235U+1/0n-->141/56 Ba + 92/36 Kr + 3 1/0n If the energy released in the reaction

Question

235U+1/0n-->141/56 Ba + 92/36 Kr + 3 1/0n If the energy released in the reaction ( nuclear mass, 234.9935 ; nuclear mass, 140.8833 ; nuclear mass, 91.9021 ) is taken as typical of that occurring in a nuclear reactor, what mass of uranium-235 is required to equal 0.20 of the solar energy that falls on Earth in 1.0 day? Ive calculated that it is 7.4305*10^15 j/day and that j=kg*m/s. I used e=mc^2 and divided so im at (7.4305*10^15)/((3.0*10^8)^2)= 0.8256kg which is needed. question is would you also count each of the three neutrons?

Explanation / Answer

yes, 3 neutrons should be counted to calculate the mass difference.
delta M = mass of (U235+n)-mass of(Ba141+Kr92+3n)

= mass of(U235)-mass of(Ba141+Kr92+2n)

then you calculate how much energy is released in fission of 1kg U.

then divide it by the total energy ie. 3.08*10^18

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