The gold foil used by Ernest Rutherford in his investigations of atomic structure was approximately 1$\mu$m thick.Given that the radius of a gold atom is approximately 130 pm, how many atoms thick was the foil?

Question

Quizplus · Accepted Answer

To find the thickness of the foil in terms of atoms, divide the thickness of the foil by the diameter of a gold atom. Given that the foil is approximately 100 nm thick and the radius of a gold atom is approximately 130 pm (or 0.13 nm), the diameter of a gold atom is 0.26 nm. Therefore, the thickness in atoms is 100 nm / 0.26 nm ≈ 384.6, which is approximately 4 x 10^2 atoms thick.

The Gold Foil Used by Ernest Rutherford in His Investigations μ\muμ

The Gold Foil Used by Ernest Rutherford in His Investigations $\mu$