Question

A certain preparation includes two beta-active components with different half-lifes. The measurment resulted in the following dependence of the natural logrithm of preparation acitivity on time `t` expressed in hours.
`{:(t,0,1,2,3,5,7,10,14,20),("In A",4.10,3.60,3.10,2.60,2.60,1.82,1.60,1.32,0.90):}`
Find the half-lifes of both componets and the ratio of radioactive nuclei of these components at the moment `t = 0`.

Answer 1

Suppose `N_(1),N_(2)` are the initial number of component nuclei whose decay constant are `lambda_(1),lambda_(2) (In("houe")^(-1))`
Then the activity at any instant is
`A=lambda_(1)N_(1)e^(-lambdat)+lambda_(2)N_(2)e^(-lambda_(2)t)`
The activity so defined is in units dis//hour. We assume that the `In A` given is of its natural logarithm. The daughter nuclei are assumed nonradioactive.
We see from the data that at large `t` the change in In A per hour of elapsed time is constant and equal to `-0.07`. Thus
`lambda_(2)= 0.07` per hour
We can then see that the best fit to data is obtained by
`A(t)= 51.1e^(-0.66t)+10.0e^(-0.07t)`
[ To get the fit we calculate `A(t)e^(0.07t)`. We see that it reaches the constant value `10.0` at `t= 7,10,14,20` very nearly. This fixes the second term. The first term is obtained by subtracting out the constant value `10.0` from each value of `A(t)e^(0.07t)` in the data for small `t`]
Thus we get `lambda_(1)= 0.66 per hour`
or,`{:(T_(1),=,1.05 hour,,),(T_(2),=,9.9 hours,,):}}` half-lives
Ratio `(N_(1))/(N_(2))= (51.1)/(10.0)xx(lambda_(2))/(lambda_(1))= 0.54`
The answer given in the book is misleading.