Final answer:
The nuclear reaction for the absorption of a neutron by a Li-6 nucleus with the emission of an alpha particle is Neutron (n) + Li-6 -> He-4 + H-3. The energy released in this reaction can be calculated using the equation E = mc^2, where E is the energy released, m is the mass difference, and c is the speed of light.
Step-by-step explanation:
The nuclear reaction for the absorption of a neutron by a Li-6 nucleus with the emission of an alpha particle (He-4) is as follows:
Neutron (n) + Li-6 -> He-4 + H-3
In this reaction, the Li-6 nucleus absorbs a neutron, resulting in the formation of a helium-4 nucleus (alpha particle) and a tritium-3 nucleus (hydrogen-3).
The energy released in this reaction can be calculated using the equation E = mc^2, where E is the energy released, m is the mass defect (difference in mass before and after the reaction), and c is the speed of light. The energy can be determined by using the equation E = Δm c^2, where Δm is the change in mass given by the mass difference between the reactants and products.