Question 12.2.3: Determine d-spacing and radius from a scattering angle. Silv......
Determine d-spacing and radius from a scattering angle.
Silver metal crystallizes in an FCC lattice. Monochromatic x-radiation from a copper target has a wavelength of 154 pm. If this radiation is used in a diffraction experiment with a silver crystal, a second-order diffracted beam is observed at a theta value of 22.18°. If the spacing between these planes corresponds to the unit cell length (d = α), what is the d-spacing between the planes that gave rise to this reflection? What is the metallic radius of a silver atom?
You are asked to calculate the d-spacing between the planes in a solid lattice and the metallic radius of an atom of a specific element.
You are given the x-ray wavelength, the theta value, and the type of crystal lattice.
Step 1. Calculate the d-spacing using Bragg’s law
d = \frac{n\lambda }{2\sin \theta } = \frac{\left(2\right) \left(154\text{ pm}\right) }{2\sin \left(22.18°\right) } = 408 pm
Step 2. Determine the metallic radius of a silver atom.
The d-spacing corresponds to the edge length of the cubic cell. Use the relationship between unit cell edge length and atomic radius for an FCC unit cell to calculate the atomic radius of silver.
\mathcal{r} = \frac{a \sqrt{2} }{4} = \frac{\left(408\text{ pm} \right)\sqrt{2} }{4} = 144 pm