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A sample of U - 238 (half-life = 4.5 xx ...

A sample of U - 238 (half-life = `4.5 xx 10^(9)`yr) ore is found to contain 23.8 g of U-238 and 20.6 g of Pb - 206. What will be the age of the ore?

A

`5.5 xx 10^(12)` years

B

`6.0 xx 10^(8)` years

C

`4.5 xx 10^(9)` years

D

`5.5 xx 10^(14)` years

Text Solution

Verified by Experts

The correct Answer is:
C
Disintegration occurs as :
`""_(92)U^(238) to ""_(82)Pb^(206) + 8 ""_(2)He^(4) + e ""_(-1)e^(0)`
Atoms of Pb present = `(20.6)/(206) = 0.1 g ` atom = U decayed
Atoms of U present = `(23.8)/(238) = 0.1 g` atom
`:. N = 0.1 g` atom
`N_0` = U present + U decayed = `0.1 + 0.1 = 0.2 g` atom
`t = (2.303 xx 4.5 xx 10^(9))/(0.693) "log"_(10)(0.2)/(0.1)`
`:. t = 4.5 xx 10^(9) ` years.
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Knowledge Check

  • By successive decay ""_(92)^(238)U changed to ""_(82)Pb^(206) , when a sample of uranium ore was analysed. It was found that it contains 1g of U^(238) and 0.1 of Pb^(206 , considering that all the Pb^(206) had accumulatd due to the decay of U^(238) Calculate the age of the ore. (Half life of U^(238)=4.5xx10^(9)yrs)

    A
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    B
    `7.099xx10^(8)` yrs
    C
    `0.154xx10^(-9)` yrs
    D
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  • Uranium ._(92)U^(238) decayed to ._(82)Pb^(206) . They decay process is ._(92)U^(238) underset((x alpha, y beta))(rarr ._(82)Pb^(206)) t_(1//2) of U^(238) = 4.5 xx 10^(9) years The analysis of a rock shows the relative number of U^(238) and Pb^(206) atoms (Pb//U = 0.25) The age of rock will be

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    B
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    C
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    D
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