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Q1. The utilized reflecting plane of a lithium fluoride (LiF) analyzing
crystal has a interplanner distance of 2.5 Å. Calculate the wavelength of
the 2nd order diffracted line which has a glancing angle of 60°.
Ans. We know 2 dhkl sin(θhkl) = n λ, n=2 in this problem, (hkl) are
the Miller indices of the utilized plane, dhkl = 2.5 Å and θhkl = 60°
λ = (2 × 2.5 Å × sin(60 °)) / (2) = 2.5 Å × sin(60 °)
= 2.5 × √3 / 2 Å = 2.165 Å
Q2. Calculate the angle at which (a) first order reflection and (b) second
order reflection will occur in a x-ray spectrometer, when x-ray of
wavelength 1.54 Å are diffracted by the atoms of a crystal given that the
interplanner distance of 4.04 Å.
Ans. We know 2 dhkl sin(θhkl) = n λ,
(a) n=1 in this problem, (hkl) are the Miller indices of the plane, dhkl =
4.04 Å and λ = 1.54 Å , then θhkl = ?
Sin(θhkl ) = (1 × 1.54 Å) / (2 × 4.04 Å) = 0.191
So, θhkl = sin-1(0.191) = 11.0°
(b) n=2 in this problem, (hkl) are the Miller indices of the plane, dhkl = 4.04
Å and λ = 1.54 Å , then θhkl = ?
Sin(θhkl ) = (2 × 1.54 Å) / (2 × 4.04 Å) = 0.381
So, θhkl = sin-1(0.381) = 22.4°
Q3. An x-ray of wavelength of 1.5 Å are diffracted by a set of atomic
planes as given below
Find the angle α for the first order diffraction.
Ans. We know 2 dhkl sin(θhkl) = n λ, n=1 in this problem, (hkl) are the
Miller indices of the plane, dhkl = 2.9 Å and λ = 1.5 Å , then θhkl = ?
So, Sin(θhkl ) = 0.259, θhkl =15°
Then, α = 90° - θhkl = 75° (Ans)
d=2.9 Å
α
θ
θ
Q4. The placing between the principal planes in a crystal of NaCl is 2.82 Å.
It is found that the first order Bragg reflection occurs at 10°.
(a) What is the wavelength of x-ray?
(b) At what angle the second order reflection occurs?
(c) What is the highest order of reflection seen?
Ans. We know 2 d sin(θ) = n λ, n=1 in this problem, (hkl) are the Miller
indices of the plane, d = 2.82 Å = 2.82 × 10-10 meter
(a) λ = 2 × 2.82 × 10-10 meter × sin(10°) / 1
= 0.979 Å
(b) For n=2, θ = sin-1(n λ / 2 d) = sin-1[ (2 × 9.79 2 × 10-11 m) / (2 × 2.82 ×
10-10 ) ] = 20.31°
(c) For the highest order, maximum value of sinθ=1
nmax = 2 d (sinθ)max / λ = 5.76 (As n is an integer, the highest order
can be seen is 5)
Q5. X rays of wavelength 1.5418 Å are diffracted by (111) planes in a
crystal at an angle of 30° in the first order. Calculate the lattice constant of
the cubic cell.
Ans. Given, λ = 1.5418 Å = 1.5418 × 10-10 m
(h k l) = (1 1 1), and θ= 30°, what is lattice constant a=?
Inter planar spacing is given by dhkl = a / √(h2+k2+l2)
According to Bragg’s law of X-ray diffraction, 2d sinθ = n λ
Then, d = n λ / (2 sinθ) = 1.5418 × 10-10 m = 1.5418 Å
Then a = dhkl √(h2+k2+l2) = 1.5418 Å × √(12+12+12) =2.67 Å
d - spacing in different crystal systems

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Braggs-Law-numerical-problems.pptx

  • 1. Q1. The utilized reflecting plane of a lithium fluoride (LiF) analyzing crystal has a interplanner distance of 2.5 Å. Calculate the wavelength of the 2nd order diffracted line which has a glancing angle of 60°. Ans. We know 2 dhkl sin(θhkl) = n λ, n=2 in this problem, (hkl) are the Miller indices of the utilized plane, dhkl = 2.5 Å and θhkl = 60° λ = (2 × 2.5 Å × sin(60 °)) / (2) = 2.5 Å × sin(60 °) = 2.5 × √3 / 2 Å = 2.165 Å
  • 2. Q2. Calculate the angle at which (a) first order reflection and (b) second order reflection will occur in a x-ray spectrometer, when x-ray of wavelength 1.54 Å are diffracted by the atoms of a crystal given that the interplanner distance of 4.04 Å. Ans. We know 2 dhkl sin(θhkl) = n λ, (a) n=1 in this problem, (hkl) are the Miller indices of the plane, dhkl = 4.04 Å and λ = 1.54 Å , then θhkl = ? Sin(θhkl ) = (1 × 1.54 Å) / (2 × 4.04 Å) = 0.191 So, θhkl = sin-1(0.191) = 11.0° (b) n=2 in this problem, (hkl) are the Miller indices of the plane, dhkl = 4.04 Å and λ = 1.54 Å , then θhkl = ? Sin(θhkl ) = (2 × 1.54 Å) / (2 × 4.04 Å) = 0.381 So, θhkl = sin-1(0.381) = 22.4°
  • 3. Q3. An x-ray of wavelength of 1.5 Å are diffracted by a set of atomic planes as given below Find the angle α for the first order diffraction. Ans. We know 2 dhkl sin(θhkl) = n λ, n=1 in this problem, (hkl) are the Miller indices of the plane, dhkl = 2.9 Å and λ = 1.5 Å , then θhkl = ? So, Sin(θhkl ) = 0.259, θhkl =15° Then, α = 90° - θhkl = 75° (Ans) d=2.9 Å α θ θ
  • 4. Q4. The placing between the principal planes in a crystal of NaCl is 2.82 Å. It is found that the first order Bragg reflection occurs at 10°. (a) What is the wavelength of x-ray? (b) At what angle the second order reflection occurs? (c) What is the highest order of reflection seen? Ans. We know 2 d sin(θ) = n λ, n=1 in this problem, (hkl) are the Miller indices of the plane, d = 2.82 Å = 2.82 × 10-10 meter (a) λ = 2 × 2.82 × 10-10 meter × sin(10°) / 1 = 0.979 Å (b) For n=2, θ = sin-1(n λ / 2 d) = sin-1[ (2 × 9.79 2 × 10-11 m) / (2 × 2.82 × 10-10 ) ] = 20.31° (c) For the highest order, maximum value of sinθ=1 nmax = 2 d (sinθ)max / λ = 5.76 (As n is an integer, the highest order can be seen is 5)
  • 5. Q5. X rays of wavelength 1.5418 Å are diffracted by (111) planes in a crystal at an angle of 30° in the first order. Calculate the lattice constant of the cubic cell. Ans. Given, λ = 1.5418 Å = 1.5418 × 10-10 m (h k l) = (1 1 1), and θ= 30°, what is lattice constant a=? Inter planar spacing is given by dhkl = a / √(h2+k2+l2) According to Bragg’s law of X-ray diffraction, 2d sinθ = n λ Then, d = n λ / (2 sinθ) = 1.5418 × 10-10 m = 1.5418 Å Then a = dhkl √(h2+k2+l2) = 1.5418 Å × √(12+12+12) =2.67 Å
  • 6. d - spacing in different crystal systems