# How many gold atoms can fit around earth's equator?

I know that earth has a radius $$R_E = 6371\,km$$ and gold has the following properties:

• $$\rho_{Au} = 1.932 \cdot 10^4 \frac{kg}{m^3}$$
• $$M_{Au} = 196.9666 \cdot u \quad (u = 1.66054*10^{-27}\,kg)$$

where $$\rho_{Au}$$ is the density of gold and $$M_{Au}$$ is the mass of a gold atom.

Assuming the equator is a perfect circle, we can determine its circumference \begin{align*} U_E &= 2\pi R_E \\ &= 2\pi \cdot 6371 \, km \\ &\approx 40\,030 \,km \end{align*} Let's take circle and cut it open at one point, so we can stretch it out to a straight line with length $$U_E$$. (please correct me here if that's wrong)

To calculate how many atoms fit around the equator, we can divide the circumference $$R_E$$ by the diameter of a single gold atom, let's call it $$D_{Au}$$.

Two problems:

(1) Is it actually possible to treat the circumference of the circle as a simple straight line, given that the ring of atoms around the equator has height $$1$$.

(2) How can I find the diameter/radius of a gold atom across a line? I tried viewing a gold atom as a simple sphere and calculating its radius using the formula $$V_{Au} = \frac{4}{3} \pi R_{Au}^3$$ Which won't give any value of the atomic radii that you can find on the internet, since they all define the atomic radius as the distance between $$2$$ atoms in a given packing in a volume or on an area. But I need to get the radius in the context of a line, right?

• Hints: 1) you know the density and molar mass of gold, so you can calculate the dimensions of a cube that contains one mole of gold atoms; 2) that cube has the same width, height, and depth. Multiplying the number of atoms in the cube's height, width, and depth gives one mole. This means that the cube root of the Avagadro number is the number of gold atoms down the side of the cube, assuming that each gold atom is just touching its neighbors. Commented Oct 19, 2021 at 19:25