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!

Find number of nuclei

  1. Feb 6, 2014 #1

    utkarshakash

    User Avatar
    Gold Member

    1. The problem statement, all variables and given/known data
    Nuclei of radioactive element A are produced at rate t^2 (where t is time) at any time t. The element A has decay constant λ. Let N be the number of nuclei of element A at any time t. At time t=t0, N is minimum. Then N at t0 is what?

    3. The attempt at a solution
    [itex]N(t) = kt^2 - \lambda t [/itex]

    For N to be minimum, dN/dt = 0. But doing this gives me the wrong answer.
     
  2. jcsd
  3. Feb 6, 2014 #2

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    If "The element A has decay constant λ" then the decay should be [itex]- \lambda N[/itex], not "[itex]-\lambda t[/itex]".
     
  4. Feb 6, 2014 #3
    Hi utkarshakash!

    Your expression for N(t) seems to be incorrect. From the first statement, can you write down an expression for dN/dt?
     
  5. Feb 6, 2014 #4

    utkarshakash

    User Avatar
    Gold Member

    [itex]\dfrac{dN}{dt} = kt^2 - \lambda N [/itex]
     
  6. Feb 7, 2014 #5

    collinsmark

    User Avatar
    Homework Helper
    Gold Member

    That looks very good to me. :approve:

    There is just one thing though about the k constant. The problem statement said, "Nuclei of radioactive element A are produced at rate t^2" I'm wondering if you meant to write
    "produced at rate k t2" or if k = 1. Putting the k in there makes the answer more general, so there isn't anything wrong with doing so. I just don't know if the problem statement wants you to assume that k = 1 or not.

    Whatever the case, moving on:

    So if we want to find out where N(t) is a minimum or maximum at some point, we set what equal to zero so we can find it?
     
  7. Feb 7, 2014 #6

    utkarshakash

    User Avatar
    Gold Member

    Should I find N by solving this differential equation and then differentiate it and set it zero or just set kt^2-λN equal to 0? Personally I don't feel setting the latter equal to 0 would be reasonable as it would still contain 'N' in it.
     
  8. Feb 7, 2014 #7

    collinsmark

    User Avatar
    Homework Helper
    Gold Member

    If it helps, imagine plotting a curve (a graph) of N vs. t. We don't know exactly what that plot looks like yet. But imagine for a moment that at some point in that plot, N (represented on the vertical axis) is a minimum at time t (on the horizontal axis).

    So we don't know everything about what the graph looks like, but one thing we do know is what the slope of the curve is. The slope of the curve is dN/dt. That's all that dN/dt is: the slope of the curve.

    So if the curve is smooth (let's assume that it is), what does the slope of the curve have to be at the point of the minimum?

    It doesn't really matter if N and t are in the equation for dN/dt insofar that dN/dt is the slope. dN/dt is the slope whether it's a function of a few variables or many variables. It is what it is. (And it's fortunate here that N is in the equation, because otherwise you wouldn't be able to solve for it at the point where the slope is zero.)

    [Edit: To be clear, at this point we don't really know what the slope of N vs. t strictly as a function of t. Nor do we know what the slope is solely as a function of N either. But we do know the slope if we were given both N and t -- and more importantly -- if we know what the slope is, we have an immediate relationship between N and t at that particular slope; and that's enough to solve the problem. :wink:]
     
    Last edited: Feb 7, 2014
  9. Feb 8, 2014 #8

    utkarshakash

    User Avatar
    Gold Member

    Setting dN/dt = 0 gives [itex]N = k t_0 ^2 / \lambda[/itex]. But this is not the correct answer.
     
  10. Feb 8, 2014 #9

    Curious3141

    User Avatar
    Homework Helper

    Did you try ##k = 1##, i.e. ##\displaystyle N_{min} = \frac{t_0^2}{\lambda}##?

    If that's not the correct answer either, then are you sure the question is not asking how many nuclei are present initially, i.e. at ##t = 0##? It is actually possible to calculate this in terms of ##\lambda## and ##t_0##.
     
    Last edited: Feb 8, 2014
  11. Feb 8, 2014 #10

    utkarshakash

    User Avatar
    Gold Member

    Yes, but it does not matches any of the options given. The question asks the number of nuclei of element A at t=t0
     
  12. Feb 8, 2014 #11

    Curious3141

    User Avatar
    Homework Helper

    What are the options? If too unwieldy for Latex, scan and attach. In fact, if you're doing that, include the entire question.
     
