Question

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=1.009u, where the atomic mass unit, u, is defined as follows: 1u=1.66×10−27kg.) 1. An electron (M=5.49×10−4u). Express your answer using four significant figures. 2. A proton (M=1.007u). Express your answer using one significant figure. 3. The nucleus of a lead atom (M=207.2u). Express your answer using four significant figures.

Answer 1

Let us consider that the v_i and v_f are the initial and final velocities of the neutron (whose mass is m) before and after the collision.

Initial velocity for the target particle (whose mass is M) is zero as it is statinoary before the collison.

Let us consider that the final velocity of the target particle is V.

So, from the momentum conservation principle,

Note: It is assumed that the neutron is continue to move in the same direction (true for small mass target particle that are ligher than neutron) as prior to the collison. In case if it bounce back then the v_f direction is going to be -Ve (This is case for heavy mass target particles that are heavier than neutron).

From the kinetic energy conservation principle,

By dividing the above two equations,

And by dividing this result equation with momentum expression, we can eliminate the V.

Hence, the ratio of final kinetic energy to the initial kinetic energy of the neutrons is given by,

in other way,

And we can prove in similar way to get the final expression as shown below, for the case of heavior target particle case, where the neutron will bounce back.

In either case, the neumerator value will still be positive as we are taking sqaure of that difference in mass values.

Now, let us use this for calculating the efficient target particle.

1. Target particle: Electron with M=5.49x10^-4 u

which means, the neutron had lost only 0.22% of its initial kinetic energy while colliding the eletron as target particle.

2. Target particle: Proton with M=1.007u

which means, the neutron had lost almost all of its initial kinetic energy while colliding the proton as target particle, meaning neutron comes to rest soon after colliding the proton by transferrring all its kinetic energy to the proton.

3. Target particle: Lead atom with M=207.2u

which means, the neutron had lost only 1.93% of its initial kinetic energy while colliding the Lead atom as the target particle, which means, after colliding this heavy target particle, the neutron bounce back with 98.07% of its initital kinetic energy.

Hence, among the given target particles, Proton is the promising target particle to slow down the neutrons.

Note: It is presumed that the aim is to finding out the slow down of netrons. But in the case of protons, they get accelerate and then triggers the stationary neutrons within the vicinity. In such case, it is better to consider the next efficient material that is Lead Atoms among the given three options.