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# In chain analysis of a rock , the mass ratio of two radioactive isotope is found to be $\;100 \%\;$ . The mean lives of two isotopes are $\;4 \times 10^{9}\;$years and $\;2\times10^{9}\;$years respectively . If it is assumed that at time of formation of rock , both isotopes were in equal proportion . Then calculate age of rock . Ratio of atomic weights of two isotope is $\;1.02 \%\;(log_{10}^{1.02}=0.0086)$

$(a)\;18.3 \times 10^{9} \;years\qquad(b)\;19.2 \times 10^{10} \;years\qquad(c)\;20.2 \times 10^{11} \;years\qquad(d)\;11.6 \times 10^{8} \;years$

Answer : $\;18.3 \times 10^{9} \;years$
Explanation :
At time of formation of rock , both isotopes have same no.of nuclei No. Let $\;\lambda_{1}\;$ & $\;\lambda_{2}\;$ be decay constants of two isotopes .
If $\;N_{1}\;$ and $\;N_{2}\;$ are no.of their Nuclei after a time t , we have
$N_{1}=N_{0} e^{-\lambda_{1} t }\;$ & $\;N_{2}=N_{0} e^{-\lambda_{2} t}\;$
$\large\frac{N_{1}}{N_{2}}=e^{-(\lambda_{1}-\lambda_{2})t}$$-----(1) Let mass of two isotopes at time t be \;m_{1}\; & \;m_{2}\; and let their respective atomic weights be \;M_{1}\; and \;M_{2}\; we have m_{1}=N_{1} M_{1}\; & \;m_{2}=N_{2} M_{2} \large\frac{N_{1}}{N_{2}}=\large\frac{m_{1}}{m_{2}}\;\large\frac{M_{2}}{M_{1}}$$----(2)$
Substituting value given in problem we get
$\large\frac{N_{1}}{N_{2}}=100\;.\large\frac{1}{1.02}=\large\frac{100}{1.02}$
Let $\;z_{1}\;$ & $\;z_{2}\;$ be mean lives of two isotopes .Then
$z_{1}=\large\frac{1}{\lambda_{1}}\;$ and $\;z_{2}=\large\frac{1}{\lambda_{2}}$
Which gives
$\;\lambda_{1}-\lambda_{2}=\large\frac{z_{1}-z_{2}}{z_{1} z_{2}}=\large\frac{2 \times10^{9}-4\times10^{9}}{2 \times10^{9}\times4\times10^{9}}$$=-0.25\times10^{-9} Setting this value in equation (1) we get \large\frac{N_{1}}{N_{2}}=e^{10.25\times10^{-9}}t or \;t=\large\frac{1}{0.25\times10^{-9}}\;log_{e} \large\frac{(100)}{1.02}$$=18.34 \times 10^{9} year\;.$
edited Mar 26, 2014