Answer to Question #96351 in General Chemistry for Brayden Angelina Oblak

Question

a person of 60kg has 120g of potassium, of which 0.0123g is potassium-40. This results in an activity of 3.2x10^3 Bq. What is the decay constant

Answer 1

The activity of potassium is$=3.2\times10^3Bq$

As $\frac{dN}{dt}=kN$, where $k=$ decay constant

$3.2\times10^3=k\times0.0123$

Hence, $k=2.60\times10^5s^{-1}$