In the chemical analysis of a rock the mass ratio of two radioactive isotopes is found to be `100:1`. The mean lives of the two isotopes are `4xx10^9` years and `2xx10^9` years, respectively. If it is assumed that at the time of formation the atoms of both the isotopes were in equal propotional, calculate the age of the rock. Ratio of the atomic weights of the two isotopes is `1.02:1`.

Answer

A, C, D

Explanation

At the time of observation `(t=t)`,
`m_1/m_2=100/1` (given)
Further it is given that `A_1/A_2=1.02/1`
Number of atoms, `N=m/A`
:. `N_1/N_2=m_1/m_2xxA_2/A_1=100/1.02` …(i)
Let N_0 be the number of atoms of both the isotopes at the time of formation, then
`N_1/N_2=(N_0e^(-lambda1t))/(N_0e^(-lambda2t))=e^((lambda_2-lambda_1)t)` ...(ii)
Eq.(i) and Eq.(ii), we have
`e^((lambda_2-lambda_1)t)=100/1.02`
or `(lambda_2-lambda_1)t=1n100-1n1.02`
:. `t=(1n100-1n1.02)/(((1)/(2xx10^9)-(1)/(4xx10^9)))`
Substituting the values, we have
`t=1.834xx10^10yr`