# 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

- Steve4PhysicsLv 78 months agoFavourite 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 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.)

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).