Anonymous asked in Science & MathematicsPhysics · 8 months 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

1 Answer

  • 8 months ago
    Favourite answer

    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

    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.)

    • ...Show all comments
    • Steve4Physics
      Lv 7
      8 months 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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