The density of a proton (hydrogen nucleus)

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Homework Help Overview

The discussion revolves around calculating the density of a proton, specifically focusing on the conversion of diameter to radius, volume calculation, and the resulting density value. Participants are examining the implications of this density in comparison to more familiar substances like water.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants are checking the validity of the density calculation, questioning the relevance of comparing it to the density of water, and exploring the implications of the calculated density in a broader context.

Discussion Status

The discussion is ongoing, with participants expressing uncertainty about the correctness of the density value and its significance. Some have provided hints and comparisons to water, while others are questioning the meaningfulness of the density of an elementary particle.

Contextual Notes

There are references to the assumptions made in the calculations, such as treating the proton as a solid sphere, and the philosophical implications of discussing the density of subatomic particles.

mrcake
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Homework Statement
A proton, which is the nucleus of a hydrogen atom, can be modeled as a sphere with a diameter of 2.4 fm and a mass of 1.67 x 10^-27 kg. Determine the density of the proton
Relevant Equations
D = m/V
First of all i want is to check if my answer is right or not because i am really not sure about my answer.
Because the length is given in the form of a diameter we will divide that by 2 so we get it in the form of radius,
2.4fm / 2 = 1.2fm = r
Then we will convert the radius from "fm" to "m" so we that'll be,
r = 1.2 x 10^-15 m.
so we now have the radius in meters and from that we can calculate the volume of the proton, by the following formula V = 4/3 x pi x r^3
V = 4/3 x pi x (1.2 x 10^-15)^3 = 7.23 x 10^-45 m^3
Since we have the mass and we calculated the volume we can plug them into formula of density so we'll get,
D = m/V
D = 1.67 x 10^-27 / 7.23 x 10^-45 = 2.3 x 10^17 kg/m^3
 
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Something is not right with your answer. Take a good look at it. Hint: Water has density 1000 kg/m3.
 
hi thanks for reply, but what does water have to do with the proton of the hydrogen atom
 
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How many protons does a glass of water (0.25 L) contain? What would be the mass of that glass of water if your number for the density is correct?
 
kuruman said:
How many protons does a glass of water (0.25 L) contain? What would be the mass of that glass of water if your number for the density is correct?
Water is mostly space. A single proton may have a huge classical density - compare with a neutron star, for example.
 
mrcake said:
D = 1.67 x 10^-27 / 7.23 x 10^-45 = 2.3 x 10^17 kg/m^3
Looks about right given the numbers.
 
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PeroK said:
Looks about right given the numbers.
Is it right?
 
mrcake said:
Is it right?
Google thinks it's right to within an order of magnitude.
 
mrcake said:
Is it right?
It seems a fairly meaningless number, if you ask me. The density of water or hydrogen gas is meaningful; but the density of an elementary particle, given that it isn't actually a solid, localised sphere seems a pointless calculation. Sorry for the philosophical answer.
 
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