we know that `N = N_(0)e^(-lambda t)`
:. (dN)/(dt) = N_(0)e^(-u) (-lambda) = -N_(0)e^(-11)`
Taking log on both sides
`(dN)/(dt) = log_(e) (-N_(0) lambda)) - lambda t `
Comparting it with the graph line
we get `lambda= (1)/(2)gamma t^(-1) [(AC)/(BC) = (1)/(2))]`
`:. T_(1//2) = ((0.693)/(lambda) = 0.693 xx 2 = 1.386 year`
Now` (N)/(N_(0)) = ((1)/(2))^(1)/(T_(1//2)) rArr (1)/(p)= ((1)/(2)) ^(4.13)/*(1.386)) = (1)/(8)`