gold atom

logistic_guy

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The radius of an atom of gold (Au)\displaystyle (\text{Au}) is about 1.35\displaystyle 1.35 A˚\displaystyle \text{\AA}. (a)\displaystyle \bold{(a)} Express this distance in nanometers (nm)\displaystyle (\text{nm}) and in picometers (pm)\displaystyle (\text{pm}). (b)\displaystyle \bold{(b)} How many gold atoms would have to be lined up to span 1.0 mm\displaystyle 1.0 \ \text{mm}? (c)\displaystyle \bold{(c)} If the atom is assumed to be a sphere, what is the volume in cm3\displaystyle \text{cm}^3 of a single Au\displaystyle \text{Au} atom?
 
The radius of an atom of gold (Au)\displaystyle (\text{Au}) is about 1.35\displaystyle 1.35 A˚\displaystyle \text{\AA}. (a)\displaystyle \bold{(a)} Express this distance in nanometers (nm)\displaystyle (\text{nm}) and in picometers (pm)\displaystyle (\text{pm}). (b)\displaystyle \bold{(b)} How many gold atoms would have to be lined up to span 1.0 mm\displaystyle 1.0 \ \text{mm}? (c)\displaystyle \bold{(c)} If the atom is assumed to be a sphere, what is the volume in cm3\displaystyle \text{cm}^3 of a single Au\displaystyle \text{Au} atom?

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

Let r\displaystyle r be the radius.

1 A˚=1010 m\displaystyle 1 \ \text{\AA} = 10^{-10}\ \text{m}

Then,

r=1.35 A˚=1.35 A˚×1010 mA˚=0.000000000135 m=0.135 nm=135 pm\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}}
 
(b)\displaystyle \bold{(b)}

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

1 mm2r=0.001 m2×0.000000000135 m=3703704 gold atoms\displaystyle \frac{1 \ \text{mm}}{2r} = \frac{0.001 \ \text{m}}{2 \times 0.000000000135 \ \text{m}} = \textcolor{blue}{3703704 \ \text{gold atoms}}
 
(c)\displaystyle \bold{(c)} If the atom is assumed to be a sphere, what is the volume in cm3\displaystyle \text{cm}^3 of a single Au\displaystyle \text{Au} atom?
V=43πr3=43π(0.000000000135)3=1.03×1029 m3\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:

V=0.0000000000000000000000000000103 m3\displaystyle V = \textcolor{red}{0.0000000000000000000000000000103 \ \text{m}^3}

😛😛

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

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