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
1. Unit Cell Parameters
- The lattice constant (length of the unit cell) is \( a \).
- Each unit cell contains two atoms:
- Atom Q at position \( x_1 = 0 \)
- Atom P at position \( x_2 = a/2 \)
- The atomic form factors for these atoms are \( f_Q \) and \( f_P \) respectively.
2. The Structure Factor Calculation
The structure factor \( S_n \) for the \( n \)-th order diffraction peak is given by the sum over the atoms in the unit cell:
$$ S_n = \sum_{j} f_j e^{i \phi_j} $$
The phase difference \( \phi \) for an atom at position \( x \) is related to the path difference \( \Delta = x \cos \theta \). Given the diffraction condition \( a \cos \theta = n\lambda \), the phase difference is:
$$ \phi = \frac{2\pi}{\lambda} (x \cos \theta) $$
Substituting the atomic positions:
- For atom Q (\( x=0 \)): \( \phi_Q = 0 \)
- For atom P (\( x=a/2 \)): \( \phi_P = \frac{2\pi}{\lambda} \left( \frac{a}{2} \cos \theta \right) = \frac{\pi}{\lambda} (a \cos \theta) = \frac{\pi}{\lambda} (n\lambda) = n\pi \)
Therefore, the structure factor is:
$$ S_n = f_Q e^{i(0)} + f_P e^{in\pi} = f_Q + f_P (-1)^n $$
3. Evaluating Intensity
The intensity \( I \) of the diffraction peak is proportional to the square of the magnitude of the structure factor: \( I \propto |S_n|^2 \).
- For even values of \( n \) (\( n = 2, 4, 6, \dots \)): $$ S_{\text{even}} = f_Q + f_P(1) = f_Q + f_P $$ $$ I_{\text{even}} \propto (f_Q + f_P)^2 $$
- For odd values of \( n \) (\( n = 1, 3, 5, \dots \)): $$ S_{\text{odd}} = f_Q + f_P(-1) = f_Q - f_P $$ $$ I_{\text{odd}} \propto (f_Q - f_P)^2 $$
Conclusion
Since \( (f_Q + f_P)^2 > (f_Q - f_P)^2 \) (assuming \( f_P \) and \( f_Q \) are positive and unequal), the intensity of the diffraction peaks is lower for odd values of \( n \) compared to even values of \( n \).
Correct Option: lower for odd values of \( n \), when compared to even values of \( n \)
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 ________.
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: $m_n = 1.675 \times 10^{-27} \text{ kg}$, $h = 6.626 \times 10^{-34} \text{ J.s}$)
- A metal with body centered cubic (bcc) structure shows the first (i.e. smallest angle) diffraction peak at a Bragg angle of $\theta=30^\circ$. The wavelength of X-ray used is 2.1 Å. The volume of the PRIMITIVE unit cell of the metal is
