The radius of a lead atom is 175 pm. How many lead atoms would have to be laid side by side to span a distance of 1.58 mm?
If the radius is 175 pm then the diameter is 350 pm.
350 pm/atom x # atoms = 1.58E9 pm
Solve for # atoms.
20
The radius of a strontium atom is 215 pm. How many strontium atoms would have to be laid side by side to span a distance of 2.56 mm?
To find out how many lead atoms would have to be laid side by side to span a distance of 1.58 mm, we need to calculate the number of atoms in a given distance.
First, let's convert the radius of a lead atom from picometers (pm) to meters (m).
1 pm = 1 x 10^-12 m
So, the radius of a lead atom (175 pm) can be expressed as:
175 pm * (1 x 10^-12 m/1 pm) = 175 x 10^-12 m
Next, let's convert the distance we want to span from millimeters (mm) to meters (m).
1 mm = 1 x 10^-3 m
So, the distance of 1.58 mm can be expressed as:
1.58 mm * (1 x 10^-3 m/1 mm) = 1.58 x 10^-3 m
Now, we need to determine how many lead atoms can fit in the given distance.
The lead atoms can be considered as spheres with the diameter equal to twice the radius:
Diameter of a lead atom = 2 * radius = 2 * 175 x 10^-12 m = 350 x 10^-12 m
To find the number of atoms, we divide the total distance by the diameter of a lead atom:
Number of lead atoms = Total distance / Diameter of a lead atom
Number of lead atoms = (1.58 x 10^-3 m) / (350 x 10^-12 m)
Simplifying the expression, we cancel out the meters:
Number of lead atoms = (1.58 x 10^-3) / (350 x 10^-12)
Number of lead atoms = 4.51 x 10^9
Therefore, approximately 4.51 billion lead atoms would have to be laid side by side to span a distance of 1.58 mm.