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Math Help - [SOLVED] help with half-life, formula with time and remaining time formula

  1. #1
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    [SOLVED] help with half-life, formula with time and remaining time formula

    Okay I don't understand any of this & if I could get someones help I would appreciate it so much. We learned before the compounding formulas. I thin some how they relate but I don't know. Please help.

    Problem:

    At the beginning of an experiment, a scientist has 216 grams of radioactive goo. After 225 minutes, her sample has decayed to 27 grams.

    What is the half-life of the goo in minutes?

    Find a formula for , the amount of goo remaining at time .

    How many grams of goo will remain after 8 minutes?

    Hint: do part 2 first, i.e. you are just finding k from given pair. Then to find half-life, set outcome G(t) to be 1/2 of starting amount and solve for t.
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  2. #2
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    Hello there,

    From my general chemistry studies, I know that exponential growth and decay can be modelled by:

     G(t) = Ae^{kt} , where A is the starting amount, k is just a constant, and t is the time.

    I will show you #2 and then you will be able to do #1.

     G(t) = Ae^{kt}

    Your information indicates terminating mass as 27 g, A = 216 g, and t = 27 min:

     27 = 216 e^{k27}

     ln\frac{1}{8} = k27

     \Rightarrow k = 0.077016 .

    I hope that this helps. Be sure to write back for more help.
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  3. #3
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    Quote Originally Posted by scherz0 View Post
    Hello there,

    From my general chemistry studies, I know that exponential growth and decay can be modelled by:

     G(t) = Ae^{kt} , where A is the starting amount, k is just a constant, and t is the time.

    I will show you #2 and then you will be able to do #1.

     G(t) = Ae^{kt}

    Your information indicates terminating mass as 27 g, A = 216 g, and t = 27 min:

     27 = 216 e^{k27}

     ln\frac{1}{8} = k27

    k = -0.077016

    I hope that this helps. Be sure to write back for more help.
    ^ your answer for k should be negative as would befit decay given your original formula

    Half life is given by:

    t_{\frac{1}{2}} = \frac{\ln(2)}{k}

    and is derived from the standard exponential decay equation:

    G(t) = G(0)e^{-kt}
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  4. #4
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    Thanksd scherz0 and e^(i*pi) but I am more lost than before. I am not saying it's both of your faults or what not. I am just not sure what is going on.
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    Quote Originally Posted by scherz0 View Post
    Hello there,

    From my general chemistry studies, I know that exponential growth and decay can be modelled by:

     G(t) = Ae^{kt} , where A is the starting amount, k is just a constant, and t is the time.

    I will show you #2 and then you will be able to do #1.

     G(t) = Ae^{kt}

    Your information indicates terminating mass as 27 g, A = 216 g, and t = 27 min:

     27 = 216 e^{k27}

     ln\frac{1}{8} = k27

     \Rightarrow k = 0.077016 .

    I hope that this helps. Be sure to write back for more help.

    wouldn't the time be 225? why 27?
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  6. #6
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    Quote Originally Posted by lsnyder View Post
    Okay I don't understand any of this & if I could get someones help I would appreciate it so much. We learned before the compounding formulas. I thin some how they relate but I don't know. Please help.

    Problem:

    At the beginning of an experiment, a scientist has 216 grams of radioactive goo. After 225 minutes, her sample has decayed to 27 grams.

    What is the half-life of the goo in minutes?

    Find a formula for , the amount of goo remaining at time .

    How many grams of goo will remain after 8 minutes?

    Hint: do part 2 first, i.e. you are just finding k from given pair. Then to find half-life, set outcome G(t) to be 1/2 of starting amount and solve for t.
    Since 216/2 = 108, 108/2 = 54 and 54/2 = 27 it should be clear that 225 minutes corresponds to 3 half lives. Therfore half life = 225/3 = 75 minutes.
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  7. #7
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    Quote Originally Posted by lsnyder View Post
    Okay I don't understand any of this & if I could get someones help I would appreciate it so much. We learned before the compounding formulas. I thin some how they relate but I don't know. Please help.

    Problem:

    At the beginning of an experiment, a scientist has 216 grams of radioactive goo. After 225 minutes, her sample has decayed to 27 grams.

    What is the half-life of the goo in minutes?
    This is an "exponential" problem but it isn't necessary to use "e" itself. If something has a half life of T minutes, then it is multiplied by 1/2 every T minutes. In t minutes there are t/T periods of T minutes so you multiply by \left(\frac{1}{2}\right)^{t/T}. If it starts at 216 grams then t minutes later you have 216\left(\frac{1}{2}\right)^{t/T}. After 225 you hav 27 so 216\left(\frac{1}{2}\right)^{225/T}= 27. Then \left(\frac{1}{2}\right)^{225/T}= \frac{27}{216}= \frac{1}{8}= \left(\frac{1}{2}\right)^3 0.125. So we must have 225/T= 3, T= 225/3= 75 minutes. That's very radioactive stuff!

    Find a formula for , the amount of goo remaining at time .
    The formula is P(t)= 216\left(\frac{1}{2}\right)^{t/75}.

    [quote]How many grams of goo will remain after 8 minutes?
    This is an ambiguous question. Does it refer to 8 minutes after the original time, when the mass was 216 grams or is it 8 minutes after the first 225 minutes? If the first, use P(8)= 216\left(\frac{1}{2}\right)^{8/75}. If the latter either set t= 225+ 8= 233 minutes so P(233)= 216\left(\frac{1}{2}\right)^{233/75} or, equivalently, "restart" at 225 minutes: 27\left(\frac{1}{2}\right)^{8/75}.

    Hint: do part 2 first, i.e. you are just finding k from given pair. Then to find half-life, set outcome G(t) to be 1/2 of starting amount and solve for t.
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  8. #8
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    wow, that made a lot more since.
    Thank you Mr. Fantastic and Halls
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  9. #9
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    HallsofIvy, it was the first one on the question about the 8 minutes. just incase you wondered which equalled to 200.6059846. Thanks again
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