Question

You have a radioactive sample which has an activity of 3.89E5 Bq at t = 0 s. Later on at t = 1540 s you measure the activity again and find it is 3.15E5 Bq.

What is the half-life of the sample? (to 3 s.f and in s)

Answer 1

Given data

Initial activity of sample is $R_0=3.89*10^5 Bq$

Final activity of sample is $R=3.15*10^5 Bq$

The activity of sample present after time t is

$R=R_0e^{-\frac{ln2 t}{T_{1/2}}}$

$3.15*10^5=(3.89*10^5)e^{-\frac{ln2 (1540)}{T_{1/2}}}$

$T_{1/2}=5058.8 s= 5060 s$

So the half life of sample is $T_{1/2}=5060 s$ .