Answer:
5.0 x 10⁹ years.
Explanation:
k = ln(2)/(t1/2) = 0.693/(t1/2).
Where, k is the rate constant of the reaction.
t1/2 is the half-life of the reaction.
∴ k =0.693/(t1/2) = 0.693/(1.251 × 10⁹ years) = 5.54 x 10⁻¹⁰ year⁻¹.
kt = ln([A₀]/[A]),
where, k is the rate constant of the reaction (k = 5.54 x 10⁻¹⁰ year⁻¹).
t is the time of the reaction (t = ??? year).
[A₀] is the initial concentration of (K-40) ([A₀] = 100%).
[A] is the remaining concentration of (K-40) ([A] = 6.25%).
∴ (5.54 x 10⁻¹⁰ year⁻¹)(t) = ln((100%)/( 6.25%))
∴ (5.54 x 10⁻¹⁰ year⁻¹)(t) = 2.77.
∴ t = 2.77/(5.54 x 10⁻¹⁰ year⁻¹) = 5.0 x 10⁹ years.
Answer:
[tex]\boxed{4.99 \times10^{9} \text{ y}}[/tex]
Explanation:
The half-life of potassium-40 (1.248 ×10⁹ y) is the time it takes for half of it to decay.
After one half-life, half (50 %) of the original amount will remain.
After a second half-life, half of that amount (25 %) will remain, and so on.
We can construct a table as follows:
[tex]\begin{array}{cccc}\textbf{No. of} & &\textbf{Fraction} &\textbf{Percent}\\ \textbf{Half-lives} & \textbf{t/(10$^{9}$ y}) &\textbf{Remaining} &\textbf{Remaining}\\0 & 0 & 1 & 100\\\\1 & 1.248 & \dfrac{1}{2} & 50\\\\2 & 2.496 & \dfrac{1}{4} & 25\\\\3 & 3.744 & \dfrac{1}{8} & 12.5\\\\4 & 4.992 & \dfrac{1}{16} & 6.25\\\\5 & 6.240 & \dfrac{1}{32} & 3.125\\\\\end{array}[/tex]
We see that after four half-lives, ¹/₁₆ or 6.25 % of the original mass remains.
Conversely, if 6.25 % of the sample remains, the age of the sample must be four half-lives.
[tex]\text{Age of rock} = 4 \times 1.248 \times 10^{9}\text{ y}= \boxed{\mathbf{4.99 \times 10^{9}} \textbf{ y}}[/tex]
