Question

Consider the nuclear fission re action  Assuming all the kinetic energy is carried away by the fast neutrons only
and total binding energies of    MeV and 780 MeV respectively, the average kinetic energy
carried by each fast neutron is (in MeV)

• A. 200
• B. 180
• C. 67
• D. 60

Answer 1

The Correct Option is D

Total binding energy of reactants = Binding energy of ${}^{235}_{92}\text{U}$ = 1800 MeV Total binding energy of products = Binding energy of ${}^{144}_{56}\text{Ba}$ + ${}^{89}_{36}\text{Kr}$ \[ = 1200 + 780 = 1980 \text{ MeV} \] The increase in binding energy (i.e., energy released) = \[ 1980 - 1800 = 180 \text{ MeV} \] This energy is carried away by 3 fast neutrons, so average energy per neutron: \[ \frac{180}{3} = 60 \text{ MeV} \] Thus, the average kinetic energy carried by each fast neutron is 60 MeV.