Heat released in neutron capture by boron

physics_dipsi

I am working on some neutron experiments and have to calculate the heat released in the boron neutron capture

10B + ¹n - > Li + Alpha particle

The reaction has a Q value of 2.31 MeV. Is this entire energy converted to heat?

Also, in the two types of reactions of neutron with Boron, the energy of the excited Li in the reaction is less than the Li in the ground state. Can someone please throw some light on that ?

Thanks

Astronuc

The Q value is the minimum possible energy released. That only comes from the differences in rest mass. One would also have to consider the kinetic energy of the neutron. If there is not internal excitation in the Li nucleus, then the difference in rest mass is manifest as kinetic energy (heat) of the reactant particles.

When one refers to two types of reactions of neutron with boron, is one consider one isotope, or two? There is B-10 and B-11, and each has a different reaction with a neutron.

physics_dipsi

Hi Astronuc, Thanks for the reply. The two reaction of Boron are 10B capturing a thermal neutron and releasing a Li_excited (94% times of reaction) or Li_ground state (6%).

Bob S

The thermal neutron capture cross sections for B¹⁰ and B¹¹ are 3850 and 5 barns respectively. See thumbnail. The thermal neutron kinetic energy is ~ 1/40 eV. The excited state of Li⁷ decays to the ground state with a 0.48 MeV gamma, and the remaining energy is the kinetic energy of the recoiling lithium nucleus (0.84 MeV) and the alpha particle (1.47 MeV). See

http://web.mit.edu/nrl/www/bnct/info...scription.html

Bob S.

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