Sign in
Please select an account to continue using cracku.in
↓ →
Join Our JEE Preparation Group
Prep with like-minded aspirants; Get access to free daily tests and study material.
The total number of octahedral void(s) per atom present in a cubic close packed structure is:
In a Cubic Close Packed (ccp) or Face-Centered Cubic (fcc) structure:
Atoms per unit cell ($$Z$$): 4 (1 corner + 3 face centers).
Octahedral voids per unit cell ($$N$$): 4 (1 at the body center and 3 at the edge centers).
$$\text{Octahedral voids per atom} = \frac{\text{Total Voids}}{\text{Total Atoms}} = \frac{4}{4} = 1 \text{}$$
Create a FREE account and get:
Predict your JEE Main percentile, rank & performance in seconds
Educational materials for JEE preparation
Ask our AI anything
AI can make mistakes. Please verify important information.
AI can make mistakes. Please verify important information.