MCQOPTIONS
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| 1. |
The edge of unit cell of FCC \[Xe\] crystal is \[620\,pm\]. The radius of \[Xe\] atom is [MP PET 2004] |
| A. | 219.25 Pm |
| B. | 235.16 Pm |
| C. | 189.37 Pm |
| D. | 209.87 Pm |
| Answer» B. 235.16 Pm | |