Nuclear Chemistry - Result Question 3
####3. ${ }^{23} Na$ is the more stable isotope of $Na$. Find out the process by which ${ } _{11}^{24} Na$ can undergo radioactive decay.
(2003, 1M)
(a) $\beta^{-}$-emission
(b) $\alpha$-emission
(c) $\beta^{+}$-emission
(d) $K$-electron capture
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Answer:
Correct Answer: 3. (a)
Solution:
- In stable isotope of $Na$, there are 11 protons and 12 neutrons. In the given radioactive isotope of sodium $\left(Na^{24}\right)$, there are 13 neutrons, one neutron more than that required for stability. A neutron rich isotope always decay by $\beta$-emission as
$$ { } _0 n^{1} \longrightarrow{ } _{-1} \beta^{0}+{ } _1 H^{1} $$
