Nuclear fission calculation of energy released

[tahmidbro]

Summary:: 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.

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?

 

[phyzguy]

You are not approaching it correctly. How many uranium atoms are in 2 kg of fuel? If 0.7% of these fission, how much energy is released? Also, the question says the final answer should be in Joules, not MeV.

 

[tahmidbro]

There are (2000/235) x 6.02 x 10^(23) = 5.1234 x 10^(24) atoms in 2kg of fuel. If 0.7% of these fission, 5.1234 x 10^(24) x 0.7%= 3.586 x 10^(22) atoms will undergo fission and if fission of each atom releases 200 MeV, then 3.586 x 10^(22) x 200 x 10^(6) x 1.6 x 10^(-19) = 1.1476 x 10^(12) Joules of energy is released. Is this correct?
Probably the answer from the solution book was wrong.

 

[mjc123]

Looks OK to me. I would calculate the number of moles as 2000/238 as most of the uranium is U-238, but that's just a small correction. The book's answer (aside from being in the wrong units) looks a factor of 1000 too small. Probably they either took the mass as 2 g or the atomic weight as 235 kg/mol.

 

[mfb]

The second approach is right and the answer looks good.

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

Here was the problem with the first approach. That value is the mass difference between the uranium atom and the fission products, not the mass of an uranium atom.

 

[phyzguy]

One minor problem is that since most of the Uranium is U238, the atomic weight of Uranium is 238.03 gms/mole. So you should divide by this number, not 235 when you calculate the number of U atoms. But this is only a ~1% correction.

 

[mfb]

The question doesn't specify if 0.7% is the mass fraction or the atomic fraction. I would assume the mass fraction, in that case OP's calculation is right.

 

[tahmidbro]

One minor problem is that since most of the Uranium is U238, the atomic weight of Uranium is 238.03 gms/mole. So you should divide by this number, not 235 when you calculate the number of U atoms. But this is only a ~1% correction.

As far as I know, Uranium 235 can absorb a neutron and can undergo fission, not U- 238. Probably this is why they have given % of uranium 235 in the question.

 

[mfb]

U-238 can do induced fission as well, but it needs neutrons with a high energy and doesn't produce enough of them to sustain a chain reaction. But that's not the point. If the 0.7% refer to the number of atoms then you need to consider the U-238 mass to find the total number of atoms and then take 0.7% of that. It's just a 1% effect, however.

 

[tahmidbro]

U-238 can do induced fission as well, but it needs neutrons with a high energy and doesn't produce enough of them to sustain a chain reaction. But that's not the point. If the 0.7% refer to the number of atoms then you need to consider the U-238 mass to find the total number of atoms and then take 0.7% of that. It's just a 1% effect, however.

m= 0.014 kg

E = 200MeV

mol of Uranium x avogadro constant = no. of mol of uranium nuclei

So , energy released = 0.014/235 x 6.02 x 10^(23) x 200 MeV
= 7.17 x 10 ^21 MeV Hurrah!
A friend of mine helped me to do it.
Anyway, Thanks to you all! :-)