in the fcc lattice of particle $A, B$ particles occupy all tetrahedral void and $C$ occupy all octahedral void. Find the correct relation for closest distance between particle $B$ and $C$ . Given radius of particle $A$ is 200 A.

r_{B} + r_{C}

= 2 pm

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r_{B} + r_{C} = \sqrt{6}

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r_{B} + r_{C} = 0.639

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r_{B} + r_{C} = 0.25

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Solution

The correct option is B $r_{B} + r_{C} = \sqrt{6}$ pm
$A \Rightarrow$ In $F C C$
$B \Rightarrow$ Tetrahedra void
$A \Rightarrow$ oetahedra voil
the $r_{B} = 0.414 r_{A}$
$r_{C} = 0.225 r_{A}$
$r_{B} + r_{C} = (0.639) r_{A}$

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in the fcc lattice of particle A, B particles occupy all tetrahedral void and C occupy all octahedral void. Find the correct relation for closest distance between particle B and C. Given radius of particle A is 200 A.

The correct option is B rB+rC=√6 pmA⇒ In FCCB⇒ Tetrahedra voidA⇒ oetahedra voilthe rB=0.414 rArC=0.225 rArB+rC=(0.639) rA

in the fcc lattice of particle $A, B$ particles occupy all tetrahedral void and $C$ occupy all octahedral void. Find the correct relation for closest distance between particle $B$ and $C$ . Given radius of particle $A$ is 200 A.

The correct option is B $r_{B} + r_{C} = \sqrt{6}$ pm
$A \Rightarrow$ In $F C C$
$B \Rightarrow$ Tetrahedra void
$A \Rightarrow$ oetahedra voil
the $r_{B} = 0.414 r_{A}$
$r_{C} = 0.225 r_{A}$
$r_{B} + r_{C} = (0.639) r_{A}$