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Alpha Particle Approaching a Gold Nucleus

  1. Jan 17, 2017 #1
    1. The problem statement, all variables and given/known data
    An alpha-particle with velocity 3.5 x 106 strikes a block of gold, atomic number 79 and mass number 197. Find the distance of the closest possible approach between the alpha particle and a gold nucelus, assuming Coulomb's law holds over such distances. Assume mp = mn = 1.67 x 10-27
    2. Relevant equations
    p = mv
    KE = 1/2mv2
    EPE = kq1q2/r2


    3. The attempt at a solution

    Considering conversation of momentum and conservation of energy I worked out p = mv = (4 x 1.67 x 10-27)(3.5 x 106) = 2.3 x 10-20.

    The particle would repel the nucleus, at the closest approach they must have the same velocity. So p = mauv' + malpha v'
    v' = p/mau + mualpha
    v' = 2.3 x 10-20 / 201 x 1.67 x 10-27
    v' = 6.4 x 104

    So the change in KE in KEv minus KEv'.

    At this point I believe I've gone wrong as I've calculated the change in KE to be negligible yet the online question pp I'm using tells me to consider conversation of momentum (here). As I understand the remainder of the question is simply equate change KE to change in EPE, rearrange to find r and plug in values.
     
  2. jcsd
  3. Jan 17, 2017 #2

    blue_leaf77

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    Homework Helper

    The gold nucleus is to be assumed as a field/potential generator and hold firm in its initial place. In this case, only conservation of energy is needed. Start from infinite distance where the alpha particle is moving with the given velocity in zero potential. Then the total energy in the closest distance will give you the answer.
     
  4. Jan 17, 2017 #3

    gneill

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    Staff: Mentor

    I don't believe that's true in this instance. If you check the question and hints at the website you'll find that the use of conservation of momentum is recommended in addition to conservation of energy.

    @FelaKuti , check your calculation for v'. Your method's okay but the numerical result looks off; maybe a finger problem on the calculator.
     
  5. Jan 17, 2017 #4

    blue_leaf77

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    I see. I only read the first hint where no mention of momentum conservation is made. So yes the momentum conservation is needed and the gold nucleus is also moving.
     
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