The diameter of a rubidium atom is \(4.95 \AA\). 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 \(1.0 \mathrm{~cm}\) on a side? (b) How many Rb atoms could be
placed on a square surface that is \(1.0 \mathrm{~cm}\) 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?