5. A 2-Me V neutron traveling in water has a head-on collision with an 160 nucleus. (a) What are the energies of the neutron and nucleus after the collision

- Final energy of neutron is 1.557 Mev

- Final energy of nucleus is 0.443 Mev

Explanation:

Given the that;

for ¹⁶O, A = 16

we know that Energy of neutron and nucleus after collision = ( A - 1 / A + 1)² E₀

so we substitute

Fe = ( 16 - 1 / 16 + 1)² × 2

= (0.8823)² × 2

= 1.557 Mev

therefore Final energy of neutron is 1.557 Mev

Final energy of nucleus will be

2 - Final energy of neutron

= 2 - 1.5569

= 0.443 Mev

therefore Final energy of nucleus is 0.443 Mev

where is the circuit diagram ?