Question

The decay constant of ${}^{238}\text{U}$ is $4.9\times {10}^{-18}{S}^{-1}$ (a) What is the average - life of ${}^{238}\text{U}$ ? (b) What is the half-life of ${}^{238}\text{U}$ ? (c) By what factor does the activity of a ${}^{238}\text{U}$ sample decrease in $9\times {10}^9$ years ?

Answer 1

$\lambda =4.9\times {10}^{-18}{s}^{-1}$

(a) $\text{Average life of}$

${}^{238}\text{U}=\frac{1}{\lambda }=\frac{1}{4.9\times {10}^{-18}}$

=$\frac{1}{4.9}\times {10}^{-18}\text{sec}$

=$\frac{{10}^{18}}{4.9\times 365\times 24\times 3600}\text{years}$

=$\frac{{10}^{16}}{4.9\times 365\times 24\times 36}\text{years}$

=$6.47\times {10}^{-7}\times {10}^{18}\text{years}$

=$6.47\times {10}^7\text{years}$

(b)$\text{Half of Uranium}$

$=\frac{0.693}{\lambda }=\frac{0.693}{4.9\times {10}^{-18}}$

= $\frac{0.693}{4.9}\times {10}^{18}\text{sec}$

=$0.1414\times {10}^{18}\text{sec}$

=$\frac{0.1414\times {10}^{18}}{365\times 24\times 3600}$

=$\frac{1414\times {10}^{12}}{365\times 24\times 36}$

=$4.48\times {10}^{-3}\times {10}^{12}$

=$4.5\times {10}^9\text{years.}$

(c)$\text{Since A}=\frac{A_0}{\frac{t}{2\frac{t_1}{2}}}$

$\Rightarrow \frac{A_0}{A}=2^{\frac{9\times {10}^9}{4.5\times {10}^9}}=2^2=4$