Question

A material having FCC structure has inter atomic distance as 3° A. which of the following are true?
1. Effective no. of atoms = 4/cell
2. APF is 74%
3. Atoms per sq. mm is [100] is 1.11 × 10¹³
4. All of these

Answer 1

Correct Answer - Option :

Concept:

FCC has 4 effective atoms / unit cell. With APF of 74%

For FCC,

\(a = \frac{{4r}}{{\sqrt 2 }}\)

Where a is plane length [100]

Number of atoms in 100 \(= 1 + 4 \times \frac{1}{4} = 2\)

Calculation:

2r = 3° A ⇒ r = 1.50° A

\(a = \frac{{4r}}{{\sqrt 2 }} = 2\sqrt 2 \;r\)

∴ a = 4.2426 × 10^-7 mm

Now,

Atom/square mm \( = \frac{{2\; \times \;1}}{{{a^2}}} = \frac{{2\; \times \;{{10}^{14}}}}{{{{\left( {4.2426} \right)}^2}}}\)

∴ Atom/square mm = 1.11 × 10¹³ atoms.