An element crystallises in a face-centred cubic (fcc) unit cell with cell edge a. The distance between the centres of two nearest octahedral voids in the crystal lattice is

Question

An element crystallises in a face-centred cubic (fcc) unit cell with cell edge a. The distance between the centres of two nearest octahedral voids in the crystal lattice is (A) a (B) √2 a (C) ( a /√2) (D) ( text a /2)

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<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/chemistry/4917338efd22180fac038f248f56ebce-.png" /> distance between nearest octahedral voids(O.V.) =√((a/2))2+((a/2))2 ⇒ =(a/√2)

Q. An element crystallises in a face-centred cubic (fcc) unit cell with cell edge a. The distance between the centres of two nearest octahedral voids in the crystal lattice is

a

$2$ a

$\frac{a}{2}$

$\frac{a}{2}$

distance between nearest octahedral voids(O.V.)

$= (\frac{a}{2})^{2} + (\frac{a}{2})^{2}$

$\Rightarrow= \frac{a}{2}$

Q. An element crystallises in a face-centred cubic (fcc) unit cell with cell edge a. The distance between the centres of two nearest octahedral voids in the crystal lattice is

a

2​ a

2​a​

2 a ​

distance between nearest octahedral voids(O.V.) =(2a​)2+(2a​)2​ ⇒=2​a​

$2$ a

$\frac{a}{2}$

$\frac{a}{2}$

distance between nearest octahedral voids(O.V.)

$= (\frac{a}{2})^{2} + (\frac{a}{2})^{2}$

$\Rightarrow= \frac{a}{2}$