SATHEE: Electronic Devices Question 172

Electronic Devices Question 172

Question: Atomic radius of fcc is

[J & K CET 2001]

Options:

A) $\frac{a}{2}$

B) $\frac{a}{2 \sqrt{2}}$

C) $\frac{\sqrt{3}}{4} a$

D) $\frac{\sqrt{3}}{2} a$

Show Answer

Answer:

Correct Answer: B

Solution:

For the fcc structure $4 r = {(a^{2} + a^{2})}^{1 / 2}$

$= a \sqrt{2}$

Therefore $r = \frac{a \sqrt{2}}{4} = \frac{a}{2 \sqrt{2}}$

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Electronic Devices Question 172

Question: Atomic radius of fcc is

Options:

Answer:

Solution:

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