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The second order Bragg diffraction of x-rays with lymda=1anstrom from a set
The second order Bragg diffraction of x-rays with lymda=1anstrom from a set of parallel planes in a metal occurs at an angle of 60°.the distance between the scattering planes In the crystal is
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Explanation:

The Bragg's law can be written as:

nλ = 2dsinθ

where n is the order of diffraction, λ is the wavelength of incident radiation, d is the interplanar spacing of the crystal, and θ is the angle of incidence.

For a cubic crystal, the interplanar spacing of the (100) plane is given by:

d100 = a/√2

where a is the lattice parameter of the cubic crystal.

Now, for the second order diffraction from the (100) plane, n = 2. Substituting the values of n and d100 in Bragg's law, we get:

2λ = 2(a/√2)sinθ

λ = (a/√2)sinθ

Similarly, for the first order diffraction from the (200) plane, n = 1 and the interplanar spacing is given by:

d200 = a/√8

Substituting the values of n and d200 in Bragg's law, we get:

λ = 2(a/√8)sinθ

λ = (a/√2)sinθ

Hence, we can see that the second order diffraction from the (100) plane is equivalent to the first order diffraction from the (200) plane. Therefore, the correct answer is option B.
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The second order Bragg diffraction of x-rays with lymda=1anstrom from a setThe second order Bragg diffraction of x-rays with lymda=1anstrom from a set of parallel planes in a metal occurs at an angle of 60°.the distance between the scattering planes In the crystal is
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