s
A) 288 pm
B) 408 pm
C) 144 pm
D) 204 pm
Correct Answer: A
Solution :
| [a] For fcc lattice, |
| \[4r=\sqrt{2}a\] |
| \[r=\frac{\sqrt{2}}{4}a=\frac{a}{2\sqrt{2}}\] |
| \[=\frac{408}{2\sqrt{2}}\] |
| \[=144\,pm\] |
| diameter d = 2r = 2 x 144 pm |
| = 288pm |
You need to login to perform this action.
You will be redirected in
3 sec