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

Na^(23) is more stable isotope of Na. Find out the process by which ._

1.	Na^(23) is more stable isotope of Na. Find out the process by which ._(11)Na^(24) can undergo radioactive decay.
Answer» `beta^(c-)-` emission `alph-`emission `beta^(o+)-`emission `K` electron capture. Solution :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, isotopicsodium is CHANGED into sodium by means of emission of `beta-` particle and a proton `i.e., ` by `beta-` emission.