Question

A monochromatic light beam of wavelength 0.1790 nm is incident on BCC iron crystal lattice, with lattice parameter 0.2866 nm. Assume that the order 
   of reflection is 1. The diffraction angle  

θ

, rounded to the nearest integer for the (220) set of planes is 124 degrees​

Answer 1

Answer:

your answer in this picture

Answer 2

The correct question is: For BCC iron, compute (a) the interplanar spacing, and (b) the diffraction   angle   for   the   (220)   set   of   planes.   The   lattice parameter    for    Fe    is    0.2866    nm.    Also,    assume    that monochromatic radiation having a wavelength of 0.1790 nm is used, and the order of reflection is 1

Answer: The interplanar scaling is 0.1013 nm and diffraction angle is 124.16

Given:

set of planes = 220

Lattice parameter for Fe = 0.2866 nm

Wavelength of monochromatic light = 0.1790nm

order of reflection = 1

To find:

(a) the interplanar scaling

(b) the diffraction angle

Solution:

With a= 0.2866 nm and h= 2, k = 2, l = 0, and considering the (220) planes, the value of the interplanar spacing is determined.

(A) Therefore,

⇒ $d_{hkl} = \frac{a}{\sqrt{h^{2}+k^{2}+l^{2}} }$

⇒ $d_{hkl} = \frac{0.2866 nm}{\sqrt{2^{2}+2^{2}+0^{2}} }$

⇒ 0.1013 nm

(B) Because this is a first-order reflection, the value of can now be computed with n=1.

⇒ $sin(theta) = \frac{n*lambda}{2d_{hkl} }$

⇒ $\frac{1*0.1790nm}{2*0.1013nm}$

⇒ 0.884 ⇔ Theta = Sin$^{-1}$(0.884)

≅ 62.13°

The diffraction angle is 2*(Theta) = 2* 62.13 ≅ 124.16°

Hence, the interplanar scaling is 0.1013 nm and diffraction angle is 124.16.

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