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What incident energy is required to ionize a helium atom?

  1. Sep 2, 2011 #1
    1. The problem statement, all variables and given/known data
    The energy required to ionize a helium atom is 24.6 eV. If helium gas is bombarded with charged particles, at what incident energy will ionization of helium just be observed if the projectiles are (a) alpha particles (b) protons (c) electrons?

    2. Relevant equations
    Most of the exercises in this chapter have to do with conservation of momentum and energy, but otherwise I really don't know what equations would be relevant.

    3. The attempt at a solution
    I need help knowing even where to begin with this. It seems like the entire kinetic energy (which I assume = "incident energy") of the incoming particle should go into the ionization, with the result that the total kinetic energy of the system after the collision = 0 and the helium atom is ionized. This would imply that the incident energy required is 24.6 eV and that the same amount is required regardless of what sort of particle is the projectile. But this is wrong. Hint please?
  2. jcsd
  3. Sep 3, 2011 #2


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    Consider the process as inelastic collision. You can assume that after leaving the helium atom, the speed of the electron is zero.

  4. Sep 3, 2011 #3
    Yeah, I figured this one out. (Sorry, I suppose I should've posted something.) The trick is that since the total energy before the collision, Ei, which is the kinetic energy of the incoming particle, should be just enough to cause the ionization of the helium atom, we can assume it is as low as it can be consistent with conservation of momentum. I.e., the collision is perfectly inelastic and Ei is low enough to preserve conservation of momentum and have just enough left over to ionize the helium, i.e., 24.6 eV.

    So we have for conservation of energy:

    Ei = Ef
    .5mVi2 = .5(m + H)Vf2 + 24.6 eV

    where m is the mass of the incoming particle and H is the mass of the helium atom.

    For conservation of momentum:

    Pi = Pf
    mVi = (m + H)Vf
    Vf = Vi(m/m + H)


    .5mVi2 = .5(m + H)[m/(m + H)]2Vi2 + 24.6eV
    .5mVi2[1 - m/(m + H)] = 24.6eV

    Solve for .5mVi2 in the three cases where:

    H = m
    H = 4m
    H + m ≈ H

    The problem I could really use help with is the one on very high speed proton-proton collisions: https://www.physicsforums.com/showthread.php?t=526305

    I tried treating it like the above only using relativistic formulas instead of nonrelativistic. But that didn't work. I doubt that any of the problems in this introductory book I'm using could really be very hard, but it may be that in this proton-proton case there's something a little out of the ordinary. The book, published in 1964, cites a reference for this problem: G. K. O'Neil, Science 141: 679 (1963). I don't have easy access to this journal or I might just look up the article.
  5. Sep 4, 2011 #4


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    Apply conservation of momentum and determine the energy after collision. You can use the relation between the momentum and energy: E2=p2c2+m02c4 (m0 is the rest mass)

  6. Sep 5, 2011 #5
    So is the idea the following?

    Compute the initial kinetic energy from T = mc2(γ - 1). So

    3 GeV = 938(γ - 1) MeV (.001 GeV/MeV)
    2.062 = .938γ
    γ = 2.1983


    1/(1 - V2/c2).5 = 2.1983
    1 - V2/c2 = .2069
    V = .89c

    Now use conservation of momentum.

    Pi = Pf
    mγV = Pf
    938(2.1983).89c = Pf
    1836.3/c = Pf

    Now if the collision is inelastic and no initial energy is absorbed into producing other particles, the initial total mass = the final total mass. So

    Ef2 = Pf2c2 + [(2)(mc2)]2
    Ef2 = (1836.3/c)2c2 + [(2)(938)]2
    Ef2 = 3,372,000 MeV + 3,519,376 MeV
    Ef = 2.625 GeV

    Ei = Ti + mc2
    Ei = 3 GeV + .938 GeV

    So the fraction of the energy of the moving proton not available for inelastic interactions in this proton-proton collision is

    2.625/3.938 = .6666

    The answer the book gives is .6, which I don't think is close enough to be due to rounding error.

    The main problem I see here is that I'm comparing Ef based on two protons with Ei based on one. But I don't know what help there is for it. Pi is based on one proton, Pf on two. And I can't just use Ei = 3 GeV + (2)(.938) GeV, since what we need is the energy of the moving proton alone.
  7. Sep 5, 2011 #6


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    Take care what you denote by m. In the formula I wrote in my post, m means the moving mass, m=m0γ. Recalculate the problem with that.

  8. Sep 5, 2011 #7
    Sorry, but I'm not getting you. Your formula was: "E2=p2c2+m02c4 (m0 is the rest mass)". And my formulas all use "m" for rest mass, multiplying by γ where needed. What am I doing wrong?

    I got that Science article by the way. I'll read through it tomorrow and see if I can see what's up.
  9. Sep 6, 2011 #8


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    I am sorry... I got confused with the notations. You are right, it is the rest mass in the formula E2=p2c2+m02c4, which you denoted by m.

    Try to derive the relation between the final KE and the generator energy instead of calculating with numbers from the beginning. You get the momentum of the accelerated proton from the formula above, using E=m0c2+E(generator). No need to calculate the speed.
    And let me know, what you learnt from the Science article.
    I got the formula from Giancoli: Physics for Scientist and Engineers.

  10. Sep 6, 2011 #9


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    I have found a mistake at the beginning of your derivation:

    0.938 γ= 3+0.938 .

