A radioactive nucleus $\frac{A}{Z}$ X undergoes spontaneous decay in the sequence

$\frac{A}{Z}X{\to }_{Z-1}B{\to }_{Z-3}B{\to }_{Z-2}D$ , where Z is the atomic number of element X. The possible decay particles in the sequence are :

The correct option is β⁺, α, β⁻.

To determine the decay particles in the sequence, we need to trace the change in the atomic number ($Z$) and mass number ($A$) during each decay step.

Let the initial radioactive nucleus be represented as:
$X_Z^A$
Here, the atomic number (number of protons) is $Z$.

Step 1: First Decay ($X_Z^A\to B_{Z-1}$)
The atomic number decreases from $Z$ to $Z-1$.
A decrease in the atomic number by 1 without a change in the mass number indicates a positron emission or beta-plus (${\beta }^{+}$) decay, where a proton converts into a neutron:
$X_Z^A\to B_{Z-1}^A+e^{+}+\nu$
Thus, the first decay particle is a β⁺ particle.

Step 2: Second Decay ($B_{Z-1}\to B_{Z-3}$)
The atomic number decreases from $Z-1$ to $Z-3$.
The net change in the atomic number is:
$(Z-3)-(Z-1)=-2$
A decrease in the atomic number by 2 is characteristic of an alpha ($\alpha$) decay, which emits a helium nucleus (${\text{He}}_2^4$):
$B_{Z-1}^A\to B_{Z-3}^{A-4}+{\text{He}}_2^4$
Thus, the second decay particle is an α particle.

Step 3: Third Decay ($B_{Z-3}\to D_{Z-2}$)
The atomic number increases from $Z-3$ to $Z-2$.
The net change in the atomic number is:
$(Z-2)-(Z-3)=+1$
An increase in the atomic number by 1 without a change in the mass number indicates a beta-minus (${\beta }^{-}$) decay, where a neutron converts into a proton and emits an electron:
$B_{Z-3}^{A-4}\to D_{Z-2}^{A-4}+e^{-}+\overline{\nu }$
Thus, the third decay particle is a β⁻ particle.

Combining the steps in sequence, the decay particles are:
β⁺, α, β⁻