Question

the concentration of the radio active isotope potassium-40 in a rock sample is found to be 6.25%. what is the age of the rock

Answer 1

`\boxed{4.99 *10^(9) \text{ y}}`

Step-by-step 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:

`\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 & (1)/(2) & 50\\\\2 & 2.496 & (1)/(4) & 25\\\\3 & 3.744 & (1)/(8) & 12.5\\\\4 & 4.992 & (1)/(16) & 6.25\\\\5 & 6.240 & (1)/(32) & 3.125\\\\\end{array}`

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.

`\text{Age of rock} = 4 * 1.248 * 10^(9)\text{ y}= \boxed{\mathbf{4.99 * 10^(9)} \textbf{ y}}`

Answer 2

5.0 x 10⁹ years.

Step-by-step explanation:

• It is known that the decay of a radioactive isotope isotope obeys first order kinetics.

• Half-life time is the time needed for the reactants to be in its half concentration.

• If reactant has initial concentration [A₀], after half-life time its concentration will be ([A₀]/2).

• Also, it is clear that in first order decay the half-life time is independent of the initial concentration.

• The half-life of K-40 = 1.251 × 10⁹ years.

• For, first order reactions:

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.

• Also, we have the integral law of first order reaction:

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.