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Fusion of protons in the Sun

  1. Jul 31, 2009 #1
    1. The problem statement, all variables and given/known data

    The sun is powered by fusion, with four protons fusing together to form a helium nucleus (two of the protons turn into neutrons) and, in the process, releasing a large amount of thermal energy. The process happens in several steps, not all at once. In one step, two protons fuse together, with one proton then becoming a neutron, to form the "heavy hydrogen" isotope deuterium. A proton is essentially a 2.4-fm-diameter sphere of charge, and fusion occurs only if two protons come into contact with each other. This requires extraordinarily high temperatures due to the strong repulsion between the protons. Recall that the average kinetic energy of a gas particle is (3/2)kbT, where kb is the Boltzmann constant.

    a) Suppose two protons, each with exactly the average kinetic energy, have a head-on collision. What is the minimum temperature for fusion to occur?

    b) Your answer to part a is much hotter than the 15 million K in the core of the sun. If the temperature were as high as you calculated, every proton in the sun would fuse almost instantly and the sun would explode. For the sun to last for billions of years, fusion can occur only in collisions between two protons with kinetic energies much higher than average. Only a very tiny fraction of the protons have enough kinetic energy to fuse when they collide, but that fraction is enough to keep the sun going. Suppose two protons with the same kinetic energy collide head-on and just barely manage to fuse. By what factor does each proton's energy exceed the average kinetic energy at 15 million K?


    2. Relevant equations

    (1/2)mv^2 = (3/2)kbT

    3. The attempt at a solution

    I isolated for T: T = (mv^2)/(3kb).

    However, I don't know how to find the speed of the particle.
     
    Last edited: Aug 1, 2009
  2. jcsd
  3. Jul 31, 2009 #2

    cepheid

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    The basic physics here is that the protons must have sufficient kinetic energy that they will come within 2.4 fm of each other and fuse before electric (Coulomb) repulsive forces send them flying apart. If they don't have enough kinetic energy, it will have all been used up before the particles reach this distance, and they'll stop and then begin to accelerate away from each other. You can calculate how much electric potential energy the system of two elementary charges separated by 2.4 fm would have in order to deduce how much initial kinetic energy each would have to have in order to get them within that distance. That's how you figure out the velocity.
     
  4. Aug 1, 2009 #3
    So:

    kq^2/r = (1/2)mv^2 + (1/2)mv^2 = mv^2

    Solving for v:

    v = sqrt(kq^2/mr) = sqrt(((9x10^9)(1.6x10^-19)^2)/((1.6x10^-27)(2.4x10^-15))) = 7.6 x 10^6 m/s

    Now plugging that into this equation:

    T = (mv^2)/(3kb)

    I obtain a value of 2.3 x 10^9 K.
     
    Last edited: Aug 1, 2009
  5. Aug 1, 2009 #4
    I checked, and that was the correct answer!

    As for problem B, I'm not sure what they're asking. I found the average kinetic energy to be: 3.11x10^-16 J at 15 million K. Do I divide this by a certain number?
     
  6. Aug 1, 2009 #5

    Redbelly98

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    They're asking you to compare two kinetic energies:
    (1) the average kinetic energy for the temperature you calculated in (a) and (2) the average kinetic energy for the sun's actual temperature.
     
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