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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 nucleus 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}( Ans: neutron energy = 0.15MeV)

**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

## Q = 2.8 Mev##

## \theta = (\vec v_B,\vec v_N) ##

Energy equation :

## E_0 = E_B + E_N + Q##

Momenta equation:

##\left\{

\begin{array}{}

\vec P_i = 4\ \vec v_{0} \\

\vec P_f = 10\ \vec v_B + \vec v_N

\end{array}

\right.

##

## The Attempt at a Solution

[/B]

Hello, I have reached a dead end with this problem and I'd greatly appreciate some help.

My idea for solving question (a) is to isolate ##E_0## and study the solutions of ##\frac{dE_0}{dE_n} = 0 ## that are positive ( a nul solution means that no neutron is produced, and a negative solution is non physical ) , and then study the sign of the second derivative.

The first steps are to write the conservation of momentum equation, and apply the right substitutions with the energy equation:

By conservation of momentum, ## \frac{|\vec P_i|^2}{4} =\frac{|\vec P_f|^2}{4} ##, so

## 4 v_{0}^2 = 25v_B^2 + \frac{1}{4} v_N^2 + 5 v_bv_n\cos(\theta) ##

In terms of energy, that previous equation is equivalent to

## 2E_0 = 5 E_B + \frac{1}{2} E_N + \sqrt{10} \sqrt{E_BE_N} \cos(\theta) ##

Using ## E_B = E_0 - E_N - Q ##, I get :

## 3E_0 -\frac{9}{2}E_n - 5Q + \sqrt{10} \sqrt{(E_0 - E_N - Q)E_N} \cos(\theta) = 0 ##

And I am stuck because I don't see any easy way to isolate ##E_0## !