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Rutherford Scattering problem with a heavier nucleus

  1. Nov 11, 2005 #1
    An alpha particle approaches a Au (gold) nucleus with atomic mass number 197 with a speed of 1.5*10^7 m/s. The alpha particle is then scattered 49 degrees from the horizontal at the slower speed of 1.47*10^7 m/s.

    In what direction does the Au nucleus recoil below the x-axis (the horizontal) in degrees?

    I don't think I'm approaching this problem in the right way. Part B of the question asks with what speed does the Au nucleus recoil, so I thought that once I had the velocity of the nucleus, I could somehow figure out the angle it recoiled at, but I'm not sure if I have to use the force of repulsion to do that.

    I also had another similar problem where they tell us the radius of an oxygen atom with atomic mass 16 has a radius of 3.0 fm. and the quesiton asked at what speed would a proton need to be fired toward the oxygen nucleus if it were to have a turning point at 1.20 fm from the surface.

    Looking through an example in my book, by using conservation of energy we get:

    final kinetic energy (K_f) + final potential energy (U_f) = initial kinetic energy (K_i) + initial potential energy (U_i)

    from that i got the expression:

    0 + ((q_alpha)(q_Au))/(r_min*4*pi*epsilon_0) = .5m(v_i)^2

    So I solve for my v_i using 1.20 fm + 3.0 fm as my r_min. 1.6*10^-19 as my charge for the proton, 16*1.6*10^-19 as the charge of my nucleus, and for the mass I used the mass of the proton (1.67*10^-27).

    After calculating it by plugging in the numbers, I got 3.24*10^7 m/s, but it's wrong and I'm not sure if I used the wrong r_min because I tried using just 1.2 fm and and 4.2 fm (converted into meters) but I'm sill not getting the right answer.

    If anyone can help, I'd be really grateful, these two problems really confused me. Thanks. I've also attached a picture of the diagram for the problem with the Au nucleus.
     

    Attached Files:

  2. jcsd
  3. Nov 11, 2005 #2

    Doc Al

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    Staff: Mentor

    conservation of momentum

    Don't forget that momentum is conserved.
    Again you seem to have forgotten that momentum is conserved. You have incorrectly assumed that when the proton gets to the turning point, the kinetic energy is zero. (That would violate momentum conservation!)
     
  4. Nov 11, 2005 #3
    I tried using conservation of momentum to solve for the velocity of the nucleus' recoil.

    I had
    m_alpha*v_alpha_i = m_alpha*v_alpha_f + m_Au*v_Au_f

    and I solved for the velocity of the nucleus and I got an answer around 3030 m/s, but that was wrong, and I don't think that I'm plugging in the numbers incorrectly either.

    A friend of mine was working on the same problem and she told me that she solved for it using conservation of energy instead, and I tried that once too but I still got the wrong answer.

    I had
    .5m_alpha*(v_alpha_i)^2 = .5m_alpha*(v_alpha_f)^2 + .5m_Au*(v_Au_f)^2

    and then I solved for the velocity of the nucleus and got around 3020 m/s. The real answer was 2.49*10^5 m/s, so I think I'm missing a conponent of energy other than the kinetic energy.
     
    Last edited: Nov 11, 2005
  5. Nov 11, 2005 #4

    Doc Al

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    Staff: Mentor

    Momentum is a vector! Treat the initial direction of the alpha particle as being along the x-axis. Treat each component of momentum separately.
     
  6. Nov 11, 2005 #5
    When I was reading an example in my book using gold and an alpha particle, it said:

    "At the moment of closest approach, just before the alpha particle is reflected, the charges are at rest and the system has only potential energy"

    But this was a case in which we assumed that a gold nucleus wouldn't recoil if a an alpha particle is shot at it. In my case, it's a proton and an oxygen nucleus.
     
    Last edited: Nov 11, 2005
  7. Nov 11, 2005 #6

    Doc Al

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    Right. But in this example (Rutherford's gold foil experiment) the gold nucleus is fixed.

    For gold, you assume it because the gold is bound to its neighbors in a lattice structure. Are you expected to assume that the oxygen atom is bound to a solid as well? (I was thinking this was a molecule of oxygen gas.)
     
  8. Nov 11, 2005 #7
    I didn't know the gold was fixed, but the example in my book says that the alpha particle is shot toward the nucleus of a gold atom, so I figured it was just a standalone nucleus and the reason they give for the gold not moving is because it's much more massive than the alpha particle. The problem doesn't say much about the oxygen nucleus, but it specifically says it's a nucleus and doesn't mention that it's an atom. This was all they gave us:

    The oxygen nucleus (atomic number 16) has a radius of 3.0 fm.
    and then they asked:
    With what speed must a proton be fired toward an oxygen nucleus to have a turning point 1.20 fm from the surface?

    In that case, since it the gold nucleus does move, how would we find a velocity for it without knowing first the initial speed of the proton?
     
  9. Nov 11, 2005 #8

    Doc Al

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    I didn't mean to mislead you: The gold's much greater mass than the alpha particle is certainly a significant factor as well.
    Assuming that the oxygen nucleus can move and that the collision is dead center, you need to find the speed at the moment of closest approach. Here's a hint: At the moment of closest approach, what can you say about the relative speeds of the proton and the oxygen nucleus?
     
  10. Nov 11, 2005 #9
    My first instict tells me that both the proton and nucleus would be stationary and some distance when they come within a certain distance of each other because only the proton is moving. Assuming that the proton starts off at a distance of inifinity to the nucleus, wouldn't the initial KE from the proton still equal the potential from the minimum distance that the two need to meet to become stationary? Or is this distance, and the distance that they want you to deflect it by have to be different?
     
  11. Nov 11, 2005 #10

    Doc Al

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    Staff: Mentor

    The key is to realize that at the distance of closest approach the proton and nucleus must have the same speed (not necessarily stationary, unless the nucleus is so heavy that you can ignore its recoil). If they still had different speeds, then they clearly haven't reached the distance of closest approach. Use conservation of momentum to relate the speed of proton/nucleus at closest approach to the initial speed of the proton. Then use conservation of energy to solve for that initial speed.
     
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