a)indium-115 has a half life of 4.5 hours. if you start with a 12.0-mg sample of indium-115, how much will remain after 13.5 hours?
b)how much of a 3.50-mg smaple of nickel-63 will remain after 368 years? the half-life of nickel-63 is 92 years?

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  1. There is a short way and a long way of doing these. The short way is easy if the numbers are right. In this case they are.
    a) Note that 13.5/4.5 = 3.00 half lives; therefore, 1/2 of 12.0 mg sample = 6.0 mg left after the first half life, 1/2 of that = 3.0 mg left after the 2nd half life and 1/2 of that = 1.5 mg left after the 3rd half life.
    b) This is done the same way if you notice that 368 years/92 years = 4.00 half lives.

    Here is the long way of doing them when the number of half lives is not an even number.
    k = 0.693/t1/2
    ln(No/N) = kt

    So, for the first one,
    k = 0.693/t1/2
    k = 0.693/4.5 = 0.154
    Then substitute into the other equation for k you've just calculated.
    ln((No/N) = kt
    ln(12.0/N) = 0.154*13.5
    ln(12.0/N) = 2.079
    Take the antilog of both sides.
    12.0/N = 8.0
    N = 12.0/8.0 = 1.5 mg

    The b part is done the same way the long way, too.

    N=12.0/1.053 =

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