Question

Potassium-40 (40K) decays to Argon-40 (40Ar) with a half-life of 1.27 billion years. If a lunar rock was collected, and it was found that the number of 40K was 12% of its original number, then how old is that rock?

Answer 1

The rock is 3.88 billion years old.

Step-by-step explanation:

Initial mass of the K-40 isotope = x

Final mass of the K-40 isotope = 12% of x = 0.12x

Half life of the K-40 =
`t_{(1)/(2)}` =1.27 billion years

Age of the sample = t

Formula used :

`N=N_o* e^(-\lambda t)\\\\\lambda =\frac{0.693}{t_{(1)/(2)}}`

where,

`N_o` = initial mass of isotope

N = mass of the parent isotope left after the time, (t)

`t_{(1)/(2)}` = half life of the isotope

`\lambda` = rate constant

`N=N_o* e^{-((0.693)/(t_(1/2)))* t}`

Now put all the given values in this formula, we get

`0.12x=x* e^{-(\frac{0.693}{1.27 \text{billion years}})* t}`

`\ln(0.12) * \frac{1.27 \text{billion years}}{0.693}=-t`

t = 3.88 billion years

The rock is 3.88 billion years old.