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Homework Help: Calculate the temperature of neutrons emerging from a reactor

  1. Jan 2, 2018 #1
    1. The problem statement, all variables and given/known data
    A collimated beam of thermal neutrons emerges from a nuclear reactor and passes through a speed selector into a detector. The number of neutrons detected in a second with speeds in the range 4000 to 4010 m s−1 is twice as large as the number per second detected with speeds in the range 2000 to 2010ms−1. What is the temperature of the moderator in the nuclear reactor?


    2. Relevant equations


    3. The attempt at a solution
    So the speed distribution is proportional to ##v^3e^{-\frac{v^2}{v_{th}^2}}## so my instinct was to write ##v^3e^{-\frac{v^2}{v_{th}^2}}|_{4000}^{4010}=2v^3e^{-\frac{v^2}{v_{th}^2}}|_{2000}^{2010}## but then i don't know how to solve this.

    Many thanks
     
    Last edited: Jan 2, 2018
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  3. Jan 2, 2018 #2

    TSny

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    The speed distribution should have a factor of ##v^2## instead of ##v^3##. However, the rate at which neutrons strike the detector would have the factor of ##v^3##.

    What does ##v_{th}## represent? If it has dimensions of speed, then note that the argument of your exponential function is not dimensionless.

    Can you explain the notation ##v^3e^{-\frac{v^2}{v_{th}}}|_{4000}^{4010}## as regards the interpretation of ##|_{4000}^{4010}## ?
     
  4. Jan 2, 2018 #3
    Sorry that is meant to be ##v_{th}^2## I have missed of the ##^2##. That notation is meant to be putting in limits from 4000 to 4010
     
  5. Jan 2, 2018 #4

    TSny

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    Limits of an integration? Did you perform an integration?
     
  6. Jan 2, 2018 #5
    no, but i probably should have!!
     
  7. Jan 2, 2018 #6

    TSny

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    Yes. However, an interval of 10 m/s is quite small compared to 2000 m/s or 4000 m/s. So, you can get a decent answer by considering the integrand as constant over the interval of 10 m/s.
     
  8. Jan 2, 2018 #7
    So can we approximate this as ##(4000)^3e^{-\frac{4000}{v_{th}^2}}\int_{4000}^{4010}{dv}=2(2000)^3e^{-\frac{2000}{v_{th}^2}}\int_{2000}^{2010}{dv}## ?

    ##T=\frac{m}{K_B}\Big(\frac{4000^2-2000^2}{\ln{\frac{4000^3}{2(2000^3)}}}\Big)^2##
     
  9. Jan 2, 2018 #8

    TSny

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    Looks good. But did you drop a factor of 2 in the relation between ##v_{th}^2## and ##T##?
     
  10. Jan 2, 2018 #9
    I did- was a typo though. Thank you for your help!! Very much appreciated
     
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