle: same mass, opposite charge
Gamma ($\gamma$)	high-energy photons (hundreds of keV or more)	neutral, massless; wavelengths shorter than X-rays

Displacement rules — [JEE/NEET Important]

• Alpha decay: $^{A}_{Z}X \to\ ^{A-4}_{Z-2}Y + {}^{4}_{2}He$ — A drops by 4, Z by 2.
• Beta-minus: $^{A}_{Z}X \to\ ^{A}_{Z+1}Y + e^- + \bar{\nu}$ — A unchanged, Z up by 1 (a neutron becomes a proton).
• Beta-plus: $^{A}_{Z}X \to\ ^{A}_{Z-1}Y + e^+ + \nu$ — A unchanged, Z down by 1.
• Gamma: no change in A or Z — an excited nucleus sheds energy as a photon, usually after an alpha or beta decay.

Stability logic (NCERT Points to Ponder): stability wants N/Z ≈ 1 for light nuclei, drifting to ~3/2 for heavy ones (extra neutrons offset proton repulsion). Nuclei off this ratio — excess neutrons or protons — are unstable; only ~10% of known isotopes are stable. Electron-positron pairs annihilate into gamma photons on meeting.

The Decay Law — [JEE/NEET Essential]

The rationalised NCERT trims the derivation, but retains $\lambda$, half-life, mean life and activity in its own summary table — and JEE/NEET test them every single year. Here is the complete kit.

Exponential decay

Radioactive decay is statistical: each nucleus has a fixed probability per unit time, $\lambda$ (the decay/disintegration constant), of decaying. For N undecayed nuclei, $\dfrac{dN}{dt} = -\lambda N$, giving

$$\boxed{N = N_0e^{-\lambda t}}$$

Half-life

Time for half the nuclei to decay:

$$T_{1/2} = \frac{\ln 2}{\lambda} = \frac{0.693}{\lambda}$$

After n half-lives: $N = \dfrac{N_0}{2^n}$ — the workhorse of most numericals.

Mean life

Average lifetime of a nucleus:

$$\tau = \frac{1}{\lambda}, \qquad T_{1/2} = 0.693\,\tau$$

After one mean life, N falls to $N_0/e$ ≈ 37% of the start.

Activity

The decay rate — what detectors actually measure:

$$R = \left|\frac{dN}{dt}\right| = \lambda N = R_0e^{-\lambda t}$$

Units: the SI becquerel (1 Bq = 1 decay/s) and the traditional curie (1 Ci = $3.7 \times 10^{10}$ Bq).

Key Point: N, R and the remaining mass all fall by the same factor $e^{-\lambda t}$ — halving every $T_{1/2}$. Given any one at two times, you can extract $\lambda$ and everything else.

[JEE Tip] The three time constants in one line: $\tau = 1/\lambda > T_{1/2} = 0.693/\lambda$. Mean life always exceeds half-life. And for 'what fraction survives 3 half-lives?' — $\left(\frac{1}{2}\right)^3 = 12.5\%$, decayed fraction 87.5%.

Solved Examples

Example 1: Identify the decays [Board Rapid]

Name the radiation: (a) helium nuclei, (b) electrons/positrons, (c) photons of hundreds of keV.

Solution:

1. (a) Alpha decay — $^4_2$He nuclei ejected.
2. (b) Beta decay — $\beta^-$ (electrons) or $\beta^+$ (positrons, same mass as electrons but opposite charge).
3. (c) Gamma decay — high-energy photons, wavelengths shorter than X-rays.
4. Takeaway: NCERT's three-decay taxonomy verbatim — a permanent one-marker.

Example 2: Tracking A and Z through a decay chain [JEE Pattern]

$^{238}_{92}$U undergoes one alpha decay, then two beta-minus decays. Find the final nuclide's A and Z.

Solution:

1. Alpha: A: 238 → 234; Z: 92 → 90 ($^{234}_{90}$Th).
2. Beta-minus #1: A unchanged; Z: 90 → 91 ($^{234}_{91}$Pa).
3. Beta-minus #2: A unchanged; Z: 91 → 92 ($^{234}_{92}$U).
4. Answer: $^{234}_{92}$U — same element as the start, four mass units lighter.
5. Takeaway: alpha moves (A, Z) by (-4, -2); each $\beta^-$ moves (0, +1). Chain arithmetic is pure addition.

Example 3: Fraction left after n half-lives [NEET Numerical]

A sample's half-life is 30 days. What fraction survives after 90 days, and what fraction has decayed?

