You are conducting experiments with your x-ray diffractometer.
(a) A specimen of molybdenum is exposed to a beam of monochromatic x-rays of wavelength set by the Kα line of silver. Calculate the value of the smallest Bragg angle, θhkl, at which you can expect to observe reflections from the molybdenum specimen.
DATA:
λKα of Mo = 0.721 Å
lattice constant of Mo, a = 3.15 Å
λKα of Ag = 0.561 Å
lattice constant of Ag, a = 4.09 Å
unanswered
(b) If you wish to repeat the experiment described in part (a) but using an electron diffractometer, across what vale of potential difference must electrons be accelerated from rest in order to give the identical value for the smallest Bragg angle, θhkl?
unanswered
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a) 7.23
b) 475 ev
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(a) To calculate the value of the smallest Bragg angle, θhkl, at which you can expect to observe reflections from the molybdenum specimen, you need to use the Bragg's Law equation:
nλ = 2dsinθ
Where:
n is the order of the reflection (usually 1)
λ is the wavelength of the incident x-rays
d is the interplanar spacing of the crystallographic planes
θ is the angle of incidence (known as the Bragg angle)
First, we need to determine the interplanar spacing, d, for the (hkl) plane in molybdenum. For cubic crystals like molybdenum, the interplanar spacing is given by:
d = a/√(h^2 + k^2 + l^2)
Given that the lattice constant of molybdenum (a) is 3.15 Å, and we're interested in the (hkl) plane, we can calculate the interplanar spacing.
For the (hkl) plane of molybdenum, let's assume the indices are (h, k, l). In this case, we'll consider the (110) plane since it is commonly used for molybdenum. Therefore, (h, k, l) = (1, 1, 0).
Using the formula for the interplanar spacing, we have:
d = 3.15 Å / √(1^2 + 1^2 + 0^2) = 3.15 Å / √2 = 2.23 Å
Next, we can calculate the smallest Bragg angle, θhkl, using the Bragg's Law equation. The wavelength of the incident x-rays can be obtained from the Kα line of silver, as the Kα line of molybdenum is not provided in the data.
Given that λKα of Ag = 0.561 Å, we can use it as the wavelength for the incident x-rays. Plugging the values into the Bragg's Law equation:
2dsinθ = nλ
2 * 2.23 Å * sinθ = λKα of Ag (0.561 Å)
Simplifying the equation, we get:
sinθ = (λKα of Ag / (2d))
sinθ = (0.561 Å / (2 * 2.23 Å))
sinθ ≈ 0.1259
To find the smallest Bragg angle, θhkl, we can take the inverse sine of the value:
θhkl ≈ sin^(-1)(0.1259)
θhkl ≈ 7.22°
Therefore, the value of the smallest Bragg angle, θhkl, at which you can expect to observe reflections from the molybdenum specimen is approximately 7.22°.
(b) To repeat the experiment using an electron diffractometer and achieve the identical value for the smallest Bragg angle, θhkl, we need to relate the behavior of electrons to that of x-rays in diffraction.
In electron diffraction, the de Broglie wavelength, λ, is given by the de Broglie equation:
λ = h / p
Where λ is the wavelength, h is the Planck's constant, and p is the momentum of the electrons.
The momentum of electrons can be calculated using the kinetic energy, KE, and mass, m, of an electron. Since the electrons are accelerated from rest, the kinetic energy is equal to the potential energy, PE. Therefore:
PE = KE = qV
Where q is the charge of an electron, and V is the potential difference across which the electrons are accelerated.
To get the identical value for the smallest Bragg angle, θhkl, in the electron diffraction, we need to match the de Broglie wavelength with the x-ray wavelength used in part (a).
Using the de Broglie equation, we have:
λ = h / p
Since h is a constant, the momentum of electrons, p, needs to be adjusted by choosing an appropriate potential difference, V, so that the de Broglie wavelength λ is equal to the x-ray wavelength used in part (a).
Therefore, we have:
λe = λx
h / pe = λx
Since the de Broglie wavelength of electrons (λe) needs to be matched with the x-ray wavelength (λx), we can rearrange the equation:
pe = (h / λx)
Now, substituting the wavelength of the x-rays, given by λKα of Ag (0.561 Å), we can solve for the required momentum of electrons:
pe = (h / λKα of Ag)
pe = (h / 0.561 Å)
Finally, to find the potential difference, V, needed to accelerate the electrons from rest to achieve the identical smallest Bragg angle, θhkl, we use the energy-momentum relationship:
pe = √(2mE) = √(2mqV)
Where E is the total energy of the electron, m is the mass of the electron, q is the charge of the electron, and V is the potential difference.
Solving for V:
√(2mqV) = (h / 0.561 Å)
V = (h^2 / (2mq * (0.561 Å)^2))
Using the known values for h, m (mass of an electron), q, and the x-ray wavelength (λKα of Ag = 0.561 Å), you can substitute these values into the equation to calculate the potential difference, V.
Please note that without the precise values for h, m (mass of an electron), and q, a specific numerical answer cannot be provided.