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

The rock is 3.75 billion years ago.

Step-by-step 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* 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* e^{-5.5452* 10^(-10)* t}`

`80e^{-5.5452* \:10^(-10)x}=10`

`\frac{80e^{-5.5452* \:10^(-10)x}}{80}=(10)/(80)`

`\ln \left(e^{-5.5452* \:10^(-10)x}\right)=\ln \left((1)/(8)\right)`

`x=(10^(10)* \:3\ln \left(2\right))/(5.5452)=3.75* 10^9`

So,

The rock is 3.75 billion years ago.