Question Details

A radioactive nucleus $\frac{A}{Z}$ X undergoes spontaneous decay in the sequence $\frac{A}{Z} X \to_{Z - 1} B \to_{Z - 3} C \to_{Z - 2} D$ , where Z is the atomic number of element X. The possible decay particles in the sequence are :

Options

β⁺, α, β⁻

β⁻, α, β⁺

α, β⁻, β⁺

α, β⁺, β⁻

Show Answer

Correct Answer :

Option A

β⁺, α, β⁻

Solution :

The correct answer is β⁺, α, β⁻.

To determine the sequence of decay particles, let us analyze the changes in the atomic number (Z) and mass number (A) at each step of the radioactive decay sequence.
Let the initial radioactive nucleus be represented as:
${}_{Z}^{A}X$

Step 1: Decay from X to B
The transition is shown as:
${}_{Z}^{A}X \to {}_{Z - 1}^{A_{1}}B$
In this step, the atomic number decreases by 1 (from Z to Z - 1). This is characteristic of a positron decay (beta-plus decay, β⁺), where a proton converts into a neutron, emitting a positron and a neutrino:
${}_{Z}^{A}X \to {}_{Z - 1}^{A}B + {}_{+ 1}^{0}e$
Thus, the first decay is a β⁺ decay.

Step 2: Decay from B to C
The transition is shown as:
${}_{Z - 1}^{A}B \to {}_{Z - 3}^{A_{2}}C$
In this step, the atomic number decreases by 2 (from Z - 1 to Z - 3). A decrease of 2 in the atomic number is characteristic of an alpha decay (α), where a helium nucleus is emitted:
${}_{Z - 1}^{A}B \to {}_{Z - 3}^{A - 4}C + {}_{2}^{4}{He}$
Thus, the second decay is an α decay.

Step 3: Decay from C to D
The transition is shown as:
${}_{Z - 3}^{A - 4}C \to {}_{Z - 2}^{A_{3}}D$
In this step, the atomic number increases by 1 (from Z - 3 to Z - 2). An increase of 1 in the atomic number is characteristic of an electron emission (beta-minus decay, β⁻), where a neutron converts into a proton, emitting an electron and an antineutrino:
${}_{Z - 3}^{A - 4}C \to {}_{Z - 2}^{A - 4}D + {}_{- 1}^{0}e$
Thus, the third decay is a β⁻ decay.