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7 May, 08:44

A sample of uranium ore contains 6.73 mg of 238U and 3.22 mg of 206Pb. Assuming all of the lead arose from the decay of the uranium and that the half-life of 238U is 4.51 x 109years, determine the age of the ore

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  1. 7 May, 09:04
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    The age of the ore is 4.796*10^9 years.

    Explanation:

    To solve this question, we use the formula;

    A (t) = A (o) (1/2) ^t/t1/2

    where;

    A (t) = 3.22mg

    A (o) = 6.731mg

    t1/2 = 4.51*70^9 years

    t = age of the ore

    So,

    A (t) = A (o) (1/2) ^t/t1/2

    3.22 = 6.73 (1/2) ^t/4.51*10^9

    Divide both sides by 6.73

    3.22/6.73 = (1/2) ^t/4.51*10^9

    0.47825 = (0.5) ^t/4.51*10^9

    Log 0.4785 = t/4.51*10^9 • log 0.5

    Log 0.4785/log 0.5 • 4.51*10^9 = t

    t = 1.0634 * 4.51*10^9

    t = 4.796*10^9

    So therefore, the age of the ore is approximately 4.796*10^9 years.
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