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Math Help - Half Life

  1. #1
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    Half Life

    The half life of plutonium-239 is 24,360 years. The maximum amount an adult human can handle without injury is .13 micrograms (.000000013 grams). suppose a researcher possesses a 1-gram sample of plutonium - 239. the amount "A" (in grams) after "t" years is A(t) = 1 x (1/2)^ t/360

    a) How much of 1-gram sample will remain after 10,000 years?

    b) How long will it take before there is .000000013 grams left?

    My issue is this: For part "A", how do I input 1/2 to the 10000/36th power on my calculator. I Have a TI-30Xa, yet I have been unable to figure out how to enter this formula.

    For part B, is it as simple as counting the number of half lives that pass until I get an amount less than .13 micrograms?

    Thanks in advance for your help!
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  2. #2
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    I dont understand the formula for A(t).
    shouldnt it be : A(t)=1 x (1/2)^(t/24360) as the half life is 24360 years, which mean, only after this time the quantity will be half it was

    so the answer would be : A(10000)= (1/2)^(2,436) which is easier to calculate

    anyway...did u find the solution for this?
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  3. #3
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    Quote Originally Posted by Hapa View Post
    The half life of plutonium-239 is 24,360 years. The maximum amount an adult human can handle without injury is .13 micrograms (.000000013 grams). suppose a researcher possesses a 1-gram sample of plutonium - 239. the amount "A" (in grams) after "t" years is A(t) = 1 x (1/2)^ t/360

    a) How much of 1-gram sample will remain after 10,000 years?

    b) How long will it take before there is .000000013 grams left?

    My issue is this: For part "A", how do I input 1/2 to the 10000/36th power on my calculator. I Have a TI-30Xa, yet I have been unable to figure out how to enter this formula.

    For part B, is it as simple as counting the number of half lives that pass until I get an amount less than .13 micrograms?

    Thanks in advance for your help!
    your equation is incorrect. it should be ...

     <br />
A(t) = \left(\frac{1}{2}\right)^{\frac{t}{24360}}<br />

    now find A(10000)


    .13 micrograms = 1.3 \times 10^{-7} grams

    solve the equation for t using logs ...

     <br />
1.3 \times 10^{-7} = \left(\frac{1}{2}\right)^{\frac{t}{24360}}<br />
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