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 answer is β⁺, α, β⁻.

Let us analyze the spontaneous radioactive decay sequence step-by-step to identify the decay particles emitted in each stage. Let the initial nucleus be represented as $X_Z^A$, where $A$ is the mass number and $Z$ is the atomic number.

Step 1: First Decay ($Z\to Z-1$)
The parent nucleus $X_Z^A$ decays into a nucleus with atomic number $Z-1$.
When a nucleus undergoes positron decay (${\beta }^{+}$ decay), a proton is converted into a neutron, releasing a positron ($e_{+1}^0$) and a neutrino. This decreases the atomic number by 1 while keeping the mass number constant:
$X_Z^A\to B_{Z-1}^A+{\beta }^{+}+\nu$
Thus, the first decay particle is a positron (β⁺).

Step 2: Second Decay ($Z-1\to Z-3$)
The nucleus $B_{Z-1}^A$ decays into a nucleus with atomic number $Z-3$. The change in atomic number is:
$\Delta Z=(Z-3)-(Z-1)=-2$
An alpha particle ($\alpha$, which is a helium nucleus ${\mathrm{He}}_2^4$) consists of 2 protons and 2 neutrons. The emission of an alpha particle decreases the atomic number of the parent nucleus by 2 and the mass number by 4:
$B_{Z-1}^A\to C_{Z-3}^{A-4}+\alpha$
Thus, the second decay particle is an alpha particle (α).

Step 3: Third Decay ($Z-3\to Z-2$)
Finally, the nucleus with atomic number $Z-3$ decays into a nucleus $D$ with atomic number $Z-2$. The change in atomic number is:
$\Delta Z=(Z-2)-(Z-3)=+1$
When a nucleus undergoes beta-minus decay (${\beta }^{-}$ decay), a neutron is converted into a proton, releasing an electron ($e_{-1}^0$) and an antineutrino. This increases the atomic number by 1:
$C_{Z-3}^{A-4}\to D_{Z-2}^{A-4}+{\beta }^{-}+\overline{\nu }$
Thus, the third decay particle is a beta-minus particle (β⁻).

Combining the steps, the sequence of emitted particles is: β⁺, α, β⁻.