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Homework Help: Alpha Particle & Gold Nucleus Collision

  1. Nov 13, 2015 #1
    1. The problem statement, all variables and given/known data
    An alpha particle (a helium nucleus, containing 2 protons and 2 neutrons) starts out with kinetic energy of 10.5 MeV (10.5x106 eV), and heads in the +x direction straight toward a gold nucleus (containing 79 protons and 118 neutrons). The particles are initially far apart, and the gold nucleus is initially at rest. Answer the following questions about the collision.

    What is the initial momentum of the alpha particle? (You may assume its speed is small compared to the speed of light).

    What is the initial momentum of the gold nucleus?

    What is the final momentum of the alpha particle, long after it interacts with the gold nucleus?

    What is the final momentum of the gold nucleus, long after it interacts with the alpha particle?

    What is the final kinetic energy of the alpha particle?

    What is the final kinetic energy of the gold nucleus?

    Assuming that the movement of the gold nucleus is negligible, calculate how close the alpha particle will get to the gold nucleus in this head-on collision.

    2. Relevant equations
    p_f = p_i
    1 eV = 1.6e-19 J
    k = 1/2 mv^2
    m_proton/neutron = 1.7e-27 kg
    q_proton = 1.6e-19
    r_proton/neutron = 1e-15 m

    3. The attempt at a solution
    k_i = 10.5e+5 eV
    10.5e+5 eV * 1.6e-19 = 1.68e-12
    k_i = 1.68e-12 J

    m_alpha = (2 * 1.7e-27) + (2 * 1.7e-27)
    m_alpha = 6.8e-27 kg

    k_i = 1/2 mv^2
    1.68e-12 = 1/2 (6.8e-27) (v)^2
    v = 2.22e+7 m/s

    p_alpha,i = mv
    p_alpha,i = <1.51e-19, 0, 0> kg * m/s

    m_au = (79 * 1.7e-27) + (118 * 1.7e-27)
    m_au = 3.349e-25 kg

    v_au,i = 0

    p_au,i = mv
    p_au,i = <0, 0, 0> kg * m/s

    r_alpha = cube root 4 * 1e-15
    r_alpha = 1.59e-15 m
    r_au = cube root (79 + 118) * 1e-15
    r_au = 5.82e-15

    q_alpha = 2 * 1.6e-19
    q_alpha = 3.2e-19
    q_au = 79 * 1.6e-19
    q_au = 1.264e-17

    u_elec = 1/4 pi E_0 * q_1 q_2 / r
    u_elec = 9e+9 * (4.04e-26 / 7.41e-15)
    u_elec = 9e+9 * 5.46e-22
    u_elec = 4.92e-12 J

    p_f = p_i
    p_alpha,f + p_au,f = p_alpha,i + p_au,i
    p_alpha,f + p_au,f = p_alpha,i + 0
    (m_alpha * v_alpha,f) + (m_au * v_au,f) = 1.51e-19
    6.8e-27 * v_alpha,f + 3.349e-25 * v_au,f = 1.51e-19

    I don't know where to go from here to get p_alpha,f .
    I calculated u_elec because I could, but I don't know how its related to the problem.
    Last edited by a moderator: Apr 16, 2017
  2. jcsd
  3. Nov 14, 2015 #2
    I am not very sure about it, but i think this is what happened:
    when the alpha particle collides with the gold nucleus, there are 2 quantities that are conserved: linear momentum and energy. The kinetic energy lost by the alpha particle will be used to overcome the electric potential energy between the particles and the kinetic energy of the gold nucleus. Meaning KHe=U + KAu The alpha particle will stop completely when it has used up all of its kinetic energy.
    From principle of conservation of linear momentum, pHeinitial = pHefinal + pAufinal , since alpha particle stops completely, its final momentum will be 0, meaning pHeinitial = pAufinal.
    However, alpha particle will be repelled because like charges repel, so the final kinetic energy of the alpha particle will be the potential energy between the particles. We can find the potential energy from the equation pHeinitial = pHefinal + pAufinal after finding the kinetic energy of the gold particle.
    The closest distance between the particles can be found easily if we use the equation U=q1q2/4piEor
    This is what i think, correct me if i am wrong.
  4. Nov 14, 2015 #3


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    Staff: Mentor

    Calculating radii for the particles is pointless (and your method is drastically flawed anyways); the particles will never physically touch in any way as the electric force will keep them separate. There's not enough KE to overcome the electric potential "barrier", so treat the two objects as point particles.

    What you have is a perfectly elastic collision between particles where the "elastic" is provided by the electric field. You'll need to use the methods for elastic collisions to work out the post-collision details. So brush up on elastic collisions.

    I suggest that you find and use more accurate values for your constants. The KE of the alpha particle was given to you with three significant figures but you're using mass and charge constants that are only two figures, and they are rounded to two figures, which isn't a great way to start. The rounding of initial values and intermediate step values in a calculation introduces what are called rounding and truncation errors that creep into significant figures as calculations proceed.

    Always try to use values for constants that have the same or (better) more figures than your given data values. Don't round intermediate results in a multi-step calculation, and keep a few extra "guard" digits on all intermediate values. Only round results to the required number of figures at the end of calculation.
  5. Nov 14, 2015 #4
    As qneill said, this is an elastic collision. When the electric potential energy of an atom is larger than the kinetic energy of an oncoming alpha particle, the particle simply bounces off it.

