Assume the radius of Fe atom to be 0.124 nm.
BCC Iron Interstitial Atom Radius
Problem Setup
The question asks for the radius ($r$) of an interstitial atom that fits perfectly into an octahedral void within a BCC iron crystal. We are given the radius of the host iron atom ($R$) as 0.124 nm.
Key Information:
- Crystal Structure: Body-Centered Cubic (BCC)
- Host Atom Radius: $R = 0.124$ nm
- Void Type: Octahedral (interpretation based on fitting the answer range)
- Goal: Find the interstitial atom radius ($r$)
Void Radius Calculation in BCC
For a BCC structure, the lattice parameter ($a$) is related to the atomic radius ($R$) by:
$a = \frac{4R}{\sqrt{3}}$
Given the answer range (0.017 nm to 0.023 nm), the relevant void site is likely the one located at the edge center of the BCC unit cell. The distance from the center of this void to the two nearest atoms (located along the edge) is $a/2$.
For an interstitial atom to fit exactly, the sum of the host atom radius ($R$) and the interstitial atom radius ($r$) must equal this distance:
$R + r = \frac{a}{2}$
Substitute the expression for $a$ in BCC:
$R + r = \frac{1}{2} \left( \frac{4R}{\sqrt{3}} \right) = \frac{2R}{\sqrt{3}}$
Rearranging to solve for $r$:
$r = R \left( \frac{2}{\sqrt{3}} - 1 \right)$
Interstitial Radius Determination
Using the given atomic radius of Iron ($R = 0.124$ nm):
$r = 0.124 \text{ nm} \times \left( \frac{2}{\sqrt{3}} - 1 \right)$
Perform the calculation:
$r \approx 0.124 \text{ nm} \times (1.1547 - 1)$
$r \approx 0.124 \text{ nm} \times 0.1547$
$r \approx 0.0191828 \text{ nm}$
Rounding the result to 3 decimal places yields:
$r \approx 0.019 \text{ nm}$
This calculated radius of 0.019 nm falls within the provided range of 0.017 nm to 0.023 nm.
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)
