Question

A radioactive nucleus `X `decay to a nucleus `Y` with a decay with a decay Concept `lambda _(x) = 0.1s^(-1) , gamma ` further decay to a stable nucleus Z with a decay constant `lambda_(y) = 1//30 s^(-1)` initialy, there are only X nuclei and their number is `N_(0) = 10^(20)` . Set up the rate equations for the population of `X , Y and Z` The population of `Y` nucleus as a function of time is given by `N_(y) (1) = N_(0) lambda_(x)l(lambda_(x) - lambda_(y))( (exp(- lambda_(y)t))` Find the time at which `N_(y)` is maximum and determine the populations `X and Z` at that instant.

Answer 1

Correct Answer - [(i) `(dN_(x))/(dt) = - lambda_(x) N_(x) (dN_(y))/(dt) = lambda_(x) N_(x) - lambda_(y) N_(y), (dN_(z))/(dt) = lambda_(y) N_(y)` (ii) 16.48 s, (iii) `N_(x) = 1.92 xx 10^(19), N_(y) = 5.76 xx 10^(19), N_(c) = 2.32 xx 10^(19)`]