Question

A 1MeV neutron is scattered through an angle of 60° in a collision with a H-2 nucleus. (a) What is the energy of the scattered neutron? (b) What is the energy of the recoiling nucleus? (c) How much of a change in lethargy does the neutron undergo in this collision? 2.

Answer 1

RELATION BETWEEN ENERGY (E') AND ENERGY (E) OF NEUTRON BEFORE AND AFTER COLLISION CAN BE WRITTEN AS:

E' E((cos + cos20(A2 - 1)/ (A1)

IN THE QUESTION ATOMIC MASS A =4 FOR HYDROGEN NUCLEUS, AND =0.5 , E= 1Mev

cos e

(A) THEREFORE ENERGY OF SCATTERED NEUTRON E' = 1 *(0.5+(0.25+(16-1))^0.5/(4+1))^2 =0.7762 Mev

(B) FROM CONSERVATION OF ENERGY PRINCIPLE, THE ENERGY OF RECOILING NUCLEUS = 1- 0.7762 =0.2237 Mev

(C) CHANGE IN LETHARGY IS GIVEN BY

Au 1 - (A 1)2/(2.A) In((A 1)/ (A 1))

1-[(4-1)^2/(2*4)*ln((4+1)/(4-1))] = 1-(9/8)*ln((5/3))

u :

=1-0.1325=0.8675