Potential barrier in nuclear fusion

In summary, the conversation discusses calculating the height of the potential barrier for a head-on collision between two deuterons. The potential energy of the system and the force on each deuteron are calculated using equations and the concept of the center of mass frame is introduced. The conversation ends with a resolution stating that the barrier potential is equal to half of the potential energy of the system.
  • #1
kihr
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Homework Statement



To calculate the height of the potential barrier for a head on collision between two deuterons given that each deuteron is a sphere of radius R

Homework Equations



Potential of the first deuteron at a distance of 2R from it =V = ke/2R where k = 9 * 10^9 ,
e= 1.6 * 10^-19 Coulombs
The potential energy of the system of 2 deuterons = V*e



The Attempt at a Solution



Since the potential energy is the work required to be done in order to bring the second deuteron in contact with the first, the value of this potential energy should be the potential barrier. However, the answer given is half the potential energy as the potential barrier. I am unable to understand as to why this should be so. Either the answer as given is incorrect, or there is some flaw in my reasoning. Would appreciate some help in this regard. Thanks.
 
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  • #2
It may be a matter of using the center of mass (COM) frame instead of the frame where one deuteron is at rest. Note that since KE is square of the velocity you get 1/4 the energy for the moving deuteron in the CM frame since it has now half the velocity but you get double KE contributions from each deuteron, hence energy in COM frame is 2/4 times the energy in the frame you're using.

It seems like this shouldn't matter since you're working with potentials.

Let's see, let x be the distance from the origin of each deuteron so r=2x. The force on each is:
[tex]F = \frac{ke}{r^2} = \frac{ke}{4x^2}[/tex]
[tex]\int^\infty_R 2F(x)dx = \int^{\infty}_R \frac{ke\,dx}{2x^2} =[/tex]
[tex] = \left.-\frac{ke}{2x}\right|^{\infty}_R = \frac{ke}{2R}[/tex]
OK that seems correct.

That puzzles me, PE is same in either frame but KE isn't. Hmmm... let me think on this.
 
  • #3
Ahhh! I think I see the problem. In the Lab frame (one deuteron initially fixed) if you let the deuteron remain fixed then you're actually doing work on the total system since the center of mass is moving. Allowing the initially stationary deuteron to move you'll note that the total kinetic energy is not zero at closest approach. Rather there will be a point where both particles are at their closest and moving with the same velocity. The momentum is conserved so:
[tex]P_{total} = \sqrt{2m T_{init}}[/tex]
Then the closest approach momenta will each be:
[tex]p_k = \frac{1}{2}P_{total} = \sqrt{mT_{init}/2}[/tex]
and thus
[tex]T_{excess} = \frac{1}{2m}(p_1^2 + p_2^2) = \frac{1}{2m}mT_{init}[/tex]
thus
[tex]T_{excess} = \frac{1}{2} T_{init}[/tex]
The initial KE necessary to reach a given closest approach will be twice the KE transferred to PE in the Lab frame. Now if the stationary deuteron is held in place, i.e. its a target for a beam then this is not the case, all the KE goes to overcoming the potential barrier.

Now in the COM frame you have as I mentioned half as much KE but all of it gets used overcoming the potential barrier since at closest approach both particles come to a hault. THAT resolves the seeming inconsistency.

Now as to your answer. The answer you got was the correct amount of KE needed to overcome the potential barrier in both Lab and COM frames. However in the Lab frame you'd need to instill on one particle twice your answer in KE to get it to fuse with the initially stationary deuteron.

The halving answer I think is a mistake in a.) not taking into account this extra KE needed in the Lab frame and yet b.) by transitioning to the COM frame they do subtract it out. However in the COM frame you also would only need half your answer per deuteron so it is correct to halve it when calculating say fusion temperatures.

I think that is the correct resolution. Check my comments with your professor and see what he says.
 
  • #4
Thanks. Let me mull over your hints and try and figure out the whole issue.
 
  • #5
Having thought over it a little more based on your inputs, I think I have the answer to my problem. It is like this:

When the two deuterons are moving towards each other, they have kinetic and potential energies which keep changing with time (I would assume that each has the same KE and PE values at a given instant of time). When they come closest to each other and thereby come to a stop, the entire energy of the deuterons is potential energy V= kq*q/2R. By the law of conservation of energy, V must be equal to twice the kinetic energy of each deuteron when it was in motion, as all the kinetic energy has now been converted to potential energy. The barrier potential, I suppose, refers to this value of kinetic energy of the deuterons which must overcome the coulomb force opposing the motion of each other. Thus the answer is barrier potential = V/2.
Please let me know whether the above line of reasoning is correct or not. Thanks.
 

1. What is a potential barrier in nuclear fusion?

A potential barrier in nuclear fusion refers to the amount of energy that is required to overcome the repulsive forces between two positively charged nuclei to fuse them together.

2. What causes the potential barrier in nuclear fusion?

The potential barrier in nuclear fusion is caused by the strong nuclear force, which is the attractive force that holds protons and neutrons together in the nucleus of an atom. This force must be overcome in order for fusion to occur.

3. How can the potential barrier be overcome in nuclear fusion?

The potential barrier can be overcome by providing enough energy to the reactant nuclei to increase their kinetic energy and overcome the repulsive forces. This is often done through extreme temperatures and pressures in a controlled environment.

4. What is the significance of the potential barrier in nuclear fusion?

The potential barrier plays a crucial role in the feasibility of nuclear fusion as a source of energy. It determines the minimum amount of energy that must be input into the system for fusion to occur, and the efficiency of the fusion reaction.

5. Can the potential barrier be completely eliminated in nuclear fusion?

While it is not currently possible to completely eliminate the potential barrier in nuclear fusion, scientists are constantly researching and developing new techniques and technologies to reduce the amount of energy required and increase the efficiency of fusion reactions.

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