A thin target of lithium is bombarded by helium nuclei of energy E0. The lithium

Question

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.
Threshold energy = 4.400 MeV

Calculate the energy of the neutron at the threshold energy.

Explanation / Answer

Recall the answer I gave you to calculate the threshold energy:

Equations:

Ei=Ef

pi=pf

EfvB=0

Ei=E0=2mv2He
Ef=E02.8MeV=5mv2B+12mv2n
pi=4mv(He)iˆ
pf=(10mvBcosB+mvncosn)iˆ+(10mvBsinB+mvncosn)jˆ

To get threshold energy,
B=n=0
which gives
pf=(10mvB+mvn)iˆ
Solve for vn from the momentum conservation and put it into the energy equation and then use
EfvB=0
to find the minimum energy.

E0=4.40 MeV

Now, to find neutron energy, you simply recall that the energy comprises of the Boron KE and the neutron KE, so to find Eneutrons, just subract the boron energy 5mv^2:

E=5mv^2B+1/2mv2neutrons

Eneutrons= 0.15 MeV

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