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InterviewSolution
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A sample of uranium mineral was found to contain `Pb^(208)` and `U^(238)` in the ratio of 0.008 : 1. Estimate the age of the mineral (half life of `U^(238)` is `4.51 xx 10^(9)` years). |
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Answer» We know that, `t = (2.303t_(1//2))/(0.693) log [1 + (.^(206)Pb)/(.^(238)U)]` Given `t_(1//2) = 4.51 xx 10^(9)` years Ratio by mass of `.^(206)Pb : .^(238)U = 0.008 : 1` Ratio by moles of `.^(206)Pb : .^(238)U = (0.008)/(206) : (1)/(238) = 0.0092` So, `t = (2.303 xx 4.51 xx 10^(9))/(0.693) log [1 + 0.0092]` `= (2.303 xx 4.51 xx 10^(9))/(0.693) xx 0.00397` `= (0.0412)/(0.693) xx 10^(9) = 0.05945 xx 10^(9)` years Hence, age of the mineral is `5.945 xx 10^(7)` years. |
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