Question

The decrease in each day in the Uranium mass of the material in a Uranium reactor operating at a power of 12 MW is (Energy released in one \(^{92}U\) fission is about 200 MeV)

• A. \( 12.64 \times 10^{-2} \) kg
• B. \( 11.50 \times 10^{-2} \) g
• C. \( 12.64 \) kg
• D. \( 12.64 \) g

Hint:

The mass loss in a nuclear reaction can be calculated using the energy released and converting it using Einstein’s equation \( E = mc^2 \).

Answer 1

The Correct Option is D

Step 1: Energy released per fission of Uranium \( ^{92} U \) is 200 MeV. Step 2: Power given as 12 MW. We convert it into joules per second: \[ P = 12 \times 10^6 \text{ J/s} \] Step 3: Convert the energy released per fission into joules: \[ 200 \text{ MeV} = 200 \times 1.6 \times 10^{-13} \text{ J} \] Step 4: Calculate the number of fissions per second: \[ \text{Number of fissions per second} = \frac{12 \times 10^6}{200 \times 1.6 \times 10^{-13}} = 3.75 \times 10^{13} \text{ fissions per second} \] Step 5: Total mass lost per second: \[ \text{Mass lost} = 3.75 \times 10^{13} \times 2.68 \times 10^{-25} \text{ kg} \quad \Rightarrow \quad \text{Mass lost} = 12.64 \times 10^{-2} \text{ kg} \] Thus, the correct answer is option (4), \( 12.64 \) g.