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Question in nuclear fusion

 
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Jun14-12, 01:22 PM   #1
 

Question in nuclear fusion


1. The problem statement, all variables and given/known data
Calculate the amount of fission energy, in joules, that can be generated from 2kg of uranium fuel, if the U-235 represents 0.7% of the metal, and every fission reaction produces 200MeV.


2. Relevant equations



3. The attempt at a solution
My attempt is irrelevant before i even know what does the energy per fission mean. I mean, in a fission of Uranium how many atoms are used???.. Anyway, i need some good replies on the qestion above!
 
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Jun14-12, 02:18 PM   #2
 
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Quote by ehabmozart View Post
1. The problem statement, all variables and given/known data
Calculate the amount of fission energy, in joules, that can be generated from 2kg of uranium fuel, if the U-235 represents 0.7% of the metal, and every fission reaction produces 200MeV.

2. Relevant equations

3. The attempt at a solution
My attempt is irrelevant before i even know what does the energy per fission mean. I mean, in a fission of Uranium how many atoms are used???.. Anyway, i need some good replies on the question above!
I'm pretty sure that they mean that each time a U-235 nucleus undergoes a fission, that's a fission reaction.
 
Jun14-12, 02:49 PM   #3
 
Upon that assumption, how can i reach to a final answer in this question. I am actaully confused in the topic itself so i need to know how to attempt such a question
 
Jun14-12, 03:22 PM   #4
 
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Question in nuclear fusion


How much mass of U-235 is there in a 2 kg sample of Uranium?

How many U-235 atoms are in such a sample?
 
Jun14-12, 03:24 PM   #5
 
You have the energy per fission event given. If you multiply this by the number iof fission events, you get the total energy. How can you calculate the # of fission events?
 
Jun14-12, 04:23 PM   #6
 
Mu current attempt is out of 2000g of Uranium fuel there are .7/100 * 2000= 14 grams... If 235 g of U-235 has 6.02x10^23 atoms, therefore in 14 grams there is 14* 6.02x10^23 / 235= 3.586x10^22 atoms.. = number of fission... Multiplying this answer by 200 MeV i get 7.173x10^24 which is ultimately wrong. Which step did i miss or which step was wrong?? Please HELP!
 
Jun14-12, 04:40 PM   #7
 
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Quote by ehabmozart View Post
Mu current attempt is out of 2000g of Uranium fuel there are .7/100 * 2000= 14 grams... If 235 g of U-235 has 6.02x10^23 atoms, therefore in 14 grams there is 14* 6.02x10^23 / 235= 3.586x10^22 atoms.. = number of fission... Multiplying this answer by 200 MeV i get 7.173x10^24 which is ultimately wrong. Which step did i miss or which step was wrong?? Please HELP!
That's 7.173×1024 MeV.

Convert that to Joules !
 
Jun15-12, 01:00 AM   #8
 
oh dude... was the MeV value correct??? Mybad, the answer in MeV in my book was 7.173x10^21 Mev.. Forget about joules and just assure me if my value of meV was right???
 
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