Answer to Question #133842 in Atomic and Nuclear Physics for Aditya

Question #133842
given that fuel grade uranium contains 4% fissile material (235U) that the fissile material is completely utilised that 200MeV (1eV=1.6× 10^-19 J) of thermal energy is released during fission of a single 235U nucleus and that the thermal to electrical conversion of efficiency is 33% the annual requirement of fuel grade uranium for round the clock operation of a 1000MW nuclear power plant would be
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Expert's answer
2020-09-21T08:31:14-0400

Let's denote MM as fuel grade uranium.

We know the mass of 235U is 0.04M0.04M. We need to calculate how many U-235 nucleus are inside of this mass.

N=0.04M235NA=0.04M2356.021023=1.0251020MN = \frac{0.04M}{235} N_A = \frac{0.04M}{235}\cdot 6.02 \cdot 10^{23}= 1.025\cdot 10^{20} M

These nucleus will produce

E=1.0251020M200  MeV=1.02510202001061.61019M    J=E= 1.025 \cdot 10^{20} M \cdot 200 \; MeV = 1.025 \cdot 10^{20} \cdot 200 \cdot 10^{6} \cdot 1.6 \cdot 10^{-19} \cdot M \; \;J =

=328107M    J= 328 \cdot 10^7 M \; \; J

This energy should be equal to ϵEt=0.331000  MW1  year=33010631  556  926  J1.0431016  J\epsilon \mathcal{E} t = 0.33 \cdot1000\; MW \cdot 1 \; year= 330 \cdot 10^6 \cdot 31\; 556\; 926 \; J \approx 1.043 \cdot 10^{16} \; J

So,

327107M=1.0431016327 \cdot 10^7 M = 1.043 \cdot 10^{16}

M=3.2106  g=3.2103  kg.M = 3.2 \cdot 10^{6} \; g = 3.2 \cdot 10^{3} \; kg.


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