The decay constant of ^ 238 U is 4.9 × 10^ -18 s^ -1 . By what...

The half-life of ^{238}\text{U}$is given by:<br><br>$T_{1/2} = \dfrac{\ln 2}{\lambda} = \dfrac{0.693}{4.9 \times 10^{-18}} \text{ s} \approx 1.414 \times 10^{17} \text{ s}$<br><br>Converting the half-life into years (using$1 \text{ year} \approx 3.14 \times 10^7 \text{ s}$):<br><br>$T_{1/2} = \dfrac{1.414 \times 10^{17}}{3.14 \times 10^7} \text{ years} \approx 4.5 \times 10^9 \text{ years}$<br><br>The given time is$t = 9 \times 10^9 \text{ years}$.<br><br>Number of half-lives,$n = \dfrac{t}{T_{1/2}} = \dfrac{9 \times 10^9}{4.5 \times 10^9} = 2$<br><br>The activity of the sample after$n$half-lives is$A = \dfrac{A_0}{2^n}$.<br><br>Substituting$n = 2$:<br><br>$A = \dfrac{A_0}{2^2} = \dfrac{A_0}{4}$<br><br>Therefore, the activity decreases by a factor of$4.