Answer
$10 $
Work Step by Step
$\text { Consider the process }^{239} \mathrm{U}+\mathrm{n} \rightarrow^{140} \mathrm{Ce}+^{99} \mathrm{Ru}+\mathrm{Ne} . \text { We have } $
$$Z_{f}-Z_{i}=Z_{\mathrm{Ce}}+Z_{\mathrm{Ru}}-Z_{\mathrm{U}}=58+44-92=10 . $$
Thus number of $(beta-decay)$ events is $10 . $
