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_(Z) = 2.32 xx 10^(19)`
(i) Let at time `t oo t`, number of nuclel of Y and Z are `N_(y)` and `N_(z)`. Then Rate equations of the population of X, Y and Z are
`((dN_(x))/(dt)) = - lambda_(x) N_(x)` ....(i)
`((dN_(y))/(dt)) = lambda_(x) N_(x) - lambda_(y) N_(y)` .....(ii)
`rArr ((dN_(z))/(dt)) = lambda_(y) N_(y)` ....(iii)
(ii) Given `N_(y) (t) = (N_(0)lambda_(x))/(lambda_(x) - lambda_(y)) [e^(-lambda_(y)t) - e^(- lambda_(x)t)]`
For `N_(y)` to be maximum `(dN_(y)(t))/(dt) = 0`
i.e. `lambda_(x) N_(x) = lambda_(y) N_(y)`....(v) (from equation (ii))
`rArr lambda_(x) N_(0)e^(-lambda_(x)t) = lambda_(y) (N_(0) lambda_(x))/(lambda_(x) - lambda_(y)) [e^(-lambda_(y)t) - e^(- lambda_(x)t)]`
`rArr (lambda_(x) - lambda_(y))/(lambda_(y)) = (e^(-lambda_(y)t))/(e^(- lambda_(x)t)) - 1 rArr (lambda_(x))/(lambda_(y)) = e^((lambda_(x) - lambda_(y))t)`
`rArr (lambda_(x) - lambda_(y)) t ln (e) = ln ((lambda_(x))/(lambda_(y))) rArr t = (1)/(lambda_(x) - lambda_(y)) ln ((lambda_(x))/(lambda_(y)))`
Substituting the values of `lambda_(x)` and `lambda_(y)`, we have
`t = (1)/((0.1 - 1//30)) ln ((0.1)/(1//30)) = 15 ln (3) rArr t = 16.48 s`.
(iii) The population of X at this moment,
`N_(x) = N_(0) e^(-lambda X t) = (10^(20)) e^(- (0.1) (16.48)) rArr N_(x) = 1.92 xx 10^(19)`
`N_(y) = (N_(x) lambda_(x))/(lambda_(y))` [From equation (iv)]
`= (1.92 xx 10^(19)) ((0.1))/((1//30)) = 5.76 xx 10^(19)`
`N_(z) = N_(0) - N_(x) - N_(y)`
`= 10^(20) - 1.92 xx 10^(19) - 5.76 xx 10^(19)`
`rArr N_(2) = 2.32 xx 10^(19)`