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 Nuclear and Particle Physics Nuclear and Particle Physics Help Forum 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    Mar 1st 2011, 02:14 PM   #2

Join Date: Apr 2008
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Posts: 2,858
 Originally Posted by fishkeeper 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-U 235- 7.2 MeV Rb 94- 8.6 MeV 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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See the forum rules here.   Mar 1st 2011, 03:52 PM #3 Forum Admin   Join Date: Apr 2008 Location: On the dance floor, baby! Posts: 2,858 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 __________________ Do not meddle in the affairs of dragons for you are crunchy and taste good with ketchup. See the forum rules here.   Mar 1st 2011, 10:58 PM #4 Physics Team   Join Date: Jun 2010 Location: Mauritius Posts: 609 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. __________________ Jerry (Got my results!) It is easier to protect your feet with slippers than to cover the earth with carpet. No one can go back and change a bad beginning; but anyone can start now and create a successful ending. If a problem can be solved, no need to worry about it. If it cannot be solved what is the use of worrying? Last edited by Unknown008; Mar 1st 2011 at 11:01 PM. Reason: Typo about 'concerved'   Mar 1st 2011, 11:37 PM #5 Forum Admin   Join Date: Apr 2008 Location: On the dance floor, baby! Posts: 2,858 Thank you. I knew I was messing something up. -Dan __________________ Do not meddle in the affairs of dragons for you are crunchy and taste good with ketchup. See the forum rules here.  Tags binding, energy, fission Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Physics Forum Discussions Thread Thread Starter Forum Replies Last Post pranimaboity2050 Nuclear and Particle Physics 0 Oct 20th 2014 11:27 PM vyomictor Nuclear and Particle Physics 0 Sep 8th 2010 02:50 AM moshe.gorin Advanced Thermodynamics 0 Apr 25th 2010 10:26 AM newer Nuclear and Particle Physics 0 Oct 28th 2009 09:29 PM s3a Energy and Work 1 May 28th 2009 06:02 PM