Topics | Particles

Question 1 - Explain why gamma rays would be emitted during an inelastic collision between a neutron and a target atomic nucleus.

During the inelastic collision the target nucleus emits alpha and beta particles and so becomes more energetic. When its energy levels decrease to their ground level the difference in energy is emitted as protons of energy, ie gamma rays are emitted.

Question 2 - Determine the speed of a 1 MeV neutron using a non-relativistic formula for kinetic energy and hence determine whether relativistic mechanics are to be used in any further working.

This speed is five hundredths of the speed of light, so I do not have to use relativistic mechanics in any further working, I will use Newtonian mechanics.

Question 3 - If a neutron of mass m₁ and incident speed u₁ collides with a stationary nucleus of mass m₂, show that for an elastic, head on collision:

For all head on collisions total momentum is conserved, hence by the principle of conservation of linear momentum:

For all elastic collisions total kinetic energy is conserved, hence by the principle of conservation of energy:

• m₁ (u₁ - v₁) (u₁ + v₁) = m₂v₂² as required

Question 4 - From these two equations show that:

Question 5 - Show that the ratio of kinetic energy lost by object 1 in the collision to its initial kinetic energy is given by:

The collision is elastic, head on so total kinetic energy is conserved, hence by the principle of conservation of energy:

This is because as stated in the question:

Question 6 - Plot a graph of the kinetic energy ratio (K_L / K_I) against the mass ratio (m₂ / m₁) for mass ratios from 0 to 7.

Below is a table of values for the graph:

Below is the graph:

Question 7 - Prove by differentiation that the kinetic energy ratio is a maximum when the ratio of the masses equals one. Now examine the graph again - is it correct ?

By differentiating using the product rule I have obtained the following first derivative of the graph:

At all of the stationary points of the graph y the gradient is zero. So if the point (1,1) is a stationary point the gradient at that point is zero.

So the point (1,1) is a stationary point. To determine the nature of this stationary point I will find the second derivative of the graph:

At all of the maximum points of the graph y the second derivative is less than zero. So if the point (1,1) is a maximum point the second derivative at that point is less than zero.

Hence, the stationary point (1,1) is a maximum point of the graph.

Therefore by differentiation I have proven that the kinetic energy lost by the incident particle is a maximum when the ratio of the masses between it and the target particle equals one. So the most effective interchange of energy occurs when incident particle and target particle have equal mass.

When re-evaluating the graph I believe that it is correct because it shows this and that when the mass ratio increases the amount of energy interchange decreases, which I have indirectly proven by differentiation.

So if we denote the number of collisions required to thermalise a 1 MeV neutron by n, we have the following function:

So 54 head on collisions with a ¹²C nucleus are needed to thermalise a 1 MeV neutron.

The Moderator in a nuclear reactor reduces the energy of and therefore the speed of neutrons, allowing them to be readily absorbed by Uranium atoms, thus inducing fission.

Protons (H⁺) would be a very good Moderator because of the near 100% maximum loss of kinetic energy of the neutron in a collision, but protons are not used because they are absorbed.

However, heavy water (D₂0), light water (H₂0), graphite (CH₃) are used because they each have small nuclei, so there is a large loss of kinetic energy in the neutron from each collision, because by the below formula:

So if m₁ @ m₂ then m₁ (neutron) will virtually stop after collision. Showing that if a Moderator is to do an effective job of slowing the neutron down, its relative atomic mass must be very close to that of the colliding neutron’s mass.

Question 11 - If a Moderator was not used how many head on collisions with a ²³⁸U nucleus would be needed to thermalise a 1 MeV neutron ?

From each head on collision with a ²³⁸U nucleus 2% loss of kinetic energy occurs in the neutron (m₁), therefore 98% is remaining.

K_INITIAL of m₁ = 1 MeV

= 1 ´ 10⁶ eV

K_DESIRED by m₁ = 2.5 ´ 10^-2 eV

So if we denote the number of collisions required to thermalise a 1 MeV neutron by n, we have the following function:

Question 12 - Explain qualitatively why the answers to question 8 are maximum values and why the answers to question 9 and 11 are minimum values.

In question 8 I assumed that the collisions elastic and head on, and that the initial speed of the Moderator (m₂) was zero. This is because the preliminary proofs of the kinetic energy ratio equation involved the principle of conservation of linear momentum and energy because I assumed no loss of kinetic energy occurring. There can not be more interchange of energy if it is an elastic, head on collision because the coefficient of restitution, e = 1. The most general form of the kinetic energy ratio equation is shown below:

So for a collision at an angle (not head on) the amount of kinetic energy lost by the neutron reduces. Therefore I have found the maximum loss of kinetic energy from the neutron in a collision, so the answers to question 8 are maximum values.

In questions 9 and 11 I have made the same assumptions as I have in question 8 because I have used the answers to question 8 (c) and (d). Therefore for each collision I have used the maximum loss of kinetic energy of the neutron, so it follows that I have found the minimum number of collisions required to reduce the energy of the neutrons a given amount.

In reality the assumptions that I have made do not hold true because the neutrons may not be perfectly spherical, or hard enough not to change shape in a collision. So if the collisions are inelastic in reality after each collision total kinetic energy reduces, so subsequent collisions take place with less velocity, and therefore force, so more collisions would be required to reduce the energy of a neutron a given amount. There may also be forces of attraction between the particles, which I have not accounted for, these may influence the loss of kinetic energy per neutron per collision. The assumption that the velocity of the Moderator (m₂) was zero is evidently not true because atoms are in constant, random motion, so in reality, since the Moderator is not stationary (increasing the likelihood of an inelastic collision), the amount of kinetic energy lost by the neutron reduces. However collisions of gaseous atoms are generally referred to as perfectly elastic, so the effects of these factors are presumably negligible.

Faysal Riaz (Yr12 student 1998-1999)