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Elastic collisions in a nuclear reactor

  1. Oct 16, 2006 #1
    In a nuclear reactor, neutrons released by nuclear fission must be slowed down before they can trigger additional reactions in other nuclei. To see what sort of material is most effective in slowing (or moderating) a neutron, calculate the ratio of a neutron's final kinetic energy to its initial kinetic energy, Kf / Ki for a head-on, elastic collision with each of the following stationary target particles. (Note: The mass of a neutron is m = 1.009 u, where the atomic mass unit, u, is defined as follows: 1 u = 1.66 10-27 kg.)
    (a) An electron (M = 5.49 10-4 u).

    I'm not quite sure how to tackle this problem. I know that kinetic energy is 1/2 mv^2, but I don't know how I can calculate the ratio of the final and kinetic energies without an initial velocity of the neutron.
  2. jcsd
  3. Oct 17, 2006 #2

    Andrew Mason

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    Assume an initial velocity of v for the neutron. Then have it collide head on with an electron at rest. The kinetic energy of the electron has to be equal to the loss of kinetic energy of the neutron. What is the energy of the electron in terms of the initial kinetic energy? (hint: you have to apply conservation of energy and momentum).

    Better still, work out (or look up) the general relationship for transfer of energy (ie. loss of energy of the incident particle) in a one dimensional collision between objects of mass M1 and M2 where M2 is at rest. Apply that to each of the particles. What is the value for M2 that results in maximum energy loss for M1?

  4. Nov 12, 2008 #3
    I am sorry to bring this up from 2 years ago (lol...), but I am having trouble on this exact problem. I can't find a way to combine the two equations to get the final and initial energies. Any help would be greatly appreciated.
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