Question

What is thr half-like of an isotope that decays to 12.5% of its original activity in 16.3 hours?

Answer 1

Answer – We are given, time t = 16.3 hours, No = 100 % , Nt = 100-12.5 = 87.5 %

First we need to calculate the decay constant

ln Nt/No = -k*t

ln 87.5 /100 = -k * 16.3 hrs

-0.133 = -k*16.3 hrs

So, k = 0.00819 hrs^-1

Now we know

Half-life t ½ = 0.693 /k

                    = 0.693 / 0.00819 hrs^-1

                    = 84.6 hours.