(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 ?

Explanation

(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.