Moderating a Neutron In a nuclear reactor, neutrons released...

Moderating a Neutron 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, $K / K_{i},$ for a head-on elastic collision with each of the following stationary target particles. (Note: The mass of a neutron is $=1.009 \mathrm{u},$ where the atomic mass unit, $\mathrm{u},$ is defined as follows: $1 \mathrm{u}=1.66 \times 10^{-27} \mathrm{kg.} )$ (a) An electron $\left(M=5.49 \times 10^{-4} \mathrm{u}\right) .$ (b) A proton $(M=1.007 \mathrm{u})$ . (c) The nucleus of a lead atom $(M=207.2 \mathrm{u}) .$

Moderating a Neutron 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, $K / K_{i},$ for a head-on elastic collision with each of the following stationary target particles. (Note: The mass of a neutron is $=1.009 \mathrm{u},$ where the atomic mass unit, $\mathrm{u},$ is defined as follows: $1 \mathrm{u}=1.66 \times 10^{-27} \mathrm{kg.} )$ (a) An electron $\left(M=5.49 \times 10^{-4} \mathrm{u}\right) .$ (b) A proton $(M=1.007 \mathrm{u})$ .
(c) The nucleus of a lead atom $(M=207.2 \mathrm{u}) .$

Key Concepts

Kinetic Energy Transfer

The fraction of kinetic energy that a neutron retains or loses after colliding with another particle can be determined through the principles of energy and momentum conservation. In head-on collisions, formulas can be derived to relate the initial and final kinetic energies of the neutron based on the masses of the colliding particles. This concept is crucial for understanding how different materials make effective moderators by comparing energy transfer efficiencies.

Mass Ratio in Collisions

The effectiveness of a collision in transferring energy between particles is strongly dependent on the ratio of their masses. In neutron moderation, if the mass of the target nucleus is much lighter or similar to that of the neutron, a greater fraction of the neutron's kinetic energy can be transferred in a collision. Conversely, when the target mass is much larger, the neutron retains most of its energy. This mass dependence is fundamental to selecting appropriate materials for moderating neutrons in a reactor.

Neutron Moderation

In a nuclear reactor, neutrons need to be slowed down (moderated) to energies where they are more likely to induce further fission events. Moderation is achieved by allowing the neutrons to collide elastically with nuclei in a moderator material, transferring some of their kinetic energy during each collision. This process is critical for maintaining a controlled chain reaction, as the efficiency of fission reactions often depends on the energy of the incoming neutrons.

Elastic Collisions

An elastic collision is one in which both momentum and kinetic energy are conserved. In the context of neutron moderation, elastic collisions are used to model the interactions between a neutron and the nuclei of the moderator material. The conservation laws provide the framework to determine how much kinetic energy is retained or lost by the neutron after a collision with a target particle.

Transcript

00:01 Question 39 says moderating a neutron in a nuclear reactor is done by slowing down and reducing the kinetic energy of the neutron.

00:09 To see what sort of material is most effective in slowing this, we calculate the ratio of the final kinetic energy to its initial kinetic energy, that is k to k to ki, for a head -on elastic collision, with each of the following stationary target particles.

00:23 And they tell us mass in amu, which is fine.

00:27 This is probably the easiest thing to do here is to drive a couple equations.

00:33 We're given a very specific case.

00:35 We have this one thing that has an m1 and an initial velocity i'm going to call it vi.

00:41 It goes through an elastic head -on collision, and it still has its mass, m2, and then it has a final velocity that i'm going to call vf.

00:50 Now, we are asked to find k to k -i, where k is kinetic energy.

00:57 And so this is of course one half uh...

01:00 Who's this is still that particle m one uh...

01:02 One half m one v f squared to one half m one v i squared which we can see the mass and the half cancel and so the ratio of their kinetic energies is really the ratio of the final velocity of that object squared to the initial uh...

01:22 Velocity of that object squared however i don't want to do a whole bunch of problems uh...

01:26 We have three three different particles to check where i've got to go through the law of conservation of momentum and the law of conservation of kinetic energy several times.

01:35 Instead, i'm going to derive a fairly straightforward equation for this.

01:40 And so it starts with the law of conservation of momentum, where i have the mass of my neutron, i'm going to call it m1 times its initial velocity, vi, and the other particle we're told is at rest.

01:51 So this is the total initial momentum.

01:53 This has to equal the final momentum, so m1 times the final velocity of our neutron plus m2, whatever particle we're smashing it into, times its velocity, which i'm going to call v2.

02:06 Now, remember, in a head -on perfectly elastic collision, we can say that the initial velocity, vi, of the first object, minus the initial velocity of the second object, which in this case is zero, it's at rest, is equal to the opposite of their differences.

02:24 So this is vf minus v2 after the collision.

02:29 And so what i get is vi is equal to v2 minus vf.

02:36 And i'm going to go ahead then and solve this for v2 v2 equals vi plus vf this means i can eliminate the v2 from this equation so i'm going to rewrite this equation without that v2 so i'm going to go ahead and say m1 vi so that our neutron times its initial velocity is equal to the mass of our neutron m1 times its final velocity vf plus m2 times and now i'm going to make this substitution vi plus vf...

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