Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Radioactive decay and finding half life

  1. Dec 3, 2006 #1
    A sample of tritium 3 decayed to 94.5% of its original amount after a year.
    (a) What is the half-life of tritium-3?
    (b) How long would it take the sample to decay to 20% of its original
    amount?


    The only equations given in my book are dm/dt = km and m(t) = m(0)e^kt but the mass is not given in the problem. Is there another equation I should be using or am I just not seeing what exactly I should be using from the problem?
     
  2. jcsd
  3. Dec 3, 2006 #2

    Kurdt

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    It doesn't matter what mass it is because you know what ratios they have to decay to.
     
  4. Dec 3, 2006 #3
    Hint: Use the second eqn. Let initial mass be at t=0, ie m(0) in th expression. So at t = 1 year, m(t) becomes 94.5% of m(0).
    Can you proceed now ?
    Of course as Kurdt said you will find the final expression independent of mass, since only the ratio that remains is what matters.
     
  5. Dec 3, 2006 #4

    AlephZero

    User Avatar
    Science Advisor
    Homework Helper

    "The sample decayed to 94.5% of it's original amout."

    So if the initial mass is M, what is the mass after 1 year?
     
  6. Dec 3, 2006 #5
    [tex]m(1)=0,945m(0)[/tex]. Do you see why?

    So that m(0) now cancel out and you have:

    [tex]0,945=e^{-k}[/tex]

    You can now find k.
     
  7. Dec 3, 2006 #6
    I understand it now and I'm pretty sure I got it I just have one question.... why is it e^(-k)? I understand t = 1 so that just leaves me with k but why is it negative?
     
  8. Dec 3, 2006 #7

    Kurdt

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Because it describes the decay of a radioactive substance thus the power to which an exponential should be raised to describe decay must be negative. A positive power would lead to growth.
     
  9. Dec 3, 2006 #8

    HallsofIvy

    User Avatar
    Science Advisor

    You don't really need the exponential form. Since the question is ask for "half-life", write
    [tex]M(t)= M_0(\frac{1}{2})^\frac{t}{T}[/tex]
    Since for t= T, that multiplies M_0 by 1/2, T is the half-life.

    When t= 1,
    [tex]M(1)= .954M_0= M_0(\frac{1}{2})^\frac{1}{T}[/tex]
    so
    [tex].954= (\frac{1}{2})^\frac{1}{T}[/tex]
    Take logarythms of both sides and solve for T.
     
  10. Dec 3, 2006 #9
    I'm really sorry but now I'm confused.
     
  11. Dec 3, 2006 #10
    Your book didn't really do you any favors on the k thing. Since the amount of radioactive material remaining decreases as time increases, the first equation should properly be written as:

    [tex]\frac{dm}{dt}=-kt[/tex]

    The minus sign on the kt represents the fact that m is decreasing as time increases (and vice-versa.)

    When you solve that equation, you'll get

    [tex]m=m_{0}e^{-kt}[/tex]

    That's good. As time increases, m decreases. k will be the decay constant, which is just ln(2)/(half-life). With the way your equations are written, k is actually the negative decay constant. It should be pretty obvious when you try to solve it, since the half-life has to be a positive value.

    Anyway, remember that k isn't actually the value you're looking for in part a. If has units of inverse time. half-life = ln(2)/k is what you want.
     
  12. Dec 4, 2006 #11

    HallsofIvy

    User Avatar
    Science Advisor

    What is it that confuses you? Do you understand that If the half life is T, then the amount is multiplied by 1/2 every T years? Do you know how to solve [tex].954= (\frac{1}{2})^\frac{1}{T}[/tex]?

    As I said, take logarithms of both sides.
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook