The present day abundances of the isotopes $⋃^{238}$ and $⋃^{235}$ are in the ratio of $128 : 1$ . They have half lives of $4.5 \times 10^{9}$ years and $7 \times 10^{8}$ years respectively and assume that the equal amount of each isotope existed at the time of formation of the earth. If age of earth is $\frac{49 X}{76} \times 10^{9}$ years, then calculate X

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Solution

128 $N = N_{0} e^{- λ_{1} t}$ ...(i)

$N$ $= N_{0} e^{- λ_{2} t}$ ...(ii)

128 $= e^{- λ_{2} t + λ_{2} t}$

$l n$ $128 = (- \frac{l n 2}{4.5 \times 10^{9}} + \frac{l n 2}{7 \times 10^{8}}) t$

7 $l n$ $2 = l n 2 [\frac{1}{7 \times 10^{8}} - \frac{1}{4.5 \times 10^{9}}] t$

$7 = [\frac{(45 - 7)}{7 \times 45 \times 10^{8}}] t = \frac{49 \times 9}{38 \times 2} \times 10^{9} = t$

Hence $X = 9$

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