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## Homework Statement

A thin target of lithium is bombarded by helium nuclei of energy

*E*. 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

_{0}*E*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 [tex]^{7}Li+^{4}He \to ^{10}B+^{1}n-2.8MeV[/tex]

_{0}**a.**What is

*E*the minimum value of

_{0,threshold}*E*for which neutrons can be produced? What is the energy of the neutrons at this threshold?

_{0}**b.**Show that if the incident energy falls in the range

*E*<

_{0,threshold}*E*<

_{0}*E*+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.)

_{0,threshold}## Homework Equations

*E*=

_{i}*E*

_{f}*p*=

_{i}*p*

_{f}[tex]\frac{\partial E_{f}}{\partial v_{B}}=0[/tex]

## The Attempt at a Solution

To solve

**a**is a snap. (Ok, it took me a few hours.) I have [tex]E_{i}=E_{0}=2mv_{He}^{2}[/tex] [tex]E_{f}=E_{0}-2.8MeV=5mv_{B}^{2}+\frac{1}{2}mv_{n}^{2}[/tex] [tex]p_{i}=4mv_{He}\hat{i}[/tex] [tex]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}[/tex]

To get a minimum energy i choose [tex]\theta_{B}=\theta_{n}=0[/tex] which gives

[tex]p_{f}=(10mv_{B}+mv_{n})\hat{i}[/tex] Now I solve out

*v*from the momentum conservation and put it into the energy equation and then use [tex]\frac{\partial E_{f}}{\partial v_{B}}=0[/tex] to find the minimum energy. Then I put it into the

_{n}*E*equation and get

_{f}*E*and the neutron energy is

_{0}=4,4 MeV*0,15 MeV*.

**b.**To try to solve this problem I use the momentum conservation to break out [tex]v_{b}=\frac{4v_{He}-v_{n}}{10}[/tex] and put it into the energy equation to get [tex]\frac{11}{20}v_{n}^{2}-\frac{2}{5}v_{He}v_{n}-\frac{6}{5}v_{He}^{2}+2.8 MeV=0[/tex] i solve out [tex]v_{n}=\frac{4}{11}v_{He} \pm \frac{1}{\sqrt{m}}\sqrt{ \frac{140}{121} E_{0}-\frac{56}{11}MeV}[/tex] 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.