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A $1.5 \times 10^{19}$ years old rock contains which disintegrates to from. Assume that there was no lead in the rock initially and it is the only stable product formed by the decay. Calculate the ratio of the number of nuclei of lead to that of uranium [Given, half life of ${}^{238}U = 4.5 x 10^{9} years, 2^{1 / 3} \approx 1.26$ ] (round off to nearest integer)

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Detailed Solution

Let No be the initial number of uranium nuclei. After time t , let NU be the number of uranium nuclei and NPB be the number of lead nuclei. The number of half-lives passed is
$n = \frac{t}{T_{half}} = \frac{1.5 x 10^{9}}{4.5 x 10^{9}} = \frac{1}{3}$

$N_{U} = N_{0} \times {(\frac{1}{2})}^{n} = N_{0} {(\frac{1}{2})}^{\frac{1}{3}}$

$N_{pb} = N_{0} - N_{0} {(\frac{1}{2})}^{\frac{1}{3}}$

$\frac{N_{pb}}{N_{U}} = 2^{\frac{1}{3}} - 1 = 0.26$

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