This is the only problem left that I can\'t solve...Step by step explanation wou

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This is the only problem left that I can't solve...Step by step explanation would be really helpful! :)

In a nuclear reactor, neutrons released by nuclear fission must be slowed down before they can trigger additional reactions in other nuclei. To see what sort of material is most effective in slowing (or moderating) a neutron, calculate the ratio of a neutron's final kinetic energy to its initial kinetic energy, Kf/Ki , for a head-on elastic collision with each of the following stationary target particles. (Note: The mass of a neutron is m=1.009u, where the atomic mass unit,u , is defined as follows: 1 u = 1.66 x 10^-27 kg.)

Part A
An electron ( M = 5.49 x 10^-4 u).
Express your answer using four significant figures.

Part B
A proton ( M = 1.007 u).
Express your answer using one significant figure.

Part C
The nucleus of a lead atom ( M = 207.2 u).
Express your answer using four significant figures.

It's asking to find Kf/Ki for all three parts.

Explanation / Answer

Let v1 be the initial speed of the neutron From elastic collision we have the velocity after the collision (of the incoming particle) is v2 = (ma - mb)/(ma + mb)*v1 Now K = 1/2*m*v^2 Since the mass of the neutron doesn't change then Kf/Ki = v2^2/v1^2 = ((ma - mb)/(ma + mb)/ma)^2 So Kf/Ki(proton) = ((1.009-1.007)/(1.009+1.007))^2 = 9.84x10^-7 and Kf/Ki(lead) = ((1.009-207.2)/(1.009+207.2))^2 = 0.981 So the proton is more effective in reducing the K of the neutron

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