Control rods nuclear reaction equation, moles liberated, pressure

[moenste]

Homework Statement

A control rod to limit the rate at which a nuclear reactor is working is made from boron which is sealed in a casing. A boron atom (¹⁰₅B) is able to capture a neutron; an atom of lithium (Li) and an alpha particle being produced in the process. As a result helium gas is produced which occupies the spaces between the atoms in the rods which may be assumed to have a crystalline structure.

Each cubic metre of the control rod can absorb 1.5 * 10²⁷ neutrons before it must be replaced.

(a) Write down an equation for the nuclear reaction which takes place in the control rods.

(b) How many moles of helium are liberated in each cubic metre of control rod?

(c) The boron atoms themselves occupy 75 % of the total volume occupied by the rods. Calculate the pressure inside the casing, at a temperature of 300 K, just before the rod is replaced.

(Molar gas constant R = 8.3 J mol^-1 K^-1, Avogadro constant N_A = 6.0 * 10²³ mol^-1.)

Answers: (b) 2.5 * 10³ mol, (c) 2.5 * 10⁷ Pa.

2. The attempt at a solution
(a) ¹⁰₅B + ¹₀n → ⁶₃Li + ⁴₂α + ⁴₂He. Is it correct?

(b) No idea where to begin.

 

[TSny]

(a) ¹⁰₅B + ¹₀n → ⁶₃Li + ⁴₂α + ⁴₂He. Is it correct?

Is electric charge conserved in your equation?
Are the total number of nucleons conserved in your equation?

How is an alpha particle related to a Helium nucleus?

 

[moenste]

Is electric charge conserved in your equation?
Are the total number of nucleons conserved in your equation?

How is an alpha particle related to a Helium nucleus?

It should be like this then: ¹⁰₅B + ¹₀n → ⁶₃Li + ⁴₂α.

 

[TSny]

It should be like this then: ¹⁰₅B + ¹₀n → ⁶₃Li + ⁴₂α.

That's close. But the number of nucleons on the left doesn't match the number of nucleons on the right.

 

[moenste]

That's close. But the number of nucleons on the left doesn't match the number of nucleons on the right.

¹⁰₅B + ¹₀n → ⁶₃Li + ⁴₂α + ¹₀He? We have helium included, as in the problem statement.

 

[TSny]

¹⁰₅B + ¹₀n → ⁶₃Li + ⁴₂α + ¹₀He? We have helium included, as in the problem statement.

What do the 1 and 0 stand for in the symbol ¹₀He?

Also, think about what an alpha particle is made of.

Does the lithium isotope have to be ⁶₃Li? Could it be a different isotope of lithium?

 

[moenste]

What do the 1 and 0 stand for in the symbol ¹₀He?

1 is nucleon number and 0 is proton number.

Also, think about what an alpha particle is made of.

2 electrons and 2 protons and 4 - 2 neutrons.

Does the lithium isotope have to be ⁶₃Li? Could it be a different isotope of lithium?

Why shouldn't it be? When an alpha-particle is added, 4 and 2 should be subtracted.

 

[TSny]

1 is nucleon number and 0 is proton number.

Does a Helium nucleus have 0 protons?

2 electrons and 2 protons and 4 - 2 neutrons.

An alpha particle does not have any electrons.

Can you fill in the blanks in the folowing?
(a) An alpha particle has __________ protons.
(b) An alpha particle has __________ neutrons.
(c) A helium nucleus has _________ protons.
(d) A helium nucleus has _________ neutrons.

 

[moenste]

Does a Helium nucleus have 0 protons?

I'm not sure what "Helium nucleus" means...

The element from the periodic table is ⁴₂He, where 4 is the nucleon number = protons + neutrons and 2 is the proton number = electrons. So He from the periodic table has 2 protons and electrons and 4 - 2 = 2 neutrons.

Can you fill in the blanks in the folowing?
(a) An alpha particle has __________ protons.
(b) An alpha particle has __________ neutrons.
(c) A helium nucleus has _________ protons.
(d) A helium nucleus has _________ neutrons.

(a) 2 protons.
(b) 4 - 2 = 2 neutrons.
(c, d) What is a helium nucleus?

