In a nuclear reactor, neutrons released by nuclear fission must be slowed down b

Question

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 = 1.009 u, where the atomic mass unit 1 u = 1.66*10^-27 kg) (a) An electron (M = 5.49*10^-4 u) (b) A proton (M = 1.007u) (c) The nucleus of a lead atom (M = 207.2 u)

I figured it out how to calculate and my answers are...

(a) 0.9978

(c)0.9807

and for part (b)....

The mass of neutron m1=1.009 u

= 1.009*1.66*10^-27

= 1.674*10^-27

The mass of a proton m2 = 1.007u

=1.007*1.66*10^-27

=1.67162*10^-27

Vlf/Vli=(m1-m2/m1+m2)

Vlf/Vli=(1.674*10^-27)-(1.67162*10^-27)/(1.674*10^-27)+(1.67162*10^-27)

and my answer came out something wrong...

2.38*10^-30 / 3.34562*10^-27=7.1137786e-58

and then I need tp apply KEf/KEi=(Vlf/Vli)^2 to 7.1137786e-58...

maybe I'm using calculator worng...

I need help with the calculation for part (b).

Thank you!

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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