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Atomic collisions must be elastic?

  1. Mar 12, 2013 #1
    1. The problem statement, all variables and given/known data

    Ground-state hydrogen atom with 12 ev kinetic energy collides head-on with another ground-state hydrogen atom at rest. Using principles of conservation of energy and momentum, show that an inelastic collision cannot occur. Therefore the collision must be elastic.

    2. Relevant equations

    Sum of energy before equals sum of energy after
    Vector sum of momentum before equals vector sum of momentum after

    3. The attempt at a solution

    If the collision is inelastic, then the total kinetic energy after the collision will be less than it was before. Conservation of energy requires that the atoms absorb some of the initial kinetic energy and change their internal state.

    Conservation of momentum requires that the vector sum of velocities be unchanged by the collision (masses cancelling out). There is no way for this to occur, as I see it, if the total kinetic energy has been reduced by an inelastic collision. Therefore the collision must be elastic.

    This logic seems to say that all collisions between like atoms are elastic. Correct? Regardless of the initial kinetic energy?
     
  2. jcsd
  3. Mar 12, 2013 #2

    TSny

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    Hello, gmark. Welcome to PF!
    I don't think your conclusion is correct. Did you actually go through the math to show that in this problem the collision must be elastic?
     
  4. Mar 13, 2013 #3
    TSny embarrasses me into working out the math. I find that my hand-waving was wrong. Inelastic collision is possible if at least half of the initial kinetic energy remains as kinetic energy after the collision. That is, up to half the initial K can change the atom's internal state if appropriate quantum levels are available.

    Here's my math. Subscripts or superscripts indicate post-collision values, plain symbols are for pre-collision. Velocity is treated as scalar, since all motion is along a line.
    v is velocity
    m is atomic mass
    K is kinetic energy

    In a fully elastic collision, particles essentially exchange momenta. K = K'. At any time except during impact one particle or the other has zero velocity. In at least some inelastic collisions (eg bullet and ballistic pendulum), both particles have non-zero velocities after collision.

    Conservation of momentum:
    mv = m(v1 + v2)
    so v = v1 + v2

    Before collision:
    K = 1/2(mv2) = 1/2(m(v1 + v2)2)

    After:
    K' = 1/2(m(v12 + v22))

    This gives

    v12 - v*v1 + (K - K')/m = 0

    where v = √(2K/m)

    Values of v1 are real for K' ≥ K/2, so inelastic collisions are possible provided less than half of the initial kinetic energy is "lost."
     
  5. Mar 13, 2013 #4
    What is the answer to the original problem then? K = 12 eV.
     
  6. Mar 13, 2013 #5
    The answer is that this collision must be elastic. The least amount of energy that a stationary ground-state H atom can absorb is 10.2ev, to change the principal quantum number n from 1 (-13.6ev) to 2 (-3.4ev). Such a change would result in a residual K' = 12.0 - 10.2 = 1.8ev, which is less than K/2. This particular inelastic collision could not occur.

    The answer is also that inelastic atomic collisions are possible under certain conditions, contrary to my original conclusion that they cannot take place at all.
     
  7. Mar 13, 2013 #6
    Very well.

    Note, however, that there are smaller quantum effects, which are known as fine and hyperfine structures, where very tiny fluctuations of energy are possible.
     
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