A) \[\frac{4m}{{{a}^{3}}}\]
B) \[\frac{2m}{{{a}^{3}}}\]
C) \[\frac{m}{{{a}^{3}}}\]
D) \[\frac{m}{4{{a}^{3}}}\]
Correct Answer: B
Solution :
Given that. Mass of single Ag-atom \[=m\] Unit cell length \[=a\] Total atoms present per unit cell of fee lattice \[=4\] Therefore, mass of unit cell becomes \[4\,m.\] Volume of unit cell \[={{a}^{3}}\] \[\text{Density}\,(\rho )=\frac{\text{Mass}\,\text{of}\,\text{unit}\,\text{cell}}{\text{Volume}\,\text{of}\,\text{unit}\,\text{cell}}=\frac{4m}{{{a}^{3}}}\]
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