### Video Transcript

Calcite crystals contain scattering planes separated by 0.300 nanometers. What is the angular separation between first- and second-order diffraction maxima when X-rays of 0.130 nanometers wavelength are used?

Letβs start by highlighting some of the vital information weβve been given. We were told that the scattering planes in the crystal weβre considering are separated by 0.300 nanometers; weβll call that distance π. Weβre also told that the wavelength of the electromagnetic radiation used is 0.130 nanometers; weβll call that value π. We want to know the angular separation between first- and second order-diffraction maxima; weβll call that Ξπ.

Since this problem involves crystals and the regular structure that crystals are composed of, this scattering is an example of Bragg scattering. For scattering with crystals, maxima are found using the equation ππ equals two π sin π, where π is the order number, π is the wavelength of the light used, π is the separation distance between layers of the crystal lattice, and π is the angle between the scattered light and the scattering plane.

We can draw a sketch of Bragg scattering where each individual blue dot represents an atom in the crystal lattice. The two layers drawn in are separated by a distance π. When light is shined on the lattice some of it reflects off the top layer and some off the second layer. Similarly to the way it does with a thin film. Depending on the phase relationships of the reflected rays of light, there can be constructive or destructive interference. Our scattering equation is for constructive interference; that is, diffraction maxima.

When we apply this relationship to our scenario for the first-order maxima, in that case π is one so π equals two times the separation distance π times the sine of an angle weβll call π. If we then write a separate equation for the second-order maxima, in that case π is two so two times π equals two times the separation distance π times the sine of a different angle, one we will call π.

Ξπ, the angular separation we want to solve for, is equal to the magnitude of π minus π. So letβs solve for π and π now. Starting with our first-order diffraction maxima equation, when we rearrange to solve for π, we find that itβs equal to the inverse sine of π over two times π. When we plug in for π and π using the given values, and being careful to use units of meters for π and π, we find that π is equal to 12.51 degrees. Thatβs the angle for the first-order maxima.

Now letβs solve for π, the angle involved in the second-order diffraction maxima. In this case, π, the angle, is equal to the arcsin of π over π. Plugging in these values as before and entering this term on our calculator, we find that π is equal to 25.68 degrees. Now that we know both π and π, we can solve for Ξπ which weβve defined as the magnitude of π minus π.

When we subtract these angles, we find that Ξπ is 13.2 degrees. Thatβs the angular separation between the first- and second-order diffraction maxima. Thatβs the angular separation between the first and second order diffraction maxima.