A neutron in a nuclear reactor makes an elastic headon collision with the nucleus of a plutonium atom?

Rye

initially at rest. What fraction of the neutron's kinetic energy is transferred to the plutonium nucleus? (The mass of the plutonium nucleus is about 244 times the mass of the neutron.) If the initial kinetic energy of the neutron is 1.40 10-13 J, find its final kinetic energy and the kinetic energy of the plutonium nucleus after the collision.



Mableton

For two particles (masses m₁ and m₂, initial velocities u₁ and u₂), which undergo a perfectly elastic head-on collision, you can find the after - collision velocities from [1]:v₁ = (u₁∙(m₁ - m₂) + 2∙m₂∙u₂)/(m₁ + m₂)v₂ = (u₂∙(m₂ - m₁) + 2∙m₁∙u₁)/(m₁ + m₂)Let subscript 1 denotes the proton and 2 the nucleus. Since the nucleus was initially at rest, i. e. u₂=0, the velocity of the nucleus after the collision is:v₂ = 2∙m₁∙u₁/(m₁ + m₂) = 2∙u₁/(1 + (m₂/m₁))So the kinetic energy of the nucleus after the collision is:E₂' = (1/2)∙m₂∙v₂² = 2∙m₂∙u₁²/(1 + (m₂/m₁))²The fraction of the neutrons kinetic energy transferred to the nucleus is:f = E₂'/E₁ = [ 2∙m₂∙u₁²/(1 + (m₂/m₁))² ] / [ (1/2)∙m₁∙u₁²) ]= 4∙(m₂/m₁) / (1 + (m₂/m₁))² = 4∙ 244 / (1 + 244)² = 0.016= 1.6%for E₁ = 1.40×10⁻¹³JE₂' = f∙E₁ = 0.16 ∙ 1.40×10⁻¹³J = 0.02×10⁻¹³JE₁' = (1 - f)∙E₁ = (1 - 0.16) ∙ 1.40×10⁻¹³J = 1.38×10⁻¹³J