MHT CET · Chemistry · Solid State
Silver crystallises in fcc structure, if edge length of unit cell is \(316.5 \mathrm{pm}\). What is the radius of silver atom?
- A \(137.04 \mathrm{pm}\)
- B \(111.91 \mathrm{pm}\)
- C \(121.91 \mathrm{pm}\)
- D \(158.25 \mathrm{pm}\)
Answer & Solution
Correct Answer
(D) \(158.25 \mathrm{pm}\)
Step-by-step Solution
Detailed explanation
(B)
\(\mathrm{a}=316.5 \mathrm{pm}\)
For fcc unit cell, \(r=\frac{a}{2 \sqrt{2}}\)
\(\therefore r=\frac{316.5 \mathrm{pm}}{2 \times 1.414}=111.91 \mathrm{pm}\)
\(\mathrm{a}=316.5 \mathrm{pm}\)
For fcc unit cell, \(r=\frac{a}{2 \sqrt{2}}\)
\(\therefore r=\frac{316.5 \mathrm{pm}}{2 \times 1.414}=111.91 \mathrm{pm}\)
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