(a) What is the distance between second nearest neighbor atoms in vanadium? Express your answer in cm.
(b) What is the interplanar spacing between (110) planes in vanadium? Express your answer in cm.
(c) At what angle must (110) planes of vanadium be exposed to a collimated beam of X-rays in order to obtain diffraction by monochromatic chromium K-alpha radiation? Express your answer in degrees.
a) 3.03⋅10−8
something wrong ..
b) 2.14*10^-8
pls. c
a)3.03*10^-8
b) 2.14*10^-8
any ideas for c)?
c) 33.2
3.2e-8
Thank you
a)3.03*10^-8
b) 2.14*10^-8
c) 33.2
a is qrong can you please teach me how to do this
To answer these questions, we need to understand the crystal structure of vanadium and use some basic formulas. Vanadium crystallizes in a body-centered cubic (BCC) structure.
(a) To find the distance between second nearest neighbor atoms in vanadium, we can use the formula:
d = (4/3) * (a / sqrt(3))
where d is the distance between the atoms and a is the lattice parameter. For vanadium, the lattice parameter is approximately 3.02 Å (angstroms). To convert this to centimeters, we can use the conversion factor 1 Å = 1 × 10^-8 cm.
So, substitute the values into the formula:
d = (4/3) * (3.02 Å / sqrt(3)) * (1 × 10^-8 cm / 1 Å)
Simplifying the expression:
d = (4/3) * (3.02 × 10^-8 cm) / sqrt(3)
Calculating the expression:
d ≈ 1.325 × 10^-8 cm
Therefore, the distance between second nearest neighbor atoms in vanadium is approximately 1.325 × 10^-8 cm.
(b) To find the interplanar spacing between (110) planes in vanadium, we can use the formula:
d = a / sqrt(h^2 + k^2 + l^2)
where d is the interplanar spacing and (hkl) represents the Miller indices of the crystallographic plane. In this case, (110) represents the (hkl) values of (1,1,0) for the (110) plane.
Substitute the values into the formula:
d = 3.02 Å / sqrt((1^2) + (1^2) + (0^2))
Simplifying the expression:
d = 3.02 Å / sqrt(2)
Calculating the expression:
d ≈ 2.137 Å
To convert this to centimeters, use the conversion factor mentioned earlier:
d = 2.137 Å * (1 × 10^-8 cm / 1 Å)
Calculating the expression:
d ≈ 2.137 × 10^-8 cm
Therefore, the interplanar spacing between (110) planes in vanadium is approximately 2.137 × 10^-8 cm.
(c) To find the angle at which the (110) planes of vanadium must be exposed to a collimated beam of X-rays for diffraction by monochromatic chromium K-alpha radiation, we can use Bragg's Law:
nλ = 2dsinθ
where n is the order of diffraction (usually 1 for primary diffraction), λ is the wavelength of the X-rays, d is the interplanar spacing (in this case, for (110) planes), and θ is the angle of incidence.
First, we need to find the wavelength of chromium K-alpha radiation. The K-alpha transition in chromium corresponds to the emission of characteristic X-rays with a wavelength of approximately 2.29 Å.
Then, substitute the values into Bragg's Law:
1 * (2.29 Å) = 2 * (d) * sin(θ)
Since we are looking for the angle θ, we need to rearrange the formula:
sin(θ) = (1 * (2.29 Å)) / (2 * (d))
Substitute the values:
sin(θ) = (1 * (2.29 Å)) / (2 * (2.137 Å))
Convert the Ångstrom units to centimeters using the conversion factor mentioned earlier.
sin(θ) = (1 * (2.29 × 10^-8 cm)) / (2 * (2.137 × 10^-8 cm))
Calculate the expression:
sin(θ) ≈ 0.536
To find the angle θ, we can take the inverse sine (arcsin) of 0.536:
θ ≈ arcsin(0.536)
Using a calculator or trigonometric tables, we find:
θ ≈ 32.86 degrees
Therefore, the (110) planes of vanadium must be exposed to a collimated beam of X-rays at an angle of approximately 32.86 degrees to obtain diffraction by monochromatic chromium K-alpha radiation.