Inelastic Collision: Threshold Energy

In summary: So I assume, that the range given will lead to a real solution of the square root.In summary, a thin target of lithium is bombarded by helium nuclei, causing a nuclear reaction that results in a boron nucleus and a neutron. The collision is inelastic and the final kinetic energy is 2.8 MeV less than the initial energy. The threshold energy for producing neutrons is 4.4 MeV, with the energy of the neutrons at this threshold being 0.15 MeV. If the incident energy falls in the range between the threshold energy and the threshold energy plus 0.27 MeV, the neutrons ejected in the forward direction will have two possible energies depending on the initial energy of the system
  • #1
a.man
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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, mass 1. The reaction can be symbolized:
7Li+4He→10B+1n −2.8MeVa. What is E0threshold 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 E0threshold<E0<E0threshold+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 = Ef + 2.8

pi=pf

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:

7vHe - 4vL = 10vB - vN = 11vCM
7vHe = 4vL
10vB = vN
E0 = 49m2vHe2 ([itex]\frac{1}{14m}[/itex]+[itex]\frac{1}{8m}[/itex]) = 1m2vN2 ([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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  • #2
Right, I figured how to do the first part:

In center of mass system, the final particle's velocities will be zero respective to the center of mass (they will be the same speed), since this is threshold energy.

So final energy is zero and initial energy is 2.8 MeV (The fact that system has the lowest energy of inertial reference frames can be used too). Using the above equations, we can find velocity of helium in the center of mass system.

Thus, we can find its velocity in the laboratory system, and therefore calculate the initial energy in this system, which is the E0threshold we need to find (4.4 MeV).

Now use the fact that the Boron and the neutron are of the same speed to substitute in a conservation of energy reaction (now in lab system) to get the energy of a neutron (which is 0.15 MeV).

I got the right answers, so I'm thinking this method is right...

For the second part, I assume I have to find an equation that expresses energy of a neutron in terms of initial energy?
 
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  • #3
Okay, I got the second answer as well, so never mind this. (Unless anyone else has the sam problem, I guess...?)

You DID have to get an equation of the velocity of neutrons in terms of initial energy, find for which values of energy will lead to a REAL, POSITIVE value of the velocity in the laboratory system.
 

1. What is the concept of "threshold energy" in inelastic collisions?

Threshold energy refers to the minimum amount of kinetic energy required for an inelastic collision to occur between two objects. This energy is necessary for the objects to overcome any attractive forces and make physical contact with each other.

2. How is threshold energy related to the coefficient of restitution?

The coefficient of restitution is a measure of the elasticity of a collision, and is directly related to the amount of kinetic energy lost during the collision. In an inelastic collision, the coefficient of restitution is less than 1, indicating that some kinetic energy is lost. The threshold energy is the minimum amount of kinetic energy required to overcome this loss and still have a non-zero coefficient of restitution.

3. Can the threshold energy be calculated for any type of inelastic collision?

Yes, the threshold energy can be calculated for any type of inelastic collision, whether it is a simple two-object collision or a more complex multi-object collision. However, the calculation may become more complicated for multi-object collisions due to the varying masses and velocities involved.

4. How does the threshold energy affect the outcome of an inelastic collision?

If the initial kinetic energy of the colliding objects is less than the threshold energy, the collision will not occur and the objects will simply pass by each other. If the initial kinetic energy is equal to or greater than the threshold energy, the collision will occur and the objects will stick together and move as a single mass.

5. Is there a connection between the threshold energy and the conservation of momentum?

Yes, the threshold energy is directly related to the conservation of momentum in an inelastic collision. In order for the collision to occur, the objects must have enough kinetic energy to overcome any attractive forces and make physical contact. This means that the objects will have a non-zero momentum before and after the collision, thus satisfying the law of conservation of momentum.

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