The diameter of a rubidium atom is 4.95 ???. We will consider two different ways of placing the atoms on a surface. In arrangement A, all the atoms are lined up with one another to form a square grid. Arrangement B is called a close-packed arrangement because the atoms sit in the 'depressions' formed by the previous row of atoms: (a) Using arrangement A, how many Rb atoms could be placed on a square surface that is 1.0 cm on a side? (b) How many Rb atoms could be placed on a square surface that is 1.0 cm on a side, using arrangement B? (c) By what factor has the number of atoms on the surface increased in going to arrangement B from arrangement A? If extended to three dimensions, which arrangement would lead to a greater density for Rb metal?
Solution : Step 1: Rubidium is a chemical element having chemical symbol Rb. It has atomic number of 37. Given, Diameter of Rb atom = 4.95 Step 2: (a) Using arrangement A, how many Rb atoms could be placed on a square surface that is 1.0 cm on a side Side of square = 1.0cm. To find the number of atoms to be placed on 1.0cm side of square, we first convert side in cm to Angstrom. Therefore, side of square in is : 1 cm = 10 8 Now we calculate number of rubidium atoms in the square of length 10 as : 8 8 Number of Rb atoms = 10 Diameter of atom = = 2.02 × 10 atoms. Therefore, number of atoms required for arrangement A, which is a square is : 2 Area of square = a . = (2.02 × 10 ) 7 14 14 2 = 4.08 × 10 atoms 4.1 × 10 atoms/cm .