# Collision and energy threshold

• geoffrey159

## 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 nucleus 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 E0,threshold the minimum value of E0 for which neutrons can be produced? What is the energy of the neutrons at this threshold?
( Ans: neutron energy = 0.15MeV)

b. Show that if the incident energy falls in the range E0,threshold<E0<E0,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

## 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## !

The hint is given in the problem statement. The threshold is 2.8 MeV in the centre of mass frame. But in the lab frame where the Lithium starts at rest the threshold is different. Transform the COM frame and do the problem there. Then transform back to the lab.

In the COM frame the two initial particles start with mirror momentums, one from the left, one from the right. They will be moving at different speeds but have equal magnitude momentums with opposite signs. After, by conservation of momentum, the resultant particles will fly away with again opposite sign momentums, but with less total energy.

Then as to the energy of forward-ejected neutrons, there is a trick. That seems to be the major point of this homework, so I'm not going to say any more.

Thanks ! Got it for part (a) !
The center of mass has constant velocity vector ##\vec V_c = \frac{4}{11} \vec v_0 ## due to momentum conservation during the collision.
In that coordinate system:
##\left\{\begin{array}{}
\vec v_{He,c} = \vec v_0 - \vec V_c = \frac{7}{11} \vec v_0 \\
\vec v_{Li,c} = \vec v_{Li} - \vec V_c = \frac{-4}{11} \vec v_0 \\
\vec v_{N,c} = \vec v_{N} - \vec V_c \\
\vec v_{B,c} = \vec v_{B} - \vec V_c
\end{array}\right.
##

Momentum in the COM system is ##\vec 0##, and the velocity vectors of particles are going in opposite directions. As a consequence,

## 10 v_{B,c} - v_{N,c} = 4 v_{He,c} - 7 v_{Li,c} = 0 ##

Now the expression of energy in the COM system is:

##\begin{array}{}
E_{0,c} := \frac{1}{2}( 4 v_{He,c}^2 + 7 v_{Li,c}^2 ) = \frac{7}{11}E_0 \\
E_{f,c} := \frac{1}{2}( 10 v_{B,c}^2 + v_{N,c}^2 ) = v_{N,c}^2
\end{array}
##

Initial energy is minimal when ##E_{f,c}=0##, so ##E_{0,c} =2.8MeV##

## \begin{align}
E_{f,c}=0 \Rightarrow& v_{N,c}^2 = 0 \\
\Rightarrow& \vec v_N = \vec V_c \\
\Rightarrow& E_N = \frac{1}{2}V_c^2 = \frac{4}{121}E_0 = \frac{4}{121}\times\frac{11}{7}E_{0,c} = \frac{4}{77}\times 2.8MeV \approx 0.15 MeV
\end{align}
##

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##\begin{array}{}
E_{f,c} := \frac{1}{2}( 10 v_{B,c}^2 + v_{N,c}^2 ) = v_{N,c}^2
\end{array}
##

There is a typo here, forgot a factor 11/20, but luckily it does not interfere with the rest.

Got (b) too thanks to you ;-)

By conservation of momentum in laboratory coordinates, if the neutron is ejected forward, then the helium nucleus and the neutron follow the same direction, which constrains the boron nucleus to follow the same direction.

Therefore, ## 4v_0 - v_N = 10 v_B##.

Squaring the left and right expression, and using conservation of total energy for substitutions, I get an expression of neutron energy depending on helium nucleus energy:

## 11 E_N - 4\sqrt{E_0E_N}+(28-6E_0) = 0 ##

Then, neutron energy is given by the positive, real solutions of a second degree equation. Its discriminant is ##\triangle = 4 (70E_0 - 308) ##.
The solutions are real when ##\triangle > 0 ## which means that E_0 must be greater than ## \frac{308}{70} = 4.4 = E_{0,threshold}##.
In that case,

## E_N = \frac{1}{11} (2\sqrt{E_0} \pm \sqrt{70E_0 - 308})##

One root is obviously positive, the other one must satisfy the following condition:
## 2\sqrt{E_0} > \sqrt{70E_0 - 308} \Rightarrow E_0 < \frac{308}{66} \approx E_{0,threshold} + 0.27 ##

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