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Classification of Solids :: Discussion
| 9. | The interplanar spacing of (220) planes of a FCC structure is 1.7458 Å. Calculate the lattice constant. | |
| A. | 4.983 Å | |
| B. | 2.458 Å | |
| C. | 0 | |
| D. | 5.125 Å | |
| Answer: Option A | |
| Explanation: | |
|
D = 1.7458 Å = 1.7458 × 10-10 m |
|
| Tutorial Link: |
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