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: $m_n = 1.675 \times 10^{-27} \text{ kg}$, $h = 6.626 \times 10^{-34} \text{ J.s}$)
Neutron Diffraction: Finding Interplanar Spacing
To determine the interplanar spacing ($d$) of a crystal using neutron diffraction, we apply Bragg's Law and the de Broglie wavelength relationship.
Calculating Neutron Momentum and Wavelength
First, calculate the momentum ($p$) of the neutrons using their given speed ($v$) and mass ($m_n$).
Momentum, $p = m_n \times v$
$p = (1.675 \times 10^{-27} \text{ kg}) \times (10^3 \text{ m/s})$
$p = 1.675 \times 10^{-24} \text{ kg m/s}$
Next, find the de Broglie wavelength ($\lambda$) using Planck's constant ($h$) and the calculated momentum ($p$).
Wavelength, $\lambda = \frac{h}{p}$
$\lambda = \frac{6.626 \times 10^{-34} \text{ J.s}}{1.675 \times 10^{-24} \text{ kg m/s}}$
$\lambda \approx 3.9558 \times 10^{-10} \text{ m}$
Applying Bragg's Law
Bragg's Law relates the wavelength ($\lambda$), interplanar spacing ($d$), Bragg angle ($\theta$), and diffraction order ($n$):
$n \lambda = 2d \sin \theta$
We need to find $d$. Rearranging the formula:
$d = \frac{n \lambda}{2 \sin \theta}$
Given values are:
- Diffraction order, $n = 1$ (first order)
- Bragg angle, $\theta = 30^\circ$
- Wavelength, $\lambda \approx 3.9558 \times 10^{-10} \text{ m}$
Substitute these values into the equation for $d$:
$d = \frac{1 \times (3.9558 \times 10^{-10} \text{ m})}{2 \times \sin(30^\circ)}$
Since $\sin(30^\circ) = 0.5$, the equation becomes:
$d = \frac{3.9558 \times 10^{-10} \text{ m}}{2 \times 0.5}$
$d = \frac{3.9558 \times 10^{-10} \text{ m}}{1}$
$d = 3.9558 \times 10^{-10} \text{ m}$
Interplanar Spacing in Ångströms
Convert the interplanar spacing from meters to Ångströms (Å), knowing that $1 \text{ m} = 10^{10} \text{ Å}$.
$d = (3.9558 \times 10^{-10} \text{ m}) \times (10^{10} \text{ Å/m})$
$d \approx 3.96 \text{ Å}$
This calculated value falls within the range of 3.91 to 4.15 Å.
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
- 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
