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Math Help - Radioactive Decay

  1. #1
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    Radioactive Decay

    I'm new here, first post.
    I honestly forgot the concept of half-life and this week we are having our final exams. Good thing that the teacher provided us with a review test. So would someone refresh my memory.

    Quote Originally Posted by Problem 1
    An isotope of technetium is used to prepare images of internal body organs. This isotope has a half-life of approximately 6h. A patient is injected with 30 mg of this isotope.

    A) What is the technetium level in the patient after 3h?
    B) How long (in hours) will it take for the technetium level to reach 20 mg?
    This is how I did A:

    1. P(t) = 30(1/2)^t/6
    2. 3 = 30(1/2)^3/6
    3. 3/30 = 1/2^3/6
    4. log (1/2)^3/30 = 3/6
    5. (log (3/30) / log (1/2) ) = 3/6
    6. (log (1/10) / log (1/2) ) = 1/2
    7. (log (1/10) / log (1/2) )*2 = 6.64


    The problem is that my teacher said that the answer for A is 21.2 mg. I got this problem totally wrong. Can someone show me the work on how they got the answer to be 21.2 mg?

    I am clueless on how to do B.


    Quote Originally Posted by Problem 2
    A sample of prothethium-147 (used in some luminous paints) weighs 25mg. One year later, the sample weighs 18.95mg. What is the half-life of promethium-147, in years?
    The answer to this problem is 2.5 years. I need to know how to setup the equation for this problem and how you get the answer, 2.5 years.

    Quote Originally Posted by Problem 3
    Francium-223 is a very rare radioactive isotope discovered in 1939 by Marguerite Percy. A 3-microgram sample of francium-223 decays to 2.54 micrograms in 5-mins. What is the half-life of francium-223, in minutes?
    The answer to this problem is 20.8 mins. I need to know how to setup the equation for this problem and you get the answer, 20.8 mins.

    Quote Originally Posted by Problem 4
    The amount C in grams of carbon-14 present in a certain substance after t years is given by C = 20 e ^ 0.0001216t . Estimate the half-life of carbon-14.
    The answer to this problem is 5700 years. How do I set this problem up and how do you get the answer, 5700 years?

    Thanks for any help!
    Last edited by akasixcon; May 17th 2008 at 12:40 PM.
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  2. #2
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    Hello,

    Quote Originally Posted by akasixcon View Post
    I'm new here, first post.


    This is how I did A:

    1. P(t) = 30(1/2)^t/6
    2. 3 = 30(1/2)^3/6
    3. 3/30 = 1/2^3/6
    4. log (1/2)^3/30 = 3/6
    5. (log (3/30) / log (1/2) ) = 3/6
    6. (log (1/10) / log (1/2) ) = 1/2
    7. (log (1/10) / log (1/2) )*2 = 6.64


    The problem is that my teacher said that the answer for A is 21.2 mg. I got this problem totally wrong. Can someone show me the work on how they got the answer to be 21.2 mg?

    I am clueless on how to do B.
    This is how I'd do :

    If the half-life is 6 hours, then :

    \frac N2=N \cdot e^{- \lambda t_{1/2}}, where t_{1/2}=6 and \lambda is a characteristic of the technicium.

    --> \frac 12=e^{-6 \lambda}

    -----> \boxed{\lambda=\frac{\ln 2}{6}}


    Now, at time t=3, how many are there ?

    \begin{aligned} N(t={\color{red}3}) &=N_0 \cdot e^{-{\color{red}3} \lambda} \\<br />
&=30 \cdot e^{-\frac{3 \ln 2}{6}} \\<br />
&=30 \cdot e^{-\frac{\ln 2}{2}} \\<br />
&=30 \cdot e^{-\ln \sqrt{2}} \\<br />
&=\frac{30}{\sqrt{2}} \end{aligned}
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  3. #3
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    For #2:

    The decay constant is given by k=\frac{-1}{T}ln(2)

    Where T=half-life.

    At y(0)=25 and y(1)=18.95

    So we have 25e^{\frac{-1}{T}ln(2)}=18.95

    Solve for T and you will see it equals 2.50168709718....
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  4. #4
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    #4:

    Similar to the other, \frac{-1}{T}ln(2)=-.0001216

    Solve for T.
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  5. #5
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    #3:

    2.54=3e^{5k}

    k=\frac{ln(127/150)}{5}\approx{-.03328964}

    \frac{-1}{T}ln(2)=\frac{ln(127/150)}{5}

    Solve for T.

    See the pattern with these problems?.
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