Answer:
The rock is 3.88 billion years old.
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 =[tex]t_{\frac{1}{2}}[/tex] =1.27 billion years
Age of the sample = t
Formula used :
[tex]N=N_o\times e^{-\lambda t}\\\\\lambda =\frac{0.693}{t_{\frac{1}{2}}}[/tex]
where,
[tex]N_o[/tex] = initial mass of isotope
N = mass of the parent isotope left after the time, (t)
[tex]t_{\frac{1}{2}}[/tex] = half life of the isotope
[tex]\lambda[/tex] = rate constant
[tex]N=N_o\times e^{-(\frac{0.693}{t_{1/2}})\times t}[/tex]
Now put all the given values in this formula, we get
[tex]0.12x=x\times e^{-(\frac{0.693}{1.27 \text{billion years}})\times t}[/tex]
[tex]\ln(0.12) \times \frac{1.27 \text{billion years}}{0.693}=-t[/tex]
t = 3.88 billion years
The rock is 3.88 billion years old.
