A reactor is developing nuclear energy at a rate of 32,000 kilowatt. How many kg of `U^235` undergo fission per second? How many kg of `U^235` would be used up in 1000 hour of operation? Assume an average energy of 200 MeV released per fission? Take Avogadro.s number as `6 xx 10^23` and 1 MeV =`1.6xx10^(-13)` joule

Explanation

To solve the problem, we need to find out how much uranium-235 (U-235) undergoes fission per second and then how much is used in 1000 hours of operation. We will follow these steps:

### Step 1: Convert Power from Kilowatts to Joules per Second
The power of the reactor is given as 32,000 kilowatts. Since 1 kilowatt = 1,000 joules/second, we can convert this to joules per second:

\[ P = 32,000 \, \text{kW} = 32,000 \times 1,000 \, \text{J/s} = 32,000,000 \, \text{J/s} = 3.2 \times 10^7 \, \text{J/s} \]
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