Question

Geologists use the decay of potassium-40 in volcanic rocks to determine their age. Potassium-40 has a half-life of 1.26 109 years, so it can be used to date very old rocks. If a sample of rock 3.15 108 years old contains 2.73 10-7 g of potassium-40 today, how much potassium-40 was originally present in the rock

Answer 1

Answer:

$m_{o} \approx 3.247\times 10^{-7}\,g$

Explanation:

The mass of radioactive isotope at a certain time is given by the following expression:

$m(t) = m_{o}\cdot e^{-\frac{t}{\tau} }$

Time constant can be calculed in terms of half-life:

$\tau = \frac{t_{1/2}}{\ln 2}$

$\tau = \frac{1.26\times 10^{9}\,yr}{\ln 2}$

$\tau = 1.818\times 10^{9}\,yr$

The initial mass is:

$m_{o} = \frac{2.73\times 10^{-7}\,g}{e^{-\frac{3.15\times 10^{8}\,yr}{1.818\times 10^{9}\,yr} }}$

$m_{o} \approx 3.247\times 10^{-7}\,g$