Find the ( A + B ) Nearest neighbor distance in NaCl-type structure Nearest neighbor distance in CsCl-type structure Assume the edge length in each case is 100 units. [Given 2 = 1.4 and 3 = 1.7 ]

Question

Find the ( A + B ) Nearest neighbor distance in NaCl-type structure Nearest neighbor distance in CsCl-type structure Assume the edge length in each case is 100 units. [Given 2 ​ = 1.4 and 3 ​ = 1.7 ]

TestHub · Accepted Answer

To find the nearest neighbor distances:

For NaCl-type structure:

The nearest neighbor distance is $a /2$ .

Given $a = 100$ units.

Nearest neighbor distance $= 100/2 = 50$ units.

For CsCl-type structure:

The nearest neighbor distance is $3 a /2$ .

Given $a = 100$ units and $3 = 1.7$ .

Nearest neighbor distance $= (1.7 \times 100) /2 = 170/2 = 85$ units.

The question asks for (A+B), which implies the sum of these two distances.

Sum $= 50 + 85 = 135$ units.

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