Molybdenum, a BCC metal has an atomic radius of 0.136 nm. (a) Calculate the d-spacing for the (211) plane.
Molybdenum, a BCC metal has an atomic radius of 0.136 nm. (a) Calculate the d-spacing for the (211) plane.
Elements Of Electromagnetics
7th Edition
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
ChapterMA: Math Assessment
Section: Chapter Questions
Problem 1.1MA
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![**Problem Statement:**
Molybdenum, a BCC metal has an atomic radius of 0.136 nm.
**Question:**
(a) Calculate the d-spacing for the (211) plane.
**Important Concepts:**
- **Body-Centered Cubic (BCC) Structure:** In a BCC structure, the unit cell consists of one atom at each corner of the cube and one atom at the center of the cube. The atomic radius \( r \) is related to the lattice parameter \( a \) by the formula \( a = \frac{4r}{\sqrt{3}} \).
- **d-Spacing Calculation:** The d-spacing (interplanar spacing) for BCC crystals can be calculated using Bragg's Law and the Miller indices (hkl) of the plane. For a cubic structure:
\[
d_{hkl} = \frac{a}{\sqrt{h^2 + k^2 + l^2}}
\]
**Step-by-Step Solution:**
1. **Calculate the Lattice Parameter \( a \):**
Given the atomic radius \( r = 0.136 \) nm, the lattice parameter \( a \) can be determined using the equation:
\[
a = \frac{4r}{\sqrt{3}}
\]
Substituting \( r = 0.136 \) nm:
\[
a = \frac{4 \times 0.136}{\sqrt{3}} \approx 0.314 \text{ nm}
\]
2. **Calculate the d-Spacing \( d_{211} \) for the (211) Plane:**
With the lattice parameter \( a \) and the Miller indices \( h = 2 \), \( k = 1 \), \( l = 1 \), we use:
\[
d_{211} = \frac{a}{\sqrt{2^2 + 1^2 + 1^2}} = \frac{a}{\sqrt{6}}
\]
Substituting \( a = 0.314 \) nm:
\[
d_{211} = \frac{0.314}{\sqrt{6}} \approx 0.128 \text{ nm}
\]
Thus, the d-spacing for the (211) plane in Molybdenum is approximately](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fe4d256a5-cbd6-4aee-ae86-020aabe91708%2Fb54a674f-afb9-42a0-b4b6-f71fda86c57f%2Fbp09ku_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Problem Statement:**
Molybdenum, a BCC metal has an atomic radius of 0.136 nm.
**Question:**
(a) Calculate the d-spacing for the (211) plane.
**Important Concepts:**
- **Body-Centered Cubic (BCC) Structure:** In a BCC structure, the unit cell consists of one atom at each corner of the cube and one atom at the center of the cube. The atomic radius \( r \) is related to the lattice parameter \( a \) by the formula \( a = \frac{4r}{\sqrt{3}} \).
- **d-Spacing Calculation:** The d-spacing (interplanar spacing) for BCC crystals can be calculated using Bragg's Law and the Miller indices (hkl) of the plane. For a cubic structure:
\[
d_{hkl} = \frac{a}{\sqrt{h^2 + k^2 + l^2}}
\]
**Step-by-Step Solution:**
1. **Calculate the Lattice Parameter \( a \):**
Given the atomic radius \( r = 0.136 \) nm, the lattice parameter \( a \) can be determined using the equation:
\[
a = \frac{4r}{\sqrt{3}}
\]
Substituting \( r = 0.136 \) nm:
\[
a = \frac{4 \times 0.136}{\sqrt{3}} \approx 0.314 \text{ nm}
\]
2. **Calculate the d-Spacing \( d_{211} \) for the (211) Plane:**
With the lattice parameter \( a \) and the Miller indices \( h = 2 \), \( k = 1 \), \( l = 1 \), we use:
\[
d_{211} = \frac{a}{\sqrt{2^2 + 1^2 + 1^2}} = \frac{a}{\sqrt{6}}
\]
Substituting \( a = 0.314 \) nm:
\[
d_{211} = \frac{0.314}{\sqrt{6}} \approx 0.128 \text{ nm}
\]
Thus, the d-spacing for the (211) plane in Molybdenum is approximately
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