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Homework Help: Questions about the Differential Equation (Dacay)

  1. Dec 6, 2009 #1
    The half-life of a radioactive isotope is the amount of time it takes for a quantity
    of radioactive material to decay to one-half of its original amount.
    i) The half-life of Carbon 14 (C-14) is 5230 years. Determine the decay-rate
    parameter  for C − 14.
    ii)The half-life of Iodine 131 (I-131) is 8 days. Determine the decay-rate param-
    eter for I − 131.

    2. Relevant equations

    dx/dt = -bx where b is a constant

    3. The attempt at a solution

    i) let x(t) be the quantity of radionactuve meterial at time t in year and dx/dt be the rate of change of the quantity.

    dx/dt = -bx where b is a constant

    I dun no how to process after doing this assumption. Please help to give some hints on doing these questions. Many thanks!
  2. jcsd
  3. Dec 6, 2009 #2
    Let's let [itex]t_{1/2}=5230\,\text{yr}[/itex] be the half-life of C-14. What does this mean? Well, if at some initial time [itex]t_0[/itex] the total quantity of C-14 is [itex]x_0=x(t_0)[/itex], then after an additional time [itex]t_{1/2}[/itex] the amount of carbon remaining is [itex]x_0/2[/itex]; algebraically, [itex]x(t_0+t_{1/2})=x_0/2[/itex].

    The decay constant [itex]\lambda[/itex], instead of specifying how long it takes for the quantity to halve, tells you the rate of change of the percent change in quantity. That's a mouthful. So, what does that mean? Well, consider the same quantities [itex]x_0[/itex] and [itex]t_0[/itex] as before. How much carbon is left after an infinitesimal time [itex]dt[/itex]? Let's go take a look at our (verbal) definition for the decay constant: "percent change in quantity" means [itex]dx/x(t)[/itex] (remember that [itex]dx=x(t+dt)-x(t)[/itex]), and the "rate of change" of this is [itex][dx/x(t)]/dt[/itex]. Thus, the decay constant is given algebraically by

    Can you think of a way of incorporating these to find [itex]\lambda[/itex] for C-14?
  4. Dec 6, 2009 #3
    t0 =0, x0 = 0
    When t = 5230, then I dun no how to find the X=???
    then can't find the lambda?
  5. Dec 6, 2009 #4


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    Starting with dx/dt = -bx, "separate" the variables by separating the derivative into differentials: dx= -bx dt so dx/x= -bdt. Integrate both sides: [itex]\int dx/x= -b\int dt[/itex].
  6. Dec 6, 2009 #5
    that means lambda is dx/dt, right?

    As u said, I do the calculation and find the follow
    ln|x| = -bt +C where C is a constant

    then what should I do?

    Can u explain more details?
  7. Dec 6, 2009 #6
    Why do you think that x0=0?
  8. Dec 7, 2009 #7
    I think if the time is zero, there is no decay so x is equal to ZERO
  9. Dec 7, 2009 #8
    Ahah! There's your problem. The value x measures the quantity remaining, not the quantity which has been removed.
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