Fission of Pb 208: Can We Sustain it?

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Discussion Overview

The discussion centers around the potential for fission of Lead-208 (Pb-208), exploring whether it can sustain a fission reaction and the energy requirements for such a process. Participants delve into the theoretical aspects of nuclear fission, energy calculations, and the characteristics of Pb-208 as a nuclear fuel.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • Some participants question the sustainability of fission in Pb-208, suggesting that the products of the reaction would not induce further reactions in adjacent Lead atoms.
  • There is uncertainty regarding the energy required for a neutron to induce fission in Pb-208, with participants seeking quantitative measures.
  • One participant notes that Pb-208 is deeply bound, making it an inefficient fuel for fission.
  • Another participant suggests that spallation might occur instead of fission when interacting with Pb-208.
  • Some participants discuss the need for a nuclear equation to calculate the Q-value and the energy needed for fission, expressing difficulty in finding this information.
  • There are references to activation energy and barrier energy calculations as potential methods to determine the kinetic energy required for a neutron to induce fission.
  • One participant mentions the average energy of prompt fission neutrons and its relevance to sustaining a reaction.
  • Questions arise about what a favorable reaction equation would look like if fission were to occur, specifically regarding the products formed.

Areas of Agreement / Disagreement

Participants generally agree that fission of Pb-208 is unlikely to be sustainable, but there is no consensus on the specifics of energy requirements or the mechanisms involved. Multiple competing views regarding the nature of the reaction and the calculations needed remain unresolved.

Contextual Notes

Participants express limitations in their understanding of the energy calculations and the nuclear equations necessary for assessing fission in Pb-208. There is a reliance on empirical values and a lack of clarity on how to derive the required energy for fission.

Who May Find This Useful

This discussion may be useful for those interested in nuclear physics, particularly in the context of fission reactions, energy calculations, and the properties of isotopes like Lead-208.

NuclearGaTech
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Does anyone know why we don't fission Lead 208? I know it is considered with magic number and stable, but is it still capable of producing fission.

How much energy would an incident neutron need in order to produce fission, and could/why could Pb208 have a sustainable fission reaction?
 
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I don't believe that the fissioning of Lead 208 is sustainable. IE the products of the reaction will NOT cause another reaction in adjacent Lead atoms. In effect, you CANT use this as a fuel.
 
Right, i figured that it would not be sustainable, but I cannot quantitatively show why, I am having difficulty understanding how much energy I would need to fission a single lead nucleus.

The neutron is without charge, so it should be able to just jump right into the nuclear well and create a excited Pb209, right? but how much energy would the neutron need to induce it's fission?
 
Lead 208 is deeply bound. That makes it one of the least effective fuels for fission. So not only is it hard, it's inefficient. Why go there?
 
I know it has the magic shells, and its tightly bound. Buty, i have a question about how much energy a neutron would need to fission this Pb208 atom.

I cannot figure out how much, i am in need of a quantitative answer. Don't know how to get it- that's what i need some help on.

At first i thought i would just need to get two atoms out of the nuclear well, so i thought it was just a force equal to the coulumb force at the radius of the two fission products, but i don't think this is the case.

It seems that all my knowledge about what makes something fissionable it based on having a valid nuclear equation from which i can make assesments. So, i try to find a reasonable nuclear equation for Pb208, if it were to fission. With an equation I can find the Q-value, but still need to know how much energy would be needed to fission, which is the problem i have.

What do you guys think of that?
 
You probably can't "fission" Pb-208 in the usual sense - what you will do is get spallation.
 
It is possible to get fission from lead isotopes, not only spallation. I haven't read the whole article though.

http://iopscience.iop.org/0305-4616/5/7/014

To figure out how much energy you will gain depends what the resulting isotopes will be. Fission with U has a maximum probability of going to an isotope of 2/5 and 3/5 the mass of the U. You can try to calculate the same for the Pb, but I'm not sure this will give a correct result.
 
Yeah, I'm trying to find some proton and neutron numbers close to the magic numbers- hopefully this would be preferable for Pb. The Q-value will be positive, and display the energy ejected, but still about the required energy of the incident neutron...?

A lot of textbooks talk about a critical energy needed for fission to occur, but all they display are empirical values and no information on how to determine.

I think the best thing to go on is: Find a favorable equation, use its Q-value versus the coulomb repulsion. I guess a neutron would have to have enough energy that, along with the Q-value of the equation, would have to be enough to overcome the coulomb force.
 
So, I'm looking at the same problem, but I feel that one can use the activation energy minus separation energy for PB-208 to find the amount of Kinetic Energy that the Neutron needs to introduce to the Pb-208 in order for it to undergo Fission.

The Class notes either #9 or #10 have a table that states that the activation energy is taken to be 0.9 MeV less than the barrier energy as determined by equation 11.3. Along with the separation energy being the Neutron separation energy from the nucleus A+1.

So if anyone knows how to calculate the Barrier energy that's probably all we need to find how much KE the neutron needs to induce fission.

As for the the amount of energy that a prompt neutron will have allowing the reaction to be self-sustaining. I am looking at page 148 in the Fundamentals of Nuclear Science and Engineering and it says, "The average energy of prompt fission neutrons is about 2MeV." So That's how I think I'm going to answer the question.
 
  • #10
So, what would be a favorable reaction equation? would it split evenly into 2 Nb-104 atoms?
 

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