What Is the Threshold Energy for Lithium-Helium Nuclear Reactions?

[kraigandrews]

Homework Statement

A thin target of lithium is bombarded by helium nuclei of energy E0. The lithium nuclei are initially at rest and are essentially unbound for this high-energy collision. A nuclear reaction can occur in which
4He + 7Li → 10B + 1n −2.8 MeV
The collision is inelastic, and the final kinetic energy is less than E0 by 2.8 MeV. (1 MeV = 106 eV = 1.6 ×10−13 J). The relative masses of the particles are: helium, mass 4; lithium, mass 7; boron, mass 10; neutron, mass 1.

(a) Determine the threshold energy, i.e., the minimum value of E0 for which neutrons can be produced. [HINT: Analyze the collision in the center-of-mass frame of reference. At the threshold energy, the final particles are produced at rest in the center-of-mass frame.]


(b) Calculate the energy of the neutron at the threshold energy.


(c) Determine the threshold energy for the reaction
4He + A(Iso1) → A+3(Iso2) + 1n − β
where β is the inelastic energy loss.
Data: A = 10; β = 3.1 MeV.

Homework Equations

Pi=Pf

KE=.5mv^2-Q; Q is the energy lost in the collision

The Attempt at a Solution

I am not sure if this remotely correct, but this is my best attempt:

m_he*Vo=(m_he+m_li)*Vo'
Eo=.5(m_he+m_li)*Vo'^2-Q

 

[Nate810]

Eo=.5*(4+7)*Vo'^2-2.8*10^6Vo'=(4+7)/(2*2.8*10^6)^.5Threshold energy=.5*(4+7)*(4+7)/(2*2.8*10^6)^.5Neutron energy at threshold energy=2.8*10^6mHe*Vo=(mHe+mA)*Vo'Eo=.5(mHe+mA)*Vo'^2-βVo'=(mHe+mA)/(2*β)^.5Threshold energy=.5*(mHe+mA)*(mHe+mA)/(2*β)^.5