Uranium ores contain one radium —226 atom for every 2.8 x 10⁶ uranium-238 atoms. Calculate the half-life of ${}_{92}U^{238}$ given that the half-life of ${}_{88}{Ra}^{226}$ is 1600 years ( ${}_{88}{Ra}^{226}$ is a decay product of ${}_{92}U^{238}$ ):

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$1.75 x 10^{3} years$

$1600 x \frac{238}{92} years$

1600 years

$4.5 x 10^{9} years$

answer is C.

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Detailed Solution

$N_{1} λ_{1} = N_{2} λ_{2}$

$T = \frac{0.693}{λ}$

Hence $2.8 x 10^{6} x \frac{0.693}{T_{1} (U)} = 1 x \frac{0.693}{T_{2} (Ra)}$

$∴ T_{1} (U) = 1600 x 2.8 x 10^{6}$

$= 4.48 x 10^{9} years ≃ 4.5 x 10^{9} years$

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