Question

An igneous rock originally has 3 grams of uranium 238 in it. When dated the rock only contains 1.8 grams. What are the parent and daughter concentrations (in percentages) and what is its age?

Answer 1

Answer :

The parent and daughter concentrations (in percentages) is, 60 % and 40 % respectively.

The age of rock is $3.32\times 10^9\text{ years}$

Explanation :

First we have to calculate the parent and daughter concentrations (in percentages).

$\text{Parent concentrations}=\frac{1.8g}{3g}\times 100=60\%$

and,

$\text{Daughter concentrations}=\frac{(3-1.8)g}{3g}\times 100=40\%$

As we know that, the half-life of uranium-238 = $4.5\times 10^9$ years

Now we have to calculate the rate constant, we use the formula :

$k=\frac{0.693}{t_{1/2}}$

$k=\frac{0.693}{4.5\times 10^9\text{ years}}$

$k=1.54\times 10^{-10}\text{ years}^{-1}$

Now we have to calculate the time passed.

Expression for rate law for first order kinetics is given by:

$t=\frac{2.303}{k}\log\frac{a}{a-x}$

where,

k = rate constant  = $1.54\times 10^{-10}\text{ years}^{-1}$

t = time passed by the sample  = ?

a = initial amount of the reactant  = 3 g

a - x = amount left after decay process = 1.8 g

Now put all the given values in above equation, we get

$t=\frac{2.303}{1.54\times 10^{-10}}\log\frac{3}{1.8}$

$t=3.32\times 10^9\text{ years}$

Therefore, the age of rock is $3.32\times 10^9\text{ years}$