An element A decays into C in two steps: A to | vedclass

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(A) Let the atomic number of $A$ be $Z$ and its mass number be $M$.In the first step: $A 	o B + _{2}^{4}He$.The atomic number of $B$ becomes $Z - 2$

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$A$ and $C$ are isotopes.

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(A) Let the atomic number of $A$ be $Z$ and its mass number be $M$.In the first step: $A 	o B + _{2}^{4}He$.The atomic number of $B$ becomes $Z - 2$ and the mass number becomes $M - 4$.In the second step: $B 	o C + 2e^{-}$.Since each beta decay $(e^{-})$ increases the atomic number by $1$ without changing the mass number,the emission of $2e^{-}$ increases the atomic number of $B$ by $2$.Therefore,the atomic number of $C = (Z - 2) + 2 = Z$.The mass number of $C$ remains $M - 4$.Since $A$ and $C$ have the same atomic number $Z$ but different mass numbers,they are isotopes.

An element $A$ decays into $C$ in two steps: $A \to B + _{2}^{4}He$ and $B \to C + 2e^{-}$. Then:

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