Answer to Question #123101 in Atomic and Nuclear Physics for serra

Question

The half life of Rn is 3.82 days. Determine the time it takes for 60% of the Rn to decay, the activity of 10 mg of Rn.

Answer 1

Given,

$$t_{1/2}=3.8\:days=3,30,048s,A_0=10^{-2}\:kg$$

Formula for decay

$$A(t)=A_0(1/2)^{t/t_{1/2}}$$

when, $60\%$ decay occurs then $A(t)=40\%$ of $A_0$ ,thus

$$0.4A_0=A_0(1/2)^{t/330048}\\ \implies t= - 330048\ln(2/5)/\ln(2)\approx4.36\times 10^5s$$