Elastic collisions in a nuclear reactor

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SUMMARY

In nuclear reactors, neutrons released from fission must be moderated to sustain further reactions. The discussion focuses on calculating the ratio of a neutron's final kinetic energy (Kf) to its initial kinetic energy (Ki) during a head-on elastic collision with stationary target particles, specifically an electron with a mass of 5.49 x 10-4 u. The conservation of energy and momentum principles are essential for determining the energy transfer in such collisions. The mass of a neutron is given as 1.009 u, and the atomic mass unit (u) is defined as 1.66 x 10-27 kg.

PREREQUISITES
  • Understanding of elastic collisions in physics
  • Knowledge of conservation of energy and momentum
  • Familiarity with kinetic energy calculations (K.E. = 1/2 mv2)
  • Basic understanding of atomic mass units (u)
NEXT STEPS
  • Study the principles of elastic collisions in one-dimensional systems
  • Learn how to apply conservation of momentum in particle collisions
  • Research the effects of different target particle masses on neutron moderation
  • Explore the mathematical derivation of energy transfer equations in collisions
USEFUL FOR

Physicists, nuclear engineers, and students studying nuclear reactions and particle physics will benefit from this discussion, particularly those focused on neutron moderation techniques in nuclear reactors.

map7s
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In a nuclear reactor, neutrons released by nuclear fission must be slowed down before they can trigger additional reactions in other nuclei. To see what sort of material is most effective in slowing (or moderating) a neutron, calculate the ratio of a neutron's final kinetic energy to its initial kinetic energy, Kf / Ki for a head-on, elastic collision with each of the following stationary target particles. (Note: The mass of a neutron is m = 1.009 u, where the atomic mass unit, u, is defined as follows: 1 u = 1.66 10-27 kg.)
(a) An electron (M = 5.49 10-4 u).

I'm not quite sure how to tackle this problem. I know that kinetic energy is 1/2 mv^2, but I don't know how I can calculate the ratio of the final and kinetic energies without an initial velocity of the neutron.
 
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map7s said:
In a nuclear reactor, neutrons released by nuclear fission must be slowed down before they can trigger additional reactions in other nuclei. To see what sort of material is most effective in slowing (or moderating) a neutron, calculate the ratio of a neutron's final kinetic energy to its initial kinetic energy, Kf / Ki for a head-on, elastic collision with each of the following stationary target particles. (Note: The mass of a neutron is m = 1.009 u, where the atomic mass unit, u, is defined as follows: 1 u = 1.66 10-27 kg.)
(a) An electron (M = 5.49 10-4 u).

I'm not quite sure how to tackle this problem. I know that kinetic energy is 1/2 mv^2, but I don't know how I can calculate the ratio of the final and kinetic energies without an initial velocity of the neutron.
Assume an initial velocity of v for the neutron. Then have it collide head on with an electron at rest. The kinetic energy of the electron has to be equal to the loss of kinetic energy of the neutron. What is the energy of the electron in terms of the initial kinetic energy? (hint: you have to apply conservation of energy and momentum).

Better still, work out (or look up) the general relationship for transfer of energy (ie. loss of energy of the incident particle) in a one dimensional collision between objects of mass M1 and M2 where M2 is at rest. Apply that to each of the particles. What is the value for M2 that results in maximum energy loss for M1?

AM
 
I am sorry to bring this up from 2 years ago (lol...), but I am having trouble on this exact problem. I can't find a way to combine the two equations to get the final and initial energies. Any help would be greatly appreciated.
 

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