Nuclear fission energy release problem

Hi!

I am stuck in this question from my exercise book :

Q. Calculate the amount of energy in joules generated from 2 kg of uranium fuel if the uranium 235 represents 0.7% of the metal and every fission releases 200 MeV.

I've already attempted but could not come up with the correct answer: which is 7.17 x 10^(21) MeV ( they did not show any working in the solution book )

My attempt:

200 x 10^(6) x 1.6 x 10^(-19) = m( 3 x 10^(8))^2

m= 3.56 x 10^(-28) kg used up in 1 fission.

0.7% of 2 = 0.014 kg

so the number of times 3.56 x 10^(-28) kg is used up ( or the number of fissions ) = 0.014/(3.56 x 10^(-28)) = 3.93 x 10^(25) fissions

Total energy released = 200 x 1.6 x 10^(-19) x 10^(6) x 3.93 x 10^(25) = 1.25 x 10^(15) MeV

Is there any way I can get the correct answer?

Reply 1

username4407536

The logic for finding number of fissions is not right
Edit: Your calculation for finding total energy released, is also not right. It should simply be number of fissions x energy released per fission

You should rather find the mass of uranium (0.014 kg)
Then the number of uranium nuclei.
This is the number of fissions.
Multiplying this by 200 MeV, gives the energy released

(edited 3 years ago)

Reply 2

tahmidbro

Original post by golgiapparatus31

its (0.014/235 ) x 6.02 x 10^(23) x 200 MeV . Now answer is 7.17 x 10^(21) MeV .

That is mole x avogadro constant = number of uranium nuclei, right?
Thanks a lot for helping me !

(edited 3 years ago)

Reply 3

username4407536

Original post by tahmidbro

its (0.014/235 ) x 6.02 x 10^(23) x 200 MeV . Now answer is 7.17 x 10^(21) MeV .

That is mole x avogadro constant = number of uranium nuclei, right?
Thanks a lot for helping me !

Question info states "...2 kg of uranium fuel if the uranium 235 represents 0.7%..."
so as you said in the first post, the fuel contains 2 x 0.07 = 0.14 kg of uranium-235.
235 g of uranium-235 is 1 mole of uranium-235 = 6.02 * 10^23 nuclei of uranium-235
so 0.014 kg is ... nuclei

or 1 nucleus has mass 235 * 1.66 * 10^-27 kg.
so 235 * 1.66 * 10^-27 kg -> 1 nucleus
so 0.014 kg is -> ... nuclei

Then once you have number of nuclei, energy released = number of nuclei x energy released per fission
because number of nuclei is equal to number of fissions