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(D) When an element $A$ emits an $\alpha$-particle $(^4_2He)$,its mass number decreases by $4$ and its atomic number decreases by $2$.Let $A$ be repre

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Isodiasphere

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(D) When an element $A$ emits an $\alpha$-particle $(^4_2He)$,its mass number decreases by $4$ and its atomic number decreases by $2$.Let $A$ be represented as $^Z_A X^M$,where $M$ is the mass number and $Z$ is the atomic number.After $\alpha$-emission,the new element $B$ is $^Z_{A-2} Y^{M-4}$.Since the mass number changes by $4$ and the atomic number changes by $2$,the difference between the number of neutrons $(N = M - Z)$ for $A$ and $B$ is:$N_A = M - Z$$N_B = (M - 4) - (Z - 2) = M - Z - 2$The difference in the number of neutrons is $(N_A - Z_A) - (N_B - Z_B) = 0$,which defines isodiaspheres (elements having the same $(N - Z)$ value).

An element $A$ emits an $\alpha$-particle and forms $B$. $A$ and $B$ are

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