Question

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

• A. $ \beta $ -rays
• B. $ \alpha $ -rays
• C. $ \beta ^{+} $ -rays
• D. $ \gamma $ -rays

Answer 1

Answer: (A)

$\underset{\text{less stable}}{_{11}Na^{24}} \to \,\underset{\text{Stable}}{_{11}Na^{23}} + \underset{\text{neutron}}{_{0}n^{1}}$
The neutron on decomposition gives a prof
and a $ \beta $ -particle $(_{-1}e^{0})\, _0n^1 \to \underset{\text{proton}}{_1H^1} + \underset{\beta-\text{particle}}{_{-1}e^0} $.
Hence, $ _{11}Na^{24} $ gets changed into $ _{11}Na^{23} $ means of $ \beta $ -emission.