Question

A $0.50\text{g}$ sample of rock was found to have $2.5\times {10}^{-6}\text{mol}$ of ${}_{19}^{40}K(t_{1/2}=1.3\times {10}^9\text{years})$ and $7.5\times {10}^{-6}{\text{mol of}}_{20}^{40}Ca$. How old is the rock?

• A. $6.5\times {10}^8\text{years}$
• B. $1.3\times {10}^9\text{years}$
• C. $2.6\times {10}^9\text{years}$
• D. $5.2\times {10}^9\text{years}$

Answer 1

The correct option is C $2.6\times {10}^9\text{years}$

${}_{19}^{40}K\to {}_{20}^{40}Ca+{}_{-1}^{0}e$
t=0 $A_0$
t=t $A_0-x$ $x$
Given
$A_0-x$ = $2.5\times {10}^{-6}$
$x$ = $7.5\times {10}^{-6}$
therefore
$A_o$ = $10\times {10}^{-6}$
$\lambda =\frac{1}{t}\ln \frac{N_o}{N_t}$
$\frac{0.693}{1.3\times {10}^9}\times t$ = $2.303\times log\frac{10\times {10}^{-6}}{2.5\times {10}^{-6}}$
$t$ = $2.6\times {10}^{9}years$