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Beryllium-9 gets hit by alpha?


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So, I decided to find out why beryllium (and aluminum) actually emits neutrons when hit by alpha particles. I was quite sure that the two extra protons and neutrons make it an isotope that's very unstable and a neutron emitter. But no.

 

Be-9 + alpha (2 neutrons 2 protons) -> C-12 + a neutron

 

It was quite interesting to see that alpha decay doesn't work in reverse (at least not in this case). Why is this? Why doesn't the nuclei take all the particles becoming C-13? Does this have something to do with preserving momentum or something?

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So' date=' I decided to find out why beryllium (and aluminum) actually emits neutrons when hit by alpha particles. I was quite sure that the two extra protons and neutrons make it an isotope that's very unstable and a neutron emitter. But no.

 

Be-9 + alpha (2 neutrons 2 protons) -> C-12 + a neutron

 

It was quite interesting to see that alpha decay doesn't work in reverse (at least not in this case). Why is this? Why doesn't the nuclei take all the particles becoming C-13? Does this have something to do with preserving momentum or something?[/quote']

 

Check the masses of Be-9 + alpha and compare it with the mass of C-13. What does that tell you?

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It tells me that it's increased by about 4 atomic mass units, which would be the case IF the nuclei would take all 2 protons and 2 neutrons, as I said.

 

No, I meant the mass of (Be-9 + alpha). That mass is 0.01143 amu larger than C-13, so there is an excess of 10.65 MeV that the C-13 would need to shed, and that's ignoring any KE the alpha has to have to overcome the Coulomb repulsion and cause the interaction. That's a lot of energy to get rid of.

 

By emitting a neutron, the C-13 sheds about 5MeV of that excitation.

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actually wouldn't it be exactly = if you had of a pure sample of Be9 then you would get a mix of be9 and C13 at least according to the masses of the samples

 

 

the mass of Be-9 + an alpha = to the mass of C13 because you are adding exactly two neutrons and two protons.

 

the mass discrepency that you seem to speak of is because of the average atomic weight that is listed in the periodic table

 

4 protons, 5 neutrons +2 protons, 2 neutrons = 6 protons, 7 neutrons

 

 

sorry for the repitition just its late and I wasn't sure that I was getting my point across

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"By emitting a neutron' date=' the C-13 sheds about 5MeV of that excitation."

 

A fast neutron then?

 

Btw, I had no idea whatsoever about the Coulomb repulsion. Can someone explain it?[/quote']

 

Not necessarily a fast neutron - the 5 MeV is solely due to mass differences. Some of the excitation energy could go into KE of the neutron, but it could also go into a gamma as the C-12 de-excites.

 

The key here is that it only takes ~5 MeV to strip a neutron from C-13. But you have 10.65 MeV of excess energy - so that neutron is not really bound to the nucleus.

 

Edit: Forgot about the Coulomb repulsion part: The protons in the nucleus and the incoming alpha repel each other. You have to have enough energy to overcome the repulsion in order to get within the range of the nuclear force. PE = kq1q2/r, where r will be of order 10-15m

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the mass of Be-9 + an alpha = to the mass of C13 because you are adding exactly two neutrons and two protons.

 

the mass discrepency that you seem to speak of is because of the average atomic weight that is listed in the periodic table

 

4 protons' date=' 5 neutrons +2 protons, 2 neutrons = 6 protons, 7 neutrons

 

 

sorry for the repitition just its late and I wasn't sure that I was getting my point across[/quote']

 

No. You can't just add those masses, because in forming bound systems energy must be released and the mass changes.

 

The mass of 6 protons + 6 neutrons is not the same as the mass of C-12.

 

The mass discrepancy is actually given by the table of nuclides, since you need the mass of the specific isotopes, rather than the averages.

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"The protons in the nucleus and the incoming alpha repel each other. You have to have enough energy to overcome the repulsion in order to get within the range of the nuclear force."

 

So, slow alpha particles are repelled by the nucleus' protons? I'd imagine that might have something to do with the bad ratio of neutrons for alpha particles (I think it was 30 neutrons / million alphas in an experiment with a beryllium sheet and an Am-241 alpha source), since alphas don't usually go too fast.

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"The protons in the nucleus and the incoming alpha repel each other. You have to have enough energy to overcome the repulsion in order to get within the range of the nuclear force."

 

So' date=' slow alpha particles are repelled by the nucleus' protons? I'd imagine that might have something to do with the bad ratio of neutrons for alpha particles (I think it was 30 neutrons / million alphas in an experiment with a beryllium sheet and an Am-241 alpha source), since alphas don't usually go too fast.[/quote']

 

Am-241 releases 5.638 MeV, so the alpha will have ~5.5 MeV of KE at the start. The barrier for an alpha on Be-9 is about half that.

