Calculating Energy Released by Fission of 1g of U-235

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The discussion focuses on calculating the energy released by the fission of 1 gram of U-235. The fission reaction is represented, and the atomic masses of the involved isotopes are provided. The calculation involves determining the mass defect, which is found to be 0.19204u, leading to an energy release of approximately 178.79 MeV. This energy is then converted to Joules, resulting in a value of about 2.86 x 10^-11 J per fission event. The final energy released from 1 gram of U-235 is estimated to be around 7.33 x 10^10 J.
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Homework Statement



Estimate the energy released, in Joules, by the fission of 1g of U-235 according to the equation:


^{235}_{92}{U} + ^{1}_{0}{n} → ^{144}_{56}{Ba} + ^{90}_{36}{Kr} + 2^{1}_{0}{n} + Q

atomic masses:

^{235}_{92}{U} = 235.04438u

^{144}_{56}{Ba} = 143.92279u

^{90}_{36}{Kr} = 89.92089u

^{1}_{0}{n} = 1.00866u

Avogadro's number = 6.023 x 1023
1u = 931MeV
1MeV = 1.6 x 10-13J

Homework Equations





The Attempt at a Solution



(235.04438u + 1.00866u) - (143.92279u + 89.92089u + 1.00866u + 1.00866u) = 0.19204u

0.19204u x 931 = 178.78924MeV
178.78924 x 1.6 x 10-13J = 2.86 x 10-11

Do I then do (6.023 x 1023/235) x 2.86 x 10-11 = 7.33 x 1010J

For the final answer?

Thanks
 
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Looks good to me.
 
barryj said:
Looks good to me.


Thanks.
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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