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Chapter 40: Problem 41

The binding energies of \(K\) -shell and \(L\) -shell electrons in copper are 8.979 and \(0.951 \mathrm{keV},\) respectively. If a \(K_{a}\) x ray from copper is incident on a sodium chloride crystal and gives a firstorder Bragg reflection at an angle of \(74.1^{\circ}\) measured relative to parallel planes of sodium atoms, what is the spacing between these parallel planes?

Short Answer

Expert verified
The spacing between the planes is approximately 1.54 脜.

Step by step solution

01

Understand K-alpha X-ray

The K-alpha x-ray emission occurs when an electron from the L-shell fills a vacancy in the K-shell. The energy of the emitted photon is the difference between the K-shell and L-shell binding energies.
02

Calculate Energy of K-alpha X-ray

The energy of the K-alpha x-ray is calculated by subtracting the L-shell binding energy from the K-shell binding energy: \[ E_{K\alpha} = E_K - E_L \] where \( E_K = 8.979 \text{ keV} \) and \( E_L = 0.951 \text{ keV} \). This gives us \( E_{K\alpha} = 8.979 \text{ keV} - 0.951 \text{ keV} = 8.028 \text{ keV} \).
03

Convert Energy to Wavelength

Use the energy-wavelength relation for photons: \[ E = \frac{hc}{\lambda} \] where \( E \) is the energy in keV, \( h \) is Planck's constant \((4.135667696 \times 10^{-15} \text{ eV}\cdot\text{s}) \), and \( c \) is the speed of light \((3 \times 10^8 \text{ m/s})\). Solve for \( \lambda \): \[ \lambda = \frac{hc}{E} \] substituting \( E = 8.028 \text{ keV} \times 10^3 \text{ eV/keV} \), we can convert to meters by recognizing \( 1 \text{ eV} = 1.602176634 \times 10^{-19} \text{ J} \).
04

Apply Bragg's Law

Bragg's Law relates the angle of reflection \( \theta \) to the wavelength \( \lambda \) and the distance \( d \) between crystal planes: \[ n\lambda = 2d\sin\theta \] For the first-order reflection \( n=1 \), plug in \( \lambda \) and \( \theta = 74.1^{\circ} \), and solve for \( d \).
05

Calculate the Plane Spacing

Substitute the values into Bragg's Law: \[ d = \frac{\lambda}{2\sin\theta} \] using the wavelength \( \lambda \) calculated in the previous step and \( \theta = 74.1^{\circ} \). Determine the exact value for \( d \) in meters or angstroms as desired.

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Key Concepts

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

K-shell and L-shell Electrons
Electrons in atoms occupy various "shells," which are essentially layers that represent different energy levels. The K-shell is the innermost shell and has the highest binding energy, as electrons are closest to the nucleus and are very tightly bound. Directly following the K-shell is the L-shell, which is further out and has a lower binding energy compared to the K-shell.

When an L-shell electron transitions to fill a vacancy in the K-shell, it releases energy equivalent to the difference in binding energies between these two shells. This process leads to the emission of K-alpha X-rays, a crucial mechanism in identifying elemental compositions via X-ray spectroscopy.
Binding Energy
Binding energy refers to the energy required to remove an electron from its shell within an atom. It plays a pivotal role in X-ray emission processes. The binding energy of the K-shell electrons is typically much higher compared to that of the L-shell electrons due to the closer proximity of the K-shell to the nucleus.

In the case of copper, the K-shell binding energy is 8.979 keV, while the L-shell is significantly lower at 0.951 keV. The difference between these energies (8.028 keV) directly influences the energy of the emitted photon during the K-alpha X-ray emission. Understanding binding energies helps in predicting and calculating the X-ray energies emitted by elements.
Energy to Wavelength Conversion
When dealing with photons like X-rays, there is a crucial relationship between energy and wavelength. This is described by the equation \[ E = \frac{hc}{\lambda} \] where:
  • \(E\) is the energy of the photon.
  • \(h\) is Planck's constant \(4.135667696 \times 10^{-15} \text{ eV}\cdot\text{s}\).
  • \(c\) is the speed of light \(3 \times 10^8 \text{ m/s}\).
  • \(\lambda\) is the wavelength.
To find the wavelength from energy, rearrange the equation to: \[ \lambda = \frac{hc}{E} \] Applying this to a K-alpha X-ray with energy 8.028 keV allows us to calculate its wavelength, which is necessary for analyzing crystal structures via Bragg's Law.
X-ray Emission
X-ray emission is a phenomenon that occurs when high-energy electrons are decelerated or when electrons transition between different energy levels within an atom. In this context, K-alpha X-rays are particularly noteworthy.

The transition of an electron from the L-shell, where it is less tightly bound, to a vacancy in the K-shell releases energy in the form of an X-ray. This specific emission is used in various analytical techniques, like X-ray fluorescence, to identify and quantify elements in a material. The precise measurement of these X-rays can reveal valuable information about the electronic structure and chemical environment of the samples.
Crystal Plane Spacing
When X-rays interact with crystalline materials, they cause diffraction patterns that depend on the spacing of the crystal planes, known as "d-spacing." Bragg鈥檚 Law is fundamental in determining this spacing. It is expressed as: \[ n\lambda = 2d\sin\theta \] where:
  • \(n\) is the order of reflection (usually 1),
  • \(\lambda\) is the wavelength of the x-ray,
  • \(d\) is the distance between crystal planes,
  • \(\theta\) is the angle of incidence that satisfies the reflection condition.
By measuring \(\theta\) and knowing \(\lambda\), one can solve for \(d\), the spacing between the planes. This analysis provides insights into the crystal structure and is an essential tool in material science, chemistry, and physics.

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