  11. Sep 6, 2011 #10
    Ahh! Much quicker technique. And it gets the same answer as I was getting with my more laborious technique (after correcting the boo-boo you found, and thanks for that).

    The Science article doesn't provide much illumination beyond explaining why one would be interested in this sort of problem. It seems that at the time the article was written, particle accelerators typically accelerated a particle toward a stationary target particle, thereby wasting a considerable amount of kinetic energy in conserving momentum. At nonrelativistic speeds, a perfectly inelastic collision wastes half the initial kinetic energy this way. At relativistic speeds, the waste is considerably greater, and that is the point of the exercise. But the article doesn't explain how to calculate the amount of relativistic kinetic energy that is taken up conserving momentum; it only gives a table showing the amounts for various accelerators that were in operation at the time. Calculating these amounts is left as an exercise for the student! Literally.

    So here's how I think it must go.

    Eproton = Ti + mc2
    Eproton = 3 + .938
    Eproton = 3.938 GeV

    Eproton2 = pi2c2 + (mc2)2
    3.9382 = pi2c2 + .9382
    pi2c2 = 14.628
    pi = 3.825/c

    The total initial energy is:

    Ei = Ti + (2)mc2
    Ei = 3 + (2)(.938)
    Ei = 4.876 GeV

    Now Ei = Ef and pi = pf. So the "Remainder Energy" (which could be rest energy of whatever particles exist after collision, radiation energy, etc.) after the collision that is not taken up conserving momentum is:

    Ef2 = (pf/c)2c2 + RE2
    4.8762 = 3.8252 + RE2
    RE = 3.024 GeV

    Therefore, the kinetic energy after the collision is:

    Tf = Ef - RE = 4.876 - 3.024 = 1.85 GeV

    As a proportion of the initial kinetic energy, this is:

    1.85/3 = .617

    The chart in the article (reproduced for this exercise in my special relativity textbook) gives a value of .60. It bothers me to think there could be this much rounding error, but I don't know what else to think.

    The values provided versus what I compute are:

    Accelerator energy / Loss due to COM (published) / Loss due to COM (computed by me)
    3 / .60 / .62
    7 / .72 / .69
    25 / .80 / .79
    200 / .92 / .91
    1000 / .96 / .96

    There is a follow-up exercise. The article suggests that the lost kinetic energy can be recovered by building an accelerator that shoots opposing beams of protons at each other instead of shooting one beam of protons at a stationary mass. Thus, for example, if each beam is accelerated at 3 GeV, the total initial kinetic energy will be 6 GeV instead of 3 GeV, and since the total pi of the opposed protons is 0, the total pf must also be 0 and there should be no energy lost conserving momentum; it can all go into inelastic interactions.

    The task we are given is to find the energy an old model, stationary target accelerator would require to produce the same amount of energy available for inelastic interactions as a new model opposing beam accelerator where each beam has an energy of (a) 3 GeV (b) 25 GeV (c) 31 GeV.

    So suppose that the energy in each beam of a colliding beam accelerator is 3 GeV. Then the total initial energy is:

    Ei = 2(Ti + mc2)
    Ei = 2(3.938) = 7.876 GeV

    And since pf = pi = 0, kinetic energy after collision is 0 and all of Ef = Ei is available for inelastic interactions. This is what we called RE in the above, stationary target case. Using the above formulas, we need to solve for the initial Ti of the single, moving proton that will produce the same Ef as we just found in the colliding beam case.

    We have:

    Ef2 = (pf/c)2c2 + RE2
    Eproton2 = pi2c2 + (mc2)2
    Eproton = Ti + mc2
    Ei = Ti + (2)mc2

    Bearing in mind that Ef = Ei and pf = pi, we can rewrite the first equation as:

    Ei2 = (pi/c)2c2 + RE2

    We know RE and mc2, so the four equations become:

    Ei2 = (pi/c)2c2 + 7.8762
    Eproton2 = pi2c2 + .9382
    Eproton = Ti + .938
    Ei = Ti + (2).938

    Thus, we have four equations and four unknowns. We solve for Ti:

    Ti(2).938 + [(2).938]2 = 7.8762
    Ti = 31.19 GeV

    The value published in the Science article is 31 GeV. The table published in the article versus the values I computed are:

    Colliding beam energy / Single beam energy (published) / Single beam energy (computed by me)
    3 / 31 / 31.19
    25 / 1360 / 1432.62
    31 / 2040 / 2173.04

    Again, the discrepancy between the published values and what I'm getting bothers me, but I don't know what to do about it. Anyway, the energy savings achieved by the colliding beam technique are very impressive!
  12. Sep 7, 2011 #11


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    Your derivation looks good. I do not know what can be the reason of the differences between your results and those of the book. I asked an other member in the Forums, and he also thinks it correct.
    There is some flaw, but only in the notation: do not write (p/c)2c2 when you mean p2c2.
    To check rounding errors, derive the "not usable amount of energy" symbolically, in terms of the generator energy and rest proton mass, and substitute data at the end.
    If you find a reliable solution, please let me know.

  13. Sep 7, 2011 #12
    Yes, I see now that I shouldn't have expressed a general formula that way.

    Another mistake is that "we need to solve for the initial Ti of the single, moving proton that will produce the same Ef as we just found in the colliding beam case" should read "...that will produce an RE of the same magnitude as the Ef we just found in the colliding beam case."

    Thanks again for your help with this!
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