(round off to 1 decimal place).
(Given: Lattice parameter of iron at room temperature = 0.287 nm)
The problem asks for the magnetic moment per iron atom ($\mu_{atom}$) given the saturation magnetization ($M_s$) and the lattice parameter ($a$) of iron.
Calculating Volume of Unit Cell
Iron has a Body-Centered Cubic (BCC) structure at room temperature. The volume ($V$) of the cubic unit cell is calculated using the lattice parameter ($a$):
Given: $a = 0.287 \text{ nm} = 0.287 \times 10^{-9} \text{ m}$
Calculation: $V = a^3$ $V = (0.287 \times 10^{-9} \text{ m})^3$ $V = (0.287)^3 \times 10^{-27} \text{ m}^3$ $V \approx 0.02364 \times 10^{-27} \text{ m}^3$
Determining Atoms per Unit Cell
For a BCC structure, the number of atoms per unit cell ($n$) is 2.
Calculating Magnetic Moment per Atom
Saturation magnetization ($M_s$) represents the maximum magnetic moment per unit volume.
Given: $M_s = 1700 \text{ kA m}^{-1} = 1700 \times 10^3 \text{ A m}^{-1} = 1.7 \times 10^6 \text{ A m}^{-1}$
The relationship is $M_s = \frac{n \times \mu_{atom}}{V}$.
Rearranging to find $\mu_{atom}$:
$ \mu_{atom} = \frac{M_s \times V}{n} $Substituting the values:
$ \mu_{atom} = \frac{(1.7 \times 10^6 \text{ A m}^{-1}) \times (0.02364 \times 10^{-27} \text{ m}^3)}{2} $ $ \mu_{atom} = \frac{1.7 \times 0.02364}{2} \times 10^{(6 - 27)} \text{ A m}^2 $ $ \mu_{atom} = \frac{0.040188}{2} \times 10^{-21} \text{ A m}^2 $ $ \mu_{atom} \approx 0.020094 \times 10^{-21} \text{ A m}^2 $Final Answer Formatting
Convert the result to the format $X \times 10^{-23} \text{ A m}^2$ and round to 1 decimal place:
$ \mu_{atom} \approx 0.020094 \times 10^{-21} \text{ A m}^2 = 2.0094 \times 10^{-23} \text{ A m}^2 $Rounding to 1 decimal place gives $2.0 \times 10^{-23} \text{ A m}^2$.
The value to fill in the blank is 2.0.
Important Questions from Crystal Structure Density Atomic Packing Factor
- Match the crystal systems in Column I with the corresponding axial lengths (a, b, c) and interaxial angles ($\alpha$, $\beta$, $\gamma$) provided in Column II
Column I Column II (P) Tetragonal (1) $a \neq b \neq c$, $\alpha = \beta = \gamma = 90^\circ$ (Q) Rhombohedral (2) $a = b \neq c$, $\alpha = \beta = \gamma = 90^\circ$ (R) Orthorhombic (3) $a \neq b \neq c$, $\alpha = \gamma = 90^\circ \neq \beta$ (S) Monoclinic (4) $a = b = c$, $\alpha = \beta = \gamma \neq 90^\circ$ - The coordination number for an octahedral site in pure copper is __________.
The lattice parameter of face-centered cubic iron ($\gamma$-Fe) is 0.3571 nm. The radius (in nm) of the octahedral void in $\gamma$-Fe is _______________
- For a bcc metal the ratio of the surface energy per unit area of the (100) plane to that of the (110) plane is ________
- Pure iron transforms from body centered cubic (BCC) to face centered cubic (FCC) crystal structure at $912 \text{ °C}$. If the lattice parameter of the BCC phase is $0.293 \text{ nm}$ and that of the FCC phase is $0.363 \text{ nm}$, the associated volume change is ________ (in % to one decimal place)
