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## Homework Statement

55.847 grams of iron contain N

_{A}=6.02E23 atoms. the density is ρ=7.87 [gr/cm

^{3}]. What is the diameter of the iron atom

## Homework Equations

## The Attempt at a Solution

The volume of 55.847 grams is: $$\frac{55.847}{7.87}=7.1[cm^3]$$

I assume the N

_{A}number of atoms is arranged in a cube with an edge consisting of $$\sqrt[3]{6.02\times 10^{23}}\cong 84,436,877$$

The diameter: $$\frac{0.071}{84,436,877}=8.4\times 10^{-8}$$

The right answer is d=2.8E-10