Question 42.42: What is the activity of a 20 ng sample of ^92Kr, which has a......

Adapting Eq. 42-21, we have

N_{40}=\left(1.17 \times 10^{-4}\right) \frac{M_{ sam } N_{ A }}{M}       (42-21)

N_{ Kr }=\frac{M_{ sam }}{M_{ Kr }} N_A=\left(\frac{20 \times 10^{-9} \,g }{92\, g / mol }\right)\left(6.02 \times 10^{23} \,\text { atoms } / mol \right)=1.3 \times 10^{14} \text { atoms } .

Consequently, Eq. 42-20 leads to

R=\frac{N \ln 2}{T_{1 / 2}}=\frac{\left(1.3 \times 10^{14}\right) \ln 2}{1.84 \mathrm{~s}}=4.9 \times 10^{13} \mathrm{~Bq}

R=\frac{N_{40} \ln 2}{T_{1 / 2}}            (42-20)