Constraining final energies of neutron in nuclei interaction

[Order]

Homework Statement

A thin target of lithium is bombarded by helium nuclei of energy E₀. The lithium nuclei are initially at rest in the target but are essentially unbound. When a Helium nucleus enters a lithium nucleus, a nuclear reaction can occur in which the compound necleus splits apart into a boron nucleus and a neutron. The collision is inelastic, and the final kinetic energy is less than E₀ by 2.8 MeV. The relative masses of the particles are: helium, mass 4; lithium, mass 7; boron, mass 10; neutron, mass1. The reaction can be symbolized $$^{7}Li+^{4}He \to ^{10}B+^{1}n-2.8MeV$$

a. What is E_0,threshold the minimum value of E₀ for which neutrons can be produced? What is the energy of the neutrons at this threshold?

b. Show that if the incident energy falls in the range E_0,threshold<E₀<E_0,threshold+0,27 MeV, the neutrons ejected in the forward direction do not all have the same energy, but must have either on or the other of two possible energies. (You can understand the origin of the two groups by looking at the reaction in the center of mass system.)

Homework Equations

E_i=E_f

p_i=p_f

$$\frac{\partial E_{f}}{\partial v_{B}}=0$$

The Attempt at a Solution

To solve a is a snap. (Ok, it took me a few hours.) I have $$E_{i}=E_{0}=2mv_{He}^{2}$$ $$E_{f}=E_{0}-2.8MeV=5mv_{B}^{2}+\frac{1}{2}mv_{n}^{2}$$ $$p_{i}=4mv_{He}\hat{i}$$ $$p_{f}=(10mv_{B}\cos\theta_{B}+mv_{n}cos\theta_{n})\hat{i}+(10mv_{B}\sin\theta_{B}+mv_{n}cos\theta_{n})\hat{j}$$
To get a minimum energy i choose $$\theta_{B}=\theta_{n}=0$$ which gives
$$p_{f}=(10mv_{B}+mv_{n})\hat{i}$$ Now I solve out v_n from the momentum conservation and put it into the energy equation and then use $$\frac{\partial E_{f}}{\partial v_{B}}=0$$ to find the minimum energy. Then I put it into the E_f equation and get E₀=4,4 MeV and the neutron energy is 0,15 MeV.

b. To try to solve this problem I use the momentum conservation to break out $$v_{b}=\frac{4v_{He}-v_{n}}{10}$$ and put it into the energy equation to get $$\frac{11}{20}v_{n}^{2}-\frac{2}{5}v_{He}v_{n}-\frac{6}{5}v_{He}^{2}+2.8 MeV=0$$ i solve out $$v_{n}=\frac{4}{11}v_{He} \pm \frac{1}{\sqrt{m}}\sqrt{ \frac{140}{121} E_{0}-\frac{56}{11}MeV}$$ Now at the threshold energy the second term is zero, so that is one limit, but the other limit at +0,27 MeV is difficult for me to see. Also, I guess that the boron goes in the other direction with the plus sign, and you don't have to look at the center of mass system to understand that.

 

[anathema]

To fully understand the two possible energies for the neutrons, we need to look at the reaction in the center of mass (CM) system. In this system, the total momentum is zero, so we can equate the initial and final momenta:

p_{i}=p_{f}=0

This means that the momenta of the boron and neutron must be equal and opposite in direction, so we can write:

p_{B}=-p_{n}

We can also write the energy equation in the CM system:

E_{i}=E_{f}

Since the initial energy is simply the kinetic energy of the helium nucleus, we can write:

E_{i}=E_{0}

For the final energy, we can use the fact that the total kinetic energy of the system must be conserved, so we can write:

\frac{1}{2}m_{B}v_{B}^{2}+\frac{1}{2}m_{n}v_{n}^{2}=\frac{1}{2}m_{He}v_{He}^{2}-2.8 MeV

Now, using the fact that p_{B}=-p_{n}, we can write:

m_{B}v_{B}=-m_{n}v_{n}

Substituting this into the energy equation, we get:

\frac{1}{2}m_{B}v_{B}^{2}+\frac{1}{2}m_{n}v_{n}^{2}=\frac{1}{2}m_{He}v_{He}^{2}-2.8 MeV

\frac{1}{2}m_{B}v_{B}^{2}+\frac{1}{2}m_{B}^{2}\left(\frac{v_{B}}{m_{n}}\right)^{2}=\frac{1}{2}m_{He}v_{He}^{2}-2.8 MeV

Solving for v_{B}, we get:

v_{B}=\frac{m_{n}}{m_{B}+m_{n}}v_{He}

Now, if we substitute this back into the energy equation, we get:

\frac{1}{2}m_{n}v_{n}^{2}=\left(\frac{m_{n}}{m_{B}+m_{n}}