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In the nuclear fission reaction 235⁹²U → 140⁵⁴Xe + 94³⁸Sr + x + y, where x and y represent two particles, the 235⁹²U nucleus is initially stationary. The kinetic energies of the resulting products are denoted as Kₓ, K_Sr, K_Xe (2 MeV), and K_y (2 MeV), respectively. The binding energy per nucleon values for 235⁹²U, 140⁵⁴Xe, and 94³⁸Sr are 7.5 MeV, 8.5 MeV, and 8.5 MeV, respectively. Considering the laws of conservation, which of the following is correct?

1. x is a neutron, y is a neutron, K_Sr = 129 MeV, K_Xe = 86 MeV
2. x is a proton, y is an electron, K_Sr = 129 MeV, K_Xe = 86 MeV
3. x is a proton, y is a neutron, K_Sr = 129 MeV, K_Xe = 86 MeV
4. x is a neutron, y is a neutron, K_Sr = 86 MeV, K_Xe = 129 MeV

Correct answer: x is a neutron, y is a neutron, K_Sr = 129 MeV, K_Xe = 86 MeV

Solution

In nuclear fission, conservation of mass-energy and momentum applies. The binding energy difference indicates two neutrons are emitted (x and y). The kinetic energy distribution aligns with K_Sr = 129 MeV and K_Xe = 86 MeV, satisfying energy conservation.

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