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Proton radius

  1. Jun 21, 2008 #1
    I often see the charge radius of a proton at .8 * 10^-15m. An acquaintance on a different forum said he would use, if asked to determine the radius of a nucleus, to be:

    R = R0 * A^(1/3) where R0 = 1.2 * 10 ^ -15 to 1.5 * 10 ^ -15 and A is number of nucleons.

    He cited the sources:

    Principles of Modern Physics by Roberth Leighton or Concepts of Modern Physics by Arthur Beuser, or Nuclear Physics by Enrico Fermi.

    I do see 1.3 * 10^-15 used in the formula above often. Is this the Compton wavelength? If not, then where is this value coming from? If it is the Compton wavelength, I have a follow up question (or 4)...

    What is the difference between the Compton wavelength and the charge radius? If someone asks me what the diameter of a proton is, what value would I quote? Does it depend on the context of the question and what situation you are applying it to? What is the significance of the value of each?

    I tried educating myself through reading several wiki articles and arxiv papers. I, somewhat, understand how you arrive at the values of each, but fail to understand the significance of each and how and to what they are applied.

    I appreciate any insight you can offer and/or online sources you can lead me to.
  2. jcsd
  3. Jun 22, 2008 #2


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    The figure R0 = 1.2 * 10 ^ -15 to 1.5 * 10 ^ -15

    is just an emperical constant; you measure the radius of nucleis as a function of A, and then you see that the radius is a function of A^(1/3), and R0 is the constant that fits the data.


    slide #5
  4. Jun 22, 2008 #3
    Ok. I think I finally have it figured out. If someone were to ask, "what is the radius of a proton", I would cite 1.2 fm.

    Would I be correct in stating the charge radius is the exchange force range between two protons? In other words, when they reach a distance of around .875 fm, the strong force overwhelms the repulsive force. If that's the case, then doubling that figure to get the diameter of the proton would make no sense.

    Thanks for the PDF... nothing like pretty pictures to help things make sense :smile:
  5. Jun 22, 2008 #4


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  6. Jun 23, 2008 #5
    If I haven't already made it obvious... I have no formal education in nuclear physics. I just like to try to figure stuff out and came across 2 completely different radii for the proton. I'm just trying to understand the distinction between the two.

    The reason this question came up was from discussing how long a "beam" of photons would be at the LHC. I know they are not really beams, but it was a hypothetical.

    I understand discussing the spatial dimensions of a proton (or any other subatomic particle) is a bit more complicated that what a simple question presents.

    If I were to make certain, unrealistic assumptions like ignoring the repulsive force and line up 10^15 protons side by side, would that string of protons be 2.4 meters in length?

    I see now that the exchange force isn't related to the charge radius, but what is the significance of it? What's the purpose of knowing the rms charge radius and what can you apply it to? It has nothing to do with the spatial dimensions of the proton, right?

    One of the reasons I'm curious about the spatial dimension is that Wikipedia states the diameter of the proton to be 1.65 fm. In the discussion section, the reasoning was simply doubling the rms charge radius to get that figure. Given that determining the radius of a nucleus uses the standard radius of 1.2 fm, wouldn't it be more appropriate to cite the diameter to be 2.4 fm?

    Given my lack of formal training, I'll understand if you can't describe in simple enough enough terms for me to understand...

    I appreciate any attempt, though.
  7. Jun 24, 2008 #6


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    Where is your source? You are first given me:

    "I often see the charge radius of a proton at .8 * 10^-15m."

    Then you writes:

    R = R0 * A^(1/3) where R0 = 1.2 * 10 ^ -15 to 1.5 * 10 ^ -15 and A is number of nucleons.

    Are you meaning that if you stick in A = 1 in that eq, you get 1.2-1.5 fm?

    Remember that R = R0 * A^(1/3) is an empirical formula, trying to fit nuclei to data, and is not a RULE. If you want to measure the radius of a nuclei, you can do it, the forumula above is just an approximative relation, there are many nuclei that is deviating from that relation.

    This is for all kind of 'fit to data', some values are below the curve, and some are under the curve. For example: http://support.sas.com/documentation/cdl/en/grstateditug/61006/HTML/default/images/ex-carlabel.gif

    The same thing you do with nuclear radii, you measure the individual radius of each nuclei and then trying to fit a curve, as a function of A.

    You got it now?

    When you have grasped this, that the relation R = R0*A^(1/3), is just an emperical estimation - I'll explain more about radii in quantum mechanical obejcts, and also what 'strong radius' is.
  8. Jun 25, 2008 #7
    I might be approaching my search for an answer the wrong way...


    "a diameter of about 1.65×10−15".

    In the discussion section:

    "It's a bit hard to say what "proton diameter" means. Most experiments seem focused on determining the "proton root-mean-squared charge radius". This value is not very well known, but is about 0.87 fm. So by that measure, the proton diameter is about a fermi and a half. -- Xerxes"

    from http://hyperphysics.phy-astr.gsu.edu/hbase/hframe.html

    "Various types of scattering experiments suggest that nuclei are roughly spherical and appear to have essentially the same density. The data are summarized in the expression called the Fermi model:


    They are saying the radius of the proton is 1.2 fm which doesn't agree with the wiki article.

    Who is correct? Or are they both correct depending on the context of what each individual is attempting to describing?
  9. Jun 25, 2008 #8
    That would be the case :smile:

    Look at that picture :
    Nuclear matter is not like a hard sphere with a well-defined radius. You need a convention, such as the root-mean-square (RMS) of the distribution, which happens to be convenient because of its relation with the form factors (FF). The FF basically are what you observe directly, they parameterize your cross-section. Under a few simple and reasonable hypothesis, the shape of the FFs will be related to the RMS of the corresponding charge distribution.

    0.87 fm would come from the nucleon FF measurements. It definitely tells you about them directly.

    2*0.6 fm would come from an approximate fit to the radius as a function of A. You certainly are courageous to take it as such, plug in A=1, and claim a disagreement with the previous number. First hypothesis : the nucleon is the same in a nucleus and in a free state. This is daring. Second (most important here) : the fit does good near the specific low A region, even down to A=1. I would like to see that :smile:
  10. Jun 26, 2008 #9


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    I guess my answer was not clear enough, and I hope Humaninos will shed more light upon your question. If not, then I try again:

    1) The 'radius' of a proton, nucleus, atom - is often defined as the R.M.S of the charge distribution, since it is related to the Form Factor, which is then related to scattering amplitud, which is the quantity we measure in lab.

    2) If one does that, the radius of the proton is 0.87fm

    3) One has found that if one plots the 'radius' of different stable nuclei as a function of A^(1/3), one can fit a linear curve to that data set, and the constant for the slope of that line is approx 1.2fm.

    4) #3 does not saying that ALL nuclei HAS radius 1.2*A^(1/3)fm, when one does a fit, some data goes above the curve, and some of the data goes below is, just as the image I showed you where a fit to data was done. #3 means that nuclei radii behaves, on AVERAGE, as 1.2*A^(1/3)fm.

    5) The fit 1.2*A^(1/3)fm is not good for low mass nuclei, as humanino pointed out, it fits "best" for intermediate nuclei.

    here is how the 1.2*A^(1/3)fm fit is done - http://www.aanda.org/articles/aa/full/2005/25/aa2003-04/img180.gif

    You can also find this in the textbook by Krane.
  11. Jun 26, 2008 #10


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    Forgot to be clear about this:

    "They are saying the radius of the proton is 1.2 fm which doesn't agree with the wiki article."

    No, they are not saying that, it is your intepretation of what that forumula is doing that gives you another proton radius.
  12. Jun 26, 2008 #11


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