Question
An element $$A$$ decays into element $$C$$ by a two step process
$$\eqalign{
& A \to B{ + _2}H{e^4} \cr
& B \to C + 2{e^ - } \cr} $$
then
A.
$$A$$ and $$C$$ are isotopes
B.
$$A$$ and $$C$$ are isobars
C.
$$A$$ and $$B$$ are isotopes
D.
$$A$$ and $$B$$ are isobars
Answer :
$$A$$ and $$C$$ are isotopes
Solution :
From equation (Ist) there is 1 $$ \propto $$-decay in which $$B$$ has atomic no. 2 less than $$A.$$ In IInd case there is 2-$$\beta $$-decay in which $$C$$ has atomic no. 2 greater than $$B,$$ Since $$A$$ and $$C$$ have same atomic no. so they are called isotopes.