AZING! Can a 56Fe Nucleus Fission into Two 28Al Nuclei Spontaneously?

[fredrick08]

Homework Statement

Could a 56Fe nucleus fission spontaneously into 2 28Al nuclei? Explain, your answer should include some calculations, should be based on the curve of binding energy.

Homework Equations

B=(Zmh+Nmn-matom)*931.49

The Attempt at a Solution

B=(26(1.00783u)+30(1.00866u)-55.9349u)*931.49=492.27MeV=>~8.8MeV per nucleon.

from the curve of binding energy, 56Fe is the most stable atom, having the maximum amount of binding energy, thus it cannot fission spontaneously since it is bound too tightly...

i Have no idea if this is correct, but I am sure it can't fission... I've looked on net and found heaps of reasons why it cant... but I am really confused about how i am meant to answer the question.. can anyone help? thanks.

 

[Andrew Mason]

Homework Statement

Could a 56Fe nucleus fission spontaneously into 2 28Al nuclei? Explain, your answer should include some calculations, should be based on the curve of binding energy.

Homework Equations

B=(Zmh+Nmn-matom)*931.49

The Attempt at a Solution

B=(26(1.00783u)+30(1.00866u)-55.9349u)*931.49=492.27MeV=>~8.8MeV per nucleon.

from the curve of binding energy, 56Fe is the most stable atom, having the maximum amount of binding energy, thus it cannot fission spontaneously since it is bound too tightly...

i Have no idea if this is correct, but I am sure it can't fission... I've looked on net and found heaps of reasons why it cant... but I am really confused about how i am meant to answer the question.. can anyone help? thanks.

All nucleons in a stable nucleus are bound "tightly". The question is whether the amount of energy released when the two halves fly apart (due largely to colomb repulsion of the protons) is much greater than the amount of energy required to get them to fly apart (ie. to overcome the strong nuclear forces ie to move the halves far enough apart so that the coulomb repulsion exceeds the short range nuclear forces).

Think of a nucleon in a nucleus sitting at the bottom of an energy "well". The greater the binding energy, the deeper the well (the more energy required to get out of the well)

The binding energy curve (which shows binding energy per nucleon) tells you whether a nucleon in a particular size nucleus has more or less binding energy than a nucleon in another sized nucleus.

If the binding energy of a nucleon is greater in nucleus 2 than in nucleus 1, you know that the bottom of the energy well in nucleus 2 is lower than the bottom of the energy well in nucleus 1. So moving from nucleus 1 to 2 releases net energy. Conversely, you can say that in order to move from nucleus 2 to 1 energy must be added.

You should be able to tell from the curve whether a nucleon in 28Al has more or less binding energy than one 56Fe. In moving from 56Fe to 28Al, does the bottom of the nucleon's energy well go up or down?

AM

 

[fredrick08]

oh ok... so 28Al has a binding energy of 233MeV so 8.3MeV per nucleon... thus in order to go from 56Fe to 28Al it must go up the well? so if u add the energies from 2*28Al nuclei = 466 which does not add to 56Fe energy of 492? thus it can't fission... is that better? or is per nucleon rather then total... but 8.31MeV is no where near half of 8.8MeV?

 

[fredrick08]

oh so is it 2*28Al=2*8.31MeV=16.62MeV should equal to the binding energy of one 56Fe nucleon, because it doesn't it can't fission?...

 

[Andrew Mason]

oh ok... so 28Al has a binding energy of 233MeV so 8.3MeV per nucleon... thus in order to go from 56Fe to 28Al it must go up the well? so if u add the energies from 2*28Al nuclei = 466 which does not add to 56Fe energy of 492? thus it can't fission... is that better? or is per nucleon rather then total... but 8.31MeV is no where near half of 8.8MeV?

You simply need to observe that the binding energy per nucleon of 56Fe is greater than the binding energy of a nucleon in 28Al. This means that the energy well of the nucleon in 56Fe is deeper than in 28Al. When it goes up out of the 56Fe energy well and falls down to the bottom of the 28Al energy well, it will be above where it started. This means it ends up with more energy.

So is net energy released or absorbed? Can this happen spontaneously (ie without the addition of significant energy?).

AM