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General angular momentum question (rutherford scattering context)

  1. Apr 16, 2008 #1
    ..... if you consider something simple momentum conservation, like in the case of a bullet, if you consider the system of the gun and the bullet, the momentum is zero before and after the shot is fired.

    same for collisions etc....

    however when you consider something like a deflection in rutherford scattering, the particle is deflected by the coulomb repulsion and follows a currved path and so has angular momentum during the deflection, however before and after the defelction, the particle is moving in a straight line, so how does the conservation of angular momentum apply in this case, when clearly the incident alpha particle, moving in a straight line, initially has no angular momentum.....
     
  2. jcsd
  3. Apr 16, 2008 #2
    sorry, i need to clarify this a little bit...

    i meant to say, if you look at simple momentum conservation for something a bullet being fired, at first appearances momentum might seem to be 'created', this is obviously not true, and a simple analysis including BOTH the gun and the bullet clears everything up. the same is true for collisions, if the collision has net zero momentum (particle A has momentum p and particle B has momentum -p ), we can see the system taken as a whole has zero momentum.

    just to clarify, im wondering how a particle like an alpha particle which is moving in a straight line, which before scattering, has no angular momentum L (r cross p), but during the scattering it clearly has some angular momentum. it then, once scattered then proceeds back on a straight path to the detector.... im just wondering how, as a conserved qunatity can be resolved. which system as a whole should i be looking at (like in the gun and bullet scenario) to see the conservation....


    cheers
     
  4. Apr 16, 2008 #3

    Dick

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    The sum of the angular momentum of the target and the projectile remains zero. So yes, the target acquires angular momentum as well. In Rutherford scattering the nucleus is more massive than the projectile and furthermore, it's also locked in a bound state with the much more massive yet target mass. So you ignore the motion of the target. This is the same reason that in gravitational trajectory problems, you ignore the motion of the earth.
     
  5. Apr 17, 2008 #4
    ah of course, cheers that cleared it up
     
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