Calculating Fission Energy from Uranium Fuel

[ehabmozart]

Homework Statement

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.

Homework Equations

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!

 

[SammyS]

Homework Statement

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.

Homework Equations

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.

 

[ehabmozart]

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

 

[SammyS]

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?

 

[daveb]

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?

 

[ehabmozart]

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!

 

[SammyS]

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 !

 

[ehabmozart]

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?