Nuclear Fission Energy of 235U

In one nuclear reactor reaction, a (235,92)U is bombarded with a neutron, temporarily forming a highly unstable (236,92)U, which promptly decays via fission into two smaller daughter nuclei (92.37)Rb and (140,55)Cs. This reaction also liberates several neutrons.
Calculate the energy released in this reaction.
Compare this with the energy released in a typical chemical reaction (for example, the combustion of a hydrogen molecule.)
Estimate how many such fission reactions are necessary in order to bring a cup of room temperature water to boiling. Calculate how much mass of uranium (in kg) this amounts to.

The nuclear reaction is
$(235,92)U = (92,37)Rb +(140,55)Cs +3*(1,0)n + energy$

atomic masses are
$M(235U) =235.0439 amu$
$M(92Rb) =91.9197 amu$       
$M(140Cs) =139.9173 amu$
$M(1n) =1.00866 amu$

mass defect is
$\Delta(M)= 235.0439 – 91.9197 -139.9173 -3*1.00866 =0.18092 amu$
(E = m*c^2)
$1 amu = 1.6605*10^{-27} kg = 931.5 MeV =1.4904*10^{-10} J$

Energy released is by a single $(235,92)U$ atom
$E = 0.18092*931.5 =168.527 MeV = 2.69*10^{-11} J$

For hydrogen the energy of combustion is
$E(H) = 286 KJ/mol =286*10^3/6.022*10^{23} =4.75*10^{-19} J$
The energy released in the fission reaction is 8 orders of magnitude higher!

A cup of water mas mass $m =200 g$.
Caloric capacity of water is $c=4.18 J/g/K$
Heat needed to increase temperature of water from 300 K to 373 K is
$Q =200*4.18*(373-300) =61028 J$

Number of fissions is
$N = Q/E =61028/2.69*10^{-11} =2.27*10^{15} atoms$

One mole of $235U$ has mass $M =235 g$
The mass of U is
$m = M*N/N_a =235*2.27*10^{15}/6.023*10^{23} =8.85*10^{-7} grams!=$
$= 8.85*10^{-10}$ kg of U
where $N_a$ is Avogadro number.


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