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^9) Y and Z` The population of `Y` nucleus as a function of time is givenby `N_(y) (1) = N_(0) lambda_(x)l(lambda_(x) - lambda_(y))) (exp(- lambda_(y)1)) Find the time at which `N_(y)` is maximum and determine the populastion `X and Y` at that instant.

Answer

A, B, C, D

Explanation

X overset(T_(1//2) = 10sec) underset(lambda_(x) = 0.1s^(-1))(rarr) Y overset(T_(1//2) = 30sec) underset(lambda_(y) = (1)/(30)s^(-1))(rarr)Z`
The rate of equation for the population of X ,Y and Z will be
`(dN_(x))/(dt) = - lambda_(x)N_(x) `….(i)
`(dN_(y))/(dt) = - lambda_(y)N_(y)+ lambda_(x)N_(x) `….(ii)
`(dN_(z))/(dt) = - lambda_(y)N_(y) `….(iii)
`rArr`On integration , we get
`N_(x_ = N_(0)e^(-lambda _(x)t)`....(iv)
Given
`N_(y) = (lambda_(x) N_(0))/(lambda_(x) - lambda_(y)) [ e^(- lambda_(y) t) - e^(lambda_(x)t)]`
To determine the maximum `N_(p)` we find
`(dN_(p))/(dt) = 0`
from(ii)
` - lambda_(y) N_(y) +lambda_(x) N_(x) = 0 `
`rArr ambda_(x) N_(x) = lambda_(y) N_(y)`...(v)
`rArr lambda_(x) (N_(0)e^(-lambda_(x) t) = lambda_(y)[(lambda_(x) N_(0))/(lambda_(x) - lambda_(y)) (e^(- lambda_(y)t - e^(lambda_(x) t)]`
`rArr (lambda_(x) - lambda_(y))/(lambda_(y))= (e^(- lambda_(x) t))/(- e^(-lambda_(x)t)) rArr (lambda_(x))/(lambda_(y)) = e^(lambda_(x) - lambda_(y)) t)`
`rArr log _(e) (lambda_(x))/(lambda_(y)) = (lambda_(x) - lambda_(y)) t`
`t = (log _(e)(lambda_(x) //lambda_(y))/(lambda_(x) - lambda_(y)) = log _(e)[0.1//((1)/(30))]/(0.1 - (1)/(30) = 15log _(e) 3 `
` :. N_(z) = N_(0)e^(-0.1(15 log_(e) 3) = N_(0) e^(log_(e) (T^(-15))`
` rArr N_(x) = N_(0) 3^(-15) = (10^(20))/(3sqrt(3))`
since `(dN_(y))/(dt) = 0 at t = 15 log_(e) , 3 , :. N_(p) =(lambda_(x)N_(x))/(lambda_(y)) = (10^(20))/(sqrt(3))`
` and N_(x) = N_(0) - N_(x) - N_(y)`
`= 10^(20) - ((10^(20))/(3 sqrt(3))) - (10^(20))/(sqrt(3)) = 10^(20) ((3 sqrt(3)-4)/(3sqrt(3)))`