The spacing of one set of crystal planes in Nacl is d = 0.282 nm. A monochromatic beam of X-Rays produces a Bragg’s maximum when glancing angle with these planes is 7°. Assuming this is the second order maximum. The wavelength of X-Rays used is (sin 7° = 0.122)

Question

Question

The spacing of one set of crystal planes in Nacl is d = 0.282 nm. A monochromatic beam of X-Rays produces a Bragg’s maximum when glancing angle with these planes is 7°. Assuming this is the second order maximum. The wavelength of X-Rays used is (sin 7° = 0.122)

1.38 nm
0.034 nm
0.069 nm
2.07 nm

Answer 1

Bragg’s Law:

Bragg's Law relates the angle θ (at which there is a maximum in diffracted intensity) to the wavelength of X-rays and the interlayer distance d between the planes of atoms/ions/molecules in the lattice.

bragg

Condition for constructive interference (maxima in the reflected).

nλ = 2d sinθ

λ = Wavelength of the X-ray

d = distance of crystal layers

θ = incident angle (the angle between the incident plane and the scattering plane)

n = integer (order)

This condition is known as Bragg’s law.

From the above equation,

\(λ = \frac {2d sin θ}{n}\)

Calculation:

Given d = 0.282 nm, θ = 7°, n = 2;

From the above relation,

\(\lambda = \frac{{2dsinθ }}{2} = dsinθ\)

= 0.282 × 0.122 = 0.034 nm

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Question

The spacing of one set of crystal planes in Nacl is d = 0.282 nm. A monochromatic beam of X-Rays produces a Bragg’s maximum when glancing angle with these planes is 7°. Assuming this is the second order maximum. The wavelength of X-Rays used is (sin 7° = 0.122)

Answer (Detailed Solution Below)

Detailed Solution

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