Question

${}_{11}^{23}Na$ is more stable isotope of $Na$. Find out the process by which ${}_{11}^{24}Na$ can undergo radioactive decay:

• A. ${\beta }^{-}$ emission
• B. ${\beta }^{+}$ emission
• C. $\alpha$ emission
• D. $K$ electron capture

Answer 1

Answer: (A)

${\beta }^{-}$ emission

Solution:

In $N{a}_{11}^{24}$
No. of protons, $p=11$

No. of neutrons, $n=24-11=13$

Thus n/p = $13/11=1.182$

Similarly, for $N{a}^{23}$

No. of neutrons, $=23-11=12$

No. of protons, $p=11$

Thus, $n/p=12/11=1.091$

$N{a}^{23}$ is the most stable isotope of sodium. The $n/p$ ratio of $Na-24$ is more than its stable isotope $Na-23$ . Thus, during radioactive decay of $Na-24$ $n/p$ ratio will decrease and $n/p$ ratio decreases only during ${\beta }^{-}decay$

Hence option A is correct