A neutron in a nuclear reactor makes an elastic head-on collision with the nucle

Question

A neutron in a nuclear reactor makes an elastic head-on collision with the nucleus of a lead atom initially at rest. (a) What fraction of the neutron's kinetic energy is transferred to the lead nucleus? (The mass of the lead nucleus is about 207 times the mass of the neutron.) ****. ANSWER IS .019**** ___________________ (b) If the initial kinetic energy of the neutron is 1.10 10-13 J, find its final kinetic energy. Also, find the kinetic energy of the lead nucleus after the collision. neutron ______ J? nucleus ________J? *****NOTE****ANSWER IS NOT 9.81e-14 . J FOR NEUTRON OR 1.90e-15 J FOR NUCLEUS

Explanation / Answer

Two things are conserved: kinetic energy, linear momentum. Write two equations, one for conservation of momentum and one for conservation of kinetic energy. For simplicity assume the neutron has mass=1 and the lead nucleus mass = 207. Then: Momentum: V1n = V2n + 207*V2pb Energy: 0.5*V1n^2 = 0.5*V2n^2 + 0.5*(207)*V2pb^2 Now you have two equations with 3 unknowns. Eliminate V2pb from the equations and solve for V2n in terms of V1n. To find the fraction of energy loss you need: [V1n^2-V2n^2]/V1n^2. If you do the math correctly you should find the answer is about 2%

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