$(a)\;6 \times 10^{10} y\qquad(b)\;6 \times 10^{9} y\qquad(c)\;6 \times 10^{8} y \qquad(d)\;6 \times 10^{7} y$

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Answer : (b) $\;6 \times 10^{9} y$

Explanation :

Let $\;N_{0}\;$ be number of atoms of each isotope at time of formation of earth (t=0) and $\;N_{1}\;$ and $\;N_{2}\;$ the no.of atoms at present $\;(t=t)\;.$ Then

$N_{1}=N_{0}\;e^{- \lambda_{1}t}---(1)$

$N_{2}=N_{0}\;e^{- \lambda_{2}t}---(2)$

$\large\frac{N_{1}}{N_{2}}=e^{(\lambda_{2}-\lambda_{1})} t---(3)$

Further it is given that

$\large\frac{N_{1}}{N_{2}}=\large\frac{99.3}{0.7}----(4)$

Equating (3) and (4) and taking log both sides , we have

$(\lambda_{2}-\lambda_{1})t=ln(\large\frac{99.3}{0.7})$

$t=(\large\frac{1}{\lambda_{2}}-\lambda_{1})\;ln(\large\frac{99.3}{0.7})$

Substituting values

$t=\large\frac{1}{\large\frac{0.693}{7.04\times10^{8}}-\large\frac{0.693}{4.47\times10^{9}}}\;ln(\large\frac{99.3}{0.7})$

$t=6\times10^{9} y$

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