Nuclear Chemistry Question 1
Question: $ _{11}^{23}Na $ is the more stable isotope of Na. Find out the process by which $ _{11}^{24}Na $ can undergo radioactive decay
[IIT Screening 2003]
Options:
A) $ {{\beta }^{-}} $ emission
B) $ \alpha $ emission
C) $ {{\beta }^{+}} $ emission
D) K electron capture
Show Answer
Answer:
Correct Answer: A
Solution:
$ _{11}^{23}Na\to \frac{n}{p}ratio=12/11 $
$ _{11}^{24}Na\to \frac{n}{p}ratio=23/11 $ so decrease in $ \frac{n}{p} $ ratio gives out $ \beta $ -particle $ n\to p+e({{\beta }^{-}}) $ .
BETA
