Calculating atomic radius from weight + density

AI Thread Summary
To calculate the atomic radius of a palladium atom with a face-centered cubic (FCC) structure, the density is given as 12.0 g/cm³ and the atomic weight as 106.4 g/mol. The calculations involve determining the weight of a single palladium atom and the total weight of atoms in a unit cell, which contains four atoms. The unit cell volume is derived from the density, and the edge length is calculated using the volume. The radius is then found by dividing the edge length by 2, but an error occurred when miscalculating the edge length due to a misunderstanding of the formula involving 2*sqrt(2). The correct atomic radius is 0.138 nm, highlighting the importance of careful formula application.
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Homework Statement


Calculate the radius of a palladium atom, given that Pd has an FCC crystal structure, a density of 12.0g/cm3, and an atomic weight of 106.4g/mol


Homework Equations


Pd contains 4 atoms per unit cell.
unit cell edge lengh = 2*Radius*sqrt(2)


The Attempt at a Solution



calculate atom weight:
106.4/6.0221418e23 = 1.7668133e-22

calculate unit cell weight:
1.7668133e-22*4 = 7.0672532e-22

calculate unit cell volume in cm3:
7.0672532e-22/12 = 5.8893777e-23

convert to nm3:
5.8893777e-23*(10^7)^3 = 0.0588938

calculate edge length:
0.0588938^(1/3) = 0.3890659

calculate radius:
0.3890659/2 = 0.194533

my answer:
0.194533nm

correct answer:
0.138nm

I can't figure out what I've done wrong here.
 
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I figured it out. I got confused and thought 2*sqrt(2) was equal to 2. So silly!
 
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