Consider, the radioactive disintegration ${ }_{82} A^{210} \rightarrow B \rightarrow C \rightarrow{ }_{82} D^{206}$, The sequence of emission can be

Given, $82^{A^{210}} \longrightarrow B \longrightarrow C \longrightarrow{ }_{82} D^{206}$ i.e. difference in mass no. between $A$ and $D$ is of 4 units i.e., mass no. is decreased by 4 units while atomic no. remains the same, i.e. 82 for $A$ and $D$. Also, (i) On emission of each $\beta$ -particle, mass no. remains the same but atomic no. is increased by one unit. (ii) On emission of each $\alpha$ -particle, mass no. is decreased by 4 units, while atomic no. is decreased by 2 units. Thus, on emission of $2-\beta$ and $1-\alpha$ particle, we get $_{82} D^{206}$ i.e. ${ }_{82} A^{210} \stackrel{-\beta}{\longrightarrow}{ }_{83} A^{210} \stackrel{-\beta}{\longrightarrow}{ }_{84} A^{210} \stackrel{-a}{\longrightarrow}{ }_{82} A^{206}$ Thus, option (d) is the correct answer.