Theoretical density = (n)(AW) (volume of cell)(N) 4r BCC unit cell N = Ne.URY Ne-/ FCC unit cell 4r V2 a=2r Simple cubic k = 1.38×10 J/(atom-K)= 8.62×10% eV/(atom- K) R=8.314 J/mol-K) = 1.986 cal/mol-K) K= °C + 273 NA = 6,023×10 I nm = 1×10 m, 1 = 1x10m 5. Calculate the planar density (in atoms/nm²) for (110) in FCC gold, which has a lattice parameter, ao, of 0.40788 nm. (14 points) 6. Below are listed the atomic weight, density and atomic radius for a hypothetical element. Determine whether its crystal structure is FCC, BCC or SC, justify your answer and show why it is not the other choices. (24 points) Alloy Atomic Weight (g/mol) Mass Density (g/cm3) Radius (8) Type of Cubic Structure A 132.91 1.892 2.654
The Correct Answer and Explanation is :
Let’s begin by solving the first problem: calculating the planar density for the (110) plane in an FCC (Face-Centered Cubic) gold crystal.
Planar Density Calculation for (110) Plane in FCC Gold
- Lattice Parameter: Given, ( a_0 = 0.40788 \, \text{nm} ).
- Face-Centered Cubic (FCC) Structure: The atoms in an FCC crystal are located at each corner and the centers of all the faces of the cube.
- (110) Plane in FCC: This plane cuts through the center of the unit cell edge and passes through the adjacent face center.
- Atoms per unit area on the (110) plane: In FCC, for the (110) plane, the pattern repeats after one unit cell length along one axis and half the diagonal of the face along the other. Therefore:
- Along the length (( a_0 )), there is 1 atom per ( a_0 ).
- Along the diagonal of the face (( \sqrt{2}a_0 )), half belongs to another repeating unit, so the real spacing for the repeating unit along the diagonal is ( \frac{\sqrt{2}a_0}{2} ).
- Area of repeating unit on (110) plane:
[
\text{Area} = a_0 \times \left(\frac{\sqrt{2}a_0}{2}\right) = \frac{\sqrt{2}a_0^2}{2}
] - Number of atoms per repeating unit: Each corner atom contributes ( \frac{1}{4} ) of an atom to the plane, and each face-centered atom contributes ( \frac{1}{2} ). Thus, there are ( 2 \times \frac{1}{4} + 1 \times \frac{1}{2} = 1 ) atom per repeating unit.
- Planar Density (( \rho_{\text{planar}} )):
[
\rho_{\text{planar}} = \frac{1 \text{ atom}}{\frac{\sqrt{2}a_0^2}{2}} = \frac{2}{\sqrt{2}a_0^2} = \frac{2}{\sqrt{2} \times (0.40788 \text{ nm})^2}
]
[
\rho_{\text{planar}} = \frac{2}{\sqrt{2} \times 0.166354} \approx \frac{2}{0.467} \text{ atoms/nm}^2 \approx 4.283 \text{ atoms/nm}^2
]
So, the planar density of the (110) plane in FCC gold is approximately ( 4.283 \, \text{atoms/nm}^2 ).
Problem 2: Determine the Crystal Structure of Hypothetical Element ‘A’
Given data:
- Atomic Weight (AW): 132.91 g/mol
- Density (( \rho )): 1.892 g/cm³
- Atomic Radius (( r )): 2.654 Å
We can use the formula for theoretical density of cubic crystals:
[
\rho = \frac{n \times \text{AW}}{V \times N_A}
]
Where:
- ( n ) is the number of atoms per unit cell (1 for SC, 2 for BCC, 4 for FCC).
- ( V ) is the volume of the unit cell (( a^3 ), ( a = 2r ) for SC, ( a = \frac{4r}{\sqrt{3}} ) for BCC, and ( a = \frac{4r}{\sqrt{2}} ) for FCC).
- ( N_A ) is Avogadro’s number ( = 6.022 \times 10^{23} \, \text{atoms/mol} ).
- Calculate ( a ):
- SC: ( a = 2 \times 2.654 \text{ Å} = 5.308 \text{ Å} )
- BCC: ( a = \frac{4 \times 2.654 \text{ Å}}{\sqrt{3}} \approx 6.125 \text{ Å} )
- FCC: ( a = \frac{4 \times 2.654 \text{ Å}}{\sqrt{2}} \approx 7.5 \text{ Å} )
- Calculate theoretical densities for each structure and compare to actual density (1.892 g/cm³).
This approach will allow us to identify which structure fits the given properties by matching the calculated densities to the provided density. If the calculated density matches closely with the actual density, that’s likely the correct structure. Each calculation will involve using the formula above and substituting the respective values for ( n ) and ( a ) based on the structure (SC, BCC, or FCC).
Would you like a detailed step-by-step calculation for each structure or proceed directly to generating an image based on this explanation?