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Solid State

Section: Chemistry

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Question : 6 of 16

Marks: +1, -0

Gold crystallises in fcc lattice. The edge length of the unit cell is

4\ \text{\AA}

. The closest distance between gold atoms is '

x

\text{\AA}

and density of gold is '

y

\text{g cm}^{-3}

x

y

(.

Molar mass of gold

=197\ \text{g mol}^{-1} ;\ N=6 \times 10^{23}\ \text{mol}^{-1})

[AP EAPCET 21 May 2025 S1]

$\sqrt{2}, 41.04$
$2\sqrt{2}, 20.52$
$2\sqrt{3}, 10.25$
$\sqrt{3}, 5.15$

Solution:

The closest distance

r

is given by

r =\; \frac{a}{\sqrt{2}} \Rightarrow r =\; \frac{4}{\sqrt{2}} = 2\sqrt{2}\ \text{\AA} = x\ \text{\AA}

The density (

d

) is given by,

d =\; \frac{Z M}{a^3 N_A}

Z = 4,\ M = 197\ \text{g mol}^{-1}

a = 4 \times 10^{-8}\ \text{cm},\ N_A = 6 \times 10^{23}\ \text{mol}^{-1}

d =\; \frac{4 \times 197\ \text{g mol}^{-1}}{(4 \times 10^{-8}\ \text{cm})^3 \times 6 \times 10^{23}\ \text{mol}^{-1}}

\approx 20.52\ \text{g cm}^{-3}=y\ \text{g cm}^{-3}

\therefore\ x=2\sqrt{2}\ \text{and}\ y=20.52

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