For Bragg’s diffraction by a crystal to occur, then which of the following is correct for the X-ray of wavelength $\lambda $ and interatomic distance $d$:
(A) $\lambda $ is greater than $2d$
(B) $\lambda $ equals$2d$
(C) $\lambda $ is smaller than or equal to $2d$
(D) $\lambda $ is smaller than $2d$

Hint: In Bragg’s diffraction, there is an occurrence of constructive and destructive interference patterns, where the path difference between two waves is given by the expression, $2d\sin \theta $, where $\theta $ is the glancing angle.

Complete step by step solution:
The Bragg’s diffraction occurs when an electromagnetic wave hits the atoms present in a crystal lattice and gets diffracted due to the electrons present in the lattice. And the effect produced in Bragg’s diffraction is amplified by the reflection of the wave by different crystal planes in the lattice. Thus the interatomic distance plays a significant role in determining the peaks produced by the diffraction. Therefore, the electromagnetic wavelength is comparable to the interatomic distance d, which generally falls under the wavelength of the X-rays.
For the interference to occur, the interatomic distance must be related to the path difference as follows:
$2d\sin \theta = n\lambda $
Where d is the interatomic distance between two crystal planes, $\theta $ is the glance angle, $\lambda $ is the wavelength of the electromagnetic wave, and n is an integer.
The range of $\lambda $ that can produce Bragg’s diffraction in a given crystal can be derived as follows:
For$n = 1$ we have,
$2d\sin \theta = \lambda $
$d\sin \theta = \dfrac{\lambda }{2}$
We know that $\sin \theta $ has the maximum value 1. So, the maximum value for $\dfrac{\lambda }{2}$ is-
$\dfrac{\lambda }{2} = d$
Or $\lambda = 2d $ (Maximum value)
The minimum value of $\sin \theta $ is $0$ (We take the modulus value because wavelength or the distance cannot have a negative value)
So we have, 0 as the starting point.
Thus the value of $\lambda $ should be less than or equal to $2d$.

Therefore option (C) is correct.

Note: This property of Bragg’s diffraction is used in X-ray crystallography which can be used to image and study the lattice structure of various crystals. The wavelength at which the diffraction peaks can be used to calculate the interatomic distance.