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Calculate density of neutron in a beam

  1. Apr 27, 2013 #1
    Problem:

    A flux of 10^12 neutrons/m^2 emerges each second from a port in a nuclear reactor. If these neutrons have a Maxwell-Boltzmann energy distribution corresponding to T=300 K, calculate the density of neutrons in the beam.

    Solution:

    The average velocity of neutrons coming from the detector is [itex]\bar{v}=\sqrt{\frac{8kT}{πm}}[/itex].

    Substituting the following,

    k=1.38x10-23 J/K
    T=300 K
    m=1.675 x 10-27 kg

    yields a velocity [itex]\bar{v}=2509 m/s[/itex]

    A flux F is given by [itex]F=\rho\bar{v}[/itex] [itex]\Rightarrow \rho=\frac{F}{\bar{v}}[/itex]

    Substituting F=10^12 m^-2 s^-1 and the obtained value above for [itex]\bar{v}[/itex] yields,

    ρ=4 x 10^9 /m^3

    However this disagrees with the value in my books which says the correct answer is 1.6 x 10^9 /m^3. I think it's a pretty straight-forward problem. Where did I go wrong?
     
  2. jcsd
  3. Apr 27, 2013 #2

    TSny

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    There's a complication due to the fact that the average velocity of the neutrons emerging through the port is not the same as the average velocity of the neutrons inside the reactor. This causes a modification in the formula ##F = \rho \overline{v}##. See the middle of page 2 of these notes.
     
  4. Apr 27, 2013 #3
    I see that the end result of the analysis yields a factor of 1/4. Using this modified formula, I then get the correct answer for the density.

    However, it's not exactly clear from where the 1/4 arises. Could you please elaborate?
     
  5. Apr 27, 2013 #4

    TSny

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    Consider all particles moving toward the opening in a certain direction. Particles with greater speed are more likely to make it through the opening during a fixed time interval dt than particles with less speed. So, the distribution of speeds of particles making it through the opening is shifted somewhat towards higher speeds compared to the Maxwellian distribution.

    Also, if you look at particles moving toward the opening from different directions, then those moving in a direction roughly perpendicular to the hole will have a greater chance of getting through than those moving almost parallel to the opening. So, the distribution of directions of the particles making it through the hole will not be isotropic.

    You cannot expect the average speed of particles moving through the hole to be the same as the average speed ##\overline{v}## of particles in the Maxwellian distribution.

    The only way that I can see why the factor is 1/4 rather than some other factor is to carry through the calculation as shown in the link.
     
  6. Apr 27, 2013 #5

    TSny

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    There is one thing that confuses me about the statement of the problem. The density ##\rho## that you calculate using the formula ##F = \frac{1}{4} \rho\overline{v}## is the number density of the neutrons while they are inside the reactor in a Maxwell-Boltzmann distribution. But the problem is asking for the density in the "beam". The particles will emerge from the opening ("port") in all polar directions from θ = 0 to θ = ##\pi/2## radians and all azimuthal directions from 0 to ##2\pi## radians. So, they will spread out and the density will vary with position.

    So, I wonder if I am understanding the statement of the problem.
     
  7. Apr 27, 2013 #6
    The answer I arrive at using [itex]F=\frac{1}{4}\rho\bar{v}[/itex] agrees with the answer listed in my textbook. Perhaps, for an infinitesimal moment after passing through the opening, the constituent neutrons of the beam are roughly of the same average speed as the neutrons in the Maxwellian distribution inside the reactor.
     
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