How far do fission products travel?

[mesa]

Hey guys, about how far do fission products travel on average in UO2? This will obviously be a fairly short distance, but does anyone have any ideas or links to information on this?

 

[berkeman]

Do you mean in a fuel rod in a reactor? Or somewhere else?

https://en.wikipedia.org/wiki/Uranium_dioxide

 

[mesa]

Do you mean in a fuel rod in a reactor? Or somewhere else?

https://en.wikipedia.org/wiki/Uranium_dioxide

Indeed, yes.

 

[berkeman]

Indeed, yes.

Have you had any luck finding info about this online with your searches? If not, @Alex A or @Astronuc may have thoughts...

 

[Alex A]

10um is commonly quoted as a rule of thumb and that is all I had in my head so I did a quick google. This rather fun paper https://www.osti.gov/servlets/purl/6868318 on fission fragment rockets gives a value of 6.2um in Uranium Carbide and 16.2um in carbon. Uranium dioxide will probably be quite close to UC. There could be some variation according to the fragment type. Energy loss is probably roughly proportional to track length. Think alpha particle like ionisation paths.

Papers on fission counters may give more information.

 

[Astronuc]

I've seen some numbers, but I can't remember the text/source.

The lighter fission products, e.g., isotopes of Br, Kr, Rb, Sr, Y, Zr . . . have higher energy and travel about 7-10 microns in UO₂. The lighter the isotope, the higher the energy. The heavier fission products, e.g., isotopes of Sb, Te, I, Xe, Cs, La, . . . . travel about 4-6 microns in UO₂. Two fission products may share between 160 and 170 MeV, and one might be 65 MeV (the heavier one) and the the other about 105 MeV, for example. The rest of the energy is carried away by neutrons and gammas, and internal to the fission products, which decay by beta decay. Neutrinos, or rather, antineutrinos carry away some energy as well, in conjunction with beta decay.

The text, Introduction to Nuclear Engineering, 3rd Edition, by John Lamarsh and Anthony J. Baratta, Prentice Hall, 2001, p. 108 shows ranges of fission products to be 6.6 microns in U and 14 microns in U₃O₈, so probably about 10 microns in UO₂. Of course, this depends on porosity of the ceramic, which could increase the range, since the pores have little stopping power, even filled with gas.
.

 

[Vanadium 50]

You can use the Bethe-Bloch formula, especially if you just want an estimate:

dE/dx for U₂O is about 12 MeV per cen times Z². So for E = 100 MeV and Z = 40, you get 50 microns. I am dropping some logarithmic terms here, so it's only good to a factor of a few, but you can use the full formula if you want better accuracy.

 

[mesa]

10um is commonly quoted as a rule of thumb and that is all I had in my head so I did a quick google. This rather fun paper https://www.osti.gov/servlets/purl/6868318 on fission fragment rockets gives a value of 6.2um in Uranium Carbide and 16.2um in carbon. Uranium dioxide will probably be quite close to UC. There could be some variation according to the fragment type. Energy loss is probably roughly proportional to track length. Think alpha particle like ionisation paths.

Papers on fission counters may give more information.

Great source, and a fun paper in general, you don't see proposals like these too often anymore.

 

[mesa]

The text, Introduction to Nuclear Engineering, 3rd Edition, by John Lamarsh and Anthony J. Baratta, Prentice Hall, 2001, p. 108 shows ranges of fission products to be 6.6 microns in U and 14 microns in U₃O₈, so probably about 10 microns in UO₂. Of course, this depends on porosity of the ceramic, which could increase the range, since the pores have little stopping power, even filled with gas.
.

Good information as always Astronuc, I'll have to dust my copy off when I get into work tomorrow. This seems inline with the paper Alex A referenced.

 

[mesa]

You can use the Bethe-Bloch formula, especially if you just want an estimate:

dE/dx for U₂O is about 12 MeV per cen times Z². So for E = 100 MeV and Z = 40, you get 50 microns. I am dropping some logarithmic terms here, so it's only good to a factor of a few, but you can use the full formula if you want better accuracy.

Nice, looking into it more.

 

[Vanadium 50]

One thing to keep in mind is that the terms I neglected tend to increase ionization at low energy (called the Bragg peak) so my estimate is too long. However, if you want to include this, you will need to integrate the equation, because the ionization changes as the fragment slows.