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Math Help - The half life of a radioactive substance.

  1. #1
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    The half life of a radioactive substance.

    The half life of a radioactive substance.-captureproblem.jpg
    The half life of a radioactive substance.-capturesolution.jpg

    I understand part a) as the equation is worked so that it is in terms of k. The solution for part b) is losing me at:

    \lambda=-\frac{ln(2)}{k}*y_0e^k^(^t^1^+ ^\lambda)

    I see that where t was before there is now the symbol lambda but the part after the * symbol is lost on me right now. Why is the equation in terms of k now being multiplied by:

    y_0e^k^(^t^1^+ ^\lambda)

    Thanks for any responses...
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  2. #2
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    Re: The half life of a radioactive substance.

    In part b), it is supposed to be a period, not a multiplication. They are using \lambda as the value of t you found in part a). So, if y_1 = y_0 e^{kt_1}, then y_0 e^{k(t_1+\lambda)} = y_0e^{kt_1}e^{k\lambda}. Now, replace y_0e^{kt_1} with y_1 since they are equal by assumption. So, y_0e^{kt_1}e^{k\lambda} = y_1e^{k\lambda}. Next, substitute in the value set for \lambda: y_1e^{k\lambda} = y_1e^{-\ln 2} = \dfrac{1}{2}y_1.
    Thanks from sepoto
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  3. #3
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    Re: The half life of a radioactive substance.

    Hey sepoto.

    Remember that you are looking at the time t1 + lambda and not t1 (and you also have the information about the half life at t1 and t1 + lambda from the question). Just plug in the information for the lambda + t1 and everything should be ok.
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  4. #4
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    Re: The half life of a radioactive substance.

    So:

    t=\frac{-ln(2)}{k}

    So t might be some number 1590 in years. So if lambda is 1590 then why this equation below. So then "t_1+\lambda" is some time t plus lambda where lambda is the half life of 1590?

    y_0e^{k(t_1+\lambda)}
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  5. #5
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    Re: The half life of a radioactive substance.

    I'm not exactly sure what you are trying to say: can you please ask again but in more detail (and clarity)?
    Thanks from sepoto
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