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. 

Answer 2

In a nuclear fission reaction, the total energy released is the difference between the initial and final total binding energies. The total initial binding energy is the binding energy of the \( ^{235}_{92}U \) nucleus, which is 1800 MeV. The total final binding energy is the sum of the binding energies of the fission products \( ^{144}_{56}Ba \) and \( ^{89}_{36}Kr \), which are 1200 MeV and 780 MeV, respectively. The energy released in the fission process is the difference between the initial and final binding energies: \[ \text{Energy released} = \text{Initial binding energy} - \text{Final binding energy} \] \[ \text{Energy released} = 1800 \, \text{MeV} - (1200 \, \text{MeV} + 780 \, \text{MeV}) = 1800 \, \text{MeV} - 1980 \, \text{MeV} = 180 \, \text{MeV}. \] This energy is carried away by the fast neutrons. Since 3 neutrons are emitted in the reaction, the average kinetic energy carried by each fast neutron is: \[ \text{Average kinetic energy per neutron} = \frac{180 \, \text{MeV}}{3} = 60 \, \text{MeV}. \] Thus, the average kinetic energy carried by each fast neutron is \( {60 \, \text{MeV}} \).