A) fcc
B) bcc
C) hcp
D) None of these
Correct Answer: A
Solution :
\[\rho =\frac{Z\times M}{{{N}_{A}}\times {{a}^{3}}}\] Here, \[M=108,\,\,{{N}_{A}}=6.023\times {{10}^{23}}\] \[a=409\text{ }pm=4.09\,\,\times \,\,{{10}^{-}}^{8}cm,\] \[p=10.5\text{ }g/c{{m}^{3}}\] Put on these values and solving we get. \[Z\text{ }=\text{ }4\text{ }=\text{ }number\text{ }of\text{ }atoms\text{ }per\text{ }unit\text{ }cell\] So, the structure of the crystal lattice is fee.You need to login to perform this action.
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