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Ch.2 - Atoms, Molecules, and Ions

Chapter 2, Problem 90a

The diameter of a rubidium atom is 495 pm 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?

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Hi everyone here we have a question asking how many cesium atoms are needed to line up with one another to form a square grid on a 1.0 millimeter by 1.0 millimeter square surface. If the diameter of a cesium atom is 0.267 nanometers. So we have one atom over 0.267 nm times one nanometer over one times 10 to the negative ninth nano meters, Times one times 10 to the negative 3rd m over one millimeter. So are nanometers are going to cancel out and our meters are going to cancel out, leaving us with 3. times 10 to the negative six atoms per millimeter. So we have 1.0 millimeter squared Equals 3.745, 3 times to the sixth Squared equals 1.40 times 10 to the 13th Adams. And that is our final answer. Thank you for watching. Bye.