Uranium ores contain one radium $$226$$ atom for every $$2.8$$ x $$106$$ uranium $$238$$ atoms. Calculate the $$half-life$$ ofU $$92238$$ given that the $$half-life$$ of R $$88226a$$ is $$1600$$ years (R$$ $$88226ais$$ a decay product of U $$92238)

Correct option is 34.5 x 109 years

Questions

Uranium ores contain one radium 226 atom for every $2.8 x 10^{6}$ uranium 238 atoms. Calculate the half-life of ${}_{92}^{238}U$ given that the half-life of ${}_{88}^{226}R a$ is 1600
years ( ${}_{88}^{226}R a$ is a decay product of ${}_{92}^{238}U$ )

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175 x106 years

1600 x23892 years

4.5 x 109 years

1600 years

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detailed solution

Correct option is C

at steady state N1λ1=N2λ2 2.8 x1061T1=11T2 T2= 1600 years T1= 4.45 x 109

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Uranium ores contain one radium 226 atom for every 2.8 x 106 uranium 238 atoms. Calculate the half-life ofU 92238 given that the half-life of R 88226a is 1600 years (R 88226ais a decay product of U 92238)

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Ready to Test Your Skills?

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Uranium ores contain one radium 226 atom for every $2.8 x 10^{6}$ uranium 238 atoms. Calculate the half-life of ${}_{92}^{238}U$ given that the half-life of ${}_{88}^{226}R a$ is 1600
years ( ${}_{88}^{226}R a$ is a decay product of ${}_{92}^{238}U$ )