The silicon wafer is coated with a layer of metallic aluminum which acts as an electrical contact. The x-ray diffraction pattern of aluminum is measured in a diffractometer with Fe Kα radiation. At what angle, θ, do you expect to observe the first reflection of aluminum, i.e., the reflection at the lowest angle? Express your answer in degrees.
Please help!!
13.2
I got 24.5
They are BOTH wrong.
wrong
i found 8.87
8.87 is right. THANKS!
To determine the angle at which the first reflection of aluminum occurs, we need to use the Bragg's law equation for x-ray diffraction. The equation is as follows:
nλ = 2dsin(θ)
Where:
- n is the order of the reflection (in this case, n = 1)
- λ is the wavelength of the x-ray radiation (Fe Kα radiation has a wavelength of approximately 1.937 Å)
- d is the spacing between atomic planes in the crystal lattice
- θ is the angle at which the reflection occurs
For aluminum, we need to find the spacing between its atomic planes. The Miller indices of aluminum's most common crystal structure (face-centered cubic or FCC) are (111). The interplanar spacing, d, for an FCC crystal structure can be calculated using the following equation:
d = a / sqrt(h^2 + k^2 + l^2),
Where:
- a is the lattice constant, which represents the distance between neighboring atoms in the crystal lattice (for aluminum, a = 4.050 Å)
- h, k, and l are the Miller indices of the desired plane (in this case, h = k = l = 1)
Plugging these values into the equation:
d = 4.050 Å / sqrt(1^2 + 1^2 + 1^2)
≈ 2.460 Å
Now we can substitute the values we have into Bragg's law and solve for the angle θ:
1.937 Å = 2 * 2.460 Å * sin(θ)
sin(θ) = (1.937 Å) / (2 * 2.460 Å)
≈ 0.3947
To find the angle θ, we can take the inverse sine (sin^(-1)) of 0.3947:
θ ≈ sin^(-1)(0.3947)
≈ 23.09°
Therefore, we expect to observe the first reflection of aluminum at an angle of approximately 23.09 degrees.