`._(11)^(23)Na` is the most stable isotope of Na. Find the process by which `._(11)^(24) Na` can undergo radioactive decay :

Options

A. `beta^(-)`-emission
B. `alpha`-emission
C. `beta^(+)`-emission
D. K-electron capture

Answer

A

Explanation

Since `._(11)^(23)Na` is the stable isotope, in `._(11)Na^(24) ` n/p ratio is higher than required. It will decompose by `beta^(-)` emission , all the other three methods (`alpha` emission, `beta^(+)` emission, k-electron capture) increase n/p ratio.
`{:(._(11)Na^(24),rarr,._(12)Mg^(24)+._(-1)e^(0),),(("unstable"),,("stable"),),(,,,):}`