# Bragg X-Ray Scattering

• MarkusNaslund19
In summary, the cubic crystal with an interatomic spacing of 0.24 nm can be used to select gamma-rays of 100 keV from a radioactive source. With the incident beam being normal to the crystal, the 100 keV gamma-rays appear at an angle of 1.48 degrees based on Bragg's Law and the equation for energy, wavelength, and angle. The angle of incidence must equal the angle of reflection, not deflection.

#### MarkusNaslund19

1. Question

A cubic crystal of interatomic spacing of 0.24 nm is used to select
gamma–rays of energy 100 keV from a radioactive source containing a continuum of energies. If the incident beam is normal to the crystal, at
what angle with respect to the incident beam do the 100 keV gamma–rays appear?

2. Equations and Variables

Bragg's Law: n(lambda) = 2dsinx

n is the order of intensity
lamba is the wavelength
d is the Bragg plane separation
x is the angle
E is energy, E=100keV=1.60x10^-14J

3. Attempt

Since the incident beam is normal to the crystal, d=0.24nm.

E=hf
E=h(c/lambda)
lambda=(hc)/E
=(6.626x10^-34)(3.00x10^8)/1.60x10^-14
=1.24x10^-11m

sinx=n(lambda)/2d
=[1(1.24x10^-11)]/[2(0.24x10^-9)]
= 0.0258

x = 1.48 degrees.

Just a bit unsure of myself. I guess it's the theory that is troubling me. I thought that the angle of incidence equals the angle of deflection. Thanks fellas.

i think you may be confusing the fact that angle of incidence must equal to the angle of REFLECTION

not deflection (refraction)

4. Response

Hello, thank you for your question. Based on your calculations, it appears that the 100 keV gamma-rays would appear at an angle of 1.48 degrees with respect to the incident beam. This is in accordance with Bragg's Law, which states that the angle of incidence is equal to the angle of deflection for constructive interference to occur. In this case, the Bragg plane separation, d, is equal to 0.24 nm, and the wavelength of the gamma-rays is 1.24x10^-11m. By plugging these values into the equation, we can see that the angle of deflection, or x, is indeed 1.48 degrees. This shows that Bragg X-Ray Scattering is a useful technique for selecting specific energies from a continuum of energies, as demonstrated in this scenario. I hope this helps clarify any confusion you may have had and thank you for your interest in scientific principles.

## 1. What is Bragg X-Ray Scattering?

Bragg X-Ray Scattering is a technique used in X-ray crystallography to study the structure of crystalline materials. It involves directing a beam of X-rays at a crystal and measuring the diffraction pattern produced by the scattered rays. This allows scientists to determine the arrangement of atoms within the crystal.

## 2. How does Bragg X-Ray Scattering work?

In Bragg X-Ray Scattering, X-rays are directed at a crystal at a specific angle. When the X-rays interact with the atoms in the crystal, they are scattered in different directions. The scattered rays produce a diffraction pattern, which can be measured and analyzed to determine the structure of the crystal.

## 3. What is the significance of Bragg X-Ray Scattering in scientific research?

Bragg X-Ray Scattering is an important tool in the study of crystal structures and has many applications in various fields of science, including chemistry, biology, and materials science. It allows scientists to understand the arrangement of atoms in crystals and how this affects their physical and chemical properties.

## 4. How is Bragg X-Ray Scattering different from other X-ray techniques?

Bragg X-Ray Scattering differs from other X-ray techniques in that it specifically focuses on the diffraction pattern produced by the scattering of X-rays from a crystal. This allows for the determination of the crystal structure, whereas other X-ray techniques may be used for other purposes, such as imaging or elemental analysis.

## 5. Are there any limitations to Bragg X-Ray Scattering?

Like any scientific technique, there are limitations to Bragg X-Ray Scattering. It is most effective for studying crystalline materials, so it cannot be used for amorphous materials. Additionally, it requires a single crystal sample, which can be difficult to obtain. It also has limitations in terms of the size and complexity of crystals that can be analyzed.