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Q. Emission of $\beta$-rays in radioactive decay results in the change of

AIIMSAIIMS 1996Nuclei

Solution:

$\beta$ -decay is a process in which a neutron is converted into a proton and an electron or a proton into a neutron and a positron:
$n \rightarrow p+e^{-}+\bar{v}$
$p \rightarrow n+e^{+}+\bar{v}$
Now as the process of $\beta$ -decay is associated with the increase of atomic number of the parent nucleus, so the charge of the nucleus changes by keeping the mass number constant as we lose a neutron and gain a proton.