a) Volume of 1.93 mg of gold: [tex]1\cdot 10^{-10} m^3[/tex]
b) Thickness: 690 angstroms
c) Approximately 138 atoms of gold
Explanation:
a)
The relationship between mass and density of a material is given by
[tex]d=\frac{m}{V}[/tex]
where
d is the density
m is the mass
V is the volume
For the 1.93 mg of gold, we have
[tex]m = 1.93 mg = 1.93\cdot 10^{-6} kg[/tex] (mass)
[tex]d=19300 kg/m^3[/tex] (density)
Solving for V, we find the volume:
[tex]V=\frac{m}{d}=\frac{1.93\cdot 10^{-6}}{19300}=1\cdot 10^{-10} m^3[/tex]
b)
In part a), we have calculated the volume of the film of gold. Now we want to calculate the thickness.
Thickness and volume are related by the equation
[tex]V=At[/tex]
where
V is the volume
A is the area
t is the thickness
Here we have:
[tex]V=1\cdot 10^{-10} m^3[/tex] is the volume
[tex]A=14.5 cm^2 = 14.5\cdot 10^{-4} m^2[/tex] is the area
Therefore, solving for the thickness,
[tex]t=\frac{V}{A}=\frac{1\cdot 10^{-10}}{14.5\cdot 10^{-4}}=6.9\cdot 10^{-8} m[/tex]
And converting into angstroms,
[tex]t=6.9\cdot 10^{-8} m=690 \cdot 10^{-10} m = 690A[/tex]
c)
The thickness of the gold foil is
t = 690 A
We are told here that the diameter of a gold atom is
d = 5 A
If we imagine the gold atoms to be in line inside the gold foil, then the total thickness is equal to the number of atoms N multiplied by the diameter of a single atom of gold:
[tex]t=Nd[/tex]
where N is the number of atoms of gold.
Re-arranging for N, we find:
[tex]N=\frac{690}{5}=138[/tex]
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