
#1
Nov1913, 07:58 AM

P: 14

The binding energy (BE) per nucleon for 235U is 7.6 Mev. The 235U undergoes a nuclear fission to produce two fragments both having a BE of 8.5Mev. What is the energy released from a complete fission of 1kg of 235U (joules)?
Here I assumed that it breaks into two 118X element. Therefore BE(product) per atom is 2*118*8.5=2006 BE(reactant) per atom is 236*7.6=1793.6 Energy released per atom= 212.4 Mev Number of atoms in 1 kg U=1/.235*6.022*[itex]10^{23}[/itex] = 2.5625*[itex]10^{24}[/itex] Total energy = 2.5625*[itex]10^{24}[/itex] *212.4 = 5.4428*[itex]10^{26}[/itex] Mev = 8.7*[itex]10^{13}[/itex] joules I want know if my answer is correct ? 



#2
Nov1913, 08:20 AM

P: 1,248

Seems about right (up to two sig. figs.)




#3
Nov1913, 05:38 PM

Mentor
P: 10,813





#4
Nov1913, 09:54 PM

P: 14

Energy of fission reaction 



#5
Nov2013, 07:28 AM

P: 733

Fission does not fragment into two equal parts, you get one heavy and one light fission product and 23 free neutrons, e.g. I135 + Tc99 + 2n.




#6
Nov2013, 03:10 PM

Mentor
P: 10,813





#7
Nov2013, 11:47 PM

P: 14





#8
Nov2113, 07:48 AM

Mentor
P: 10,813

If an apple costs 30 Cents and you have to buy 100 apples by going to the shop 2 times, it does not matter how many apples you buy each time (like 50+50 or 20+80 or whatever), the total cost is independent of that. 


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