Question

A domain in ferromagnetic iron is in the form of a cube of side length 1μm. Estimate the number of iron atoms in the domain. The molecular mass of iron is $55g.mole$ and its density is $7.9g/c{m}^3$. Assume that each iron atom has a dipole moment of $9.27\times {10}^{-24}A{m}^2$.

• A. $8.65\times {10}^{10}atoms$
• B. $10.65\times {10}^{10}atoms$
• C. $18.65\times {10}^{10}atoms$
• D. $20.75\times {10}^{10}atoms$

Answer 1

The correct option is A

$8.65\times {10}^{10}atoms$

Step 1: Given data

The molecular mass of iron $55g/mole$ and its density is $7.9g/c{m}^3$

The dipole moment of each iron atom is $9.27\times {10}^{-24}A{m}^2$

Step 2: Formula used

$Massofcube=volume\times density$

Step 3: Calculating the number of iron atoms

$$\begin{gathered} Massofcube=volume\times density \\ Massofcube={\left({10}^{-4}\right)}^3\times (7.9) \\ Massofcube=7.9\times {10}^{-12} \end{gathered}$$

An Avogadro number $(6.023\times {10}^{23})$ of iron atoms has mass $55g$

$$\begin{gathered} No.ofatomsinthecube=\frac{mass}{molecularmass}\times N_A[N_A=Avogadronumber] \\ No.ofatomsinthecube=\frac{7.9\times {10}^{-12}}{55}\times 6.023\times {10}^{23} \\ No.ofatomsinthecube=8.65\times {10}^{10}atoms \end{gathered}$$

The number of irons atoms in the domain is $8.65\times {10}^{10}atoms$.

Hence, option A is the correct option.