In the table given below we have listed values of `R(MB_(q)) " and in " ((R)/(R_(0)))`
(i) When we plot the graph of R versus r. we obtain an expoential curve as shown.
from the graph we can say that activity R reduces to `50% " in " t =OB = 40` min
`"So " t_(1//2) =40 " min"`
(ii) The adjacent figure shows the graph of in `(R//R_(0))` versus t
Slope of this graph `=-lambda`
`"from the graph"" "lambda=- ((-4.16 -3.11)/(1)) rArr =1.05h^(-1)`
`"Half-life"" "T_(1//2) =(0.693)/(lambda) =(0.0=693)/(1.05)=0.66h`
`=39.6 " min" = 40 " min"`