Question

Suppose you find a rock that contains 10 micrograms of radioactive potassium-40, which has a half-life of 1.25 billion years. By measuring the amount of its decay product (argon-40) present in the rock, you conclude that there must have been 80 micrograms of potassium-40 when the rock solidified. How old is the rock?

Answer 1

Answer:

The rock is 3.75 billion years ago.

Explanation:

Given that:

Half life = 1.25 billion years

1 billion years = 10⁹ years

So,

Half life = 1.25 × 10⁹ years

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

Where, k is rate constant

So,  

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

$k=\frac {ln\ 2}{1.25\times 10^9}\ year^{-1}$

The rate constant, k = 5.5452 × 10⁻¹⁰ years⁻¹

Initial amount [A₀] = 80 micrograms

Final amount $[A_t]$ = 10 micrograms

Using integrated rate law for first order kinetics as:

$[A_t]=[A_0]e^{-kt}$

So,  

$10=80\times e^{-5.5452\times 10^{-10}\times t}$

$80e^{-5.5452\times \:10^{-10}x}=10$

$\frac{80e^{-5.5452\times \:10^{-10}x}}{80}=\frac{10}{80}$

$\ln \left(e^{-5.5452\times \:10^{-10}x}\right)=\ln \left(\frac{1}{8}\right)$

$x=\frac{10^{10}\times \:3\ln \left(2\right)}{5.5452}=3.75\times 10^9$

So,

The rock is 3.75 billion years ago.