A neutron is absorbed by a 3 6 L i nucleus with the subsequent emission of an alpha particle. L 3 6 i + n 0 1 → H 2 4 e + H 1 3 + Q Calculate the energy released, in MeV , in this reaction. [Given: ma

Question

A neutron is absorbed by a 3 6 L i nucleus with the subsequent emission of an alpha particle. L 3 6 i + n 0 1 → H 2 4 e + H 1 3 + Q Calculate the energy released, in MeV , in this reaction. [Given: mass 3 6 Li = 6.015126 u ; mass (neutron) = 1.0086654 u ; Mass (alpha particle) = 4.0026044 u and Mass (tritium) = 3.0100000 u . [ Take 1 u = 931 MeV c - 2 ]

TestHub · Accepted Answer

L 3 6 i + n 2 4 → H 2 4 e + H 1 3 + Q Total initial mass = 6.015126 + 1.0086654 = 7.0237914 amu Total final mass = 4.0026044 + 3.01 = 7.0126044 amu Mass defect, Δ m = 7.0237914 − 7.0126044 = 0.0111870 amu Energy released, Q = 0.0111870 × 931 = 10 . 415 MeV .

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