Fission of Pb 208: Can We Sustain it?

[NuclearGaTech]

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?

 

[Drakkith]

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.

 

[NuclearGaTech]

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?

 

[Vanadium 50]

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?

 

[NuclearGaTech]

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?

 

[Vanadium 50]

You probably can't "fission" Pb-208 in the usual sense - what you will do is get spallation.

 

[eXorikos]

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.

 

[NuclearGaTech]

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.

 

[Vriggchan]

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.

 

[GaTechNuclear]

So, what would be a favorable reaction equation? would it split evenly into 2 Nb-104 atoms?