Question

What is the fraction of surface atoms in a golden sphere with a 1 cm radius?

Avogadro number - 6•1023

Atomic weight of gold - 197

Density of gold – 19.3 g/cm3

Answer 1

Atomic weight of gold = 197 g/mole

Avagadro number = 6x10²³

Mass of 1 gold atom = 197 / 6x10²³

= 32.83x10^-23 g

Density of gold = 19.3 g/cm3

Volume of 1 gold atom = 32.83x10^-23 / 19.3

= 1.70x10^-23 cm3

4/3**r³ = 1.70x10^-23

r³ = 4.05x10^-24

r = 1.59x10^-8 cm

radius of golden sphere, R = 1 cm

The surface atoms form a spherical shell of thickness "2r". The shell's outer radius is R, and its inner radius is (R-2r).

Fraction of surface atoms = Volume of shell / Volume of sphere

= [ 4/3**R³ - 4/3**(R - 2r)³ ] / (4/3**R³)

= [ R³ - (R - 2r)³ ] / R³  

Putting values in avove expression:

Fraction of surface atoms = [1³ - (1 - 2*1.59x10^-8)³] / 1³

= 9.54x10^-8