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Niobium has a BCC crystal structure, an atomic radius of 0.143 nm and an atomic weight of 92.91 g/mol. Calculate the theoretical density for Nb. i g/cm³

Niobium has a BCC crystal structure, an atomic radius of 0.143 nm and an atomic weight of 92.91 g/mol. Calculate the theoretical density for Nb. i g/cm³

Elements Of Electromagnetics
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 Description **Niobium** has a **BCC (Body-Centered Cubic) crystal structure**, an **atomic radius** of **0.143 nm**, and an **atomic weight** of **92.91 g/mol**. Calculate the **theoretical density** for Nb. ### Solution To calculate the theoretical density of a BCC structure, use the formula: \[ \text{Density} = \frac{n \cdot A}{V_c \cdot N_A} \] Where: - \( n \) = Number of atoms per unit cell (for BCC, \( n = 2 \)) - \( A \) = Atomic weight - \( V_c \) = Volume of the unit cell - \( N_A \) = Avogadro's number (\(6.022 \times 10^{23} \, \text{atoms/mol}\)) #### Step-by-step: 1. **Calculate the Volume of the Unit Cell (\(V_c\)):** - In BCC, the relation between atomic radius (\( r \)) and lattice parameter (\( a \)) is: \[ a = \frac{4r}{\sqrt{3}} \] - Substitute \( r = 0.143 \, \text{nm} \): \[ a = \frac{4 \times 0.143}{\sqrt{3}} \, \text{nm} \] \[ a \approx 0.3294 \, \text{nm} \approx 3.294 \, \text{Å} \] - **Volume of unit cell (\(V_c\)):** \[ V_c = a^3 = (3.294 \, \text{Å})^3 = 35.68 \, \text{Å}^3 \] - Convert to \( \text{cm}^3 \) (\(1 \, \text{Å}^3 = 10^{-24} \, \text{cm}^3\)): \[ V_c = 35.68 \times 10^{-24} \, \text{cm}^3 \] 2. **Substitute the Values into the Density Formula:** \[ \text{Density} = \frac{2 \times
Transcribed Image Text:### Problem Description **Niobium** has a **BCC (Body-Centered Cubic) crystal structure**, an **atomic radius** of **0.143 nm**, and an **atomic weight** of **92.91 g/mol**. Calculate the **theoretical density** for Nb. ### Solution To calculate the theoretical density of a BCC structure, use the formula: \[ \text{Density} = \frac{n \cdot A}{V_c \cdot N_A} \] Where: - \( n \) = Number of atoms per unit cell (for BCC, \( n = 2 \)) - \( A \) = Atomic weight - \( V_c \) = Volume of the unit cell - \( N_A \) = Avogadro's number (\(6.022 \times 10^{23} \, \text{atoms/mol}\)) #### Step-by-step: 1. **Calculate the Volume of the Unit Cell (\(V_c\)):** - In BCC, the relation between atomic radius (\( r \)) and lattice parameter (\( a \)) is: \[ a = \frac{4r}{\sqrt{3}} \] - Substitute \( r = 0.143 \, \text{nm} \): \[ a = \frac{4 \times 0.143}{\sqrt{3}} \, \text{nm} \] \[ a \approx 0.3294 \, \text{nm} \approx 3.294 \, \text{Å} \] - **Volume of unit cell (\(V_c\)):** \[ V_c = a^3 = (3.294 \, \text{Å})^3 = 35.68 \, \text{Å}^3 \] - Convert to \( \text{cm}^3 \) (\(1 \, \text{Å}^3 = 10^{-24} \, \text{cm}^3\)): \[ V_c = 35.68 \times 10^{-24} \, \text{cm}^3 \] 2. **Substitute the Values into the Density Formula:** \[ \text{Density} = \frac{2 \times
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