Any Nuclear Scientist here?

Question

if a nuclear reactor produces energy at the rate of 'x' watts also assume energy released per fission to be 'Y' Mev then how many atoms are fissioned per sec of U-235?

Answer 1

If you want the mathematical answer:

X watts means X joules/sec of energy released. This is the same as X / (1.602*10^(-13)) MeV/sec.

If each U-235 fission yields Y MeV of energy, then the number of fissions per second is:

X / (Y*1.602*10^(-13)) fissions/sec

That assumes that no energy from the fissions is lost, which it certainly will be, but you can easily find the average efficiency of a nuclear reactor somewhere if you care about that.

Answer 2

If a fission event releases Y MeV then we first need to convert this to energy into Joules. 1 MeV = 1,000,000 ev and a eV = 1.602e-19 J so an MeV is 1.602e-13J.
Now the total power of the reactor is x watts which is x joules/sec
Two find the number of fissions per second simply take
x/(y*1.602e-13).
A typical value of Y is 200 MeV and lets assume we have 1000MW thermal reactor.
1000,000,000/(200 *1.602e-13)=3.12e19 fissions per second.
At first this seems like a lot but remember that mole is 6.022e23 atoms. At the rate above it would take 5hr and 20min to burn a mole of U-235. So really 3e19 fissions/s is not that big.

Compare that with the amount of coal required for a 1000MW thermal coal power plant.

Answer 3

Well, it all starts with one atom being split. It depends upon how far along the chain reaction has gotten and how the control rods are absorbing stray particles. Not really enough info in the question. But, a split U-235 atom will yield approx 200 meV.

Answer 4

That all depends on where your power range is... it'd be like close to 1x10^15 (or so) reactions per second. It's an enormous amount of reactions occuring.

Answer 5

Yeeeaaahhhhh an enormous amout of reaction