Question

Each nuclear fission of \( {}^{235}\text{U} \) releases 200 MeV of energy. If a reactor generates 1 MW power, then the rate of fission in the reactor is

• A. \( 3.125 \times 10^6 \)
• B. \( 3.125 \times 10^8 \)
• C. \( 3.125 \times 10^{10} \)
• D. \( 3.125 \times 10^{16} \)

Hint:

- Power = Energy / Time. Rate of energy production = Power. - If each event (fission) releases \(E_{event}\) energy, and the rate of events is \(R_{event}\) (events/sec), then Power \(P = R_{event} \times E_{event}\). - Conversion factors: \( 1 \, \text{MeV} = 10^6 \, \text{eV} \) \( 1 \, \text{eV} = 1.602 \times 10^{-19} \, \text{J} \) (use \(1.6 \times 10^{-19}\) if context suggests) \( 1 \, \text{MW} = 10^6 \, \text{W} = 10^6 \, \text{J/s} \).

Answer 1

The Correct Option is D

Energy released per fission \( E_{fission} = 200 \, \text{MeV} \).
Power generated by the reactor \( P = 1 \, \text{MW} = 1 \times 10^6 \, \text{Watts} = 10^6 \, \text{J/s} \).
First, convert energy per fission to Joules: \( 1 \, \text{MeV} = 10^6 \, \text{eV} \).
\( 1 \, \text{eV} = 1.
6 \times 10^{-19} \, \text{J} \).
So, \( E_{fission} = 200 \times 10^6 \times 1.
6 \times 10^{-19} \, \text{J} \).
\[ E_{fission} = 2 \times 10^2 \times 10^6 \times 1.
6 \times 10^{-19} \, \text{J} \] \[ = 2 \times 1.
6 \times 10^{2+6-19} \, \text{J} = 3.
2 \times 10^{8-19} \, \text{J} = 3.
2 \times 10^{-11} \, \text{J} \] The rate of fission \( R_{fission} \) (number of fissions per second) is related to the power P by: \( P = R_{fission} \times E_{fission} \).
So, \( R_{fission} = \frac{P}{E_{fission}} \).
\[ R_{fission} = \frac{10^6 \, \text{J/s}}{3.
2 \times 10^{-11} \, \text{J/fission}} \] \[ R_{fission} = \frac{1}{3.
2} \times \frac{10^6}{10^{-11}} \, \text{fissions/s} \] \[ R_{fission} = \frac{1}{3.
2} \times 10^{6 - (-11)} = \frac{1}{3.
2} \times 10^{17} \, \text{fissions/s} \] Calculate \( \frac{1}{3.
2} \): \( \frac{1}{3.
2} = \frac{10}{32} = \frac{5}{16} \).
\( 5 \div 16 \): \( 5.
0 \div 16 = 0.
3 \).
(163=48, rem 2) \( 20 \div 16 = 1 \).
(161=16, rem 4) \( 40 \div 16 = 2 \).
(162=32, rem 8) \( 80 \div 16 = 5 \).
So, \( \frac{1}{3.
2} = 0.
3125 \).
\[ R_{fission} = 0.
3125 \times 10^{17} \, \text{fissions/s} \] To match the options format \( 3.
125 \times 10^X \): \[ R_{fission} = 3.
125 \times 10^{-1} \times 10^{17} = 3.
125 \times 10^{16} \, \text{fissions/s} \] This matches option (4).