/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 63 The K-shell and L-shell ionizati... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 30: Problem 63

The K-shell and L-shell ionization energies of a metal are \(8979 \mathrm{eV}\) and \(951 \mathrm{eV},\) respectively. Concepts: (i) How is a \(K_{\alpha}\) photon produced, and how much energy does it have? (ii) What must be the minimum voltage across the X-ray tube to produce a \(K_{\alpha}\) photon? (iii) What is meant by the phrases "K-shell ionization energy" and "L-shell ionization energy"? (iv) What does the difference between the K-shell and L-shell ionization energies represent? Calculations: (a) Assuming that there is a vacancy in the \(\mathrm{L}\) shell, what must be the minimum voltage across an X-ray tube with a target made from this metal to produce \(K_{\alpha}\) X-ray photons? (b) Determine the wavelength of a \(K_{\alpha}\) photon.

Short Answer

Expert verified
A \(K_{\alpha}\) photon has 8028 eV of energy, requires at least 8028 V, and has a wavelength of approximately 1.54 脜.

Step by step solution

01

Concept of K_alpha Photon

A \(K_{\alpha}\) photon is produced when an electron transitions from the L-shell (second energy level) to fill a vacancy in the K-shell (first energy level) of an atom. The energy of this photon is the difference between the ionization energy of the K-shell and the L-shell.
02

Energy of K_alpha Photon

The energy of a \(K_{\alpha}\) photon is calculated by subtracting the L-shell ionization energy from the K-shell ionization energy. Thus, \(E = 8979 \mathrm{eV} - 951 \mathrm{eV} = 8028 \mathrm{eV}\).
03

Minimum Voltage for K_alpha Photon

The minimum voltage required to produce a \(K_{\alpha}\) photon is determined by setting \(V = E/e\), where \(E\) is the photon energy and \(e\) is the charge of an electron (\(1.602 \times 10^{-19} C\)). Therefore, \(V = 8028 \mathrm{eV} = 8028 \mathrm{V}\).
04

Ionization Energy Definitions

The K-shell ionization energy is the energy required to remove an electron from the K-shell (innermost shell) of an atom, while the L-shell ionization energy is the energy required to remove an electron from the L-shell (second shell).
05

Difference in Ionization Energies

The difference between the K-shell and L-shell ionization energies represents the energy released when an electron transitions from the L-shell to the K-shell, producing a \(K_{\alpha}\) X-ray photon.
06

Minimum Voltage for X-ray Tube

If there is a vacancy in the L-shell, the minimum voltage required is equal to the \(K_{\alpha}\) photon energy, which is \(8028 \mathrm{V}\), as calculated previously.
07

Wavelength of K_alpha Photon

The wavelength \(\lambda\) of the \(K_{\alpha}\) photon is calculated using the formula \(\lambda = \frac{hc}{E}\), where \(h\) is Planck's constant \((6.626 \times 10^{-34} \mathrm{Js})\), \(c\) is the speed of light \((3 \times 10^{8} \mathrm{m/s})\), and \(E\) is the energy of the photon, which is \(8028 \mathrm{eV}\). Converting \(E\) to joules, \(E = 8028 \times 1.602 \times 10^{-19} \mathrm{J}\). Thus, \(\lambda = \frac{6.626 \times 10^{-34} \times 3 \times 10^{8}}{8028 \times 1.602 \times 10^{-19}} \approx 1.54 \times 10^{-10} \mathrm{m}\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

K-shell ionization energy
The concept of K-shell ionization energy revolves around the removal of an electron from the innermost shell of an atom, known as the K-shell. When we talk about ionization energy in this context, it refers to the amount of energy required to overcome the electrostatic attraction between the electron and the nucleus to remove the electron from the atom. For the K-shell, this energy is particularly high because the electrons are closest to the nucleus and are thus tightly bound.

Understanding this is crucial in fields like X-ray spectroscopy, where knowing the ionization energy helps in identifying elements based on their unique energy levels. Because each element has a unique atomic structure, the K-shell ionization energy will be different for different elements, and it is often used in applications such as chemical analysis and identifying unknown substances.
  • The K-shell is the closest shell to the nucleus and has the strongest bond.
  • K-shell ionization energy is often higher than that of outer shells because electrons are more tightly bound.
  • This energy is essential to understand when analyzing element composition using X-rays.
L-shell ionization energy
L-shell ionization energy is linked to the removal of an electron from the second shell of an atom, the L-shell. This shell is a bit further from the nucleus compared to the K-shell, which generally means that the ionization energy required to remove an electron from the L-shell is less than from the K-shell.

