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  1. #1
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    can u solve the problem???

    radium decomposes at a rate propotional to the amount present.if p percent of the original amount disappears in l years how much will remain at the end of 2l years?
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  2. #2
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    Let m be the percentage mass of radium present at time t.

    Let $\displaystyle m_0$ be the mass present at time $\displaystyle t=0$.

    $\displaystyle \frac{dm}{dt} \propto m$

    $\displaystyle \frac{dm}{dt}=km$ where k is a real constant (k is actually negative since this is "exponential decay". You can write k=-b or something to show that you know it is negative. I just left it because it doesn't change the overall answer at the end.)

    Integrating gives:

    $\displaystyle \int \frac{1}{m} dm=\int k dt$

    $\displaystyle ln m=kt + C$

    $\displaystyle m=e^{kt+C}$

    $\displaystyle m=Ae^{kt}$ where $\displaystyle A=e^C$

    using the intial conditions:

    $\displaystyle m=m_0 e^{kt}$

    when t=I years:

    $\displaystyle m_I=m=m_0e^{Ik}$

    At t=2I years:

    $\displaystyle m=m_0e^{2Ik}$

    $\displaystyle m=m_0(e^{Ik})^2$

    $\displaystyle m=m_0(e^{Ik})(e^{Ik})$

    $\displaystyle m=m_I(e^{Ik})$

    and I think we're done! =D

    P.S: I did change notation. I used m instead of p. If you like you can work it through again with m=p or just write this at the end.
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  3. #3
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    Hello, padmini priya!

    Radium decomposes at a rate propotional to the amount present.
    If $\displaystyle p$ percent of the original amount disappears in $\displaystyle L$ years.
    how much will remain at the end of $\displaystyle 2L$ years?
    Let $\displaystyle Q$ = quantity of radium at time $\displaystyle t.$

    We have: .$\displaystyle \frac{dQ}{dt} \:=\:-kQ \quad\Rightarrow\quad \frac{dQ}{Q} \:=\:-kt $

    Integrate: .$\displaystyle \ln(Q) \:=\:-kt + C \quad\Rightarrow\quad Q \:=\:e^{-kt+c} \quad\Rightarrow\quad Q \:=\:Ce^{-kt}$


    When $\displaystyle t = 0,\;Q = Q_o$ (initial amount)
    We have: .$\displaystyle Q_o \:=\:Ce^0 \quad\Rightarrow\quad C \:=\:Q_o$

    The function (so far) is: .$\displaystyle Q \;=\;Q_o\,e^{-kt}$


    When $\displaystyle t = L,\;Q \:=\:(1-p)Q_o$

    So we have: .$\displaystyle (1-p)Q_o \;=\;Q_o\,e^{-kL} \quad\Rightarrow\quad e^{-kL} \:=\:1-p \quad\Rightarrow\quad -kL \:=\:\ln(1-p) $

    . . Hence: .$\displaystyle k \:=\:-\frac{\ln(1-p)}{L} $

    The function is: .$\displaystyle Q \;=\;Q_o\,e^{\text{-}\left[\text{-}\frac{ln(1-p)}{L}\right]t} \quad\Rightarrow\quad Q \;=\;Q_o\,e^{\frac{\ln(1-p)}{L}\,t}$


    When $\displaystyle t = 2L\!:\;\;Q \;=\;Q_o\,e^{\frac{\ln(1-p)}{L}\cdot2L} \;=\;Q_o\,e^{2\ln(1-p)} \;=\;Q_o\,e^{\ln(1-p)^2}$

    We have: .$\displaystyle Q \;=\;Q_o(1-p)^2$


    Therefore: .$\displaystyle \frac{Q}{Q_o} \;=\;(1-p)^2\quad _{\Leftarrow\;\;\text{Percent of radium after 2L years}}$

    . .$\displaystyle \overbrace{_{\text{Fraction of radium after 2L years}}}^{\Uparrow} $

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