 

I think the low number is due to the fact that you need pretty close to a head-on collision, and the target nucleus is only a few 10-15 meters across, and spaced at ~10-10 m apart. Getting an efficiency of a few parts in 105 seems about right.

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That barrier chart-thing is quite cool. :P Do you happen to know where to find a similar chart for neutrons in the ionization aspect, as their risk as ionizing particles seems to drop when they exceed 2 MeV (or something).

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That barrier chart-thing is quite cool. :P Do you happen to know where to find a similar chart for neutrons in the ionization aspect, as their risk as ionizing particles seems to drop when they exceed 2 MeV (or something).

 

Not sure what you mean. Neutrons can always be absorbed because they have no charge. They cause ionization directly, by scattering or absorption recoil, and indirectly because neutron absorption usually generates gammas.

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"how dangerous are neutron emissions compared to say gamma rays?"

 

When calculating the equivalence dose of radiation, all ionization-capable photons (including gamma) have a Q factor of 1. Neutrons, as you can see from the list, can have a Q factor of 20. Alphas, fission fragments and other similar heavy particles have a Q factor of 20. So if I had to choose between 0.2 Gray of 1 MeV neutrons or 1 Gray of gamma, I'd probably go with the gamma. :P Btw, it seems that I forgot the 2 MeV - 20 MeV part from the neutron list. Fixing...

 

Still, I don't know why for example 21 MeV neutrons aren't as dangerous as 1 MeV neutrons. Probably has something to do with neutrons not being absorbed by the bombarded atoms (no matter what swansont says :> ).

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Still, I don't know why for example 21 MeV neutrons aren't as dangerous as 1 MeV neutrons. Probably has something to do with neutrons not being absorbed by the bombarded atoms (no matter what swansont says :> ).

 

I would agree. Radiation isn't a problem if the energy isn't deposited. Slow neutrons tend to have bigger absorption cross sections - they have a larger deBroglie wavelength and spend more time interacting. Most of the energy they deposit is because of the change in nuclear structure, not due to KE. Using the nuclei under discussion: a C-12 nucleus capturing a neutron with ~0 KE releases ~5 MeV of energy.

 

While neutrons with a large KE deposit that much more energy per interaction, the interaction probability can drop, and the overall danger goes down.

 

 

(What I was confused about was you asking for a "similar chart" for neutrons, in reference to the Coulomb barrier - neutrons don't see a Coulomb barrier)

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Oops, I meant similar in the way of having MeV at the y-axle and a probability of ionization on the x. Yeah well, guess we've solved that now. :P

 

Heh, I also tried finding an answer for "how can you know if a nucleus is fissionable?".

 

http://www.physlink.com/Education/AskExperts/ae659.cfm

 

I trust that's you with the first answer there, swansont? :)

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Hey wait a sec...

 

"While neutrons with a large KE deposit that much more energy per interaction, the interaction probability can drop, and the overall danger goes down."

 

Yes but the equivalence dose is calculated using the absorbed dose as a base value. :o So it isn't about the probability of interactions in this case. Although, that's the answer to "why should I use borated paraffin in neutron source experiments". :P

 

Hmmh... Or did you mean that if a person receives a 1 Gray dose of 1 MeV neutrons or a 1 Gray dose of 21 MeV neutrons, the person's atoms absorb about four times less neutrons in the case of the 21 MeV neutrons?

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Hey wait a sec...

 

"While neutrons with a large KE deposit that much more energy per interaction' date=' the interaction probability can drop, and the overall danger goes down."

 

Yes but the equivalence dose is calculated using the absorbed dose as a base value. :o So it isn't about the probability of interactions in this case. Although, that's the answer to "why should I use borated paraffin in neutron source experiments". :P

 

Hmmh... Or did you mean that if a person receives a 1 Gray dose of 1 MeV neutrons or a 1 Gray dose of 21 MeV neutrons, the person's atoms absorb about four times less neutrons in the case of the 21 MeV neutrons?[/quote']

 

I sort of misread the question. My answer was in terms of flux rather than dose, as far as absorption probability goes.

 

The Q factor accounts for the energy deposited, combined with how localized the energy is. My guesss is that the higher energy neutrons do mare scattering while they slow down, so much of the KE is deposited over a larger volume as compared to the lower energy neutrons. Moderate energy neutrons will have a nuclear recoil in addition to the mass-energy change, which will ionize the atom in question, and cause local energy deposition.

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Ok, I think I got it. But could you point me to a site that has some basic information about the bit advanced nuclear physics, since the nuclear recoil things and the like are starting to get over my head. :P

 

Hyper Physics Concepts is a good place for basic stuff. So is Eric Weisstein's World of Physics

 

 

The recoil issue is just conservation of momentum. A neutron with some KE gets captured and there will be some KE left over in the resulting nucleus because momentum is conserved. This may be large compared to the ionization energy of the atom, in which case you ionize the atom - this causes more damage to surrounding cells.

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