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Chapter 3: Problem 54

(a) Derive planar density expressions for FCC (100) and (111) planes in terms of the atomic radius \(R\). (b) Compute and compare planar density values for these same two planes for nickel.

Short Answer

Expert verified
In a nickel FCC crystal structure, the planar densities for the (100) and (111) planes are approximately \(6.43\times10^{15}\) atoms/cm\(^2\) and \(3.52\times10^{15}\) atoms/cm\(^2\), respectively. Comparing these values, we find that the (100) plane has a higher planar density than the (111) plane in nickel.

Step by step solution

01

Part (a): Deriving Planar Density Expressions for FCC (100) and (111) Planes

Planar density is defined as the number of atoms per unit area of a certain plane in a crystal lattice, which can be calculated by taking the ratio of the number of atoms in the plane \(\textit{(N)}\) to the area of that plane \(\textit{(A)}\): Planar Density = \(\dfrac{N}{A}\) For an FCC crystal structure, the basis atoms are located at the corners and face centers of the unit cell. Let's derive planar density expressions for the (100) and (111) planes in terms of atomic radius R.
02

FCC (100) Plane

In the (100) plane, there are 4 atoms at the corners of the unit cell, and one atom at the center of each face. Each corner atom is shared by 4 adjacent unit cells and each face atom is shared by 2 adjacent unit cells. Number of atoms in the (100) plane, \(N_{100}\) = \(\dfrac{1}{4}(4)+\dfrac{1}{2}(4)=2\) The side length of the unit cell in FCC, \(a\) = \(2\sqrt{2}R\) Area of the unit cell, \(A_{100}\) = \(a^2 = 8R^2\) Planar Density \(\textit{(PD)}\) for the (100) plane, \(PD_{100}\) = \(\dfrac{N_{100}}{A_{100}} =\dfrac{2}{8R^2} =\dfrac{1}{4R^2}\)
03

FCC (111) Plane

In the (111) plane, there are 3 atoms at the corners and one atom at the center. Each corner atom is shared by 6 adjacent unit cells and the center atom is shared by 3 adjacent unit cells. Number of atoms in the (111) plane, \(N_{111}\) = \(\dfrac{1}{6}(3)+\dfrac{1}{3}(1)=1\) The distance between adjacent layers in the (111) plane, \(d\) = \(\dfrac{a}{\sqrt{3}}=\dfrac{2\sqrt{2}R}{\sqrt{3}}\) Area of the unit cell, \(A_{111}\) = \(\sqrt{3}d^2=2\sqrt{6}R^2\) Planar Density \(\textit{(PD)}\) for the (111) plane, \(PD_{111}\)= \(\dfrac{N_{111}}{A_{111}}=\dfrac{1}{2\sqrt{6}R^2}\)
04

Part (b): Computing and Comparing Planar Density Values for Nickel

Nickel has an FCC crystal structure and an atomic radius of \(1.245 \times 10^{-8}\) cm. We can use the expressions derived in part (a) to find and compare the planar density values for (100) and (111) planes in nickel. For the (100) plane, \(PD_{100}\) = \(\dfrac{1}{4R^2}=\dfrac{1}{4(1.245\times10^{-8} cm)^2}=6.43\times10^{15}\) atoms/cm\(^2\) For the (111) plane, \(PD_{111}\) = \(\dfrac{1}{2\sqrt{6}R^2}=\dfrac{1}{2\sqrt{6}(1.245\times10^{-8} cm)^2}=3.52\times10^{15}\) atoms/cm\(^2\) Comparing the planar density values, we can see that \(PD_{100}>PD_{111}\), which means the (100) plane has a higher planar density than the (111) plane in nickel.

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