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Unclear fusion Pratt 3

Discussion in 'Electronic Design' started by Genome, Apr 10, 2004.

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  1. Genome

    Genome Guest

    Consider deuterium. Take away its electron. It becomes a boson. I
    mentioned bosons in 'It's a fundamental universe'. Bosons like to get
    together and occupy the same energy/space.

    Why doesn't it go bang and turn into helium?

    Look at the energy per nucleon curve. If that wasn't there the universe
    would be something different.

    Things do depend on uncertainty in position though.

    DNA
     
  2. John Larkin

    John Larkin Guest

    The electrostatic repulsion between two protons is fierce, so you need
    a lot of pressure (or velocity) to push them close enough that the
    strong nuclear force kicks in and glues them together. If you fire
    them at each other in an accelerator, you'll get a lot of near misses,
    billiard-ball elastic collisions where they almost stick but at the
    last attosecond slime past each other because of the huge
    electrostatic repulsion. I think you'll get inelastics too, with
    photons or something kicked out and wasting energy.

    Room-temperature neutrons can just sort of casually saunter into a
    nucleus and blow it up. Protons and antiprotons snuggle up like
    bunnies.

    John
     
  3. Dave VanHorn

    Dave VanHorn Guest

    Room-temperature neutrons can just sort of casually saunter into a
    Exploding bunnies, that is.

    Interesting point though, leaving the electron on probably helps, since it
    masks the positive charge till a bit later.. But it probably dosen't help
    much, because the nucleus is so tiny.
     
  4. Rich Grise

    Rich Grise Guest

    What's smaller than Nanotech? Picotech? Just build sub-nanoscopic
    manipulators
    out of neutronium, and grab four individual hydrogen atoms by the protons,
    and hold them together until they release the appropriate amount of energy.
    One of the engineering challenges would then, of course, be how to capture
    that energy. I think if you design the picomanipulators properly, you could
    have the alpha particle get kicked out the same direction every time,
    making a nifty torch ship motor. Probably do some pretty decent MHD, too.

    Anybody know enough nuclear physics to calculate out how hot (i.e. fast)
    the helium/alpha exhaust would be going? (clearly, you have to dump the
    electrons as well - pointing that stream could be interesting as well.)

    Cheers!
    Rich
     

  5. Greetings,

    Since this post is fairly long the answer is in the last two paragraphs for
    those that might be too put off by length to read the whole thing. That
    said...

    The first step in making this calculation would be to find the "Q" value for
    this reaction. Of course this reaction (four protons coming together to
    form an alpha particle) doesn't really occur in real life, so this is just a
    theoretical exercise...

    The Q value tells us how much energy is released by this reaction. We use
    Einstein's E=mc^2 formula along with the mass defect (difference in masses
    between the four individual separate protons and the mass of the single
    alpha particle).

    mass proton = 1.0072764669 amu
    mass neutron = 1.0086649233 amu
    mass electron = 5.48579911E-4 amu
    Helium 4 atom = 4.0026032497 amu
    Approximate alpha particle mass = mass(He-4 atom) - 2*mass(electron)
    Approximate alpha particle mass = 4.001506089878 amu

    Note: MeV stands for mega electron volts.

    So mass defect between four protons and one alpha particle ~= 4*1.0072764669
    amu - 4.001506089878 amu = 0.02759977772 amu. To find the Q value simply
    multiply the mass defect by the conversion factor 931.5MeV/amu.

    Theoretical Q for this reaction = 0.02759977772 amu * 931.5 MeV/amu = 25.71
    MeV.

    This is how much energy would theoretically be liberated by this reaction.
    A small amount of mass was converted into this much energy.

    For comparisons sake a single fission of some atom such as U-235 releases
    approximately 200MeV, while the oxidation of say carbon liberates only a few
    electron volts.

    Okay, so now how fast is the alpha moving? Well in theory all that 25.71
    MeV is transferred into kinetic energy of the product(s) (in this case an
    alpha particle). There is a slight problem here in that an alpha particle
    is composed of 2 neutrons and 2 protons, not 4 protons. From the
    information given above we see a neutron is actually slightly more massive
    than a proton. In practice the kinetic energy of the alpha particle in this
    theoretical reaction would probably be approximately 1.2MeV less than the
    calculated 25.71MeV figure since some of that energy would have to get used
    up creating something like a couple of positrons or something so that a
    couple of the protons can convert into neutrons. So as a decent
    approximation the alpha particle's kinetic energy would be around 24.5MeV.

