Semiconducting Devices Solids And Crystals Question 18
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 $
Answer:
Correct Answer: B
Solution:
For the fcc structure $ 4r={{(a^{2}+a^{2})}^{1/2}} $ $ =a\sqrt{2} $
Þ $ r=\frac{a\sqrt{2}}{4}=\frac{a}{2\sqrt{2}} $
BETA
