1
MCQ (Single Correct Answer)

AIPMT 2015

The vacant space in bcc lattice unit cell is
A
48%
B
23%
C
32%
D
26%

Explanation

Packing efficiency of bcc lattice = 68% Hence, empty space = 32%.
2
MCQ (Single Correct Answer)

AIPMT 2014

If $$a$$ is the length of the side of a cube, the distance between the body centered atom and one corner atom in the cube will be
A
$${2 \over {\sqrt 3 }}a$$
B
$${4 \over {\sqrt 3 }}a$$
C
$${{\sqrt 3 } \over 4}a$$
D
$${{\sqrt 3 } \over 2}a$$

Explanation

The distance between the body centered atom and one corner atom is $${{\sqrt 3 } \over 2}a$$.
3
MCQ (Single Correct Answer)

NEET 2013

A metal has a fcc lattice. The edge length of the unit cell is 404 pm. The density of the metal is 2.72 g cm$$-$$3. The molar mass of the metal is
(NA Avogadro's constant = 6.02 $$ \times $$ 1023 mol$$-$$1)
A
27 g mol$$-$$1
B
20 g mol$$-$$1
C
40 g mol$$-$$1
D
30 g mol$$-$$1

Explanation

for fcc Z = 4.

d = $${{ZM} \over {{N_A}{a^3}}}$$

$$ \Rightarrow $$ M = $${{d{N_A}{a^3}} \over Z}$$

= $${{2.72 \times 6.023 \times {{10}^{23}} \times {{\left( {404 \times {{10}^{ - 10}}} \right)}^3}} \over 4}$$

= 27 g mol–1
4
MCQ (Single Correct Answer)

NEET 2013

The number of carbon atoms per unit cell of diamond unit cell is
A
6
B
1
C
4
D
8

Explanation

Total number of carbon atoms per unit cell

= $$8 \times {1 \over 8} + 6 \times {1 \over 2} + 4$$ = 8

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