Question

If the number of uranium nuclei required per hour to produce a power of 64 kW is \(7.2 \times 10^{18}\), then the energy released per fission is

• A. \(0.64 \times 10^{-10}\) J
• B. \(3.2 \times 10^{-13}\) J
• C. \(0.32 \times 10^{-10}\) J
• D. \(3.2 \times 10^{-10}\) J

Hint:

Always ensure your units are consistent before performing calculations. Convert power to Watts (J/s) and time to seconds to get the energy in Joules. This avoids common errors in magnitude.

Answer 1

The Correct Option is C

Step 1: Formula for Power: Power (\( P \)) is energy per unit time. \[ P = \frac{E_{total}}{t} \] Total energy \( E_{total} \) produced by \( N \) fissions, where each fission releases energy \( E_{fission} \), is: \[ E_{total} = N \times E_{fission} \] So, \[ P = \frac{N \times E_{fission}}{t} \]
Step 2: Substitute Given Values: \( P = 64 \, \text{kW} = 64 \times 10^3 \, \text{J/s} \) Number of nuclei \( N = 7.2 \times 10^{18} \) Time \( t = 1 \, \text{hour} = 3600 \, \text{s} \) Rearranging for \( E_{fission} \): \[ E_{fission} = \frac{P \times t}{N} \] \[ E_{fission} = \frac{64 \times 10^3 \times 3600}{7.2 \times 10^{18}} \]
Step 3: Calculation: \[ E_{fission} = \frac{64 \times 36 \times 10^5}{7.2 \times 10^{18}} \] \[ E_{fission} = \frac{64 \times 360}{72} \times 10^{5-18} \] \[ E_{fission} = \frac{64 \times 5}{1} \times 10^{-13} \] \[ E_{fission} = 320 \times 10^{-13} \] \[ E_{fission} = 3.2 \times 10^{-11} \, \text{J} \] Wait, let's re-check the options and calculation. \( 320 \times 10^{-13} = 32 \times 10^{-12} = 0.32 \times 10^{-10} \). This matches Option (C).