    How fast is that?

    Well use the formula 0.5mv^2 = kinetic energy. Where m is mass in kg, and v
    is the velocity of the particle in meters per second. You have to be
    careful in this step to make sure you don't use classical mechanics if the
    particle is moving at relativistic velocities. In this case 24.5MeV for an
    alpha particle is quite "slow", so classical mechanics will yield a fairly
    accurate result.

    So first convert MeV into joules. 24.5MeV (1.6022E-19 J/1 eV) = 3.93E-12 J.
    Now find the mass of an alpha particle in kilograms.

    4.001506089878 amu * (1.6605387E-27 kg / amu) = 6.6446557205E-27 kg

    Solving for velocity v = sqrt(2*kinetic energy / mass) = 34373216
    meters/second
    Or in other words 3.44E7 m/s

    So this "exhaust" for this theoretical reaction is boogieing along with
    pretty high velocity. For comparisons sake room temperature (ie: 300 K)
    particles have about 0.025eV of kinetic energy. Of course I want to
    reemphasize that this reaction does not occur in real life. Even single
    proton-proton --> deuterium fusion reactions are exceedingly "slow" at
    occurring, much less four protons all perfectly coming together all at once.
     
  6. used...


    Whoops... I made a slight error here. The kinetic energy of the alpha
    should be about 1.02MeV less than the theoretically calculated value of
    25.71MeV.

    Why 1.02MeV?

    Well a positron (an antimatter electron with a positive charge) has a mass
    of basically an electron. And by multiplying the mass of an electron by the
    conversion factor 931.5MeV/amu we see that a rest electron has effectively
    511keV of rest energy (in the form of matter). So a proton also has that
    much rest energy (in the form of matter). Thus two positrons has a rest
    energy of 2 * 511keV = 1.02MeV. In my head I did this calculation earlier
    on the last post and thought it should be 1.2MeV, but that isn't right.

    The cool thing about antimatter is that if it ever meets of up with regular
    matter they annihilate each other and convert all of their mass into energy.
    In the case of the positrons they eventually meet up with regular electrons
    and produce two 511keV gamma rays.
     
  7. Rich Grise

    Rich Grise Guest

    Oh, ok - Forgot about that part. Well, since we're in thoughtspace, just
    absorb
    a couple of the electrons into their protons and make the two neutrons
    that way. :)


    So as a decent
    That's pretty fast!

    Then, lessee - you've got specific impulse, thrust, etc., etc., -
    I read somewhere a millennium or so ago, an article called "Torchships
    Now!" which made claims like at 1G continuous accel/decel, you could
    get to the moon in like a couple of hours, and Pluto in a few days or
    weeks. He had all the math worked out, but I remember equations about
    as well as I remember dates. %-]

    Cheers!
    Rich
     
  8. I read in sci.electronics.design that Rich Grise <>
    1g is quite a high acceleration. Let's see. Distance at constant
    acceleration = 0.5 x acc. x time ^2. 0.5 x 9.81 (g in m/s) x (24 x
    3600)^2 = 3.66 x 10^10 metres. The earth-moon distance is 3.8 x 10^8
    metres, so it wouldn't take a day to get there but about 2.5 hours.
    You' be going past very quickly, though: 9.81 x 2.5 x 3600 = 88290 m/s.
     
  9. It might be more useful to accelerate at 1g to the half-way point and
    then de-accelerate at 1g until you get there. More like 3.5 hours, if
    I did the arithmetic right.

    Not quite the same thing, but if we built a tunnel to China that went
    through the center of the earth I could get there in something like
    45min and (assuming the tunnel was evacuated) with zero energy. Hmm..
    in that case you'd be weightless for the entire trip.

    Best regards,
    Spehro Pefhany
     
  10. The 'lovely' thing about this one, is that for a tunnel that goes straight
    between _any_ two places on the Earth's surface, assuming you free fall all
    the way, and the system is frictionless, the journey time is basically
    constant!...

    Best Wishes
     
  11. So how come I'm not in free fall whenever I'm in a tunnel?
     
  12. I read in sci.electronics.design that Walter Harley
    It's not a geodesic tunnel: there aren't any.
     
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