Question 3.2: Using the classical expression for the neutron kinetic energ......
Using the classical expression for the neutron kinetic energy E (see Chapter 2), calculate the velocity v of a 2 MeV neutron before it collides with the nucleus of a nearly stationary atom. What percentage of the speed of light is the neutron traveling at in this case?
In Chapter 2, we learned that the classical expression for the kinetic energy E = 1/2 mv^2 turns out to be relatively accurate for neutron energies up to about 100 MeV, because even at 100 MeV, the kinetic energy of a neutron is only about one-tenth of its rest mass energy (939.55 MeV). If the velocity is measured in meters per second, and the kinetic energy E is measured in MeV, the velocity is related to the energy by v = 1.383 × 10^7 \sqrt{E} . Since E = 2 MeV, v = 1.955 × 10^7 m/s. Since the speed of light is 3 × 10^8 m/s, the 2 MeV neutron is only traveling at 0.1955 × 10^8/3 × 10^8 = 6.5% of the speed of light.