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Potential barrier in nuclear fusion

  1. Jan 23, 2010 #1
    1. The problem statement, all variables and given/known data

    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

    2. Relevant 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

    3. 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.
  2. jcsd
  3. Jan 23, 2010 #2


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    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.
  4. Jan 23, 2010 #3


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    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]
    [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.
  5. Jan 23, 2010 #4
    Thanks. Let me mull over your hints and try and figure out the whole issue.
  6. Jan 23, 2010 #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.
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