Question

Consider the nuclear fission reaction\( \text{n}^{1} + \text{U}^{235} \rightarrow \text{Ba}^{144} + \text{Kr}^{89} + 3 \text{n}.\) Assuming all the kinetic energy is carried away by the fast neutrons only and total binding energies of \( \text{U}^{235} \), \( \text{Ba}^{144} \), and \( \text{Kr}^{89} \) to be 1800 MeV, 1200 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

The energy released in a fission reaction is the difference between the total binding energies of the products and reactants.

The energy carried by each fast neutron is calculated as:
\(\ \ K_E = \frac{\text{B.E of products} - \text{B.E of reactants}}{3} \ \ K_E = \frac{1980 - 1800}{3} = \frac{180}{3} = 60 \text{MeV}\)