The gold foil Rutherford used in his scattering experiment had a thickness of approximately 4×10^−3 mm. If a single gold atom has a diameter of 2.9 x10^-8cm , how many atoms thick was Rutherford's foil? Express your answer using one significant figure.
To find out how many atoms thick Rutherford's foil was, we can divide the thickness of the foil by the diameter of a single gold atom.
First, let's convert the given values to the same unit. We can convert the thickness of the foil from millimeters (mm) to centimeters (cm) by dividing it by 10:
Thickness of the foil = 4 × 10^(-3) mm = (4 × 10^(-3)) / 10 cm = 4 × 10^(-4) cm
Now, let's divide the thickness of the foil by the diameter of a single gold atom:
Number of atoms thick = (Thickness of the foil) / (Diameter of a single gold atom)
Number of atoms thick = (4 × 10^(-4) cm) / (2.9 × 10^(-8) cm)
To divide these two values, we divide the coefficients and subtract the exponents:
Number of atoms thick = 1.38 × (10^(-4) / 10^(-8))
Simplifying the exponent division, we subtract the exponents:
Number of atoms thick = 1.38 × 10^(-4 - (-8))
Number of atoms thick = 1.38 × 10^(-4 + 8)
Number of atoms thick = 1.38 × 10^4
Since we are asked to express our answer using one significant figure, we round the number to the nearest one:
Number of atoms thick ≈ 1.4 × 10^4
Therefore, Rutherford's foil was approximately 1.4 × 10^4 atoms thick.
assuming rectangular packing of atoms, just divide by the diameter of a gold atom; the sheet was
4*10^-3mm / 2.9*10^-7mm = 13793 atoms thick