(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

A thin target of lithium is bombarded by helium nuclei of energy E_{0}. 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 E_{0}by 2.8 MeV. The relative masses of the particles are: helium, mass 4; lithium, mass 7; boron, mass 10; neutron, mass 1. The reaction can be symbolized:

^{7}Li+^{4}He→^{10}B+^{1}n −2.8MeV

a. What is E_{0threshold}the minimum value of E_{0}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_{0threshold}<E_{0}<E_{0threshold}+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.)

2. Relevant equations

E = E_{f}+ 2.8

p_{i}=p_{f}

3. The attempt at a solution

I've tried solving the equations through the laboratory system, but for that I have to make an assumption that the angles of scattering are zero (and even then I'm not sure how to do it). Using the center of mass system, I get:

7v_{He}- 4v_{L}= 10v_{B}- v_{N}= 11v_{CM}

7v_{He}= 4v_{L}

10v_{B}= v_{N}

E_{0}= 49m^{2}v_{He}^{2}([itex]\frac{1}{14m}[/itex]+[itex]\frac{1}{8m}[/itex]) = 1m^{2}v_{N}^{2}([itex]\frac{1}{20m}[/itex]+[itex]\frac{1}{2m}[/itex])

All the above equations were made from my knowledge of calculations in the system. However, I'm not sure how these would lead to getting the threshold energy...

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# Homework Help: Inelastic Collision: Threshold Energy

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