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:

  1. 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

$$ { } _0 n^{1} \longrightarrow{ } _{-1} \beta^{0}+{ } _1 H^{1} $$



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