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At what glancing angle would the first order diffraction occur, when copper radiation (wavelength = 154 pm) interacts with lattice planes that are 154 pm apart?
- a)30°
- b)60°
- c)45°
- d)43.5°
Correct answer is option 'A'. Can you explain this answer?
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To find the glancing angle at which first order diffraction occurs, we can use the equation for the Bragg condition:
nλ = 2d sin(θ)
where n is the order of the diffraction, λ is the wavelength of the radiation, d is the distance between the lattice planes, and θ is the glancing angle.
In this case, n = 1, λ = 154 pm (or 154 x 10^-12 m), and d = 154 pm (or 154 x 10^-12 m).
Plugging these values into the equation, we have:
1 * (154 x 10^-12 m) = 2 * (154 x 10^-12 m) * sin(θ)
Simplifying, we get:
sin(θ) = 1/2
To find the glancing angle, we take the inverse sine of both sides:
θ = arcsin(1/2)
Using a calculator, we find:
θ ≈ 30°
So the glancing angle at which first order diffraction occurs is approximately 30°.
nλ = 2d sin(θ)
where n is the order of the diffraction, λ is the wavelength of the radiation, d is the distance between the lattice planes, and θ is the glancing angle.
In this case, n = 1, λ = 154 pm (or 154 x 10^-12 m), and d = 154 pm (or 154 x 10^-12 m).
Plugging these values into the equation, we have:
1 * (154 x 10^-12 m) = 2 * (154 x 10^-12 m) * sin(θ)
Simplifying, we get:
sin(θ) = 1/2
To find the glancing angle, we take the inverse sine of both sides:
θ = arcsin(1/2)
Using a calculator, we find:
θ ≈ 30°
So the glancing angle at which first order diffraction occurs is approximately 30°.
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At what glancing angle would the first order diffraction occur, when copper radiation (wavelength = 154 pm) interacts with lattice planes that are 154 pm apart?a)30°b)60°c)45°d)43.5°Correct answer is option 'A'. Can you explain this answer?
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