Question

Potassium-40 has a half-life of$\text{1.26}\times {\text{10}}^{\text{9}}$ yr, decaying to calcium-40 and argon-40 in a ratio of 8.54 to 1. If a rock sample contained no argon when it formed a solid but now contains one argon-40 atom for every potassium-40 atom, how old is the rock?

Short answer

Rock is $\text{4.28}\times {\text{10}}^{\text{9}}\text{y}$old.

Step-by-step solution

Given data and Concept

It’s giventhat8.54+1+1 = 10.54 atoms are the initial number of nuclei of K.

${\text{T}}_{\frac{\text{1}}{\text{2}}}$=${\text{1.26.10}}^{\text{9}}$y

In order to solve this problem, we will be applying an equation that determines the number of daughter nuclei as:

role="math" localid="1658425803971" $N=N_0{e}^{-\lambda t}$

Where N = number of daughter nuclei

${\text{N}}_0$= initial number of nuclei

$\lambda$= decay constant

t = time interval that shows how old is the rock

Where decay constant $\lambda$is given as:

$\lambda =\frac{In2}{T_{\frac{1}{2}}}$

Where: ${\text{T}}_{\frac{\text{1}}{\text{2}}}$= Half-life

Therefore, we can express time interval t as:

$t=\frac{2.36}{\lambda }$

And, when we put the relation that determines decay constant into this equation, we get:

$t=\frac{2.36T_{\frac{1}{2}}}{In2}$

Calculation

Now, we will put known values into the previous equation and calculate how old the rock as:

$\begin{array}{rcl}t& =& \frac{2.36\times 1.26\times {10}^{9}y}{In2}\\ & =& 4.28\times {10}^{9}y\\ & & \end{array}$

Therefore, the rock is $\text{4.28}\times {\text{10}}^{\text{9}}\text{y}$old.