Question

Calculate the energy released in MeV in the following reaction: ²₁H + ³₁H → ⁴₂He + n Given: m(²₁H) = 2.014102 u m(³₁H) = 3.016049 u m(⁴₂He) = 4.002603 u mₙ = 1.008665 u

Imagine the fission of a ⁵⁶₂₆Fe into two equal fragments of ²⁸₁₃Al nucleus. Is the fission energetically possible? Justify your answer by working out Q value of the process. Given: m(⁵⁶₂₆Fe) = 55·93494 u, m(²⁸₁₃Al) = 27·98191 u.

In a fission event of ²³⁸₉₂U by fast moving neutrons, no neutrons are emitted and final products, after the beta decay of the primary fragments, are ¹⁴⁰₅₈Ce and ⁹⁹₄₄Ru. Calculate Q for this process. Neglect the masses of electrons/positrons emitted during the intermediate steps.

Calculate the binding energy of an alpha particle in MeV. Given : mass of a proton = 1.007825 u mass of a neutron = 1.008665 u mass of He nucleus = 4.002800 u 1u = 931 MeV/c²