Question

A nucleus with atomic number Z and neutron number N undergoes two decay processes. The result is a nucleus with atomic number Z-3 and neutron number N-1. Which decay processes took place?

• A. Two ${\beta }^{-}$ decays
• B. Two ${\beta }^{+}$ decays
• C. An $\alpha -decay$ and a ${\beta }^{-}$ decay
• D. An $\alpha -decay$ and a ${\beta }^{+}$ decay

Answer 1

The correct option is D An $\alpha -decay$ and a ${\beta }^{+}$ decay

$\alpha \to 2n+2p$
$\beta +p\to n$
neutron no.
decrease of 2 by $\alpha$
decrease of 1 by $\beta$
Atomic No.
decrease of 2 by $\alpha$
increase of 1 by $\beta$
.
Let no. of $alpha$ decay be $n_1$
Let no. of $\beta$ decay be $n_2$
then
$Z-2n_1-n_2=Z-3$
and
$N-2n_1+n_2=Z-1$
Solving
$n_1=1$
$n_2=1$