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    The number of particles at time t of a certain radioactive substance is N.
    The substance is decaying in such a way that DN/dt = -N/3
    Given that at time t=0 the number of particles is No, find the time when the number of particles remaining is 0.5No
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    (Original post by Claxxy)
    The number of particles at time t of a certain radioactive substance is N.
    The substance is decaying in such a way that DN/dt = -N/3
    Given that at time t=0 the number of particles is No, find the time when the number of particles remaining is 0.5No
    So what hav u dun so far?
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    Ok, so from the information given we know:

    \frac{dN}{dt} = \frac {-N}{3} and when t = 0, N=N_0

    By separating the variables you get:

    \frac{dN}{N} = \frac{-dt}{3}

    Now integrate both sides:

    \int \frac{1}{N}\ dN = -1/3\int 1\ dt

    Giving:

    \ln N = (-1/3)t + k

    e both sides:

    N = e^{-t/3+k} = e^{-t/3} \times e^k =Ae^{-t/3}

    Since we know that N=N0 when t = 0, we know that our constant, A, must equal N0.

    N = N_0e^{-t/3}

    Now we have our equation, we can solve it! So let's set N = (1/2)N0:

     (1/2)N_0 = N_0e^{-t/3}



1/2 = e^{-t/3}



ln (1/2) = -t/3



t= -3ln(1/2)



t= 2.08

    Hope that helps you!
 
 
 
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