This difference in ionization energy levels between shells is significant in X-ray spectroscopy. When an electron is removed from the L-shell, the resulting energy change can also predict transitions that lead to the emission of X-ray photons. Though an atom's electrons are not as tightly bound in the L-shell as in the K-shell, they still require a significant amount of energy to ionize, particularly noticeable in heavier elements.
  • The L-shell is located further from the nucleus, leading to lower ionization energies than the K-shell.
  • Ionization energies of L-shell electrons are critical for predicting and analyzing X-ray emissions.
  • Used alongside K-shell values, L-shell ionization energies assist in calculating key X-ray parameters.
K_alpha photon
A \(K_{\alpha}\) photon represents a type of X-ray photon that is produced during specific atomic electron transitions. Specifically, when an electron from the L-shell moves down to fill a vacancy in the K-shell, a \(K_{\alpha}\) photon is emitted due to the energy difference between these two shells.

This photon has significant practical implications, especially in material characterization and medical imaging. The energy of a \(K_{\alpha}\) photon can be calculated by subtracting the L-shell ionization energy from the K-shell ionization energy. This energy is then used to determine characteristics of the emitted \(K_{\alpha}\) X-ray. In practice, this transition provides a clear spectral line that can be used to identify elements in a sample.
  • Occurs during electron transition from L-shell to K-shell.
  • Produced by the difference in ionization energies of the K and L-shells.
  • Essential for X-ray spectroscopy, aiding in the identification of elemental composition.
  • Calculated energy can assist in defining setup requirements for X-ray equipment.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Write down the fourteen sets of the four quantum numbers that correspond to the electrons in a completely filled \(4 \mathrm{f}\) subshell.

The nucleus of the hydrogen atom has a radius of about \(1 \times 10^{-15} \mathrm{m}\). The electron is normally at a distance of about \(5.3 \times 10^{-11} \mathrm{m}\) from the nucleus. Assuming that the hydrogen atom is a sphere with a radius of \(5.3 \times 10^{-11} \mathrm{m},\) find (a) the volume of the atom, (b) the volume of the nucleus, and (c) the percentage of the volume of the atom that is occupied by the nucleus.

A hydrogen atom \((Z=1)\) is in the third excited state, and a photon is either emitted or absorbed. Concepts: (i) What is the quantum number of the third excited state? (ii) When an atom emits a photon, is the final quantum number \(n_{\mathrm{f}}\) of the atom greater than or less than the initial quantum number \(n_{\mathrm{i}} ?\) (iii) When an atom absorbs a photon, is the final quantum number \(n_{\mathrm{f}}\) of the atom greater than or less than the initial quantum number \(n_{\mathrm{i}} ?\) (iv) How is the wavelength of a photon related to its energy? Calculations: Determine the quantum number \(n_{\mathrm{f}}\) of the final state and the energy of the photon when the photon is (a) emitted with the shortest possible wavelength, (b) emitted with the longest possible wavelength, and (c) absorbed with the longest possible wavelength.

In the ground state, the outermost shell \((n=1)\) of helium (He) is filled with electrons, as is the outermost shell \((n=2)\) of neon (Ne). The full outermost shells of these two elements distinguish them as the first two so-called "noble gases." Suppose that the spin quantum number \(m_{\mathrm{s}}\) had three possible values, rather than two. If that were the case, which elements would be (a) the first and (b) the second noble gases? Assume that the possible values for the other three quantum numbers are unchanged, and that the Pauli exclusion principle still applies.

Molybdenum has an atomic number of \(Z=42 .\) Using the Bohr model, estimate the wavelength of the \(K_{\alpha}\) X-ray.
See all solutions

Recommended explanations on Physics Textbooks

Magnetism

Read Explanation

Physical Quantities And Units

Read Explanation

Measurements

Read Explanation

Conservation of Energy and Momentum

Read Explanation

Kinematics Physics

Read Explanation

Fluids

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.