Question
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 , where the atomic mass unit, m=1.009u , is defined as follows: 1 u = 1.66 x 10^(-27) kg.
A)An electron- M= 5.49 x 10^(-4)u .
B)A proton- M= 1.007u .
C)The nucleus of a lead atom M= 207.2u .
A)An electron- M= 5.49 x 10^(-4)u .
B)A proton- M= 1.007u .
C)The nucleus of a lead atom M= 207.2u .
Answers
Use the formula for head-on elastic collisions that you will find at:
http://hyperphysics.phy-astr.gsu.edu/hbase/elacol2.html
For the colliding neutron of mass m1,
Vf = Vi * (m1 - m2)/(m1 + m2)
m2 is the mass of the particle it collides with.
The largest mass loss will be in collisions with protons.
http://hyperphysics.phy-astr.gsu.edu/hbase/elacol2.html
For the colliding neutron of mass m1,
Vf = Vi * (m1 - m2)/(m1 + m2)
m2 is the mass of the particle it collides with.
The largest mass loss will be in collisions with protons.
saSAsS
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