$t_{1/2}$ of $^{232}Th$ is $1.39 \times 10^{10} \ \text{years}$. Calculate the number of $\alpha$-particles emitted by $1.0 \ \text{g}$ of $^{232}Th$ per second.

(N/A) The decay constant $\lambda$ is given by $\lambda = \frac{0.693}{t_{1/2}}$.
First,convert $t_{1/2}$ to seconds: $t_{1/2} = 1.39 \times 10^{10} \ \text{years} \times 365 \times 24 \times 3600 \ \text{s/year} \approx 4.38 \times 10^{17} \ \text{s}$.
$\lambda = \frac{0.693}{4.38 \times 10^{17} \ \text{s}} \approx 1.58 \times 10^{-18} \ \text{s}^{-1}$.
The number of atoms $N$ in $1.0 \ \text{g}$ of $^{232}Th$ is $N = \frac{\text{mass}}{\text{molar mass}} \times N_A = \frac{1.0 \ \text{g}}{232 \ \text{g/mol}} \times 6.022 \times 10^{23} \ \text{atoms/mol} \approx 2.596 \times 10^{21} \ \text{atoms}$.
The rate of emission (activity) is $A = \lambda N = (1.58 \times 10^{-18} \ \text{s}^{-1}) \times (2.596 \times 10^{21} \ \text{atoms}) \approx 4.10 \times 10^{3} \ \text{particles/s}$.