Calculating Temp. required for nuclear fusion

[Zain Syed]

Hey guys, I'm working on a homework problem about nuclear fusion in stars and am..stuck on the first step: calculating the temperature needed for protons to come within 2 fm of one another and overcome Coulomb repulsion.

1. Homework Statement
Given that the protons have an average kinetic energy (3/2)k_bT, and in the Boltzmann distribution there will be some protons of 4 times that energy, show that a temperature of about a billion degrees Kelvin is needed in order for the protons to overcome the Coulomb repulsion and approach each other within 2 fm.

Homework Equations

avg KE = (3/2)k_bT
Coulomb Barrier: U = ke²/r

The Attempt at a Solution

I calculated U to be 1.4 * 10⁶eV. I then plugged this in as an inequality, where the avg KE > U and solved for T, which I found to be 1.08*10¹⁰K.

I'm certain I missed something, or a lot of something, but I don't know what that is. I'm meant to use the finite-structure constant α = 1/137, right?[/B]

 

[BvU]

Hi Zain, welcome to PF !

We're not all that good in telepathy, so if you want us to help find where it goes wrong, show your calculation in detailed steps. Who knows, perhaps you find an error somewhere, and if not, maybe PF helpers will !

 

[mfb]

Don't take the average energy, take 4 times the average as the problem statement suggests.
Maybe assume that both protons have that energy. A factor of 2 doesn't really matter for such a rough estimate.

 

[Zain Syed]

Sure! Here we go.
U = k(e²/r)
= (8.99*10^9)*(1.6*10^-19)²/(2.0*10^-15) ( found a mistake here as I was typing it out; I'd originally used 1 fm instead of 2..)
= 4.6 * 10^-13 J
= 2.8 * 10⁶ eV

I solved the KE formula for T to get
T = (2/3)*(1/k_b)*KE, where k_b = 8.62*10^-5 eV/K
T = (2/3)*(2.8*10⁶ eV)/(8.62*10^-5 eV/K)
= 2.2*10¹⁰ K

If I take 4 times the average kinetic energy I can multiply my result by 4, and then divide by 2 to account for both protons having that energy?
Am I on the right track here?

Appreciate the help :)

 

[BvU]

I read the problem statement differently (mfb means the same thing): at T the average energy may be 1/4 of what is required; there will be plenty protons that then have enough energy to get the reaction going.

Oh, and check your calculator: 9 * 1.6^2 /2 should be in the neighborhood of 10

 

[Zain Syed]

Hey guys; I checked over my work several times and I think my original answer, even though I got to it incorrectly, is actually right.
My calculation ends in 2.2 * 10¹⁰ K as being the temperature required for a proton to cross the Coulomb Barrier, so I divide this by 2 to account for their being two protons interacting with equal energy, so the temperature required is then 1.1*10¹⁰ K.

I don't know that accounting for some protons having four times the average energy makes a difference. Even those protons won't cross the Coulomb barrier classically. Only through quantum tunneling does this reaction occur.

 

[BvU]

I don't know that accounting for some protons having four times the average energy makes a difference. Even those protons won't cross the Coulomb barrier classically. Only through quantum tunneling does this reaction occur

There is no tunneling. It's a matter of relative probabilities for the various kind of possible interactions. You were thinking in the right direction mentioning $\alpha$. Matt Strassler has a nice article on this.

 

[mfb]

I don't know that accounting for some protons having four times the average energy makes a difference.

It makes a difference of a factor of 4. The energy required to reach this distance stays the same, but the average kinetic energy is just 1/4 this value, which also lowers the temperature by a factor of 4.

Sure! Here we go.
U = k(e²/r)
= (8.99*10^9)*(1.6*10^-19)²/(2.0*10^-15) ( found a mistake here as I was typing it out; I'd originally used 1 fm instead of 2..)
= 4.6 * 10^-13 J
= 2.8 * 10⁶ eV

Google disagrees

Fix that, and fix the factor of 4, and the result will be close to a billion.

 

[Zain Syed]

It makes a difference of a factor of 4. The energy required to reach this distance stays the same, but the average kinetic energy is just 1/4 this value, which also lowers the temperature by a factor of 4.

Google disagrees

Fix that, and fix the factor of 4, and the result will be close to a billion.

Thank you! I made the adjustments and got 1.38*10^9 K. I'd talked myself into thinking I had it down instead of double checking my calculator.

There is no tunneling. It's a matter of relative probabilities for the various kind of possible interactions. You were thinking in the right direction mentioning $\alpha$. Matt Strassler has a nice article on this.

Here I am confused. The next part of the question starts with "Since the core of the sun has a temperature of only about ten million degrees Kelvin, the fusion reaction can only proceed via quantum mechanical tunneling". I don't understand the material too well

 

[BvU]

The next part of the question starts with "Since the core of the sun has a temperature of only about ten million degrees Kelvin, the fusion reaction can only proceed via quantum mechanical tunneling"

Well, there I am with egg on my face ! Or not ? What temperature was it that you calculated ?

 

[Zain Syed]

Well, there I am with egg on my face ! Or not ? What temperature was it that you calculated ?

Happens to everyone! The temperature I have is 1.4*10⁹K, so a little over the 10 million degrees Kelvin in the sun.

 

[SteamKing]

Happens to everyone! The temperature I have is 1.4*10⁹K, so a little over the 10 million degrees Kelvin in the sun.

Isn't 10⁹ = 1 billion?

 

[BvU]

Happens to everyone! The temperature I have is 1.4*10⁹K, so a little over the 10 million degrees Kelvin in the sun.

Think again !

 

[Zain Syed]

1.4 billion degrees K, made a bit of an understatement.

 

[BvU]

So how can the sun do its work if the temperature is a factor 500 short of overcoming the 15 fm coulomb barrier for average speed protons on head-on collision course ?

( $1.4 \times 10^9 * 4 \ / \ 10 \times 10^3$ )

 

[Zain Syed]

So how can the sun do its work if the temperature is a factor 500 short of overcoming the 15 fm coulomb barrier for average speed protons on head-on collision course ?

( $1.4 \times 10^9 * 4 \ / \ 10 \times 10^3$ )

Quantum tunneling? I think that's the implication of the problem.

 

[mfb]

Yes it is.

(4 times the average kinetic energy is quite conservative, there are particles with more energy and fusion is rare, but even 40 times would not give enough energy for a classical contact).

 

[BvU]

Heh, I'm back with an apoloy. My 'no tunneling' is obviously wrong (I had a wrong association with 'potential hill to overcome' but Coulomb keeps going up; the crux of tunneling however is that the particles can be where they can't be classically).
Exciting material; can't help but be curious how you are doing with the remainder of the exercise !

The next part of the question starts ...