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Size of an atomic nucleus

  1. Mar 20, 2013 #1
    I've been noticing that for expressions finding the nuclear radius there is a constant term [itex]r_0[/itex], that I can't seem to find an explanation for. The full term is
    [tex]R=r_{0}A^{1/3}[/tex]
    That the atomic radius is roughly the 3rd root of the number of nucleons. But where is the [itex]r_0[/itex] coming from? I'm using Cottingham and Wong as sources but both of them fail to mention it's derivation. Plus, they both give [itex]{slightly}[/itex] different values, without looking I think Wong assigns it a value of 1.7 fm and Cottingham 1.1 fm. Can anyone explain this to me?


    Thanks.
     
  2. jcsd
  3. Mar 20, 2013 #2

    TSny

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    Nuclei are sort of fuzzy and don't have definite radii (they're not even necessarily spherical). The formula gives an approximate value of the radius and the value of ro in the formula is determined from various experiments. See http://en.wikipedia.org/wiki/Nuclear_size.
     
  4. Mar 23, 2013 #3
    This is I think an expression found by experimental or at least phenomenological means. As there ##A## is the number of nucleons, you would have that ##r_0## is the size of a single nucleon (and has to be determined experimentally of course). It is needed in this expression formally to give the right dimension to ##R##.

    Also notice that (something I think is interesting) if you compute the volume of the nucleus (interpreting really ##r_0## as the radius of the single nucleon), you will have that the total volume is ##A## times the volume of a nucleon, which is clearly consistent with what one would imagine.
     
  5. Mar 23, 2013 #4
    Looking in the Wikipedia article to the values given for the experimental sizes of neutron and proton, indeed ##r_0## is a constant determined empirically and seems to me a sort of mean value of the two.
     
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