Moderating Neutrons: Calculating Kf/Ki Ratio

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In nuclear reactors, neutrons must be moderated to trigger further reactions, and the Kf/Ki ratio is crucial for understanding this process. The discussion focuses on calculating this ratio for head-on elastic collisions with various particles, including electrons, protons, and lead nuclei. Initial velocity is not necessary for the calculation, as the ratio simplifies to a comparison of final and initial kinetic energies. Participants emphasize using conservation of momentum and energy to derive the necessary relationships, ultimately leading to a straightforward solution. The problem, initially perceived as complex, reveals itself to be simpler upon proper analysis and application of the relevant equations.
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Homework Statement


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, 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.

An electron (M = 5.49*10-4 u).

A proton (M = 1.007 u).

The nucleus of a lead atom (M = 207.2 u).

Homework Equations


Pf=Pi
m1u1 = m1v1,f2 + m2v2,f2
Kf=Ki
(1/2)m1u2 = (1/2)m1v1,f2 + (1/2)m2v2,f2

The Attempt at a Solution


Honestly, I do not even know how to begin this problem set due to lacking an initial velocity. I have been searching for a few hours of how to properly derive an equation and have not come up with anything ground breaking. Any tips are greatly appreciated, I wish I had more information to give.
 
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You don't need the initial velocity since you're only asked for a ratio Ki/Kf which reduces to a ratio of velocities. =)
 
I'm assuming that by

m1u0 = m1v1f^2 = m2v2f^2

You meant m_1u_0=m_1v_{1f}^2-m_1v_{1f}^2 where you're assuming that the colliding particle continues in the same direction, while allowing a negative velocity to prove you wrong.

That is completely correct.

All that's left for you now is to find a relationship between the momentum of a particle and its kinetic energy. Once you have the mass and the energy, you also have the velocity. But there's no need for you to go through that whole derivation. Simply find a way to express the momentum using the energy, and rephrase your equations in terms of the energy only, you should then find it easy to derive the ratio of the kinetic energies.
 
Matterwave said:
You don't need the initial velocity since you're only asked for a ratio Ki/Kf which reduces to a ratio of velocities. =)

I was under the impression you need initial velocity to be able to properly set up a conservation of momentum and kinetic energy formula.

@Royal: That was an error on my part it was meant to be addition between the two final velocities.

In terms of your response; how would one go about setting up a relationship between the two? This is my first time dealing with a question such as this and all I know if manipulating conservation formulas and substituting them.
 
Notice that Kf/Ki=mvf^2/mvi^2

See if you can manipulate your equations to get something like that.
 
Matterwave said:
Notice that Kf/Ki=mvf^2/mvi^2

See if you can manipulate your equations to get something like that.
You are very correct, sir. I could not see how something that simple was the actual work for this problem. I completely over complicated the problem, plus I was trying to solve for the entire system rather than the Kinetic Energy of the Neutron.

I ended up asking my TA about this today because I was frustrated. You end up setting up how you showed and then cancel the (1/2) and m1 leaving you with vf2/vi2. You plug in the formula for head on collisions between a mobile and stationary object which is v1,f=[(m1-m2)/(m1+m2)]. Needless to say you square that as the order of operation states and the vi cancels out leaving you to just plug masses into the equation.

Thank you very much Matterwave, wish I had been able to understand the problem earlier. It turned out be one of the easiest of all.

Very Respectfully,
Darius
 
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