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mrh881
mrh881 asked in Science & MathematicsChemistry · 1 decade ago

How to solve this half life problem?

The question is that the half-life of cesium-137 is 30.2 years. What fraction of cesium remains after 3 years.

I understand that to get the time that passed I divide 3 over 30.2. t=3/30.2

I round off to get 1/10.

So my formula looks like Nn/No=(1/2)^1/10

Afterwards my work goes to a certain extent and I get confused.

Thanks in advance for the help!

Update:

I also need to be able to solve this problem without a calculator.

2 Answers

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  • Dr.A
    Lv 7
    1 decade ago
    Favorite Answer

    t1/2 = ln 2 / k

    30.2 = ln2 /k

    k = 0.0230

    ln X = - 0.0230 x 3 = - 0.0689

    X = e^-0.0689 =0.933

  • a2gal
    Lv 7
    1 decade ago

    N = No * (.5) ^ t/h

    N is amount remaining

    No is original amount (we'll use 100 to make life easy)

    t is elapsed time

    h is half-life

    N = 100 * (.5) ^ (3/30.2)

    N = 93.35

    93.35 grams remain out of 100 (93.35/100 = 93.35%)

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