If the volume of fcc unit cell is 6.4 × 10^ -23 cm^3 find the radius of an atom in it.

Correct answer: A. 141.4 pm Volume of the unit cell, V = a^3 = 6.4 × 10^ -23 cm^3 a^3 = 64 × 10^ -24 cm^3 a = (64 × 10^ -24 )^ 1/3 cm = 4 × 10^ -8 cm Converting to picometers: a = 4 × 10^ -8 × 10^ 10 pm = 400 pm For an fcc unit cell, the relation between edge length a and radius r is given by: 4r = 2 a r = 2 a 4 = a 2 2 r = 400 2 2 = 100 2 pm r = 100 × 1.414 = 141.4 pm Answer: 141.4 pm

MHT CET 202622 AprilMorning Shift

Exam: MHT CET
Year: 2026
Date: 22 April
Shift: Morning Shift
Subject: Chemistry
Chapter: Solid State

MHT CET 2026 · 22 April · Morning Shift · Chemistry · Solid State

If the volume of fcc unit cell is 6.4 × 10^ -23 cm^3 find the...

If the volume of fcc unit cell is

6.4 \times 10^{-23}

cm^3 find the radius of an atom in it.

Options

A $141.4$ pm
B $159.0$ pm
C $180.2$ pm
D $124.2$ pm

Correct answer

A. $141.4$ pm

Step-by-step solution

Volume of the unit cell,

V = a^3 = 6.4 \times 10^{-23}

cm^3

<br><br>

a^3 = 64 \times 10^{-24}

cm^3

a = (64 \times 10^{-24})^{1/3}

= 4 \times 10^{-8}

cm

Converting to picometers:

a = 4 \times 10^{-8} \times 10^{10}

= 400

pm

For an fcc unit cell, the relation between edge length

a

and radius

r

is given by:

4r = \sqrt{2}a

r = \dfrac{\sqrt{2}a}{4} = \dfrac{a}{2\sqrt{2}}

r = \dfrac{400}{2\sqrt{2}} = 100\sqrt{2}

r = 100 \times 1.414 = 141.4

pm

Answer:

141.4

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