---

Maybe the equation is like this: ¹⁰₅B + ¹₀n → ^?_?Li + ⁴₂α + ⁴₂He. We have 11 as the nucleon number and 5 as the proton number on the left with 8 as the nucleon number and 4 as the proton number respectively. So we are lacking 3 and 1. So: ¹⁰₅B + ¹₀n → ³₁Li + ⁴₂α + ⁴₂He.

 

[mfb]

I'm not sure what "Helium nucleus" means...

The nucleus of a helium atom.

So He from the periodic table has 2 protons and electrons and 4 - 2 = 2 neutrons.

A neutral helium atom has 2 electrons, 2 protons and 2 neutrons, right. Which particles are in the nucleus, which particles are not?

 

[moenste]

A neutral helium atom has 2 electrons, 2 protons and 2 neutrons, right. Which particles are in the nucleus, which particles are not?

Nucleus has neutrons and protons, while electrons are flying around the nucleus.

 

[TSny]

The purpose of asking you how many protons and neutrons are in an alpha particle and in a $_2^4$He nucleus was to have you compare the alpha particle and the nucleus of the $_2^4$He atom.

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

 

[moenste]

The purpose of asking you how many protons and neutrons are in an alpha particle and in a $_2^4$He nucleus was to have you compare the alpha particle and the nucleus of the $_2^4$He atom.

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

Hm, it says that they are identical.

¹⁰₅B + ¹₀n → ⁷₃Li + ⁴₂He?

 

[TSny]

Hm, it says that they are identical.

¹⁰₅B + ¹₀n → ⁷₃Li + ⁴₂He?

Yes. That's good.

 

[moenste]

Yes. That's good.

Great!

How to approach (b)?

(b) How many moles of helium are liberated in each cubic metre of control rod?

I know that in ⁴₂He -- 4 g of He have 6 * 10²³ atoms (N_A number). But how to find moles I'm not sure.

 

[TSny]

I know that in ⁴₂He -- 4 g of He have 6 * 10²³ atoms (N_A number). But how to find moles I'm not sure.

How many alpha particles will be produced in 1 m³ of the control rod material before the rod must be replaced?

 

[moenste]

How many alpha particles will be produced in 1 m³ of the control rod material before the rod must be replaced?

Each cubic metre of the control rod can absorb 1.5 * 10²⁷ neutrons before it must be replaced.

1 m³ can absorb 1.5 * 10²⁷ neutrons and an α-particle has 4 - 2 = 2 neutrons.

1.5 * 10²⁷ / 2 = 7.5 * 10²⁶ α-particles.

 

[TSny]

1 m³ can absorb 1.5 * 10²⁷ neutrons and an α-particle has 4 - 2 = 2 neutrons.

1.5 * 10²⁷ / 2 = 7.5 * 10²⁶ α-particles.

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.

 

[moenste]

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 alpha particle is produced for each neutron?

 

[TSny]

One alpha particle is produced for each neutron?

Right.

 

[moenste]

Right.

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

 

[TSny]

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

Yes.

 

[moenste]

Yes.

How many moles does an alpha particle have?

In that case multiply that number by 1.5 * 10²⁷, the total number of alpha particles in 1 m³, and that will be the answer.

 

[TSny]

How many moles does an alpha particle have?

A better question would be how many alpha particles does a mole of alpha particles have?

 

[moenste]

A better question would be how many alpha particles does a mole of alpha particles have?

1 mol = number of atoms in 4 g of ⁴₂He.

So since we have 1.5 * 10²⁷ He, therefore we have 1.5 * 10²⁷ moles.

 

[TSny]

1 mol = number of atoms in 4 g of ⁴₂He.

Yes. But this is not helpful for this problem.

So since we have 1.5 * 10²⁷ He, therefore we have 1.5 * 10²⁷ moles.

No. I don't see how you came to that conclusion. A mole of particles has a specific number of particles. The number of carbon atoms in a mole of carbon is the same as the number of helium atoms in a mole of helium. What is the value of this number?

 

[moenste]

Yes. But this is not helpful for this problem.

No. I don't see how you came to that conclusion. A mole of particles has a specific number of particles. The number of carbon atoms in a mole of carbon is the same as the number of helium atoms in a mole of helium. What is the value of this number?

