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3 October, 13:44

How old is a bone in which the Carbon-14 in it has undergone 3 half-lives? A) 11,400 years old Eliminate B) 17,100 years old C) 34,200 years old D) 57,000 years old

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  1. 3 October, 15:18
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    For this problem, we use the formula for radioactive decay which is expressed as follows:

    An = Aoe^-kt

    where An is the amount left after time t, Ao is the initial amount and k is a constant.

    We calculate as follows:

    at hal-life equal to 5700,

    An = Aoe^-kt

    0.5 = e^-k (5700)

    k = 1.22x10^-4

    An = Aoe^-kt

    1.5 = e^-1.22x10^-4 (t)

    t = 3323 years
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