Kinetic energy of nuclear reaction

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SUMMARY

The discussion centers on the kinetic energy distribution of the nuclear reaction B10(n,alpha)Li7, which releases 2.79 MeV. The contributors calculate the initial kinetic energies of the He4 and Li7 nuclei, determining that the total energy is distributed among them as 0.48 MeV for He4 and 0.84 MeV for Li7, with the remaining energy attributed to the alpha particle. Additionally, they explore the specific ionization produced by the fragments using the formula for stopping power, incorporating the geometric-mean excitation potential of the medium, I=100 eV.

PREREQUISITES
  • Understanding of nuclear reactions and energy conservation principles.
  • Familiarity with kinetic energy calculations in particle physics.
  • Knowledge of specific ionization and stopping power concepts.
  • Proficiency in using formulas related to energy distribution and ionization.
NEXT STEPS
  • Study the principles of nuclear reaction mechanics, focusing on energy release and distribution.
  • Learn about specific ionization and stopping power calculations in various media.
  • Investigate the use of the dE/dS formula in practical applications of particle physics.
  • Explore advanced topics in nuclear physics, such as compound nucleus formation and decay processes.
USEFUL FOR

This discussion is beneficial for nuclear physicists, students studying nuclear reactions, and researchers focusing on particle interactions and energy distribution in nuclear processes.

jackjoo87
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reaction B10(n,alpha)Li7, Q=2.79Mev. the He4 and Li7 are oppositely directed away from the site of the compound nucleus and form one straight track.
a) what is the initial kinetic energy (KE) of He4?
b) what is the initial KE of Li7 nucleus?
c) estimate the ratio of the specific ionization produced by the 2 fragments, if the geometric-mean excitation potential of the medium is I= 100eV

my work:
a) & b)(Ek)gamma + (Ek)Li + (Ek)alpha = 0.48 MeV + 0.84 MeV + 1.47 MeV= 2.79 Mev but i don't know how the energy is distributed.
c) By using dE/dS= (4πZ2e4/moV2)NZ ln2moV2/I

(question from R. D. Evans. The Atomic Nucleus. page 646 problem 1)


Specific ionization= stopping power
 
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