Question

A fission reaction is given by ²³⁶₉₂U ➝ ¹⁴⁰₅₄Xe + ⁹⁴₃₈Sr + x + y, where x and y are two particles. Considering
²³⁶₉₂U to be at rest, the kinetic energies of the products are denoted by Kхₑ, Kѕᵣ, Kₓ(2MeV) and K????(2MeV),
respectively. Let the binding energies per nucleon of ²³⁶₉₂U, ¹⁴⁰₅₄Xe, ⁹⁴₃₈Sr be 7.5 MeV, 8.5 MeV and
8.5 MeV respectively. Considering different conservation laws, the correct option(s) is(are)
(A) x = n, y = n, Kѕᵣ = 129MeV, Kхₑ = 86 MeV
(B) x = p, y = e⁻, Kѕᵣ = 129 MeV, Kхₑ = 86 MeV
(C) x = p, y = n, Kѕᵣ = 129 MeV, Kхₑ = 86 MeV
(D) x = n, y = n, Kѕᵣ = 86 MeV, Kхₑ = 129 MeV

Answer 1

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