Question: 14. A radioactive nucleus {}_{Z}^{A}X undergoes spontaneous decay in the sequence
{}_{Z}^{A}X \rightarrow Z_{-1}B \rightarrow Z_{-3}C \rightarrow Z_{-2}D, where Z is the atomic number of element X . The possible decay particles in the sequence are :
(1) \alpha, \beta^{-}, \beta^{+}
(2) \alpha, \beta^{+}, \beta^{-}
(3) \beta^{+}, \alpha, \beta^{-}
(4) \beta^{-}, \alpha, \beta^{+}
Answer: Option (3)
Explanation:
In radioactive decay, different particles cause characteristic changes in the atomic number.
In the first step, the atomic number changes from Z to Z-1.
This decrease of 1 in atomic number occurs in \beta^{+} decay.
In the second step, the atomic number changes from Z-1 to Z-3,
which is a decrease of 2. This change is characteristic of \alpha decay.
In the third step, the atomic number changes from Z-3 to Z-2,
which is an increase of 1. This increase in atomic number occurs in \beta^{-} decay.
Thus, the sequence of decay particles is \beta^{+}, \alpha, \beta^{-}.
Therefore, the correct answer is Option (3).