Question

The half-life for the process 238u→206pb is 4.5×109 yr. a mineral sample contains 61.0 mg of 238u and 15.5 mg of 206pb. part a what is the age of the mineral

Answer 1

Answer : The age of the mineral is, $8.89\times 10^9\text{ years}$

Explanation :

First 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.

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

t = time passed by the sample  = ?

a = initial amount of the reactant  = 61.0 mg

a - x = amount left after decay process = 15.5 mg

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

$t=\frac{2.303}{1.54\times 10^{-10}\text{ years}}\log \frac{61.0mg}{15.5mg}$

$t=8.89\times 10^9\text{ years}$

Therefore, the age of the mineral is, $8.89\times 10^9\text{ years}$