Control rods nuclear reaction equation, moles liberated, pressure

[moenste]

So, how many He-atoms are produced in each m3 of the control rod by the time the rod needs to be replaced?

So 1.5 * 10²⁷ α-particles are produced in 1 m³?

Yes.

So we have 1.5 * 10²⁷ number of He-4 in 1 m³. Millions of millions He-4 are in 1 m³.

Each He-4 has N_A atoms.

Since we have 1.5 * 10²⁷ He-4, their total number of atoms in 1 m³ is 1.5 * 10²⁷ * 6 * 10²³.

1 mole = N_A

1.5 * 10²⁷ * 6 * 10²³ / N_A = 1.5 * 10²⁷ atoms in 1 m³.

 

[TSny]

So we have 1.5 * 10²⁷ number of He-4 in 1 m³. Millions of millions He-4 are in 1 m³.

When you say we have 1.5 * 10²⁷ number of He-4 , do you mean we have 1.5 * 10²⁷ number of He-4 atoms? If not, what do you mean by 1.5 * 10²⁷ number of He-4?

 

[moenste]

When you say we have 1.5 * 10²⁷ number of He-4 , do you mean we have 1.5 * 10²⁷ number of He-4 atoms? If not, what do you mean by 1.5 * 10²⁷ number of He-4?

Nevermind, 1 m³ has 1.5 * 10²⁷ atoms so the number of moles is 1.5 * 10²⁷ / N_A = 2500 moles.

Thank you!

 

[TSny]

Nevermind, 1 m³ has 1.5 * 10²⁷ atoms so the number of moles is 1.5 * 10²⁷ / N_A = 2500 moles.

OK

 

[moenste]

OK

Sorry if that sounded rude.

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.

 

[TSny]

Sorry if that sounded rude.

Not at all.

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"?

 

[moenste]

Not at all. 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 N_A atoms, so the total number of atoms is their quantity on the number atoms in one ⁴₂α or ⁴₂He particle.

 

[TSny]

I thought that a particle (one particle) is ⁴₂α or ⁴₂He. ... and each ⁴₂α or ⁴₂He has N_A atoms, ...

I don't understand why you say, "each ⁴₂α or ⁴₂He has N_A atoms".

An alpha particle consists of 2 protons and 2 neutrons bound together.

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.

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.

 

[moenste]

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?

 

[TSny]

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?

 

[moenste]

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?

 

[TSny]

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.

 

[moenste]

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 N_A [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" .

 

[TSny]

But if one alpha particle is one He-4 atom so each 4 g of He-4 atoms have N_A [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?

 

[moenste]

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 N_A number.

 

[TSny]

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 N_A number.

Yes, that's right.

 

[TSny]

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.

 

[moenste]

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.