How to calculate the density of an iron nucleus

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SUMMARY

The density of an iron nucleus can be calculated using the formula for density (Density = Mass/Volume) and the volume of a sphere (Volume = 4/3πr³). Given an iron nucleus with a radius of 5.4x10^-15 meters and a mass of 9.3x10^-26 kilograms, the calculated density is approximately 1.4x10^17 kg/m³. However, this value is significantly higher than expected due to the scale of atomic structures, as the nucleus is about 26,000 times smaller than the atom itself, leading to a volume reduction of approximately 1.7x10^13 times.

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Homework Statement



An iron nucleus has a radius of 5.4x10^-15 and a mass of 9.3x10^-26 what is its mass per unit volume in kg/m^3

Homework Equations



Area of a sphere = 4/3(pi)r^3
Density= M/V


The Attempt at a Solution



i basically plugged everything into the equations so calculated the volume and put it into the density equation but my answer is far to high i think

im getting 1.4x10^17 kgm^-3

just need some clarification on how to complete it or if I am right or not
 
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your calculation is Ok.
the size of an iron atom is 1.4*10^-10 m. This means that the radius of the core is about 26000 times smaller, and the volume is 26000^3 smaller, so about 1.7*10^13 times smaller.
If you compress iron until all the empty space between the cores is filled in it will become 1.7*10^13 times more dense.
 
ok thanks just wanted to check =]
 

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