Control rods nuclear reaction equation, moles liberated, pressure

In summary: No. The neutrons that get absorbed by the boron nucleus do not necessarily end up in the alpha particle. They could end up in the lithium nucleus.According to the reaction equation, how may alpha particles are produced for each neutron that gets absorbed.One.
  • #36
moenste said:
Sorry if that sounded rude.
Not at all. :smile:

I probably misunderstood you in post # 21. I thought that 1.5 * 1027 is number of alpha / helium particles in 1 m3.

But I guess by particles you understood atoms, so that's why the confusion was created.
Yes, I was taking "particle" to mean "atom". How were you interpreting "particle"?
 
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  • #37
TSny said:
Not at all. :smile:Yes, I was taking "particle" to mean "atom". How were you interpreting "particle"?
I thought that a particle (one particle) is 42α or 42He. And since such a large number of them (1.5 * 1027 of 42α or 42He particles) is produced in 1 m3 and each 42α or 42He has NA atoms, so the total number of atoms is their quantity on the number atoms in one 42α or 42He particle.
 
  • #38
moenste said:
I thought that a particle (one particle) is 42α or 42He. ... and each 42α or 42He has NA atoms, ...
I don't understand why you say, "each 42α or 42He has NA atoms".

An alpha particle consists of 2 protons and 2 neutrons bound together.
upload_2016-11-2_11-41-28.png

An alpha particle does not even contain 1 atom, much less Avogadro's number of atoms. However, when an alpha particle is produced in the control rod, it will capture 2 electrons from the environment. These electrons go into orbit around the alpha particle. This makes one He-4 atom.
upload_2016-11-2_11-42-29.png

So, in the control rod, each alpha particle that is produced becomes one He-4 atom.

Since each neutron that is absorbed by the control rod produces one alpha particle which, in turn, becomes one He-4 atom, the number of He-4 atoms produced is the same as the number of neutrons absorbed.
 
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  • #39
TSny said:
Since each neutron that is absorbed by the control rod produces one alpha particle which, in turn, becomes one He-4 atom, the number of He-4 atoms produced is the same as the number of neutrons absorbed.
I do understand what is said before this. But why 1 absorbed neutron produces 1 α-particle which in turns becomes 1 He-4 atom? Isn't it 1 absorbed neutron is 1 α-particle which is 1 He-4 element?
 
  • #40
moenste said:
I do understand what is said before this. But why 1 absorbed neutron produces 1 α-particle which in turns becomes 1 He-4 atom? Isn't it 1 absorbed neutron is 1 α-particle which is 1 He-4 element?
Can you explain what you mean by 1 He-4 element? In my previous post I showed simplified pictures of an alpha particle and a He-4 atom. Could you show or describe a picture of 1 He-4 element?
 
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  • #41
TSny said:
Can you explain what you mean by 1 He-4 element? In my previous post I showed simplified pictures of an alpha particle and a He-4 atom. Could you show or describe a picture of 1 He-4 element?
An alpha particle has no electrons, it has 2 protons and 2 neutrons.

When it decays (? or something else happens to it) it becomes a Helium particle which has two protons, 2 neutrons and also two electrons which are flying around it.

So one He-4 element is 42He which is one 42α-particle.

Or one alpha particle is just 1 atom of He-4?
 
  • #42
moenste said:
An alpha particle has no electrons, it has 2 protons and 2 neutrons.
Right.

When it decays (? or something else happens to it) it becomes a Helium particle which has two protons, 2 neutrons and also two electrons which are flying around it.
The alpha particle doesn't decay. The alpha particle is positively charged. So, it attracts electrons from the surroundings which go into orbit around the alpha particle. Once there are 2 electrons in orbit, the net charge of the alpha particle and the 2 electrons is zero. So, it no longer attracts any more electrons. You now have one He-4 atom.

So one He-4 element is 42He which is one 42α-particle.
We would not use the word "element" here. All that happens is that each alpha particle becomes one He-4 atom after it captures 2 electrons.

Or one alpha particle is just 1 atom of He-4?
Yes, one alpha particle becomes one He-4 atom.
 
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  • #43
TSny said:
We would not use the word "element" here. All that happens is that each alpha particle becomes one He-4 atom after it captures 2 electrons.
But if one alpha particle is one He-4 atom so each 4 g of He-4 atoms have NA [6 * 1023 number] of atoms? So one atom has 6 * 1023 atoms? To me it looks like "a chair of a chair is a chair" o_O.
 
  • #44
moenste said:
But if one alpha particle is one He-4 atom so each 4 g of He-4 atoms have NA [6 * 1023 number] of atoms?
Yes, one alpha particle (plus two electrons) will make one He-4 atom. Yes, Avogadro's number of He-4 atoms will have a total mass of 4 g.

So one atom has 6 * 1023 atoms?
How do you come to this conclusion?
 
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  • #45
TSny said:
Avogadro's number of He-4 atoms will have a total mass of 4 g.
I think I get it.

The thing with Avogadro is not relevant for this problem, right? We don't want to know the number of atoms in 4 g of He-4, but we need to know the total number of atoms. One alpha particle makes one He-4 atom. But if we need to know the number of atoms in 4 g of He-4 atom, we'll use the NA number.
 
  • #46
moenste said:
I think I get it.

The thing with Avogadro is not relevant for this problem, right? We don't want to know the number of atoms in 4 g of He-4, but we need to know the total number of atoms. One alpha particle makes one He-4 atom. But if we need to know the number of atoms in 4 g of He-4 atom, we'll use the NA number.
Yes, that's right.
 
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  • #47
Once you know the number N of He atoms formed, you can find the pressure using PV = NkT where k is Boltzmann's constant. Or you can use N and Avogadro's number to find the number n of moles of He formed. Then use PV nRT.
 
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  • #48
TSny said:
Once you know the number N of He atoms formed, you can find the pressure using PV = NkT where k is Boltzmann's constant. Or you can use N and Avogadro's number to find the number n of moles of He formed. Then use PV nRT.
Yes, the (c) part is clear. I used p = n R T / V.

Mostly was confused on the theory part with atoms.
 
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