Radioactive decay series — alpha, beta, gamma emission and transmutation rules

Question

What happens to the atomic number (Z) and mass number (A) during alpha, beta, and gamma decay? How do you trace a radioactive decay series to find the final product?

(NEET 2024 asked how many alpha and beta decays; JEE Main tests decay chain calculations)

Solution — Step by Step

An alpha particle is $^{4}_{2}\text{He}$ . When emitted:

Mass number (A) decreases by 4
Atomic number (Z) decreases by 2

^{A}_{Z}\text{X} \rightarrow ^{A-4}_{Z-2}\text{Y} + ^{4}_{2}\text{He}

The parent moves 2 places back (left) in the periodic table.

A neutron inside the nucleus converts to a proton, emitting an electron ( $\beta^-$ ) and an antineutrino.

Mass number (A) stays the same
Atomic number (Z) increases by 1

^{A}_{Z}\text{X} \rightarrow ^{A}_{Z+1}\text{Y} + ^{0}_{-1}e + \bar{\nu}

The parent moves 1 place forward (right) in the periodic table.

Gamma emission is the release of energy from an excited nucleus. The nucleus drops to a lower energy state by emitting a high-energy photon.

A stays the same, Z stays the same
Only the energy state of the nucleus changes

Given parent $^{A_1}_{Z_1}\text{X}$ and daughter $^{A_2}_{Z_2}\text{Y}$ :

Number of alpha decays: $n_\alpha = \frac{A_1 - A_2}{4}$

Number of beta decays: $n_\beta = 2n_\alpha - (Z_1 - Z_2)$

(Each alpha reduces Z by 2, each beta increases Z by 1)

flowchart TD
    A["Parent nucleus<br/>mass A, charge Z"] --> B{Type of decay?}
    B -->|Alpha| C["A → A−4<br/>Z → Z−2<br/>Emits He-4"]
    B -->|Beta minus| D["A → A<br/>Z → Z+1<br/>Emits electron"]
    B -->|Gamma| E["A → A<br/>Z → Z<br/>Emits photon only"]
    C --> F["Daughter product"]
    D --> F
    E --> F

Why This Works

Radioactive decay happens because the nucleus is unstable — it has too many protons, too many neutrons, or too much energy. Alpha decay reduces both, making heavy nuclei lighter. Beta decay converts excess neutrons to protons (or vice versa), adjusting the neutron-to-proton ratio. Gamma decay releases surplus energy without changing composition.

The formulas for counting decays work because alpha decay is the only process that changes A (by 4 each time), so dividing the total A change by 4 gives the number of alpha decays exactly. Beta decays then account for any Z discrepancy.

Alternative Method

For quick calculations: $^{238}_{92}\text{U} \rightarrow ^{206}_{82}\text{Pb}$ . Change in A = 238 − 206 = 32, so $n_\alpha = 32/4 = 8$ . Eight alphas would reduce Z by 16, from 92 to 76. But the final Z is 82, so we need $82 - 76 = 6$ beta decays to increase Z back up. Answer: 8 alpha and 6 beta decays.

Common Mistake

Students forget that gamma decay does NOT change A or Z and mistakenly count gamma emissions when calculating the number of alpha and beta decays. Gamma rays are just energy — they do not change the identity of the nucleus. When a question asks “how many alpha and beta particles are emitted,” ignore gamma completely. Only use the A and Z changes of the parent and final daughter.

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

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