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salters psa5(ii) fission question

in prep for thursdays PSA5(ii) exam i've been doing past exam papers. however, i have come across a question where i do not know where to start.
its from june 2003's paper.

"When a uranium-235 nucleus undergoes fission, two smaller nuclei are formed, each with about half the number of nucleons.

Show that the energy released in the fission of one nucleus of uranium-235 is about 4x10E-11J"

the previous part to the question is using the binding energy per nucleon against nucleon number graph to work out the binding energy of U235 in eV, which i have done.

any help greatly appreciated!

Alan
Reply 1
Avatar for Jarife
Jarife
Can you try working out the binding energy per nucleon of the two products (barium and krypton) and then, because the binding energy of the products combined is more than the uranium, energy has been released. (you can figure out how much with the values).

Perhaps you values for some masses on a data sheet, you could try and do it by finding the mass defect and then use E = mc^2
Reply 2
Avatar for supercacti
supercacti
hey i dont have this question on me but il tell you what to do. You have to calculate the mass defect (Mp*no of p - Mn*no of n) which should give you a smallish number. From there you use E=mc^2 and you get your energies! Hope that was of use to you :smile:
Reply 3
Avatar for teachercol
teachercol
9
Caluclate the binding energy of the fragemnts ( A (= 117) x BE per nuc which you get from the graph) and the U nucleus.(235 * BE per nuc)

Subtract.
Reply 4
Avatar for remember
remember
batesy71
in prep for thursdays PSA5(ii) exam i've been doing past exam papers. however, i have come across a question where i do not know where to start.
its from june 2003's paper.

"When a uranium-235 nucleus undergoes fission, two smaller nuclei are formed, each with about half the number of nucleons.

Show that the energy released in the fission of one nucleus of uranium-235 is about 4x10E-11J"

the previous part to the question is using the binding energy per nucleon against nucleon number graph to work out the binding energy of U235 in eV, which i have done.

any help greatly appreciated!

Alan


What did you get for the first part? 1.7*10^9eV

Then for the second part

8.25*117*2 = 1930.5MeV

= 1930500000eV
=1.93*10^9eV

So the change in energy is

1.93*10^9 - 1.7*10^9 = 2.4*10^8ev

2.4*10^8 * 1.6*10^-19 = 3.8*10^-11J

which is about 4*10^-11J

I think thats right, but please correct me if it isn't. I made a stupid mistake on the first part and it had me stuck for ages.

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