Calorimetry

Question :

In the chemical analysis of a rock,the mass ratio of the two radioactive isotopes A and B is found to be 100:1. The mean lives of the two isotopes are 4X10⁹ years and 2X10⁹ years respectively. If it is assumed that ,at the time of formation of the rock,the atoms of the two isotopes were in equal proportion.

Given the ratio of the atomic weight of the two isotopes is 1.02:1

Find the age of the rock

a)1.83X10⁹ years

b)1.9X10¹⁰ years

c)1.7X10¹⁰ years

d)1.83X10¹⁰ years

Solution

Given

M_A:M_B= 1.02:1

N₁(0):N₂(0)=1:1

Let N₁ be the no of moles of isotope A in the rock

Let N₂ be the no of moles of isotope B in the rock

Let m_A(t) and m_B(t) be the masses of the two isotopes respectively

Now it is given

m_A(t): m_B(t) =100:1

or

$\frac{m_a}{m_b}=\frac{N_1}{N_2}x\frac{M_a}{M_B}$

As we know

m_A(t): m_B(t) =100:1

M_A:M_B= 1.02:1

So

N₁:N₂=100:1.02

Now from the equation of radioactivity for two isotopes

$N_1\left(t\right)=N_1\left(0\right)e^{-t/{\tau }_1}$

${{}_{}}_{}{N}_2{\left(t\right)=N}_2{\left(0\right)e}^{-t/{\tau }_2}$

Dividing these two equation and substituting the values from above

$\frac{100}{1.02}=e^{-t/(\frac{1}{{\tau }_1}-\frac{1}{{\tau }_2})}$

or

$\frac{-t}{(\frac{1}{{\tau }_1}-\frac{1}{{\tau }_2})}=\log \left(\frac{100}{1.02}\right)$

Putting values for mean lives,we get

t=1.83X10¹⁰ years