  13. Feb 8, 2014 #12

    utkarshakash

    User Avatar
    Gold Member

    [itex]
    \dfrac{2t_0 - \lambda t_0 ^2}{\lambda ^2} \\
    \dfrac{t_0 - \lambda t_0 ^2}{\lambda ^2} \\
    \dfrac{2t_0 + \lambda t_0 ^2}{\lambda} \\
    \dfrac{t_0 - \lambda t_0 ^2}{\lambda}
    [/itex]
     
  14. Feb 8, 2014 #13

    Curious3141

    User Avatar
    Homework Helper

    Wow. OK. I'm not getting any of these either.

    Going purely by dimensional analysis, I'd say the answer they expect is one of the first two. Even dimensional analysis is difficult here because the constant you called ##k## is not dimensionless (it has dimensions of ##T^{-3}##). But because they assigned a numerical value to it, there are "hidden" dimensions in the constant(s) of the final expression.

    Sorry, can't help you further, I'm afraid. Either I'm missing something or the question is wrong.
     
  15. Feb 9, 2014 #14

    Curious3141

    User Avatar
    Homework Helper

    A quick update: I got confirmation from another Homework Helper that he's getting the same result.

    The question as it is posed looks incorrect. However, it would be good if you could check with your Prof/TA and post back here. Apart from our satisfaction at getting a resolution, it's nice to close the loop for students who search for this topic in future.
     
  16. Feb 9, 2014 #15

    utkarshakash

    User Avatar
    Gold Member

    There is a high probability that this question may be incorrect. But currently it is not possible for me to contact my teachers and I do not have a solution to this question. However, if I manage to get solution I will post it as soon as possible. But for now it is better to end this thread.
     
  17. Feb 9, 2014 #16
    Hi utkarshakash!

    This problem is from one of the past test papers of your coaching institute. At the first sight, it hit me that I have seen the problem somewhere. Seeing at the arousing confusion, I tried searching the past papers and found the exact same problem. The solution goes along these lines:

    $$\frac{dN}{dt}=t^2-\lambda N$$
    For ##\frac{dN}{dt}## to be minimum: ##\frac{d^2N}{dt^2}=0##.
    $$\Rightarrow \frac{d^2N}{dt^2}=2t-\lambda \frac{dN}{dt}=2t-\lambda(t^2-\lambda N)=0$$
    $$\Rightarrow N=\frac{2t_0-\lambda t_0^2}{\lambda^2}$$

    Honestly, I don't see what the authors meant here. Clearly, the problem statement mentions that N is minimum at ##t=t_0## but they seem to be minimising the rate instead.
     
  18. Feb 9, 2014 #17

    Curious3141

    User Avatar
    Homework Helper

    Ah, thanks for the clarification. The question is indeed incorrect. I've seen quite a few of the papers from coaching institutes in India. While most of the questions are good and challenging, there are some which are poorly stated, or simply wrong, as in this case. Caveat emptor.
     
  19. Feb 9, 2014 #18

    utkarshakash

    User Avatar
    Gold Member

    Hey. Thanks for posting the solution. You are correct. This question is from FULL TEST V. I also can't understand why they have minimised the rate of change of N.

    Can you please post the solution of this:

    https://www.physicsforums.com/showthread.php?t=736847&page=1

    This is also from the same test paper.
     
  20. Feb 9, 2014 #19

    Curious3141

    User Avatar
    Homework Helper

    And in fact, Bruce pointed out (in the HH forum) that even this expression is incorrect!

    It should go:

    $$2t_0-\lambda (t_0^2 - \lambda N)=0$$

    $$\frac{2t_0}{\lambda} = t_0^2 - \lambda N$$

    $$\lambda N = t_0^2 - \frac{2t_0}{\lambda}$$

    $$N = \frac{t_0^2 - \frac{2t_0}{\lambda}}{\lambda}= \frac{\lambda t_0^2 - 2t_0}{\lambda^2}$$

    Which is not one of the provided options either. Sorry, I hadn't checked it myself until now.

    Bruce was also the one who helped corroborate my earlier work, I'm acknowledging him now.
     
  21. Feb 9, 2014 #20

    utkarshakash

    User Avatar
    Gold Member

    I also didn't notice it. I thank Bruce for pointing out the mistake. I think this question is useless as the question itself is wrongly formulated and the correct option is also missing.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted