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Inelastic Collision: Threshold Energy

  1. Oct 16, 2012 #1
    1. The problem statement, all variables and given/known data

    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 −2.8MeV


    a. 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.)

    2. Relevant equations

    E = Ef + 2.8

    pi=pf


    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:

    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...
     
    Last edited: Oct 16, 2012
  2. jcsd
  3. Oct 16, 2012 #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?
     
    Last edited: Oct 16, 2012
  4. Oct 16, 2012 #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.
     
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