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

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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\;.$

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