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Can I solve this as a differential equation? (radioactive particles)

  1. Mar 3, 2010 #1
    I was wondering how one might be able to use differential equations when two opposite things are happening at the same time. For example, you have a sample of 1000 radioactive particles. The number of atoms (A) doubles every two hours. At the same time, 10 particles disappear every hour.

    I initially thought I could get a DE involving the integrating factor, but I can't seem to get my head around the logic of the problem. Does anyone know how I could go about approaching this problem?

    Last edited: Mar 3, 2010
  2. jcsd
  3. Mar 3, 2010 #2


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    It is difficult to understand just what you mean, since a quantity cannot both double and diminish over time. Also, are the growth and/or disappearance of particles (essentially) continuous phenomena, or do you mean that for example every hour, 10 particles disappear only after the hour has passed?
    Further complicating this is the fact that radioactive particles normally decay, rather than grow. Do you mean that some other type of particle decays to produce more of the particles of interest?
  4. Mar 4, 2010 #3
    I have the exact same question as Sci-Fry so I'll try to clarify.

    Both the growth and disappearance of the particles are in fact, continuous. The only difference is that growth is dependent on the amount of particles at hand, whereas disappearance isn't (it simply is a reduction of 10 particles by the end of 60 minutes). And instead of radioactive particles, let's say they are bacteria particles because they are replicating (for the most part).

    So my question is, how can you put the growth and disappearance together in order to formulate a differential equation for the problem?
  5. Mar 4, 2010 #4


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    Generally the differential equation for something like that is

    rate of change = (total number of elements put into system at time t) - (total number elements removed at time t).

    It may help to start not with a differential equation but with a difference equation, since really you're talking about a discrete process and approximating it as continuous. For the bacteria example, suppose at time t you have [itex]n_t[/itex] bacteria. At the next time step the number of bacteria increases by a multiplicative factor r + 1, but then some number b of the bacteria die, so in total the number of bacteria at time t+1 is

    [tex]n_{t+1} = (r + 1)n_t - b[/tex]

    Note that this depends on the fact that I said the bacteria multiply first, then die. I could have said some die, then multiply, in which case the right hand side would look like (r+1)(n - b), but this still looks like (r+1)n - constant, so we'll just roll with the above form. To get a differential equation, let's write the above equation as

    [tex]n_{t+1}-n_t = rn_t - b[/tex]

    The left hand side is

    [tex]\frac{n_{t+1}-n_t}{t+1 - t} \sim \frac{dn(t)}{dt},[/tex]
    where this is approximated as a derivative when we're looking at "time" scales much larger than the unit time step. Note that t, as I've used it, is actually dimensionless - in the resulting differential equation which I'm about to write down, the time is measured in units of some constant time parameter which hasn't explicitly been introduced in the difference equation. Anyways, in this approximation, [itex]n_t \sim n(t)[/itex] and so

    [tex]\frac{dn(t)}{dt} = rn(t) - b[/tex]
    which is what I said earlier: the rate of change is equal to the amount of bacteria we added to the system (which was rn_t) minus the amount of bacteria taken away (b).
  6. Mar 4, 2010 #5
    Thanks Mute for the help, but I've gotten another couple of questions.

    Why does the bacteria increase in a multiplicative factor of r+1? Or in this case, since the bacteria is doubling every 2 hours, does that mean r = 2?

    How can one introduce a time parameter into the difference equation? Do you mean something along the lines of t = hours?

    How/why did you approximate [tex]\frac{n_{t+1}-n_t}{t+1 - t} \sim \frac{dn(t)}{dt},[/tex]? Like, could you just clarify the steps on how you did it?

    And lastly, since [tex]\frac{dn(t)}{dt} = rn(t) - b[/tex] is the differential equation for this problem, is it all possible to compute data and put it into Excel to graph with this?

    Sorry about all the question, but I've just started learning diff. eq. and this is one of the problems I've encountered; thanks!
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