Question

The average energy of a neutron produced in the fission of \( ^{235}_{92}U \) is

• A. \( 160 \times 10^{-13} \) J
• B. \( 320 \times 10^{-15} \) J
• C. \( 320 \times 10^{-13} \) J
• D. \( 160 \times 10^{-15} \) J

Hint:

Remember standard values: Fission neutron energy \(\approx 2\) MeV. Thermal neutron energy \(\approx 0.025\) eV.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
This is a factual question based on nuclear physics constants. The average kinetic energy of neutrons produced in nuclear fission of Uranium-235 is approximately 2 MeV.
Step 2: Key Formula or Approach:
Convert Energy in MeV to Joules. \( 1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{J} \). \( 1 \, \text{MeV} = 1.6 \times 10^{-13} \, \text{J} \).
Step 3: Detailed Explanation:
Average energy \( E \approx 2 \, \text{MeV} \). \( E = 2 \times 1.6 \times 10^{-13} \, \text{J} \) \( E = 3.2 \times 10^{-13} \, \text{J} \) Converting to match options: \( 3.2 \times 10^{-13} = 320 \times 10^{-15} \, \text{J} \).
Step 4: Final Answer:
The energy is \( 320 \times 10^{-15} \) J.