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Calculating the radius of a nucleus of a Pd-118 atom

  1. Jun 9, 2006 #1
    Here is the question I am attempting to solve :

    A proton or neutron has a radius r of roughly 110-15 m, and a nucleus is a tightly packed collection of nucleons. Therefore the volume of the nucleus, (4/3)R3, is approximately equal to the volume of one nucleon, (4/3)r3, times the number N of nucleons in the nucleus: (4/3)R3 = N(4/3)r3. So the radius R of a nucleus is about N1/3 times the radius r of one nucleon. More precisely, experiments show that the radius of a nucleus containing N nucleons is (1.310-15 m)N1/3. What is the radius of a palladium nucleus

    So, in order to arrive at an answer I listed variables ,constants and equations that were known:
    R= (N^1/3)*r
    N=(1.3e-15)* (N^1/3)

    and so, in order decrease the number of variables given , I plugged in variables into variables:

    R=(N^1/3)*r ---> R/r =(N^1/3)
    N=(1.3e-15) * (N^1/3) ---> N=(1.3e-15)*R/r

    so I substituted the new expression for N into:

    4/3*pi*r^3 =4/3*pi*R^3 * N ----> 4/3*pi*r^3= 4/3*pi*R^3*((1.3e-15)*R/r):

    pi and 4/3 cancel , show I should be left with:

    r^4/(1.3e-15) = R
    since r = 1e-15 , then R should equal :


    And my answer is no where near the actually answer : Can you tell me what I am doing wrong
  2. jcsd
  3. Jun 10, 2006 #2

    Andrew Mason

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    So wouldn't the radius of the Pd-118 would be [itex]1.31e-15 * N^{1/3}[/itex] where N is the number of nucleons in the Pd-118 nucleus?

    Last edited: Jun 10, 2006

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