JEE Advanced · 2022 · 28 Aug 2022 · Paper 2 · Solid State

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Atom X occupies the fcc lattice sites as well as alternate tetrahedral voids of the same lattice. The packing efficiency (in %) of the resultant solid is closest to

(A)25
(B)35✓ Correct
(C)55
(D)75

Answer: (B)

Solution

In an fcc lattice, there are 4 atoms per unit cell at lattice points and 8 tetrahedral voids. If alternate tetrahedral voids are occupied, 4 additional atoms are present.

Total atoms per unit cell = 4 (fcc) + 4 (tetrahedral) = 8.

For fcc with tetrahedral void occupation, the relationship between edge length

a

and atomic radius

r_{X}

is:

a = \frac{8 r _{X}}{3}

Packing efficiency:

PE = \frac{8 \times \frac{4}{3} π r _{X}^{3}}{a ^{3}} \times 100 = \frac{8 \times \frac{4}{3} π r _{X}^{3}}{( \frac{8 r _{X}}{3} ) ^{3}} \times 100 = \frac{3 π}{16} \times 100 \approx 34%

Closest to 35%.

JEE Advanced · 2022 · 28 Aug 2022 · Paper 2 · Single correct option

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JEE Advanced · 2022 · 28 Aug 2022 · Paper 2 · Solid State

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