gold atom

[logistic_guy]

The radius of an atom of gold \(\displaystyle (\text{Au})\) is about \(\displaystyle 1.35\) \(\displaystyle \text{\AA}\). \(\displaystyle \bold{(a)}\) Express this distance in nanometers \(\displaystyle (\text{nm})\) and in picometers \(\displaystyle (\text{pm})\). \(\displaystyle \bold{(b)}\) How many gold atoms would have to be lined up to span \(\displaystyle 1.0 \ \text{mm}\)? \(\displaystyle \bold{(c)}\) If the atom is assumed to be a sphere, what is the volume in \(\displaystyle \text{cm}^3\) of a single \(\displaystyle \text{Au}\) atom?

 

[khansaheb]

The radius of an atom of gold \(\displaystyle (\text{Au})\) is about \(\displaystyle 1.35\) \(\displaystyle \text{\AA}\). \(\displaystyle \bold{(a)}\) Express this distance in nanometers \(\displaystyle (\text{nm})\) and in picometers \(\displaystyle (\text{pm})\). \(\displaystyle \bold{(b)}\) How many gold atoms would have to be lined up to span \(\displaystyle 1.0 \ \text{mm}\)? \(\displaystyle \bold{(c)}\) If the atom is assumed to be a sphere, what is the volume in \(\displaystyle \text{cm}^3\) of a single \(\displaystyle \text{Au}\) atom?

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[logistic_guy]

\(\displaystyle \bold{(a)}\)

Let \(\displaystyle r\) be the radius.

\(\displaystyle 1 \ \text{\AA} = 10^{-10}\ \text{m}\)

Then,

\(\displaystyle r = 1.35 \ \text{\AA} = 1.35 \ \text{\AA} \times \frac{10^{-10} \ \text{m}}{\text{\AA}} = 0.000000000135 \ \text{m} = \textcolor{blue}{0.135 \ \text{nm}} = \textcolor{blue}{135 \ \text{pm}}\)

 

[logistic_guy]

\(\displaystyle \bold{(b)}\)

Divide \(\displaystyle 1 \ \text{mm}\) by the diameter of the gold atom.

\(\displaystyle \frac{1 \ \text{mm}}{2r} = \frac{0.001 \ \text{m}}{2 \times 0.000000000135 \ \text{m}} = \textcolor{blue}{3703704 \ \text{gold atoms}}\)

 

[logistic_guy]

\(\displaystyle \bold{(c)}\) If the atom is assumed to be a sphere, what is the volume in \(\displaystyle \text{cm}^3\) of a single \(\displaystyle \text{Au}\) atom?

\(\displaystyle V = \frac{4}{3}\pi r^3 = \frac{4}{3}\pi (0.000000000135)^3 = \textcolor{blue}{1.03 \times 10^{-29} \ \text{m}^3}\)

Or just for fun the answer is:

\(\displaystyle V = \textcolor{red}{0.0000000000000000000000000000103 \ \text{m}^3}\)



This is one reason why you cannot see the gold atom by eyes!