# Density of a proton

I was doing some exercises the other day, when I came across this question in my book:

A proton weighs about 1.66 x 10-24 g and has a diameter of about 10-15 m. What is its density in g/cm3?

As you can see, a really simple and standard question, but... does it even make sense to say that a proton has a mass? And calculate its density?! If so, do I consider it a sphere?

It seems to me, subatomic particles are so small it doesn't really make sense to talk about mass, volume and density as if we were talking about... tennis balls!

• so, where do you think the mass comes from, if not from elementary particles.. ? from the dark matter? – mykhal Aug 30 '12 at 12:02
• You need to consider it as sphere. – BigSack Aug 30 '12 at 13:04
• I bet you are not surprised a hydrogen atom to have a mass. A hydrogen atom consists of a proton and an electron. Most of its mass is due to proton. And a mass of a hydrogen molecule $\text H_2$ is almost a mass of two protons in it. – Yrogirg Aug 30 '12 at 16:37
• The "diameter" comes from the root mean square charge radius, which is basically a measure of the "average" extent of the charge from a center point. – Snowball Aug 30 '12 at 16:41

A proton is a bound state of three quarks. The quarks themselves are (as far as we know) pointlike, but because you have the three of them bound together the proton has a finite size. It doesn't have a sharp edge any more than an atom has a sharp edge, but an edge is conventionally defined at a radius of 0.8768 femtometres. Protons are spherical in the same way that atoms are spherical even though they're made up of discrete electrons.

The three quarks have a mass, but actually the proton is a lot heavier than the combined mass of the three quarks. That's because the binding energy of the quarks is very high, and that energy increases the mass in line with Einstein's famous equation $E = mc^2$.

So, yes, it does make sense to calculate a density for the proton just as it makes sense to calculate a density for an atom.

• John, binding energy is negative. A strongly bound body will mass less than its separated components. That is why atomic nuclei mass less than their summed constituent protons and neutrons. Proton mass is about 1% Higgs interaction, or less. The balance is kinetic energy mass equivalent of its contained quarks, gluons, and all the associated virtual stuff. – Uncle Al Feb 21 '14 at 2:30
• @UncleAl: true in general but quarks are something of a special case because of confinment. – John Rennie Feb 21 '14 at 7:58

D=m/v, (with rounded numbers) Sphere for volume= (4/3)(pi)(radius^3)

r (cm)=(1.0x10^-13)cm mass (g)= (1.7x10^-24)

=(1.7x10^-24)/((4/3)pi(1.0x10^-13)^3)

=(1.7x10^-24)/(4.1887x10^39)

= 4.0585x10^14

Consider a neutron star. It's a gravitationally bound atomic nucleus. Contrary to exotic particle hopes, a very stiff equation of state (J0348+0432, 2.04 solar masses, AP4 model) suggests nothing other than neutrons, protons plus electrons. That nucleus has density, about 2.2×10^14 g/cm^3. Perhaps WFF1 is a better equation of state.

i would say that if $\rho (density) = \frac{m}{V}$ and the volume must be proportional to $V= m |\Psi (x,t)|^{2}$ then $\rho = \frac{1}{|\Psi(x,t)|^{2}}$