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]

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

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]

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.

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/description/description.html

Bob S.

 

[alphy]

for your question. I can provide some insight into the heat released in neutron capture by boron.

Firstly, in the reaction you have mentioned, the energy released is not entirely converted to heat. Some of the energy is also released in the form of kinetic energy of the products (Li and alpha particle). This is due to the conservation of energy principle, where the total energy of the system (neutron and boron) must remain constant. Therefore, the heat released in this reaction would be slightly less than the Q value of 2.31 MeV.

Secondly, the difference in energy between the excited Li state and the ground state can be explained by the concept of nuclear binding energy. When a neutron is captured by boron, the resulting nucleus (Li) is in an excited state. This means that the nucleus has excess energy, which it releases in the form of gamma rays or particles (such as alpha particles) to reach a more stable state. This release of energy results in the difference in energy between the excited state and the ground state of Li.

I hope this helps to clarify your questions about the heat released in neutron capture by boron and the difference in energy between the excited and ground states of Li. Keep up the good work with your neutron experiments!