Prior to advances in x-ray scattering technology, creativity and mathematics were the tools used to estimate the size of atoms. As it turns out, the accuracy of such estimates was best when information about solids was used in calculations. Since solids and liquids can only be compressed to a negligible extent, we are at liberty to assume that no space exists between the neighboring atoms of a solid.
First, let’s consider a rudimentary experiment involving a known volume ( V ) of oil. When this volume of oil is poured into water, it spreads out and forms a surface layer that can be measured. It is reasonable to assume that such an oil slick has a thickness of 1 molecule, because it is no longer within the confines of a container. For this reason, the thickness of an oil slick will be the approximate diameter ( D ) of its molecules:
( V / A ) = t = D
Another such technique requires usage density ( ρ ) values of various materials:
ρ = ( mass / Volume )
If we multiply the top and bottom of the density equation by any number, the value of density is not altered in any way:
( 1 kg / L )( 100 / 100 ) = ( 100 kg / 100 L ) = 1 kg / L
Rather than the number 100, what if the top and bottom of a density expression were multiplied by the number of atoms contained in a mole??? The mass expression would be the molar mass ( Mm ) of an atom or molecule measured in grams ( g ). Since the molar mass contains 6.022 x 1023 atoms, the bottom half of the density equation, expressed as ( L3 ), would be multiplied by Avogadro’s number number ( NA ):
Note: Volume is usually expressed in cubic meters ( m3 ), centimeters ( cm3 or cc ), or in liters ( L ≠ L3 ). The ( L ) in the ( L3 ) term represents length, and it was chosen for convenience. In physics, we sometimes run out of variables to use, and this is the case here 🙂.
ρ = ( Mm / V ) = [ Mm / ( NA )( L3 ) ]
By solving the ( L ) variable, we will have solved the approximate length of a cubic-shaped atom:
L3 = [ Mm / ( NA )( ρ ) ]
L = [ Mm / ( NA )( ρ ) ]1/3
A water molecule has a molar mass of ( 18 g ) and a density of ( 1.0 g / cm3 ). Substitution of the appropriate values gives us an approximation for a water molecule’s length:
L = [ 18 g / ( 6.022 x 1023 )( 1.0 g / cm3 ) ]1/3
L = 3.0 x 10-8 cm = 3.0 x 10-10 m = 0.3 nm
Q: If we assume the diameter of a typical atom to be 0.2 nm, how big of a cube does a mole of almost any solid occupy?
A: The volume of a cube is a product of its length ( l ), width ( w ), and height ( h ):
V = length x width x height
Thus, a cube with 3.0 meter dimensions will have a volume of 27.0 m3. Going in reverse, taking the cube-root of 27.0 m3 gives us the number of meters along one side of the cube ( 3.0 m ). A mole of anything contains 6.022 x 1023 parts, whether the parts are molecules, electrons, protons, neutrons, neutrinos, quarks, leptons, photons, etc., etc., etc. For this reason, taking the cube root of of Avogadro’s number will give us the number of atoms along each edge of the cube:
( 6.022 x 1023 atoms )1/3 = 8.45 x 107 atoms
If each atom has a diameter of 0.2 nm, so the length along the side of a cubic mole of substance is as follows:
L = ( 0.2 x 10-9 m )( 8.45 x 107 atoms ) = 16.8 mm ≅ 2.0 cm
Since the length, width, and height are approximately 2.0 cm, volume of a mole of any solid will be approximately 8.0 cm3. Let’s now compare this value to the value derived from the density and mass of a mole of gold:
ρ = ( m / V )
V = ( m / ρ )
V = [ 197 g / ( 19.3 g / cm3 ) ]
V = 10 cm3
RENAISSANCE PHYSICS 