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Finding Density of a Neutron

  1. Feb 24, 2010 #1

    FeDeX_LaTeX

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    Hi, I did some calculations and I worked out the density of 1 neutron to be about 3.19887549*10^57 kg/m^3. However, I want to know if the method I used is correct.

    Density = mass/volume, correct?

    If this is true, then

    Density = [mass of neutron]/[volume of neutron]

    I found the volume of a neutron by modelling it as a sphere, with volume (4/3)pi*r^3. The calculation I ended up doing was;

    (1.67492729*(10^27))/(4/3 * pi*((10^-10)/2)^3) = 3.19887549*10^57 kg/m^3

    Where the 1.67492729 is the mass of a neutron (source: wikipedia).

    Is my working correct?

    Cheers
     
  2. jcsd
  3. Feb 24, 2010 #2

    mathman

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    The method is correct, but where did you get r?
     
  4. Feb 24, 2010 #3
    It really doesn't matter, since Warner Heisenberg already showed us that when the quantity of one non-commuting variable is known in the subatomic particle, the other variable of that particle becomes uncertain (Heisenberg's uncertainty principle).
     
  5. Feb 24, 2010 #4
    I personally think the neutron's radius is unknown yet
    And since we always regard neutron as a point,its density is useless
    We need only its mass
     
  6. Feb 24, 2010 #5
    Agreed. I'd also like to add that we need know also the neutron's energy (in electron-volts, of course), as well as its spin.
     
  7. Feb 24, 2010 #6

    Vanadium 50

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    1.67492729 x 10^27 is most certainly not the mass of the neutron. Check again,.

    No we don't. The neutron has a measured radius. Of course, it's boundary is not sharp, but you could say that about many things that have a published radius: a gold atom, the planet Jupiter, the asteroid belt.
     
  8. Feb 25, 2010 #7

    Matterwave

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    10^-10 is more like the radius of an atom (where the electron likes to hang out). It's not the radius of the Neutron...but I suppose you could use that as an upper bound. (Nuclei are more on the order of 10^-15m in radius)
     
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