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Old Mar 1st 2011, 11:48 AM   #1
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Binding energy + fission energy help please

Sorry if this is in the wrong section- Its A-Level

Calculate the energy released when a U 235 (92 protons) nucleus undergoes fission producing nuclei of Rb 94 (37 protons) and Cs 142 (55 protons).
There is a graph for this question, but I shall give all the data (binding energy per nucleon) linked below.

Binding energy per nucleon-
  1. U 235- 7.2 MeV
  2. Rb 94- 8.6 MeV
  3. Cs 142- 8.4 MeV
(Read from the graph as I see it)

Ok, my calculations:

Atomic Mass unit 'u'= 1.661x10^-27

(94+142)*1.661 x10^-27 = 3.91996 x10^-25
235*1.661 x10^-27= 3.93441 x10^-25

3.93441 x10^-25 - 3.91996 x10^-25 = 1.445 x10^-27

E= MC^2 (deltas included)

E= ((1.445 x10^-27)*(3 x10^8)^2)= 1.3005 x10^-10 J

(1.3005 x10^-27)/(1.6 x10^-19) = 8.13 x10^8 eV ~ 813 MeV

Is this correct? Im really stuck and I cannot find an answer. Im worried that since i havent used any MeV values per nucleon I haven't got the correct answer. Sorry if it is hard to read :/

Thanks


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Old Mar 1st 2011, 02:14 PM   #2
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Originally Posted by fishkeeper View Post
Sorry if this is in the wrong section- Its A-Level



There is a graph for this question, but I shall give all the data (binding energy per nucleon) linked below.

Binding energy per nucleon-
  1. U 235- 7.2 MeV
  2. Rb 94- 8.6 MeV
  3. Cs 142- 8.4 MeV
(Read from the graph as I see it)

Ok, my calculations:

Atomic Mass unit 'u'= 1.661x10^-27

(94+142)*1.661 x10^-27 = 3.91996 x10^-25
235*1.661 x10^-27= 3.93441 x10^-25

3.93441 x10^-25 - 3.91996 x10^-25 = 1.445 x10^-27

E= MC^2 (deltas included)

E= ((1.445 x10^-27)*(3 x10^8)^2)= 1.3005 x10^-10 J

(1.3005 x10^-27)/(1.6 x10^-19) = 8.13 x10^8 eV ~ 813 MeV

Is this correct? Im really stuck and I cannot find an answer. Im worried that since i havent used any MeV values per nucleon I haven't got the correct answer. Sorry if it is hard to read :/

Thanks


Hmmm.... I'm having a slight problem here. A reaction will not proceed unless it is energetically favorable. To determine this you need to find out how much mass energy the nucleus holds. You have done that. Now you need to find the binding energy of the nucleus (number of nucleons times the binding energy per nucleon), the you subtract this from the mass energy. That gives you the total energy available for the fission to occur.

The problem is that I'm getting that the total energy of U235 is actually less than the total energy of the products. So the reaction should not happen.

Working on it.

-Dan
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Old Mar 1st 2011, 03:52 PM   #3
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I just noticed something which gave me a bit of a laugh. It seems there is one extra neutron that found its way into the daughter products. (142 + 94 = 236, not 235.) If we add that one neutron to the U-235 then the decay is energetically favorable.

You might want to ask about that.

-Dan
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Old Mar 1st 2011, 10:58 PM   #4
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Um, I'm afraid you got the wrong procedure. The mass of each element will not be an exact multiple of u, since there is some of that mass conserved into binding energy.

What you need to do, is find the difference in binding energies.

Binding energy in U-235 = 235*7.2 = ?

Binding energy in Cs-142 + Rb-94 = 8.4*142 + 8.6*94 = ?

Then work out the difference. Since the products are more stable, you'll have them having greater binding energy. The difference is the energy released.

Remember that:
Relative atomic mass = Total mass of each nucleon - Binding energy

Hence why the energy released will appear to be negative.

And Dan, the reaction is ideally:

U-235 + neutron ---> Rb-94 + Cs-142

That's where the extra neutron comes from.
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Last edited by Unknown008; Mar 1st 2011 at 11:01 PM. Reason: Typo about 'concerved'
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Old Mar 1st 2011, 11:37 PM   #5
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Thank you. I knew I was messing something up.

-Dan
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