The condition for X-ray diffraction to occur for a given plane $(hkl)$ at wavelength $\lambda$ is given by Bragg's Law:
$$2 d_{hkl} \sin(\theta) = n \lambda$$
For first-order diffraction ($n=1$), a reflection can only be observed if $\sin(\theta) \le 1$. Therefore, the absolute physical constraint for observing any peak is:
$$2 d_{hkl} \ge \lambda$$
For an FCC structure, reflections are allowed only if the indices $h, k, l$ are all odd or all even. The five given observed peaks and their corresponding $D$ values (where $D = \sqrt{h^2 + k^2 + l^2}$) are:
The d-spacing is related to $D$ by $d_{hkl} = a/D$, where $a$ is the lattice parameter.
The initial experiment used $\lambda_1 = 1.54 \text{ Å}$ and observed all five peaks. This means the smallest d-spacing observed, $d_{222}$, must satisfy $2 d_{222} \ge 1.54 \text{ Å}$.
To establish the maximum constraint on the observable planes for the second experiment, we assume the initial observation of the highest index peak (222) occurred at the limit ($\theta = 90^\circ$), which sets a minimum value for $a$:
$$2 d_{222} = \lambda_1 \quad \Rightarrow \quad 2 \frac{a}{\sqrt{12}} = 1.54 \text{ Å}$$ $$a = \frac{1.54 \cdot \sqrt{12}}{2} \approx 2.668 \text{ Å}$$
The new X-ray wavelength is $\lambda_2 = 3 \text{ Å}$. The condition for observing a peak is $2 d_{hkl} \ge 3 \text{ Å}$:
$$2 \frac{a}{D} \ge 3 \quad \Rightarrow \quad D \le \frac{2a}{3}$$
Using the minimum lattice parameter derived from the previous step, $a \approx 2.668 \text{ Å}$:
$$D_{\max} = \frac{2 \cdot 2.668 \text{ Å}}{3 \text{ Å}} \approx 1.778$$
We check how many of the allowed FCC reflections (and specifically the five observed ones) satisfy $D \le 1.778$:
Only the (111) reflection is observed when using the $3 \text{ Å}$ X-rays under this critical constraint.
The number of observed peaks will be 1.
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 = $1.67 \times 10^{-27}$ $\text{ Kg}$)
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: $m_n = 1.675 \times 10^{-27} \text{ kg}$, $h = 6.626 \times 10^{-34} \text{ J.s}$)