Question

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 $3.14$ \AA$^{-1}$. What is the diffraction angle in degrees (rounded off to the nearest integer) that $\vec{k}$ makes with the plane? (Use mass of neutron = $1.67 \times 10^{-27}$ Kg)

• A. 15
• B. 30
• C. 45
• D. 60

Hint:

The diffraction angle can be calculated using Bragg's Law by first finding the wavelength of the incident wave and then using the reciprocal lattice vector to determine the angle.

Answer 1

The Correct Option is B

- The diffraction condition for a neutron is given by the Bragg's Law: \[ n\lambda = 2d \sin(\theta), \] where $n$ is the order of diffraction, $\lambda$ is the wavelength of the incident beam, $d$ is the spacing between the planes of atoms, and $\theta$ is the diffraction angle.
- The energy of the neutron is related to its wavelength by: \[ E = \dfrac{h^2}{2m\lambda^2}, \] where $E$ is the energy of the neutron, $h$ is Planck's constant, and $m$ is the mass of the neutron.
- Rearranging for $\lambda$, we get: \[ \lambda = \sqrt{\dfrac{h^2}{2mE}}. \] Substitute the values for $h = 6.626 \times 10^{-34}$ J·s, $m = 1.67 \times 10^{-27}$ kg, and $E = 20.4 \times 10^{-3}$ eV to find the wavelength.
- Then, using the reciprocal lattice vector magnitude $|\vec{G}| = 3.14$ \AA$^{-1}$ and applying Bragg's Law, we calculate the diffraction angle $\theta$.
Thus, the diffraction angle is found to be 30 degrees.