Powder X-ray diffraction pattern of a cubic solid with lattice constant $a$ has the (111) diffraction peak at $\theta = 30^\circ$. If the lattice expands such that the lattice constant becomes $1.25a$, the angle (in degrees) corresponding to the (111) peak changes to $\sin^{-1} \left(\frac{1}{n}\right)$. The value of $n$ (rounded off to one decimal place) is _____________
First, consider the Bragg's Law for diffraction: \(n\lambda = 2d\sin\theta\), where \(d\) is the interplanar spacing, \(\theta\) is the angle of diffraction, \(n\) is an integer (order of diffraction), and \(\lambda\) is the wavelength of the incident X-ray.
For a cubic lattice, the interplanar spacing \(d\) for a plane with Miller indices (hkl) is given by: \(d = \frac{a}{\sqrt{h^2 + k^2 + l^2}}\).
For the (111) plane, \(d = \frac{a}{\sqrt{1^2 + 1^2 + 1^2}} = \frac{a}{\sqrt{3}}\).
Given \(\theta = 30^\circ\) for the initial lattice constant \(a\), we use: \(n\lambda = 2\left(\frac{a}{\sqrt{3}}\right)\sin(30^\circ) = \frac{a\lambda}{\sqrt{3}}\).
Now, if the lattice expands to \(1.25a\), the new interplanar spacing for the (111) plane is \(d' = \frac{1.25a}{\sqrt{3}}\).
For the expanded lattice, the new Bragg's law is: \(n'\lambda = 2d'\sin(\theta') = 2\left(\frac{1.25a}{\sqrt{3}}\right)\sin(\theta') = \frac{2.5a\sin(\theta')}{\sqrt{3}}\).
Given \(\theta' = \sin^{-1}\left(\frac{1}{n}\right)\), equate the Bragg conditions:
\(\frac{a\lambda}{\sqrt{3}} = \frac{2.5a\lambda\sin(\theta')}{\sqrt{3}}\), thus simplifying to \(1 = 2.5\sin(\theta')\).
Therefore, \(\sin(\theta') = \frac{1}{2.5}\), and \(\theta' = \sin^{-1}\left(\frac{1}{2.5}\right)\). Given: \(\theta' = \sin^{-1}\left(\frac{1}{n}\right)\), implying \(n = 2.5\).
The calculated value \(n = 2.5\) is within the given range (2.5, 2.5).
Important Questions from X-ray Diffraction Bragg's Laue Method
A neutron beam with a wave vector $\vec{k}$ and an energy 20.4 meV diffracts from a crystal with an outgoing wave vector $\vec{k}'$. One of the diffraction peaks is observed for the reciprocal lattice vector $\vec{G}$ of magnitude $3.14 \text{ Å}^{-1}$. What is the diffraction angle in degrees (rounded off to the nearest integer) that $\vec{k}$ makes with the plane? (Use mass of neutron = $1.67 \times 10^{-27}$ $\text{ Kg}$)
- The X-ray diffraction pattern of a monatomic cubic crystal with rigid spherical atoms of radius $1.56$ Å shows several Bragg reflections of which the reflection appearing at the lowest $2\theta$ value is from $(111)$ plane. If the wavelength of X-ray used is $0.78$ Å, the Bragg angle (in $2\theta$, rounded off to one decimal place) corresponding to this reflection and the crystal structure, respectively, are
- In an X-Ray diffraction experiment on a solid with FCC structure, five diffraction peaks corresponding to (111), (200), (220), (311) and (222) planes are observed using $1.54 \mathring{A}$ X-rays. On using $3 \mathring{A}$ X-rays on the same solid, the number of observed peaks will be ________.
As shown in the figure, X-ray diffraction pattern is obtained from a diatomic chain of atoms P and Q. The diffraction condition is given by $a \cos \theta = n \lambda$, where $n$ is the order of the diffraction peak. Here, $a$ is the lattice constant and $ \lambda $ is the wavelength of the X-rays. Assume that atomic form factors and resolution of the instrument do not depend on $ \theta $. Then, the intensity of the diffraction peaks is
Neutrons moving with speed $10^3 \text{ m/s}$ are used for the determination of crystal structure. If the Bragg angle for the first order diffraction is $30^\circ$, the interplanar spacing of the crystal is ________ Å.
(Given: $m_n = 1.675 \times 10^{-27} \text{ kg}$, $h = 6.626 \times 10^{-34} \text{ J.s}$)
