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First-order reflection from the reflection planes shown occurs when an x-ray beam of wavelength \(0.260 \mathrm{nm}\) makes an angle \(\theta=63.8^{\circ}\) with the top face of the crystal. What is the unit cell size \(a_{0} ?\)

Short Answer

Expert verified
The unit cell size \(a_0\) is approximately 0.585 nm.

Step by step solution

01

Understand the Problem

We need to determine the unit cell size, \(a_0\), using the given angle \(\theta\), the wavelength of x-ray, and the conditions for first-order reflection from the reflection planes. The problem involves using Bragg's Law.
02

Bragg's Law Formula

Bragg's Law relates the angle of reflection, the wavelength of the x-rays, and the distance between reflection planes in the crystal: \[ n\lambda = 2d\sin\theta \]For first-order reflection, \(n = 1\). We need to calculate the distance \(d\) using this formula.
03

Substitute Known Values

Substitute the known values into Bragg's Law to solve for \(d\):\[ 1 \times 0.260 \text{ nm} = 2d \sin(63.8^{\circ}) \]
04

Calculate \(d\)

Solve for \(d\) by rearranging the equation:\[ d = \frac{0.260 \text{ nm}}{2 \sin(63.8^{\circ})} \] Calculate \(d\) using a calculator.
05

Relate \(d\) to Unit Cell Size

Assuming the reflection comes from a plane with maximum reflection corresponding to the cube face (since this context is common without further details), and that the distance between planes \(d\) can relate to the unit cell by: \(d = \frac{a_0}{2}\) or \(d = \frac{a_0}{\sqrt{2}}\) depending on the crystal structure. Here, let's take the usual case of cubic structure involving simple, body-centered, or face-centered cubic crystals.
06

Solve for \(a_0\)

Since we lack specific detail about the crystal structure but are given to solve for a general case, assume simple cubic and thus \(d = \frac{a_0}{2}\). Then: \[ a_0 = 2d \] Substitute the value of \(d\) calculated from Step 4 to find \(a_0\).
07

Verify and Conclude

Calculate \(a_0\) using the calculated \(d\) value to ensure correctness. This will yield the final unit cell size in nanometers.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

X-ray Diffraction
Discovering the arrangement of atoms within a crystal requires some fascinating techniques. One of the most important methods is **x-ray diffraction**. It allows us to study the crystal structure by observing how x-rays are diffracted, or bent, as they pass through the crystal. X-rays are a type of electromagnetic radiation with a wavelength on the order of nanometers, which makes them ideal for probing structures that are similarly small, like atoms in a crystal.

Here's how it works: When an x-ray beam strikes a crystal, parts of the beam get scattered by the electrons in the atoms. The scattered waves could then reinforce or cancel each other, creating a pattern that tells scientists how atoms are arranged in the crystal. This pattern is called an X-ray diffraction pattern.
  • By studying this pattern, scientists can determine the distances between planes of atoms.
  • This distance is often denoted by 'd' in Bragg's Law, which we use to calculate unit cell dimensions.
Crystal Structure
**Crystal structure** refers to the orderly geometric spatial arrangement of atoms in the interior of a mineral. To visualize, imagine the atoms as tiny spheres tightly packed together in an organized manner. This arrangement creates repeating patterns, leading to a crystalline appearance when viewed under the correct magnification.

The ways atoms pack depend on multiple factors, like the nature of the atoms themselves and the conditions under which the crystal forms. There are several known crystal structures, but common types include:
  • **Simple Cubic**: Each corner atom in a cube shares the eight corners of the cube.
  • **Body-Centered Cubic (BCC)**: In addition to corner atoms, there's one atom right in the center.
  • **Face-Centered Cubic (FCC)**: Along with corner atoms, this has one atom at the center of each face of the cube.
Interestingly, the arrangement dictates physical properties like density and melting points, even influencing how the material forms bonds with others.

Understanding crystal structure is crucial because it helps us relate the 'd' value from x-ray diffraction results back to real-world dimensions, like the 'unit cell size' in the problem we're tackling.
Unit Cell Size
Measuring the size of the **unit cell** can seem challenging, but knowing how it fits within the context of crystal and diffraction makes it simpler. The unit cell is the smallest part of a crystal that, when repeated in all directions, recreates the entire crystal's structure.

In order to find our unit cell size, we use a special equation derived from Bragg's Law, which involves the 'd' value we calculate. Depending on the type of crystal structure, we relate 'd' to the unit cell size (often denoted as \(a_0\)).
  • For **Simple Cubic Structures**, the relationship is straightforward: \(d = \frac{a_0}{2}\).
  • But for more complex structures, like **Face-Centered Cubic**, it can change to: \(d = \frac{a_0}{\sqrt{2}}\).
  • This makes determining the crystal type essential in solving for \(a_0\).

Once calculated, knowing the unit cell size offers insights into material's properties and aids in the design of devices, from the smallest nanotechnology to everyday electronics.

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Most popular questions from this chapter

A diffraction grating with a width of \(2.0 \mathrm{~cm}\) contains 1000 lines/cm across that width. For an incident wavelength of \(600 \mathrm{nm},\) what is the smallest wavelength difference this grating can resolve in the second order?

If, in a two-slit interference pattern, there are 8 bright fringes within the first side peak of the diffraction envelope and diffraction minima coincide with two-slit interference maxima, then what is the ratio of slit separation to slit width?

Monochromatic light with wavelength \(538 \mathrm{nm}\) is incident on a slit with width \(0.025 \mathrm{~mm}\). The distance from the slit to a screen is \(3.5 \mathrm{~m} .\) Consider a point on the screen \(1.1 \mathrm{~cm}\) from the central maximum. Calculate (a) \(\theta\) for that point, (b) \(\alpha,\) and (c) the ratio of the intensity at that point to the intensity at the central maximum.

Light of wavelength \(600 \mathrm{nm}\) is incident normally on a diffraction grating. Two adjacent maxima occur at angles given by \(\sin \theta=0.2\) and \(\sin \theta=0.3 .\) The fourth-order maxima are missing. (a) What is the separation between adjacent slits? (b) What is the smallest slit width this grating can have? For that slit width, what are the (c) largest, (d) second largest, and (e) third largest values of the order number \(m\) of the maxima produced by the grating?

If you look at something \(40 \mathrm{~m}\) from you, what is the smallest length (perpendicular to your line of sight) that you can resolve, according to Rayleigh's criterion? Assume the pupil of your eye has a diameter of \(4.00 \mathrm{~mm}\), and use \(500 \mathrm{nm}\) as the wavelength of the light reaching you.

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