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`Na^(23)` is more stable isotope of `Na`. Find out the process by which `._(11)Na^(24)` can undergo radioactive decay.A. `beta^(c-)-` emissionB. `alph-`emissionC. `beta^(o+)-`emissionD. `K` electron capture.

Answer» Correct Answer - A
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.


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