(a) Deduce the expression, N= N_(0) e^(-lambdat) for the law of radioactive decay. (b) (i) Write symbolically the process expressing the beta^(+) decay of " "_(11)^(22)Na. Also write the basic nuclear process underlying this decay. (ii) Is the nucleus formed in the decay of the nucleus " "_(11)^(22)Na, an isotope or isobar ?

(a) Deduce the expression, N= N_(0) e^(-lambdat) for the law of radioa

1.	(a) Deduce the expression, N= N_(0) e^(-lambdat) for the law of radioactive decay. (b) (i) Write symbolically the process expressing the beta^(+) decay of " "_(11)^(22)Na. Also write the basic nuclear process underlying this decay. (ii) Is the nucleus formed in the decay of the nucleus " "_(11)^(22)Na, an isotope or isobar ?
Answer» SOLUTION : (b) (i) The process of `beta^(+)` decay of `" "_(11)^(22)Na` is: `" "_(11)^(22)Na to " "_(10)^(22)Ne + UNDERSET(beta^(+)" particle")(" "_(+1)^(0)e) + underset("neutrino")(nu)` The BASIC nuclear process underlying this decay is : `" "_(+1)^(1)p to " "_(0)^(1)n+" "_(+1)^(0)e+nu` (ii) The nucleus `" "_(10)^(22)Ne` formed due to decay of `" "_(11)^(22)Na` is an isobar.

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