For $Ra^{226}$,the $t_{1/2}$ is $1620 \ years$. From $0.001 \ g$ of $Ra$,how many $\alpha$-particles are released per $min$?

(N/A) The decay constant $\lambda$ is given by $\lambda = \frac{0.693}{t_{1/2}}$.
Given $t_{1/2} = 1620 \ years = 1620 \times 365 \times 24 \times 60 \ min = 8.51472 \times 10^{8} \ min$.
$\lambda = \frac{0.693}{8.51472 \times 10^{8}} \ min^{-1} = 8.1389 \times 10^{-10} \ min^{-1}$.
The number of atoms $N$ in $0.001 \ g$ of $Ra^{226}$ is $N = \frac{0.001}{226} \times 6.022 \times 10^{23} = 2.6646 \times 10^{18} \ atoms$.
The rate of decay (activity) is $A = \lambda N = (8.1389 \times 10^{-10} \ min^{-1}) \times (2.6646 \times 10^{18}) = 2.1686 \times 10^{9} \ min^{-1} \approx 2.17 \times 10^{9} \ min^{-1}$.