Finding Density of a Neutron

FeDeX_LaTeX

Gold Member
436
13
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
 

mathman

Science Advisor
7,688
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The method is correct, but where did you get r?
 
The method is correct, but where did you get r?
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).
 
193
0
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
 
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
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.
 

Vanadium 50

Staff Emeritus
Science Advisor
Education Advisor
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1.67492729*(10^27)
1.67492729 x 10^27 is most certainly not the mass of the neutron. Check again,.

And since we always regard neutron as a point
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.
 

Matterwave

Science Advisor
Gold Member
3,951
324
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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