Answer to Question #118607 in Atomic and Nuclear Physics for Bob

Question

What is the activity of a 0.05 mol sample of barium-137? (to 2 s.f and in Bq)

Answer 1

Number of nuclei in  0.05 mol sample of barium-137 is

$N=0.05 mol(\frac{6.022*10^{23}}{1 mol})=0.3011*10^{23}$

Half life of barium-137 is $T_{1/2}=153 s$

The activity of sample of barium-137 is

$R=\lambda N=(\frac{ln2}{T_{1/2}})N$

$=(\frac{ln2}{T_{1/2}})N=(\frac{ln2}{153 s})(0.3011*10^{23})=1.4*10^{20} Bq$