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.
(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.]

Explanation / Answer

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

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