First let us calculate for the rate constant k from the formula:
k = ln(2) / t0.5
where t0.5 is the half life
k = ln(2) / 1.3x10^9 years
k = 5.33x10^-10 years-1
Then we use the formula:
A/Ao = e^-kt
where A/Ao is the amount remaining = 25% = 0.25, t is time
Rearranging to get t:
t = ln(A/Ao) / -k
t = ln(0.25) / (-5.33x10^-10 years-1)
t = 2.6x10^9 years
Answer : The age of the sample of granite is, 2.6 billion years
Solution : Given,
As we know that the radioactive decays follow the first order kinetics.
First we have to calculate the rate constant.
Formula used : [tex]t_{1/2}=\frac{0.693}{k}[/tex]
[tex]1.3\text{ billion years}=\frac{0.693}{k}[/tex]
[tex]k=0.533(\text{billion years})^{-1}[/tex]
Now we have to calculate the age of the sample of granite.
The expression for rate law for first order kinetics is given by :
[tex]k=\frac{2.303}{t}\log\frac{a}{a-x}[/tex]
where,
k = rate constant = [tex]0.533[/tex]
t = time taken for decay process = ?
a = initial amount of the reactant = 100 g
a - x = amount left after decay process = 100 - 75 = 25 g
Putting values in above equation, we get the age of the sample of granite.
[tex]0.533=\frac{2.303}{t}\log\frac{100}{25}[/tex]
[tex]t=2.6\text{ billion years}[/tex]
Therefore, the age of the sample of granite is, 2.6 billion years
