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Dale
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First, you need to learn the right formulas. E=½mv² is the non-relativistic kinetic energy not the relativistic kinetic energy. 470 MeV of KE accelerates a neutron to about .75 c (gamma factor 1.5).Cosmos2001 said:Well, the minimum amount of energy required to accelerate a neutron close to speed of light:
neutron=1.67493×10-27 kg
c=299792458 m/s
E=½mv²
EJ=½(1.67493×10-27)(299792458)² =75.26760×10-12J
EeV=75.26760×10-12/1.60218×10-19 = 469.78242×106 = 470MeV
The average energy released in fission of one Pu-239 atom is 210MeV, emitted neutrons 5.9MeV which is very far from 470MeV, there is not enough energy to emit neutron close to speed of light.
http://en.wikipedia.org/wiki/Plutonium-239#Nuclear_properties
My question is: are there atomic nucleuses that emit neutrons having kinetic energy exceeding 470MeV?
Second, that amount of energy would result in approximately a 50% deviation from classical behavior. Well designed experiments can easily be made to detect deviations from classical behavior of less than 1% which would correspond to a KE of less than about 9 MeV.
Third, if you have a high KE particle which undergoes a nuclear reaction and releases a neutron then the KE of the neutron will be more determined by the high KE of the original particle than by the energy released from the nuclear reaction. This is how most high energy neutrons are actually produced in the lab.