A radioactive nucleus Z A X undergoes spontaneous decay in the sequence Z A X → Z − 1 □ B → Z − 3 □ C → Z − 2 □ D where Z is the atomic number of element. The possible decay particles in the

Question

A radioactive nucleus Z A ​ X undergoes spontaneous decay in the sequence Z A ​ X → Z − 1 □ ​ B → Z − 3 □ ​ C → Z − 2 □ ​ D where Z is the atomic number of element. The possible decay particles in the sequence are

TestHub · Accepted Answer

In the first step, the atomic number changes from $Z$ to $Z - 1$ . A decrease of $1$ in the atomic number indicates $β^{+}$ decay (positron emission).

In the second step, the atomic number changes from $Z - 1$ to $Z - 3$ . A decrease of $2$ in the atomic number indicates $α$ decay (emission of a helium nucleus).

In the third step, the atomic number changes from $Z - 3$ to $Z - 2$ . An increase of $1$ in the atomic number indicates $β^{-}$ decay (electron emission).

Thus, the sequence of emitted particles is $β^{+}$ , $α$ , $β^{-}$ .

Answer: $β^{+}, α, β$

Physics - Modern Physics Question with Solution | TestHub

Options:

Doubts & Discussion