I think you are talking about the N_A.

1 mol = N_A He-4.

Since we have 1.5 * 10²⁷ He-4, so the total number of atoms is 1.5 * 10²⁷ * N_A. Now we need to put the number from atoms into moles, where 1 mol is N_A. So divide the number by N_A and get 1.5 * 10²⁷ number of atoms.

 

[TSny]

I think you are talking about the N_A.

1 mol = N_A He-4.

Yes, 1 mole of He-4 contains N_A atoms of He-4.

Since we have 1.5 * 10²⁷ He-4, so the total number of atoms is 1.5 * 10²⁷ * N_A.

Doesn't the phrase "Since we have 1.5 * 10²⁷ He-4" mean that we have 1.5 * 10²⁷ atoms of He-4?
Why change this number to 1.5 * 10²⁷ * N_A?

Now we need to put the number from atoms into moles, where 1 mol is N_A. So divide the number by N_A and get 1.5 * 10²⁷ number of atoms.

Did you mean the last word to be "moles" rather than "atoms"? If so, you have the right method for converting number of atoms to moles. You just need to start with the right number of He-4 atoms.

 

[moenste]

Yes, 1 mole of He-4 contains N_A atoms of He-4.Doesn't the phrase "Since we have 1.5 * 10²⁷ He-4" mean that we have 1.5 * 10²⁷ atoms of He-4?
Why change this number to 1.5 * 10²⁷ * N_A?

Did you mean the last word to be "moles" rather than "atoms"? If so, you have the right method for converting number of atoms to moles. You just need to start with the right number of He-4 atoms.

One mole of He-4 has N_A number of atoms (one rock weights 2 kg).

But since we want to know the total number of atoms (total weight of rocks), we need to multiply this number by the number of He-4 particles (number of rocks). We have 1.5 * 10²⁷ He-4 particles * 6 * 10²³ = 9 * 10⁵⁰ atoms (we have 25 rocks, each weights 2 kg, total weight 25 * 2 = 50 kg).

Then we want to know the number of moles of helium that are liberated in each cubic metre of control rod. Since we know 1 mole = N_A, we then divide the 9 * 10⁵⁰ by 6 * 10²³ and get 1.5 * 10²⁷ moles.

Update
(b) 1 m³ has 1.5 * 10²⁷ atoms of helium. Number of moles liberated = 1.5 * 10²⁷ / 6 * 10²³ = 2500 moles. I do get the answer, but I don't get the solution. I stated my reasoning above.

(c) p = n R T / V = 2500 * 8.3 * 300 / 0.25 = 24 900 000 Pa or 2.5 * 10⁷ Pa. I get the correct answer but I don't understand: (i) what is casing?, (ii) if boron atoms occupy 75 %, we also need to know the percentage for the neutron and Li, in order to use the 2500 moles solely of helium.

If you could please explain the theory part on the (b) and (c) parts which I don't understand above, I would appreciate it very much.

 

[TSny]

One He-4 has N_A number of atoms (one rock weights 2 kg).

This is ambiguous. One mole of He-4 has N_A atoms.

But since we want to know the total number of atoms (total weight of rocks), we need to multiply this number by the number of He-4 particles (number of rocks).

I don't follow this, especially the "rocks".

We have 1.5 * 10²⁷ He-4 particles * 6 * 10²³ = 9 * 10⁵⁰ atoms (we have 25 rocks, each weights 2 kg, total weight 25 * 2 = 50 kg).

10⁵⁰ atoms is about the number of atoms in the entire earth.

You correctly stated previously that each neutron that gets absorbed results in the creation of one alpha particle. The alpha particle is the same as a nucleus of He-4. The alpha particle quickly "captures" two electrons to make a neutral He-4 atom. So, for each neutron that gets absorbed, you get one He-4 atom.

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

I don't understand: (i) what is casing?, (ii) if boron atoms occupy 75 %, we also need to know the percentage for the neutron and Li, in order to use the 2500 moles solely of helium.

The neutron gets absorbed, so it does not exist as a separate particle after the reaction. The question apparently wants you to assume that the He-4 atoms that are produced form a gas that occupies the space between the boron atoms. You can apparently neglect the Li that is produced.

 

[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.