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Chapter , Problem 90AE(choose chapter or problem)
These exercises are not divided by category, although they are roughly in the order of the topics in the chapter. They are not paired.
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 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?
Equation Transcription:
Å
Text Transcription:
4.95 AA
Questions & Answers
QUESTION:
These exercises are not divided by category, although they are roughly in the order of the topics in the chapter. They are not paired.
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 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?
Equation Transcription:
Å
Text Transcription:
4.95 AA
ANSWER:Step 1 of 4
(a)
From the given,
The length of the side of the square surface = 1.0 cm