Question

The decay constant of ²³⁸U is 4.9 × 10⁻¹⁸ S⁻¹. (a) What is the average-life of ²³⁸U? (b) What is the half-life of ²³⁸U? (c) By what factor does the activity of a ²³⁸U sample decrease in 9 × 10⁹ years?

Answer 1

Given:
Decay constant, $\lambda$ = 4.9 × 10⁻¹⁸ s⁻¹

(a) Average life of uranium $\left(\tau \right)$ is given by
$$\begin{gathered} \tau =\frac{1}{\lambda } \\ =\frac{1}{4.9\times {10}^{-18}} \\ =\frac{1}{4.9}\times {10}^{18}\mathrm{s} \\ =\frac{{10}^{16}}{4.9\times 365\times 24\times 36}\mathrm{years} \\ =\frac{{10}^{16}}{4.9\times 365\times 24\times 36}\mathrm{years} \\ =6.47\times {10}^{-7}\times {10}^{16}\mathrm{years} \\ =6.47\times {10}^9\mathrm{years} \end{gathered}$$

(b) Half-life of uranium $\left(T_{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\right)$ is given by
$$\begin{gathered} T_{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}=\frac{0.693}{\lambda }=\frac{0.693}{4.9\times {10}^{-18}} \\ =\frac{0.693}{4.9}\times {10}^{18}\mathrm{s} \\ =0.1414\times {10}^{18}\mathrm{s} \\ =\frac{0.1414\times {10}^{18}}{365\times 24\times 3600} \\ =\frac{1414\times {10}^{12}}{365\times 24\times 36} \\ =4.48\times {10}^{-3}\times {10}^{12} \\ =4.5\times {10}^9\mathrm{years} \end{gathered}$$

(c) Time, t = 9 × 10⁹ years
Activity (A) of the sample, at any time t, is given by
$A=\frac{A_0}{{\displaystyle 2^{\frac{t}{T_{1/2}}}}}$
Here, A₀ = Activity of the sample at t = 0
$\therefore \frac{A_0}{A}=2^{\frac{9\times {10}^9}{4.5\times {10}^9}}=2^2=4$