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

Options

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

Answer

A

Explanation

Isotopic `_(11)Na^(24)` is less stable than `._(11)Na^(23)` because it shows radioactive decay Less stability of `Na^(24) w.r.t Na^(23)` also based upon `13//11(n/p)` ratio. Higher the value higher will be unstability. So it is disintegrated to attain stability).
`underset(Less stabl e)(._(11)Na^(24)) rarr underset(Stabl e)(._(11)Na^(23))+ underset("Neutron")(._(0)n^(1)`
This neutron on decomposition gives proton and `beta-` particle`(._(-1)e^(0))`
`._(0)n^(1) rarr underset(Prot on)(._(1)H^(1)) ` or `._(1)P^(1)+underset(beta-partic l e)(._(-1)e^(0)`
Hence, isotopic sodium is changed into sodium by means of emission of `beta-` particle and a proton `i.e., ` by `beta-` emission.