Finding the density of a hydrogen proton

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SUMMARY

The discussion focuses on calculating the density of a proton, modeled as a sphere with a diameter of 2.4 femtometers (fm) and a mass of 1.67e-27 kilograms. The correct formula for density, D = m/v, was applied, but the user initially miscalculated the volume by using the diameter instead of the radius. The correct volume calculation should utilize a radius of 1.2 fm, leading to a density of approximately 2.3e17 kg/m³, which is significantly higher than the user's initial result.

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AryRezvani
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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 1.67e-27 kg. Determine the density of the proton.

Homework Equations



As far as I know, the only required formula is D = m/v

The Attempt at a Solution



Well my attempt looks like this:

(2.4)^3*pi*(4/3) = 57.90583579 fm^3

Used google to convert that into kilometers, and got 57.905835791 (femtometer^3) = 5.79058358e-44 meter^3.

Divided 1.67e-27 kg by 5.79058358e-44 meter^3.

Got 2.883992566e16 but apparently, it's wrong.

Any suggestions guys?
 
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Hello, AryRezvani.

Note they give you the diameter, not the radius of the proton.
 
TSny said:
Hello, AryRezvani.

Note they give you the diameter, not the radius of the proton.

Ahh... Feeling rather stupid now.

Thanks.
 
That's not stupid, just a little oversight...(I'm doing things like that a lot in my old age :redface:).
 

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