Q113 / 200 · MHT CET - Medical 2026 (22 April Shift 1) · Chemistry
If the volume of fcc unit cell is $6.4 \times 10^{-23}$ cm$^3$ find the radius of an atom in it.
- Correct answerA$141.4$ pm
- B$159.0$ pm
- C$180.2$ pm
- D$124.2$ pm
Solution
Volume of the unit cell, $V = a^3 = 6.4 \times 10^{-23}$ cm$^3$
$a^3 = 64 \times 10^{-24}$ cm$^3$
$a = (64 \times 10^{-24})^{1/3}$ cm $= 4 \times 10^{-8}$ cm
Converting to picometers:
$a = 4 \times 10^{-8} \times 10^{10}$ pm $= 400$ pm
For an fcc unit cell, the relation between edge length $a$ and radius $r$ is given by:
$4r = \sqrt{2}a$
$r = \dfrac{\sqrt{2}a}{4} = \dfrac{a}{2\sqrt{2}}$
$r = \dfrac{400}{2\sqrt{2}} = 100\sqrt{2}$ pm
$r = 100 \times 1.414 = 141.4$ pm
Answer: $141.4$ pm
$a^3 = 64 \times 10^{-24}$ cm$^3$
$a = (64 \times 10^{-24})^{1/3}$ cm $= 4 \times 10^{-8}$ cm
Converting to picometers:
$a = 4 \times 10^{-8} \times 10^{10}$ pm $= 400$ pm
For an fcc unit cell, the relation between edge length $a$ and radius $r$ is given by:
$4r = \sqrt{2}a$
$r = \dfrac{\sqrt{2}a}{4} = \dfrac{a}{2\sqrt{2}}$
$r = \dfrac{400}{2\sqrt{2}} = 100\sqrt{2}$ pm
$r = 100 \times 1.414 = 141.4$ pm
Answer: $141.4$ pm
Hence, the correct answer is option: 'A'.