Question

A sample contains 25% parent isotope and 75% daughter isotopes. If the half-life of the parent isotope is 72 years, how old is the sample?

Answer 1

Answer: The sample is 144 years old.

Explanation: This is a type of radioactive decay and all radioactive decays follow first order kinetics.

We are given half life which is 72 years.  

To calculate the rate constant, we use the formula:

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

$k=\frac{0.693}{72years}$

$k=0.0096years^{-1}$

Also 25 % of parent isotope is left, that means 75 % has been decomposed

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

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

where,

k = rate constant

t = time taken for decay process

a = initial amount of the reactant  = parent isotope= 100

a - x = amount left after decay process  = 25

Putting values in above equation, we get:

$0.0096years^{-1}=\frac{2.303}{t}\log\frac{100}{25g}$

$t=144years$

Answer 2

The radioactive decay obeys first order kinetics

the rate law expression for radioactive decay is

$ln\frac{[A_{0}]}{[A_{t}]}=kt$

Where

A0 = initial concentration

At = concentration after time "t"

t = time

k = rate constant

For first order reaction the relation between rate constant and half life is:

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

Let us calculate k

k = 0.693 / 72 = 0.009625 years⁻¹

Given

At = 0.25 A0

$ln(\frac{A0}{0.25A0})=0.009625 X time$

time = 144 years

So after 144 years the sample contains 25% parent isotope and 75% daughter isotopes**

Simply two half lives