Question

A domain in ferromagnetic iron is in the form of a cube of side length 2 $\mu$m then the number of iron atoms in the domain are (Molecular mass of iron = 55 g mol-1 and density = 7.92 g cm-3)

• A. 6.92 × 1012 atoms
• B. 6.92 × 1011 atoms
• C. 6.92 × 1010 atoms
• D. 6.92 × 1013 atoms

Answer 1

Answer: (B)

6.92 × 1011 atoms

Answer 2

: The volume of the cubic domain

$= 8 \times 10 ^ { - 18 } \mathrm {~m} ^ { 3 } = 8 \times 10 ^ { - 12 } \mathrm { Cm } ^ { 3 }$

and mass = volume × density

$= 8 \times 10 ^ { - 12 } \mathrm {~cm} ^ { 3 } \times 7.9 \mathrm {~g} \mathrm {~cm} ^ { - 3 }$

$= 63.2 \times 10 ^ { - 12 } \mathrm {~g}$

Now the avaradro number $\left( 6.023 \times 10 ^ { 23 } \right)$of iron atoms have a mass of 55 g. Hence the number of atoms in the domain are

$\mathrm { N } = \frac { 63.2 \times 10 ^ { - 12 } \times 6.023 \times 10 ^ { 23 } } { 55 }$

$= 6.92 \times 10 ^ { 11 }$ atoms