Question

A neutron in a reactor makes an elastic headon collision with the nucleus of an atom initially at rest. Assume: The mass of the atomic nucleus is about 11.3 the mass of the neutron. What fraction of the neutron’s kinetic energy is transferred to the atomic nucleus?

Answer 1

m = mass of neutron

M = mass of nucleus = 11.3 m

v_i = initial velocity of neutron before collision = ?

V_i = initial velocity of nucleus before collision = 0 m/s

v_f = final velocity of neutron after collision

V_f = final velocity of nucleus after collision

using conservation of momentum

$mv_i + MV_i = mv_f + MV_f \\ (m) v_i + (11.3 m) (0) = m v_f + (11.3 m) V_f \\ v_i = v_f + (11.3) V_f \\ v_i - v_f = (11.3) V_f \;\;\; eq-1$

using conservation of kinetic energy

$m v^2_i + M V^2_i = m v^2_f + M V^2_f \\ (m) v^2_i + (11.3 m) (0)^2 = m v^2_f + (11.3 m) V^2_f \\ v^2_i = v^2_f + (11.3) V^2_f \\ v^2_i - v^2_f = (11.3) V^2_f \\ ( v_i - v_f )( v_i + v_f ) = (11.3) V^2_f \\ (13.7) V_f ( v_i + v_f ) = (11.3) V^2_f \\ v_i + v_f = V_f \;\;\;\; eq-2$

adding eq-1 and eq-2

$( v_i - v_f ) + ( v_i - v_f ) = (11.3) V_f + V_f \\ 2 v_i = 12.3 V_f \\ v_i = 6.15 V_f$

fraction of energy transferred is given as

$fraction = \frac{(0.5) MV^2_f }{(0.5) m v^2_i } \\ fraction = \frac{M V^2_f }{m v^2_i } \\ fraction = \frac{(11.3 m) V^2_f }{m (6.15 V_f )^2 } \\ fraction = \frac{11.3}{(6.15)^2} = 0.298 ≈ 0.3$