Question

A neutron in a nuclear reactor makes an elastic, head-on collision with the nucleus of a carbon atom initially at rest. (a) What fraction of the neutron's kinetic energy is transferred to the carbon nucleus? (The mass of the carbon nucleus is about 12.0 times the mass of the neutron.) (b) The initial kinetic energy of the neutron is 1.30 10-13 J. Find its final kinetic energy and the kinetic energy of the carbon nucleus after the collision. neutron J carbon nucleus J

Answer 1

(a) Mass of neutron = m1

mass of carbon nucleus, m2 = 12*m1

Now, for two particles (masses m? and m?, initial velocities u? and u?), which undergo a perfectly elastic head-on collision, you can find the after- collision velocities as -  
v? = (u??(m? - m?) + 2?m??u?)/(m? + m?)
v? = (u??(m? - m?) + 2?m??u?)/(m? + m?)

Let subscript 1 denotes the proton and 2 the nucleus.

Since the nucleus was initially at rest, i.e. u?=0, the velocity of the nucleus after the collision is:
v? = 2?m??u?/(m? + m?) = 2?u?/(1 + (m?/m?))

So the kinetic energy of the nucleus after the collision is:
E?' = (1/2)?m??v?² = 2?m??u?²/(1 + (m?/m?))²

The fraction of the neutrons kinetic energy transferred to the nucleus is:
f = E?'/E?
= [ 2?m??u?²/(1 + (m?/m?))² ] / [ (1/2)?m??u?²) ]
= 4?(m?/m?) / (1 + (m?/m?))²
= (4*12) / (1 + 12)²
= 0.284
= 28.40 %

(b) Now, for E? = 1.30×10?¹³J

E?' = f?E? = 0.284 * 1.30×10?¹³J = 0.369 × 10?¹³ J

E?' = (1 - f)?E? = (1 - 0.284) * 1.30×10?¹³J = 0.931 × 10?¹³ J