Distance between two Protons? Does any body know?

[fusion]

Does anyone know the exact distance between a neutron and a proton or the distance between particles in the nucleus of an atom. Icouldn't find it from a search through google.

I've tried using planks lenght as the space between protons in a nuclues in coloumbs equation for the strong nuclear force, but then the energy to break a atoms nucleus would be equivalent to a neutron shot at it over the speed of light!

;)

If anyone knows please tell me. ;) ;)

[fusion]

Does any body know or doesn't anyone want to tell me! ;)

[Tormod]

I thought the distance wasn't fixed but that the nuclear particles move about.

[fusion]

What would cause them to move around? Wouldn't they be at an equilibrium between the Strong nuclear force and electrostatic force. ;)

[Tormod]

Sorry, they are not moving around like marbles...I was thinking of them as pockets of quarks which wobble. But this is probably not a correct picture.

 

I have not been able to find any info on proton distance. I found some interesting reads on the interaction of protons and neutrons, though - like at the bottom of this page:

 

http://www.emsb.qc.ca/laurenhill/science/quark.html

[fusion]

Interesting article. It said the strong force oly operates at 10^-15 and smaller distance I guess that is around the distance between two Protons. ;)

[Buffy]

We're all deceived by those billiard ball pictures. Protons have no edges, quarks have no edges. In fact if quarks are strings, they're wiggling around all over the place. Quantum world is strange... I'm not really sure there's an answer to this question other than possibly the "centers" of the protons and neutrons must have some average distance that's very small, and must be (based on fusion's latest post) less than 10^-15 in order for the strong force to hold them together...

 

Cheers,

Buffy

[TeleMad]

Most people also often mentally misrepresent atoms as microscopic solid spheres, forgetting that they are mostly empty space. Here's some math on the density of the atomic nucleus of carbon vs. that of the carbon atom as a whole.

 

GIVENS

Mass of C12 atom = 12u

Mass of electron = 5.486 x 10^-4u

Mass of C12 nucleus = mass of C12 atom – mass of 6 electrons = 11.9967084u

Volume of sphere: v = (4/3) pi * r^3

(4/3) pi = 4.1888

v = 4.1888 * r^3

Density = mass/volume: d = m/v

Radius of nucleus = roA^(1/3)

ro = 1.2 x 10^-15m; A = mass number

1 pm = 1 x 10^-12m

1u = 1.66 x 10^-27 kg = 1.66 x 10^-24g

 

CARBON ATOM

Radius = 77 pm (chemistry reference) = 7.7 x 10^-11m

Volume = 4.1888 * (7.7 x 10^-11m)^3 = 4.1888 * 460 x 10^-33m^3 = 1.9 x 10^-30m^3

Density = 12u / 1.9 x 10^-30m^3 = 6.3 x 10^30u / m^3

 

C12 NUCLEUS

Radius = 1.2 x 10^-15m * 12^(1/3) = 1.2 x 10^-15m * 2.29 = 2.75 x 10^-15m

Volume = 4.1888 * (2.75 x 10^-15m)^3 = 4.1888 * 20.8 x 10^-45m^3 = 8.71 x 10^-44m^3

Density = 12u / 8.71 x 10^-44m^3 = 1.38 x 10^44u / m^3

 

A college physics text (Raymond A. Serway & Jerry S. Faugh, College Physics:Fifth Edition, Harcourt College Publishers, 1999, p960) states that all nuclei have about the same density, and lists the value as being 2.3 x 10^17kg / m^3. Let’s check to make sure my calculated value matches theirs.

