1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Radioactive Decay

  1. Mar 22, 2013 #1
    This isnt a direct question and answer problem, it relates to an essay I am doing.

    I am a bit confused on the equations given by my tutor for radioactive decay, I suspect there is a mix up of symbols used where two symbols are used for the same think (such as N and R for the normal force).

    On the powerpoint relating to it, she stated that the decay rate, R, is as below

    However I also have the formula below

    Are they the same thing, I understand that the second one is the activity, measured in Bq, but also the first one?

    Also if I were to calculate the half-life of something, which formula would I use?

    Thanks :)
  2. jcsd
  3. Mar 22, 2013 #2
    They are the same thing. Radioactivity is a manifestation of decay, any decayed atom emits some particle that is registered as radioactivity. So one is proportional to another, and the coefficient depends on the units used to measure these two quantities.
  4. Mar 22, 2013 #3
    The activity is defined as the variation in time of the number of nuclei, and is also proportional to the number itself
    $$ A=-\frac{\mathrm{d}N}{\mathrm{d}t}=\lambda N $$
    Solving this equation you have immediately
    $$ N(t)=N_0e^{-\lambda t}=N_0e^{-t/\tau} $$
    where ##\tau=1/\lambda## is the mean lifetime (you find it tabulated).
    Now clearly then
    $$ A(t)=\lambda N(t) $$
    and your result is there.

    As for the ##R## I think it is more or less the same thing, eventually expressed as different units, but essentially has the same meaning.

    Last of all, to compute the half-life, compute ##t## for which you have ##N(t)=N/2##
  5. Mar 22, 2013 #4
    Yeah I have seen that one as well with N, I know N stands for the number of Nuclei though is it the same as well then?

    I need to derive it and end up with the one using A, would the below be ok? (i dont think it is)

    A=\frac{dN}{dt}=-λt \\
    ∫\frac{1}{N}dN=∫-λdt \\
    e^{logN}=e^{-λt} \\

    Any help is appreciated.
  6. Mar 22, 2013 #5
    First solve
    $$ \frac{\mathrm{d}N}{\mathrm{d}t}=-\lambda N $$
    exactly as you did
    $$ \int_{N_0}^{N(t)}\frac{\mathrm{d}N}{N}=-\int_{0}^{t}\lambda \mathrm{d}t $$
    $$ \ln\left[ \frac{N(t)}{N_0} \right]=-\lambda t $$
    $$ N(t)=N_0 e^{-\lambda t} $$

    Then after that you have by definition ##A=\lambda N## (see http://en.wikipedia.org/wiki/Radioactive_decay#Radioactive_decay_rates) and therefore
    $$ A(t)=\lambda N_0 e^{-\lambda t}=A_0 e^{-\lambda t} $$
    calling ##A_0=\lambda N_0## (by definition).
  7. Mar 22, 2013 #6
    Many thanks, appreciate it.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted

Similar Discussions: Radioactive Decay
  1. Radioactive decay (Replies: 9)

  2. Radioactive Decay (Replies: 4)

  3. Radioactive Decay? (Replies: 1)

  4. Radioactive decay (Replies: 5)