Question

X-rays with a wavelength of 1.10 Åscatter at an angle of 34.5 ∘ from a crystal....

X-rays with a wavelength of 1.10 Åscatter at an angle of 34.5 ∘ from a crystal.
If n=1, what is the distance between planes of atoms in the crystal that give rise to this scattering?

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Answer #1

The problem can be solved by Bragg's equation.

That is where n= an integer

   = wavelength of incident wave

d = distance between the planes of atoms OR the spacing between the planes of atomic lattice.

   = angle between the incident ray and the scattering plane of atomic lattice

Given details are, 1.10 A0 = 1.10 * 10-10 m

   34.50

  n = 1 and d=?

Therefore,

= ( 1 * 1.10*10-10 ) / ( 2*sin(34.5) )

= 1.1*10-10 / 1.133

d = 9.7*10-11 m OR 0.97 A0

  

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