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

Question

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

Solution — Step by Step

An alpha particle is a helium-4 nucleus ( $^{4}_{2}\text{He}$ ). When emitted:

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

A decreases by 4, Z decreases by 2. The element moves two places to the left in the periodic table.

Example: $^{238}_{92}\text{U} \to ^{234}_{90}\text{Th} + ^{4}_{2}\text{He}$

In beta-minus ( $\beta^-$ ) decay, a neutron converts to a proton, emitting an electron and an antineutrino:

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

A stays the same, Z increases by 1. The element moves one place to the right in the periodic table.

In beta-plus ( $\beta^+$ ) decay (positron emission), a proton converts to a neutron: $Z$ decreases by 1.

Gamma rays are high-energy photons emitted when an excited nucleus drops to a lower energy state:

^{A}_{Z}\text{X}^* \to ^{A}_{Z}\text{X} + \gamma

No change in A or Z. Gamma emission often follows alpha or beta decay — the daughter nucleus is produced in an excited state and emits gamma to reach the ground state.

graph TD
    A["Parent nucleus (A, Z)"] --> B{Type of decay?}
    B -->|Alpha| C["Daughter: A-4, Z-2"]
    B -->|Beta minus| D["Daughter: A, Z+1"]
    B -->|Beta plus| E["Daughter: A, Z-1"]
    B -->|Gamma| F["Same: A, Z (deexcitation)"]
    C --> G["Lost: He-4 nucleus"]
    D --> H["Lost: electron + antineutrino"]
    E --> I["Lost: positron + neutrino"]

Why This Works

Each decay type conserves charge and mass number (baryon number).

For decay series problems (like the uranium-238 series), track A and Z through successive decays:

Decay Type	Change in A	Change in Z
Alpha ( $\alpha$ )	$-4$	$-2$
Beta-minus ( $\beta^-$ )	$0$	$+1$
Beta-plus ( $\beta^+$ )	$0$	$-1$
Gamma ( $\gamma$ )	$0$	$0$

To find the number of alpha and beta decays in a series from $^{A_1}_{Z_1}\text{X}$ to $^{A_2}_{Z_2}\text{Y}$ :

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

Then: $n_\beta = 2n_\alpha - (Z_1 - Z_2)$

Alternative Method

For the common exam question “How many alpha and beta particles are emitted when $^{238}_{92}\text{U}$ decays to $^{206}_{82}\text{Pb}$ ?”:

$n_\alpha = \frac{238 - 206}{4} = 8$ alpha decays

Change in Z from alpha alone: $8 \times 2 = 16$ (would give $Z = 92 - 16 = 76$ )

But final Z = 82, so we need $82 - 76 = 6$ beta-minus decays to increase Z back up.

Answer: 8 alpha and 6 beta particles. This is one of the most frequently asked numericals in CBSE boards and NEET.

Common Mistake

Students often forget that gamma emission does NOT change A or Z, so they include gamma in the count of “particles emitted.” When a question asks “how many alpha and beta particles,” gamma is irrelevant to the A and Z calculation. Also, some students count beta-plus and beta-minus as the same thing — they are not. Beta-minus increases Z by 1, beta-plus decreases Z by 1. In natural radioactivity, beta-minus is far more common.

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

Try These Next