Answer
a. $ 4.9\times10^{-18}s^{-1}$
b. $ \approx 6.0 hr$
Work Step by Step
a. The decay constant and the half-life are related by equation 30–6.
$$T_{1/2}=\frac{ln\;2}{\lambda}$$
$$\lambda =\frac{ln\;2}{ T_{1/2}}=\frac{ln\;2}{4.5\times10^9years}$$
$$ =1.5\times10^{-10}yr^{-1}=4.9\times10^{-18}s^{-1}$$
b. The decay constant and the half-life are related by equation 30–6.
$$T_{1/2}=\frac{ln\;2}{\lambda}$$
$$=\frac{ln\;2}{3.2\times10^{-5}s^{-1}}$$
$$ =2.2\times10^{4}s\approx 6.0 hr$$
