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Question 48 The value of 6 at which first order peak in x-ray (A 1.53 A) diffraction corresponds to (111) plane of a se structure with the lattice constant 2.65 Åis approxi- mately.
a. 15°
b. 30°
с. 45°
d. 60°?
a. 15°
b. 30°
с. 45°
d. 60°?
Most Upvoted Answer
Question 48 The value of 6 at which first order peak in x-ray (A 1.53 ...
Understanding X-ray Diffraction
X-ray diffraction is a powerful technique used to determine the structure of crystalline materials. The angle of diffraction is related to the spacing between the crystal planes and the wavelength of the X-rays used.
Bragg's Law
Bragg's Law governs the relationship between the wavelength of X-rays, the angle of diffraction, and the interplanar spacing in the crystal structure:
- nλ = 2d sin(θ)
Where:
- n = order of the peak (usually 1 for the first peak)
- λ = wavelength of X-rays (1.53 Å)
- d = interplanar spacing
- θ = angle of diffraction
Calculating Interplanar Spacing
For a face-centered cubic (FCC) structure, the d-spacing for the (111) plane is given by:
- d = a / √(h² + k² + l²)
Where:
- a = lattice constant (2.65 Å)
- (h, k, l) = Miller indices (1, 1, 1 for the (111) plane)
Calculating d:
- d = 2.65 Å / √(1² + 1² + 1²) = 2.65 Å / √3 ≈ 1.53 Å
Finding the Angle θ
Using Bragg's Law with n = 1 and λ = 1.53 Å:
- 1.53 Å = 2(1.53 Å) sin(θ)
This simplifies to:
- 1 = 2 sin(θ)
- sin(θ) = 0.5
Now, solving for θ:
- θ = sin⁻¹(0.5) = 30°
Conclusion
Therefore, the value of θ corresponding to the first-order peak for the (111) plane in the given structure is approximately:
- b. 30°
X-ray diffraction is a powerful technique used to determine the structure of crystalline materials. The angle of diffraction is related to the spacing between the crystal planes and the wavelength of the X-rays used.
Bragg's Law
Bragg's Law governs the relationship between the wavelength of X-rays, the angle of diffraction, and the interplanar spacing in the crystal structure:
- nλ = 2d sin(θ)
Where:
- n = order of the peak (usually 1 for the first peak)
- λ = wavelength of X-rays (1.53 Å)
- d = interplanar spacing
- θ = angle of diffraction
Calculating Interplanar Spacing
For a face-centered cubic (FCC) structure, the d-spacing for the (111) plane is given by:
- d = a / √(h² + k² + l²)
Where:
- a = lattice constant (2.65 Å)
- (h, k, l) = Miller indices (1, 1, 1 for the (111) plane)
Calculating d:
- d = 2.65 Å / √(1² + 1² + 1²) = 2.65 Å / √3 ≈ 1.53 Å
Finding the Angle θ
Using Bragg's Law with n = 1 and λ = 1.53 Å:
- 1.53 Å = 2(1.53 Å) sin(θ)
This simplifies to:
- 1 = 2 sin(θ)
- sin(θ) = 0.5
Now, solving for θ:
- θ = sin⁻¹(0.5) = 30°
Conclusion
Therefore, the value of θ corresponding to the first-order peak for the (111) plane in the given structure is approximately:
- b. 30°
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Question 48 The value of 6 at which first order peak in x-ray (A 1.53 A) diffraction corresponds to (111) plane of a se structure with the lattice constant 2.65 Åis approxi- mately.a. 15°b. 30°с. 45°d. 60°?
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Question 48 The value of 6 at which first order peak in x-ray (A 1.53 A) diffraction corresponds to (111) plane of a se structure with the lattice constant 2.65 Åis approxi- mately.a. 15°b. 30°с. 45°d. 60°? for Physics 2024 is part of Physics preparation. The Question and answers have been prepared according to the Physics exam syllabus. Information about Question 48 The value of 6 at which first order peak in x-ray (A 1.53 A) diffraction corresponds to (111) plane of a se structure with the lattice constant 2.65 Åis approxi- mately.a. 15°b. 30°с. 45°d. 60°? covers all topics & solutions for Physics 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Question 48 The value of 6 at which first order peak in x-ray (A 1.53 A) diffraction corresponds to (111) plane of a se structure with the lattice constant 2.65 Åis approxi- mately.a. 15°b. 30°с. 45°d. 60°?.
Question 48 The value of 6 at which first order peak in x-ray (A 1.53 A) diffraction corresponds to (111) plane of a se structure with the lattice constant 2.65 Åis approxi- mately.a. 15°b. 30°с. 45°d. 60°? for Physics 2024 is part of Physics preparation. The Question and answers have been prepared according to the Physics exam syllabus. Information about Question 48 The value of 6 at which first order peak in x-ray (A 1.53 A) diffraction corresponds to (111) plane of a se structure with the lattice constant 2.65 Åis approxi- mately.a. 15°b. 30°с. 45°d. 60°? covers all topics & solutions for Physics 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Question 48 The value of 6 at which first order peak in x-ray (A 1.53 A) diffraction corresponds to (111) plane of a se structure with the lattice constant 2.65 Åis approxi- mately.a. 15°b. 30°с. 45°d. 60°?.
Solutions for Question 48 The value of 6 at which first order peak in x-ray (A 1.53 A) diffraction corresponds to (111) plane of a se structure with the lattice constant 2.65 Åis approxi- mately.a. 15°b. 30°с. 45°d. 60°? in English & in Hindi are available as part of our courses for Physics.
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