    I tried out a practice problem with slightly different values and the alpha particle's final momentum was negative (not 0), and only slightly smaller a number than its initial momentum.

    I just wrote down those values because I didn't want to type out all the digits for each. I keep all the numbers with more than 5 significant figures stored on my calculator.

    I know that for elastic collisions, Kf = Ki and internal energy is 0.
    Kf = Kalpha, f + KAu, f
    Ki = 1.68*10^-12 J
    1/2 (6.8e-27 kg) valpha, f^2 + 1/2 (3.349e-25 kg) vAu, f^2 = 1.68*10^-12 J

    I still have two unsolved variables, so I feel like I'm stuck in the same rut though.
  6. Nov 14, 2015 #5


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    Staff: Mentor

    What other quantity is conserved in a collision?

    If you take the time to work through a generic elastic collision using symbols only then you can apply the resulting expressions for final velocities to any such problem and you won't have to go through the agony of deriving results every time. Take a look at the Hyperphysics site's treatment of elastic collisions., and perhaps at their online calculator for head-on elastic collisions.
  7. Nov 14, 2015 #6
    v'1 = valpha, f
    valpha, f = (malpha - mAu) / (malpha + mAu) * valpha, i
    valpha, f = -2.134403055e+7 m/s

    palpha, f = malpha * valpha, f
    palpha, f = -1.451394078e-19 = -1.45e-19 kg * m/s

    v'2 = vAu, f
    vAu, f = 2malpha / (malpha + mAu) * valpha, i
    vAu, f = 8.84726651e+5 m/s

    pAu, f = mAu * vAu, f
    pAu, f = 2.962949568e-19 = 2.96e-19 kg * m/s

    kalpha, f = 1/2 malpha * valpha, f^2
    kalpha, f = 1.54892977e-12 = 1.55e-12 J

    kAu, f = 1/2 mAu * vAu, f^2
    kAu, f = 1.310700218e-13 = 1.31e-13 J

    For the final question, rclosest , I know I'm suppose use to the Energy Principle somehow but I don't know how.
  8. Nov 14, 2015 #7


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    Staff: Mentor

    What two energies are being "traded" during the collision?
  9. Nov 14, 2015 #8
    Kinetic and electrical potential energy.
  10. Nov 14, 2015 #9


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    Staff: Mentor

    Right. Assuming that the gold nucleus is essentially stationary (as proposed in the problem statement), what's the KE of the alpha particle when it's at its closest approach? Where's all the energy?
  11. Nov 14, 2015 #10
    The KE of the particle is 0 at closest approach, right?

    The energy would all be potential then.
  12. Nov 14, 2015 #11


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    Staff: Mentor

    Yup. Continue...
  13. Nov 14, 2015 #12
    Eclosest = kalpha, i + Uelec ?
  14. Nov 14, 2015 #13


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    Staff: Mentor

  15. Nov 14, 2015 #14
    Eclosest = 1.68e-12 + 4.915333452e-12
    Eclosest = 6.595333452e-12 = 6.60e-12

    I'm confused, the answer is wrong?
    Does Eclosest = rclosest ? Aren't they different units?
    Also, you said my method for calculating radii was bad, so maybe I miscalculated Uelec ?
  16. Nov 14, 2015 #15


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    Staff: Mentor

    I don't understand what you're calculating. I can spot your initial KE value (1.683-12) but the 4.9...e-12 value is a puzzle. Where does it come from? It can be hard to figure out where numbers come from. Symbols are much better for "debugging".
    Yes they are different things altogether. E is presumably an energy, r a distance.
    Probably. You can't calculate it without knowing the actual separation of the charges, and the separation at closest approach depends upon the KE available to begin with.

    Above you agreed that all the initial KE ends up as electrical PE when the alpha is at its closest approach. What's the formula for electrical PE?
  17. Nov 14, 2015 #16
    Uelec = 1/4 pi E0 * q1 q2 / r
    Uelecl = 9e+9 * q1 q2 / r

    q1 = 2 * 1.6e-19 = 3.2e-19
    q2 = 79 * 1.6e-19 = 1.264e-17

    r = ralpha + rAu
    r = (cube root 4 * 1e-15) + (cube root (79 + 118) * 1e-15)
    r = 7.40e-15

    Uelec = 9e+9 * 3.2e-19 * 1.264e-17 / 7.40e-15
    Uelec = 4.915333452e-12

    I don't get how I'm suppose to get rclosest from the formula Eclosest = kalpha, i + Uelec if rclosest isn't a part of it?

    Is there a way to convert J to m?

    This is the formula:
    Uelec = 1/4 pi E0 * q1 q2 / r
    Uelecl = 9e+9 * q1 q2 / r

    I don't get how the separation at closest approach depends upon available KE.
  18. Nov 14, 2015 #17


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    Staff: Mentor

    No, forget about the supposed radii of the particles. Even if your expressions for them were valid, the particles never get close enough to "touch". The two particles approach each other until the available KE is exhausted, it being converted to electric PE.

    Your formula for electric PE is good. Equate the initial available KE to the PE formula. Solve for r.
  19. Nov 14, 2015 #18

    1.68e-12 = 9e+9 * q1 q2 / r
    1.68e-12 = 3.64032e-26 / r
    r = 2.166857143e-14 = 2.17e-14

    Thanks a lot for the help: I appreciate it.
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