A 1/4-inch-thick lead sheet is used for protection from medi-cal X rays. If a single lead atom has a diameter 350 pm, how many atoms thick is the lead sheet?

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Step 1: Convert the thickness of the lead sheet from inches to picometers. There are 2.54 cm in an inch and 1 cm is equal to 1 x 10^10 picometers. So, the thickness of the lead sheet in picometers is 1/4 inch * 2.54 cm/inch * 1 x 10^10 pm/cm.

Step 2: Now, we need to find out how many lead atoms can fit into this thickness. To do this, we divide the thickness of the lead sheet in picometers by the diameter of a single lead atom.

Step 3: The diameter of a single lead atom is given as 350 pm. So, divide the thickness of the lead sheet in picometers (calculated in step 1) by 350 pm.

Step 4: The result from step 3 will give you the number of lead atoms that can fit into the thickness of the lead sheet. This is the number of atoms thick the lead sheet is.

Step 5: Remember that this is an approximation, as atoms are not perfect spheres and do not pack together perfectly. However, this calculation gives a good estimate of the number of atoms thick the lead sheet is.

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Atomic Diameter

The atomic diameter refers to the size of an individual atom, typically measured in picometers (pm). In this context, the diameter of a lead atom is given as 350 pm, which is essential for calculating how many atoms fit into a given thickness. Understanding atomic size is crucial for visualizing the arrangement of atoms in a material.

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Atom Structure

Thickness Conversion

The thickness of the lead sheet is provided in inches, which must be converted to the same unit as the atomic diameter for accurate calculations. Since 1 inch equals 2.54 centimeters and 1 centimeter equals 10 million picometers, converting the thickness from inches to picometers allows for a direct comparison with the atomic diameter.

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Conversion Factors

Calculating Number of Atoms

To determine how many lead atoms are stacked in the thickness of the lead sheet, one must divide the total thickness (in picometers) by the diameter of a single lead atom. This calculation provides the number of atoms that can fit across the specified thickness, illustrating how atomic scale relates to macroscopic measurements.

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