Anonymous
Anonymous asked in Science & MathematicsPhysics · 3 weeks ago

physics help?

Suppose that a neutron in a nuclear reactor initially has an energy of 4.8 × 10^−13 J. How many head-on collisions with carbon nuclei at rest must this neutron make before its energy  is reduced to 1.6 × 10^-19, assuming collisions are elastic

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  • 3 weeks ago
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    Mass of carbon nucleus (6 neutrons, 6 protons) is 12u.  Mass of neutron is 1u.

    For a head-on collision, the velocity of the neutron is changed by a factor (1-12)/(1+12) = -0.84615   (the minus sign means direction is reversed.)  E.g. see http://hyperphysics.phy-astr.gsu.edu/hbase/elacol2...

    You might need to include the derivation if not given the formula.

    Since kinetic energy is proportional to speed squared, the kinetic energy reduces by a factor

    (1/0.84615)² = 1.397

    After n collisions the kinetic energy of the neutron reduces by a factor (4.8*10^-13) / (1.6*10^-19) = 3*10^6

    Since the kinetic energy is reduced by a factor 1.397 for each collision:

    1.397ⁿ = 3*10^6

    nlog(1.397) = log(3*10^6)

    n = log(3*10^6) / log(1.397) = 45 (to 2 sig. figs.)

    • Steve4Physics
      Lv 7
      3 weeks agoReport

      Proportional to what?  Neutron’s speed after each collision (Vf) is proportional to its velocity before the collision (Vi).

      Vf/Vi = 0.84615


      The collision changes the velocity by a factor 0.84615
      This is a reduction by a factor 1/0.84615 (= 1.182).

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