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MHT CET 202621 AprilEvening Shift
Exam
MHT CET
Year
2026
Date
21 April
Shift
Evening Shift
Subject
Chemistry
Chapter
Solid State

MHT CET 2026 · 21 April · Evening Shift · Chemistry · Solid State

Calculate the atomic radius of an element having edge length 3.61 ×...

Calculate the atomic radius of an element having edge length 3.61×10−8 cm3.61 \times 10^{-8}\ \text{cm} for fcc structure.

Options

  1. A1.43×10−8 cm1.43 \times 10^{-8}\ \text{cm}
  2. B1.28×10−8 cm1.28 \times 10^{-8}\ \text{cm}
  3. C0.90×10−8 cm0.90 \times 10^{-8}\ \text{cm}
  4. D1.02×10−8 cm1.02 \times 10^{-8}\ \text{cm}

Correct answer

B. 1.28×10−8 cm1.28 \times 10^{-8}\ \text{cm}

Step-by-step solution

For a face-centered cubic (fcc) unit cell, the atoms touch each other along the face diagonal.

The relationship between the edge length aa and the atomic radius rr is given by:

4r=2a4r = \sqrt{2}a

r=a22r = \dfrac{a}{2\sqrt{2}}

Given a=3.61×10−8 cma = 3.61 \times 10^{-8}\ \text{cm}.

Substituting the value of aa:

r=3.61×10−82×1.414r = \dfrac{3.61 \times 10^{-8}}{2 \times 1.414}

r=3.61×10−82.828r = \dfrac{3.61 \times 10^{-8}}{2.828}

r=1.276×10−8 cm≈1.28×10−8 cmr = 1.276 \times 10^{-8}\ \text{cm} \approx 1.28 \times 10^{-8}\ \text{cm}

Answer: 1.28×10−8 cm1.28 \times 10^{-8}\ \text{cm}
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