Why does a U-238 tamper increase nuclear yield of nuclear bomb?

[Flexwheeler]

TL;DR

Why does U-238 tamper increase nuclear yield of nuclear bomb? I mean it is not fissile.

Why does U-238 tamper increase nuclear yield of nuclear bomb? I mean it is not fissile.

 

[mfb]

It is not fissile - it cannot sustain a chain reaction - but it is still fissionable: It can split when hit by a fast neutron. The inner nuclear explosion produces a lot of high energy neutrons, they can split additional uranium-238 nuclei in the tamper.

 

[DBO]

The tamper has sufficient inertia that it holds the core together for a tiny greater fraction of time allowing more neutron multiplication thus increasing yield.

 

[snorkack]

The tamper has sufficient inertia that it holds the core together for a tiny greater fraction of time allowing more neutron multiplication thus increasing yield.

That would also be true in case of a lead tamper.
One fission neutron has energy of about 2 MeV. If it causes a fission event, it gains about 200 MeV. Even a small fraction, like 1 %, of nuclei undergoing fission with fast neutrons would add a lot of energy. In U-238, it is actually several hundred %, because U-238 is capable of undergoing convergent chain reaction.

 

[mfb]

Even a small fraction, like 1 %, of nuclei undergoing fission with fast neutrons would add a lot of energy.

Not that much. Remember that these neutrons come from fission reactions, releasing ~80 MeV per neutron. If 1% of them induce another fission reaction you increase your yield by ~2 MeV per neutron leaving the core, or ~3% overall.

 

[Flexwheeler]

Do fissionable (here I don't mean fissile)materials when hit by a nuetron split and release energy?

 

[mfb]

There is a chance of this happening at least. In some cases the neutron needs some minimal energy.

 

[snorkack]

What cross-section for fission, through the neutron spectrum all the way to 14 MeV, qualifies a nucleus as "fissionable"?
As you rightly pointed out, though fission neutrons themselves have about 2 MeV energy, their production released about 80 MeV energy per neutron mainly in fission fragments. Fusion neutrons carry 14 MeV energy of just 18 released in producing them. A low fission cross-section would add a lot of energy here.
There are a number of isotopes fully capable of sustaining a chain reaction with fast neutrons, though not with slow ones. Definitely the low even isotopes of Pu - 238, 240, 242. Pu-244 is less well known, but probably too. U-236 probably not, U-234 and U-232 possibly (can anyone confirm?). Does that make them fissile, or merely fissionable?
U-238 definitely will not sustain a chain reaction with fast neutrons - but will have convergent chain reaction. With multiplication factor, I don´ t remember the exact value, but somewhere near 0,75. Taking 2,5 neutrons per fission and 0,75 multiplication factor as example (you can substitute real numbers when you find them), 100 neutrons into U-238 would cause 30 fissions in first generation, releasing 75 neutrons - by the time the neutrons run out, the original 100 neutrons will have caused 120 fissions.

 

[willem2]

I looked up the fusion cross section for U238 here: https://wwwndc.jaea.go.jp and I get 1.136 barn.
A square meter of a 1 cm layer of U238 would weigh, 191 Kg and contain 8.0* 10² moles, so would contain 4.8 * 10²⁶ atoms/m², the total cross section of those atoms is 0.054 m². This would mean there would be a 5.4% chance of a fusion for a 14Mev Neutron crossing a 1 cm layer. There will be more because of elastic collisions , and (n,2n) and other reactions.

The cross section for neutrons with a fission spectrum is 0.31 barn, so I don't think the neutrons produced by fissions will add much, but I think a 1 cm layer already does enough, considering that a fusion produces 18 MeV and a fission about 200MeV

 

[mfb]

Not all neutrons move orthogonal to the layer, most will see a slightly thicker effective layer.
The total cross section is 6 barn, scattering and then capture increases the chance a lot, too.

 

[shjacks45]

I looked up the fusion cross section for U238 here: https://wwwndc.jaea.go.jp and I get 1.136 barn.
A square meter of a 1 cm layer of U238 would weigh, 191 Kg and contain 8.0* 10² moles, so would contain 4.8 * 10²⁶ atoms/m², the total cross section of those atoms is 0.054 m². This would mean there would be a 5.4% chance of a fusion for a 14Mev Neutron crossing a 1 cm layer. There will be more because of elastic collisions , and (n,2n) and other reactions.

The cross section for neutrons with a fission spectrum is 0.31 barn, so I don't think the neutrons produced by fissions will add much, but I think a 1 cm layer already does enough, considering that a fusion produces 18 MeV and a fission about 200MeV

Interesting, but misses the point. Neutron cross section is determined by neutron energy. U235 (Th232, U233, Pu239, ...) only has a large cross section for thermal or very low energy neutrons. A fission weapon has no moderator to lower neutron energy so literally ALL fissions occur due to fast neutrons. Fissions also release atomic fragments at high energy enough to fragment or fuse with other nuclei. AND you neglect the contribution of fusion (tampers are in thermonuclear weapons) gamma rays. The energy necessary to split an atom does not only come from particles. And heavy elements are good gamma absorbers. The gamma from fusion can be 40 MeV. Rem fusion nuclei weigh 1/20 U235 so 20x particles for same weight.

 

[mfb]

The gamma from fusion can be 40 MeV.

Where would such a high energy come from?

