Semiconducting Devices Question 253

Question: Atomic radius of fcc is

[J & K CET 2001]

Options:

A) a2

B) a22

C) 34a

D) 32a

Show Answer

Answer:

Correct Answer: B

Solution:

For the fcc structure 4r=(a2+a2)1/2

=a2

Therefore r=a24=a22



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