The gold foil used by Ernst Rutherford in his investigations 1 um thick. Given that the radius of gold is approximately 130 pm, how many atoms thick was the foil?

Question

aaronq · Answer

How many atoms fit across the foil (assuming they're liked up side by side)

$1~\mu m=1.0*10^{6}~pm$

marihelenh · Answer

$$\frac{ 130p m }{ 1 } * \frac{ 1\mu m }{ 1.0*10^{6}p m }$$

marihelenh · Answer

Is this how I would set that part up?

marihelenh · Answer

@aaronq

aaronq · Answer

yes that would give you the size of one atom in micrometers, then you divide 1 micrometer (the thickness of the foil) by the size of 1 atom (in micrometers)