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MHT CET 202625 AprilMorning Shift
Exam
MHT CET
Year
2026
Date
25 April
Shift
Morning Shift
Subject
Chemistry
Chapter
Solid State

MHT CET 2026 · 25 April · Morning Shift · Chemistry · Solid State

An element has bcc structure with atomic radius 200 pm. Find the...

An element has bcc structure with atomic radius 200 pm. Find the volume of unit cell.

Options

  1. A5.625×10−23 cm35.625 \times 10^{-23}\ \text{cm}^3
  2. B7.513×10−23 cm37.513 \times 10^{-23}\ \text{cm}^3
  3. C8.41×10−23 cm38.41 \times 10^{-23}\ \text{cm}^3
  4. D9.848×10−23 cm39.848 \times 10^{-23}\ \text{cm}^3

Correct answer

D. 9.848×10−23 cm39.848 \times 10^{-23}\ \text{cm}^3

Step-by-step solution

For a bcc unit cell, the relation between edge length aa and atomic radius rr is given by:

3a=4r\sqrt{3}a = 4r

a=4r3a = \dfrac{4r}{\sqrt{3}}

Substituting r=200 pm=2×10−8 cmr = 200 \text{ pm} = 2 \times 10^{-8} \text{ cm}:

a=4×2×10−81.732≈4.618×10−8 cma = \dfrac{4 \times 2 \times 10^{-8}}{1.732} \approx 4.618 \times 10^{-8} \text{ cm}

Volume of the unit cell, V=a3V = a^3

V=(4.618×10−8)3≈9.848×10−23 cm3V = (4.618 \times 10^{-8})^3 \approx 9.848 \times 10^{-23} \text{ cm}^3

Answer: 9.848×10−23 cm39.848 \times 10^{-23}\ \text{cm}^3
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