A radioactive nucleus ${}_Z^AX$ undergoes spontaneous decay in the sequence ${}_Z^AX \to {}_{Z – 1}B \to {}_{Z – 3}C \to {}_{Z – 2}D$ 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) $\beta^-,\alpha,\beta^+$

Solution

We have the following possible nuclear reactions:

${}_Z^AX \to {}_{Z – 1}^AB + {}_1^0{\beta ^ + }$
${}_{Z – 1}^AB \to {}_{Z – 3}^{A – 4}C + {}_2^4H{e^{2 + }}$
${}_{Z – 3}^{A – 4}C \to {}_{Z – 2}^{A – 4}D + {}_{ – 1}^0{\beta ^ – }$

Answer: (3)