Solution:

1. Half-lives elapsed: n = 90/30 = 3.
2. Surviving: $\left(\dfrac{1}{2}\right)^3 = \dfrac{1}{8}$ = 12.5%.
3. Decayed: 1 - 1/8 = 7/8 = 87.5%.
4. Takeaway: count half-lives, halve repeatedly; 'decayed' is the complement — the single most common trap.

Example 4: Decay constant from half-life [Board Numerical]

The half-life of $^{14}$C is 5730 years. Find $\lambda$ and the mean life.

Solution:

1. Decay constant: $\lambda = \dfrac{0.693}{T_{1/2}} = \dfrac{0.693}{5730} \approx 1.21 \times 10^{-4}$ per year.
2. Mean life: $\tau = \dfrac{1}{\lambda} = \dfrac{5730}{0.693} \approx 8270$ years.
3. Takeaway: $\tau > T_{1/2}$ always (by the factor 1/0.693 ≈ 1.44).

Example 5: Activity of a sample [JEE Numerical]

A sample contains $10^{20}$ atoms of a nuclide with $T_{1/2}$ = 693 s. Find its activity in becquerel and curie.

Solution:

1. Decay constant: $\lambda = \dfrac{0.693}{693} = 10^{-3}$ s$^{-1}$.
2. Activity: $R = \lambda N = 10^{-3} \times 10^{20} = 10^{17}$ Bq.
3. In curie: $\dfrac{10^{17}}{3.7 \times 10^{10}} \approx 2.7 \times 10^{6}$ Ci.
4. Takeaway: $R = \lambda N$ — activity needs BOTH the decay constant and the population; 693-type numbers are chosen to cancel 0.693.

Example 6: Time to drop to 1/16 [NEET Numerical]

How many half-lives until activity falls to 1/16 of its initial value? If $T_{1/2}$ = 2 hours, how long is that?

Solution:

1. Halvings: $\dfrac{1}{16} = \left(\dfrac{1}{2}\right)^4$ → 4 half-lives.
2. Time: 4 × 2 = 8 hours.
3. Takeaway: express the fraction as a power of 1/2; the exponent counts the half-lives.

Example 7: Why the neutrino had to exist [JEE/NEET Extra Insight]

In beta decay the emitted electrons show a continuous energy spectrum. Why did this demand a third particle?

Solution:

1. A two-body decay (nucleus → daughter + electron) forces a unique electron energy by energy-momentum conservation.
2. Observed: electrons carry a continuous range of energies up to a maximum.
3. Resolution (Pauli): a third, nearly undetectable neutral particle — the (anti)neutrino — shares the energy randomly: $n \to p + e^- + \bar{\nu}$.
4. Takeaway: the antineutrino in the free-neutron decay (Section 1) is the same particle; conservation laws forced its prediction decades before detection.

Example 8: Annihilation arithmetic [NEET Numerical]

An electron and positron at rest annihilate. Find the total photon energy released.

Solution:

1. Mass destroyed: $2m_e = 2 \times 0.511$ MeV/$c^2$.
2. Energy: E = 1.022 MeV, shared as (at least) two gamma photons of 0.511 MeV each (momentum conservation forbids a single photon).
3. Takeaway: NCERT Points to Ponder 7 — particle-antiparticle pairs annihilate to gamma rays; 0.511 MeV per electron mass is a number worth owning.

Example 9: Which nuclei are unstable? [Conceptual]

Using the N/Z stability logic, predict the decay mode of (a) a nucleus with excess neutrons, (b) one with excess protons.

Solution:

1. (a) Neutron-rich: convert a neutron to a proton — beta-minus decay ($n \to p + e^- + \bar{\nu}$), raising Z toward the stability ratio.
2. (b) Proton-rich: convert a proton to a neutron — beta-plus decay ($p \to n + e^+ + \nu$), lowering Z.
3. Takeaway: decay is the nucleus steering itself back to the stability line; heavy nuclei may also shed bulk via alpha decay. Only ~10% of known isotopes sit stably on the line.

Example 10: Becquerel's controls [Board Conceptual]

Why was the blackening through black paper AND silver so significant in 1896?

Solution:

1. Black paper blocks all visible/UV light — so the plate wasn't exposed by ordinary light or fluorescence.
2. Silver sheet blocks weakly penetrating radiation — yet the plate still blackened.
3. Conclusion: the uranium compound emits penetrating radiation spontaneously, without needing prior illumination — a wholly new phenomenon (later traced to unstable nuclei).
4. Takeaway: the accidental discovery preceded any understanding of the nucleus by 15 years — radioactivity was data waiting for a theory.