1.38 x 10^44 (1.66 x 10^-24g) / m^3 = 2.3 x 10^20g / m^3 = 2.3 x 10^17kg / m^3

 

C12 NUCLEUS vs CARBON ATOM

1) Mass:

In a C12 atom, electrons make up 0.02743% of mass

C12 nucleus makes up 99.97257% of atom’s mass

 

2) Radius:

Atom / nucleus = 7.7 x 10^-11m / 2.75 x 10^-15m = 2.8 x 10^4

Carbon atom’s radius is about 28,000 times the radius of its nucleus

 

3) Volume:

Atom / nucleus = 1.9 x 10^-30m^3 / 8.71 x 10^-44m^3 = 2.2 x 10^13

Carbon atom’s volume is about 22 trillion times the volume of its nucleus

 

4) Density:

Nucleus / atom = (1.38 x 10^44u / m^3) / (6.3 x 10^30u / m^3) = 2.2 x 10^13

Carbon nucleus density is about 22 trillion times that of the carbon atom

[TeleMad]

And although this deals with the weak force, it does show how the effective range of forces is sometimes determined.

 

 

 

The only input needed is the mass of the particle. For the W bosons that mediate the weak force it is 80.4 Gev/c^2 (found it in a book).

 

*************

 

First we need to figure out how long a virtual W can exist. The Heisenberg equation of interest is:

 

[delta]t = h / (4[pi]E)

 

Looking up the value of h we find h = 6.63 x 10^-34 Js, so substituting we get:

 

[delta]t = (6.63 x 10^-34 Js) / (4[pi]E)

 

pi is approximately 3.14 so 4 times that is about 12.6 (keeping it to three sig figs). Substituting that we get:

 

[delta]t = (6.63 x 10^-34 Js) / (12.6E)

 

******************

 

Second, we need to calculate the energy of a W boson, in Joules. We use Einstein's energy-mass equivalence equation:

 

E = mc^2

 

The mass of a W is 80.4 Gev/c^2, so substituting we get:

 

E = (80.4Gev/c^2)c^2

 

The c^2's cancel out, leaving:

 

E = 80.4Gev

 

The conversion factor we need is 1Gev = 1.60 x 10^-10 J. Substituting we get:

 

E = 80.4(1.60 x 10^-10 J) = 1.29 x 10^-8 J

 

************************

 

Third, we substitute the value for E from step 2 into the final equation from step 1.

 

[delta]t = (6.63 x 10^-34 Js) / [12.6(1.29 x 10^-8 J)]

 

The joules on top and bottom cancel out, (and the two decimal values in the denominator can be multiplied together) leaving:

 

[delta]t = (6.63 x 10^-34 s) / (16.3 x 10^-8 )

 

Doing the division we end up with:

 

[delta]t = 4.07 x 10^-27s

 

Thus, a virtual W boson can exist for no longer than 4.07 x 10^-27 seconds.

 

***************************

 

Next, we figure out how far a virtual W boson can travel before it "disappears". We use c for the speed since that is the maximum speed a particle with rest mass can achieve.

 

Here's the simple distance formula:

 

d = vt

 

We know the velocity, v, is the speed of light in a vacuum, or c.

 

d = ct

 

d = (3.00 x 10^8 m/s)t

 

And we know t from out first set of calculations.

 

d = (3.00 x 10^8 m/s)(4.07 x 10^-27s)

 

The second cancel out, leaving:

 

d = (3.00 x 10^8 m)(4.07 x 10^-27)

 

Doing the multiplication gives us:

 

d = 1.22 x 10^-18 m

 

Thus, the maximum distance a virtual W boson can travel before "disappearing" is 1.22 x 10^-18 m (or 1.22 x 10^-16 cm).

 

Therefore, the maximum range of the weak force is 1.22 x 10^-18 m. Just to get an idea, compare that to the diameter of a proton and you will find that it is much less.

 

As you can see, once you plug a particle's mass in, the rest follows: the lifespan and maximum distance are functions of mass.

 

 

PS: I later found a confirmation for my calculation:

 

"... the weak interaction is an extremely short range force having an

interaction distance of approximately 10^-18 m." (Raymond A. Serway &

Jerry S. Faughn, College Physics: Fifth Edition, Harcourt College Publishers,

1999, p101)