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(A) For a nucleus to be stable,it must have an appropriate ratio of neutrons $(N)$ to protons $(Z)$.If the $N/Z$ ratio is higher than the stability re

Vedclass · Accepted Answer

It emits $\beta^{-}$

Vedclass · Answer

(A) For a nucleus to be stable,it must have an appropriate ratio of neutrons $(N)$ to protons $(Z)$.If the $N/Z$ ratio is higher than the stability requirement,the nucleus has an excess of neutrons.To achieve stability,the nucleus converts a neutron into a proton,an electron,and an antineutrino through $\beta^{-}$ decay: $n ightarrow p + e^{-} + \bar{
u}_{e}$.This process increases the number of protons $(Z)$ and decreases the number of neutrons $(N)$,thereby lowering the $N/Z$ ratio towards the stability line.Therefore,the correct option is $A$.

If a heavy nucleus has an $N/Z$ ratio higher than that required for stability,then:

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