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Uranium ores contain one radium `-226 `atom for every `2.8 xx 10^(6)` uranium `-238` atoms. Calculate the half-life of `._(88)Ra^(226)` is `1600` years `(._(88)Ra^(226)` is a decay product of `._(92)U^(238))`.
A. `1.75 xx 10^(3) years`
B. `1600 xx (238)/(92)years`
C. `4.5 xx 10^(9) years`
D. `1600 xx 238 years`

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Best answer
Correct Answer - c
`N_(1 )lambda_(1)=N_(2) lambda_(2)`
`T=(0.693)/(lambda)`
Hence,
`2.8xx10^(6)xx(0.693)/(T_(1)(U))=1xx(0.693)/(T_(2)(Ra))`
`:. T_(1)(U)=1600 xx 2.8 xx 10^(6)`
`4.48 xx10^(9)` years
`~~4.5 xx 10^(9)` years .

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