Discussion
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 \(G\) of magnitude 3.14 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 × 10-27 Kg)
Answer: B
The angle is 30°.
Find the neutron's wave number from its energy:
\[ \begin{aligned} E &= 20.4\ \text{meV} = 3.27 \times 10^{-21}\ \text{J} \\ p &= \sqrt{2mE} = 3.30 \times 10^{-24}\ \text{kg m/s} \\ k &= p/\hbar \approx 3.13\ \mathring{\mathrm{A}}^{-1} \end{aligned} \]
Elastic diffraction requires \(\vec{k}' - \vec{k} = \vec{G}\), which with \(|\vec{k}'| = |\vec{k}|\) is Bragg's law in the form \(G = 2k\sin\theta\), where \(\theta\) is the angle between \(\vec{k}\) and the reflecting planes:
\[ \begin{aligned} \sin\theta &= \frac{3.14}{2 \times 3.13} \approx 0.50 \\ \theta &= 30^{\circ} \end{aligned} \]
60° is the total scattering angle \(2\theta\), not the angle with the planes; 15° and 45° do not satisfy the Bragg condition.
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