Calculating Particle Distance from Density and Number

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To calculate the distance between particles given their density and number, one can determine the mean number density of particles, such as hydrogen atoms in the sun, which is calculated to be 8.4 x 10^29 atoms per cubic meter. Each atom occupies a volume of approximately 1.19 x 10^-30 cubic meters, leading to a calculated radius of about 6.57 x 10^-11 meters for the sphere around each atom. The discussion emphasizes that the typical distance between two atoms is effectively twice this radius, and it is suggested to consider the radius of a hydrogen atom in the final calculations. The importance of precise unit handling is noted, along with the recommendation to present both the calculated distance and the comparison to atomic radii for clarity. This approach provides a comprehensive understanding of particle spacing in a dense medium.
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[SOLVED] Desnity and Distibution

Homework Statement



I need to calculate the distance between particles. I know the density, and I know how many particles there are, but I am not sure how to calculate the distance between them.

Homework Equations



Not Sure

The Attempt at a Solution



Any advice would be greatly appreciated,

TFM
 
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Hi TFM,

Could you please post the full question verbatim, as it is in your textbook/homework sheet.
 
Assume that the sun is made of pure Hydrogen, and take the Hydrogen mass ass being 1.67x10^-27 kg. If the mean mass density of the sun is 1400 kg/m^3, what is the mean number density.

I have calculated this to be 8.4 x 10^29 atoms per cubic meter

hence estimate the typical inter-particle distance

TFM
 
So you know that in one meter there is 8.4 x 10^29 hydrogen atoms. So what volume does each atom occupy? Next, assume that each hydrogen atom is a particle at the centre of a sphere.
 
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Each Hydrogen Atom will occupy a volume of \frac{1}{8.4X10^{29}} which is 1.19 x 10^{-30} metres cubed

TFM
 
TFM said:
Each Hydrogen Atom will occupy a volume of \frac{1}{8.4X10^{29}} which is 1.19 x 10^{-30} metres cubed

TFM
Correct, so what is the radius of the sphere with such a volume?
 
Volume of a sphere: Vol = \frac{4}{3}\pi r^{3}

So 1.19x10^{-30} = \frac{4}{3} \pi r^{3}

so the radius is r = \sqrt[3]{\frac{3*Vol}{4* \pi }}

Giving the radius: \sqrt[3]{2.84*10^{-31}} = 6.57*10^-11 metres Cubed

TFM
 
TFM said:
Volume of a sphere: Vol = \frac{4}{3}\pi r^{3}

So 1.19x10^{-30} = \frac{4}{3} \pi r^{3}

so the radius is r = \sqrt[3]{\frac{3*Vol}{4* \pi }}

Giving the radius: \sqrt[3]{2.84*10^{-31}} = 6.57*10^-11 metres Cubed

TFM
Spot on, but watch your units :wink:
 
Do I now just have to take away the radius of a Hydrogen Atom?

TFM
 
  • #10
TFM said:
Do I now just have to take away the radius of a Hydrogen Atom?

TFM
Personally, I would have left the answer as it is since once you get down to such small distances the concept of classical radii doesn't really apply. However, you could put both answers to be safe, it depends very much on what your tutor wants.
 
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  • #11
Well, the question then asks you to compareit to the radius of a Hydrogen Atom and a Hydrogen Nuclei.

TFM
 
  • #12
TFM said:
Well, the question then asks you to compareit to the radius of a Hydrogen Atom and a Hydrogen Nuclei.

TFM
Then I definitely would subtract the hydrogen radius from your answer.
 
  • #13
Thanks foy all your assistance, Hootenanny :smile:

TFM
 
  • #14
TFM said:
Thanks foy all your assistance, Hootenanny :smile:

TFM
It was a pleasure TFM :smile:
 
  • #15
Is the typical distance between 2 atoms not twice the radius of this sphere? First from one atom to where the spheres meet, then to the other atom.
 
  • #16
kamerling said:
Is the typical distance between 2 atoms not twice the radius of this sphere? First from one atom to where the spheres meet, then to the other atom.
Indeed it is, I assumed that TFM would have realized that.
 
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