 

[shjacks45]

It is not fissile - it cannot sustain a chain reaction - but it is still fissionable: It can split when hit by a fast neutron. The inner nuclear explosion produces a lot of high energy neutrons, they can split additional uranium-238 nuclei in the tamper.

Tamper is used in thermonuclear weapon ((US Ivy Mike and Russian Tsar bombs) hence small fission trigger big fusion bomb. Per unit weight Lithium Deuteride (fusion fuel) generates 20x neutrons and very high energy gamma rays than U235. The energy to force split atoms can come from momentum or photon not just from absorbing a neutron. In Wikipedia one test was for theory 50 megaton but with tamper gave 85 megaton.

 

[shjacks45]

Where would such a high energy come from?

Wikipedia on ‘thermonuclear weapon’ gives 17Mev. Still more than 6 MeV causing nuclear instability in U238.

 

[snorkack]

Fusion is not an efficient producer of gamma rays. Strong processes are faster than electromagnetic.
By my estimate, the weak gamma rays from fusion should end around 24 MeV.

 

[mfb]

And where would 24 MeV come from? The reaction doesn't release that much energy in total!

 

[snorkack]

And where would 24 MeV come from? The reaction doesn't release that much energy in total!

α binding energy: 28,30 MeV
d binding energy: 2,22 MeV
Therefore:
d+d=α+γ
total energy 28,30-4,45=23,85 MeV. Minus recoil of α - not much because γ lacks rest mass. Plus kinetic collision energy of deuterons - relatively not much. That´ s the around 24 MeV.
As I mentioned, that radiation is weak, because strong processes
d+d=t+p
d+d=³He+n
are much faster.

 

[mfb]

d+d fusion in general is a side reaction so irrelevant that I haven't seen anyone discuss it for modern weapons. Ivy Mike had it as significant reaction I think.

 

[le berger des photon]

I"m not really replying to anything specific here. I'm making the mistake of looking at this logically and not believing anything that I don't already understand simply because some "experts" have said some thing.

For me, regardless of whatever additional technologies may exist inside of these black boxes known as atom bombs, once the black box fails to remain a strong container, the reaction would stop instantly when this container fails. Logically I don't see why the reaction would proceed far enough to release more than just a bit more than necessary to destroy the box the device is delivered in. Once it's lost its integrity, any chain reaction has to stop instantly. The way everything else in nature works, it would only make enough to do the job. It wouldn't release orders of magnitude more energy than that required to destroy the integrity of the case.

I don't know if logic has any value here in this forum. I'm anticipating that any response I get will violate Einstein's rule that says that if you can't explain a thing to a six year old then you don't understand it yourself.

Then, naturally it should proceed to telling me that I'm incapable of understanding things that six year olds can understand.

And then I'll remind you that six year olds believe in Santa Claus.

And I'll perhaps mention that I can't prove that Santa doesn't exist and that you can't prove that he does, but I'm clearly getting ahead of myself.

 

[vanhees71]

Obviously you have a wrong idea about "logic". The natural sciences are investigating objective facts about nature by observing phenomena (experimental physics) and trying to find mathematical models and natural laws to describe them (theoretical physics). The results are not to please your philosophical prejudices but to objectively describe Nature.

It might be a bit hard to explain to a six-year old the details of nuclear reactions in general and nuclear-fusion processes in the Sun at detail, and indeed a full understanding of physics without the only appropriate language, which is math including calculus, linear algebra, and some group theory, is impossible.

In the natural sciences you also cannot prove anything as in mathematics. There is only a certain degree of evidence for the general laws, which are found in the above mentioned process of quantitative observations (measurements) and theoretical reasoning (model/theory building).

Applying the scientific method to the question about the existence of Santa is simple: To all our experience Santa is just a nice invention to give children some fun. There's no sighting of an old man coming from the north pole to bring presents yet and also no theoretical deduction from known physical laws that such a man should exist. That's why we conclude that to the best of our knowledge Santa doesn't exist.

It's a characteristics of the natural sciences that usually you have to change your believes when better information and results of further research occurs, telling you that your previous believes are inaccurate of even wrong. Sometimes it also demystifies nice fairy tails like the idea of Santa.

 

[Nugatory]

Once it's lost its integrity, any chain reaction has to stop instantly.

Your intuition is misleading you here. Even after the box fails, it takes a fair number of microseconds for the cloud of vaporized bomb internals inside of it to disperse and end the reaction. With most devices there's no practical difference between "a fair number of microseconds" and "instantly", but in a fission explosion one additional microsecond is time for another few hundred doublings... and $2^{100}$ is a very big number indeed.

 

[Vanadium 50]

but none of you have ever observed a fission chain reaction

I most certainly have.

 

[le berger des photon]

I most certainly have.

was it a controlled chain reaction in a reactor? Or was it a chain reaction in an atom bomb?

If it was, tell me how you measured the output?

 

[Vanadium 50]

was it a controlled chain reaction in a reactor?

Reactor.

If it was, tell me how you measured the output?

Thermometers and neutron counters.

 

[gmax137]

are the prompt neutrons I was just reading about being liberated by collisions? or created out of thin air?

prompt neutrons are released when the uranium nucleus splits. as opposed to the delayed neutrons, which are released when the fission products decay.

 

[mfb]

Several posts full of wrong and/or unscientific statements have been removed from the thread. I also removed some replies to these posts if they wouldn't make sense without the deleted posts.

 

[berkeman]

Thread is locked temporarily for Moderation and fixing some misinformation...

UPDATE -- thread will remain closed.