The density of gold metal is 18.9 g/cm^3. (a) What is the volume occupied by one

Question

The density of gold metal is 18.9 g/cm^3. (a) What is the volume occupied by one gold atom? (b) If the diameter of a gold nucleus is 1/100 000 times that of a gold atom, What is the volume of one gold nucleus? (a) Use the volume occupied by a gold atom to estimate the thickness of one atomic layer of gold. Estimate how many layers of gold atoms there are in a strip of gold foil that is 1.0 mu m thick (1 mu m = 10^-6). The most useful tool for studying the structure of atoms is electromagnetic What we call light is one from of this radiation. We need to the properties of light in order to understand what electromagnetic reveals about atomic structure.

Explanation / Answer

4.1.1. The density of gold is 18.9 g/cm^3. Molar mass of gold is 197 g/mol. If one mol is 6.02 x 10^23 atoms, one atom of gold will weigh:

197 g/mol * 1 mol/6.02x10^23 atoms = 3.27x10^-22 g/atom

That mass equals a volume in cm^3:

3.27 x 10-22 g / 18,9 g/cm^3 = 1.73 x 10^-23 cm^3

The above number is related to the radius (and in turn, to the diameter) of the spheric atom of gold:

4/3*pi*r^3 = 1.73 x 10^-23 cm^3 -----> r = cbcroot(3/4 x 1/pi x 1.73 x 10^-23) = 1.60 x 10^-8 cm

D = 2*r = 3.20 x 10^-8 cm. If the diameter of a gold nucleus is 100000 times smaller than that of a gold atom:

D* = 3.20x10^-8 cm / 100000 = 3.20 x 10^-13 cm, r* = D*/2 = 1.60 x 10^-13 cm

V* = 4/3 * pi * r*^3 = 4/3 * 3,14159 * 1.60 x 10^-13 = 1.72 x 10^-38 cm^3 is the volume of a gold nucleus.

4.1.2 The thickness of an atomic layer of gold would be, precisely, equal to the diameter calculated above, i.e. 3.20 x 10^-8 cm. To determine how many layers of gold atoms are there in a 1 micrometer thick gold foil, first we use the same units:

3.20 x 10^-8 cm x 1 m/100 cm = 3.20 x 10^-10 m.

Then we divide the thickness of the foil by the thickness of the gold layer:

# layers: 1 x 10^-6 m / 3.20 x 10^-10 m ---> approximately 3 x 10^3 layers i.e. 3000 layers of gold atoms.

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