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Chapter 3: Problem 59

(a) Derive linear density expressions for BCC [110] and [111] directions in terms of the atomic radius \(R\). (b) Compute and compare linear density values for these same two directions for iron \((\mathrm{Fe})\).

Short Answer

Expert verified
Answer: The linear density for iron (Fe) in the [110] direction is approximately 2.52 nm\(^{-1}\), while in the [111] direction it is approximately 4.03 nm\(^{-1}\). The [111] direction has a higher linear density compared to the [110] direction.

Step by step solution

01

Determine the cell edge and body diagonal lengths

In a BCC unit cell, the cell edge length 'a' can be related to the atomic radius R using Pythagorean theorem. The body diagonal of the unit cell connects two vertices along [111] direction passing through the central atom. For the cell edge length 'a': $$a^2 + a^2 = (4R)^2$$ Solving for 'a', we get: $$a = 2R\sqrt{2}$$ For the body diagonal length 'd': $$a^2 + a^2 + a^2 = d^2$$ Substitute the value of 'a', and solve for 'd': $$d = 4R$$
02

Define the number of atoms along [110] and [111] directions

For the [110] direction, the linear density (LD) can be calculated by considering the number of atoms (_n_) distributed along the face diagonal. Since the face diagonal has one atom at each end and shares with other unit cells, we get: $$n_{110} = \frac{1}{2} + \frac{1}{2}$$ For the [111] direction, the linear density (LD) can be calculated by considering the number of atoms (_n_) distributed along the body diagonal. The body diagonal has one atom at each end and one in the middle, as it passes through the central atom: $$n_{111} = \frac{1}{2} + \frac{1}{2} + 1$$
03

Compute the linear densities for [110] and [111] directions

Now we can calculate the linear density for each direction by dividing the number of atoms along each direction by the edge length (a) for [110] and body diagonal length (d) for [111]. Linear Density for the [110] direction is: $$LD_{110} = \frac{n_{110}}{a} = \frac{1}{2R\sqrt{2}}$$ Linear Density for the [111] direction is: $$LD_{111} = \frac{n_{111}}{d} = \frac{2}{4R}$$
04

Calculate linear densities for iron (Fe)

Now, we will use the atomic radius of iron (R = 0.124 nm) to calculate the linear densities in the [110] and [111] directions. Linear Density for [110] direction in iron (Fe): $$LD_{110}^{Fe} = \frac{1}{(2)(0.124\mathrm{nm})\sqrt{2}} \approx 2.52 \mathrm{nm}^{-1}$$ Linear Density for [111] direction in iron (Fe): $$LD_{111}^{Fe} = \frac{2}{(4)(0.124\mathrm{nm})} \approx 4.03 \mathrm{nm}^{-1}$$ Comparing the linear density values for iron in [110] and [111] directions, we can conclude that the [111] direction has a higher linear density (4.03 nm\(^{-1}\)) compared to the [110] direction (2.52 nm\(^{-1}\)).

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Most popular questions from this chapter

The metal rhodium (Rh) has an FCC crystal structure. If the angle of diffraction for the (311) set of planes occurs at \(36.12^{\circ}\) (first-order reflection) when monochromatic \(\mathrm{x}\)-radiation having a wavelength of \(0.0711 \mathrm{~nm}\) is used, compute the following: (a) The interplanar spacing for this set of planes (b) The atomic radius for a Rh atom

Sketch an orthorhombic unit cell, and within that cell indicate locations of the \(0 \frac{1}{2} 1\) and \(\frac{1}{3} \frac{1}{4}\) point coordinates.

Determine the expected diffraction angle for the first-order reflection from the (310) set of planes for BCC chromium (Cr) when monochromatic radiation of wavelength \(0.0711 \mathrm{~nm}\) is used.

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(a) Derive linear density expressions for FCC [100] and [111] directions in terms of the atomic radius \(R\). (b) Compute and compare linear density values for these same two directions for copper (Cu).
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