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Ch 41: Quantum Mechanics II: Atomic Structure
Young & Freedman Calc - University Physics 15th Edition
Young & Freedman Calc15th EditionUniversity PhysicsISBN: 9780135159552Not the one you use?Change textbook
Chapter 40, Problem 38

The energies for an electron in the KK, LL, and MM shells of the tungsten atom are 69,500-69,500 eV, 12,000-12,000 eV, and 2,200-2,200 eV, respectively. Calculate the wavelengths of the KαK_{\(\alpha\)} and KβK_{\(\beta\)} x rays of tungsten.

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Step 1: Understand the problem. The Ka and Kb x-rays are emitted when electrons transition to the K shell from higher energy shells. Specifically, Ka corresponds to the transition from the L shell to the K shell, and Kb corresponds to the transition from the M shell to the K shell. The energy difference between these shells determines the energy of the emitted x-rays.
Step 2: Calculate the energy of the Ka x-ray. Use the formula for energy difference: \( E_{Ka} = E_{K} - E_{L} \), where \( E_{K} \) and \( E_{L} \) are the energies of the K and L shells, respectively. Substitute the given values: \( E_{K} = -69,500 \, \text{eV} \) and \( E_{L} = -12,000 \, \text{eV} \).
Step 3: Calculate the energy of the Kb x-ray. Use the formula for energy difference: \( E_{Kb} = E_{K} - E_{M} \), where \( E_{K} \) and \( E_{M} \) are the energies of the K and M shells, respectively. Substitute the given values: \( E_{K} = -69,500 \, \text{eV} \) and \( E_{M} = -2200 \, \text{eV} \).
Step 4: Convert the x-ray energies to wavelengths using the equation \( \lambda = \frac{hc}{E} \), where \( h \) is Planck's constant \( (6.626 \times 10^{-34} \; \text{J·s}) \), \( c \) is the speed of light \( (3.00 \times 10^{8} \; \text{m/s}) \), and \( E \) is the energy in joules. To convert eV to joules, use the conversion factor \( 1 \; \text{eV} = 1.602 \times 10^{-19} \; \text{J} \).
Step 5: Perform the calculations for \( \lambda_{Ka} \) and \( \lambda_{Kb} \) using the respective energies \( E_{Ka} \) and \( E_{Kb} \). Ensure units are consistent throughout the calculation, and express the wavelengths in meters or nanometers as appropriate.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Energy Levels in Atoms

In an atom, electrons occupy specific energy levels or shells, denoted as K, L, M, etc. Each shell has a distinct energy associated with it, which is negative due to the attractive force between the negatively charged electrons and the positively charged nucleus. The energy levels determine the electron's binding energy and influence the emission of photons when electrons transition between these levels.
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X-ray Emission

X-rays are produced when high-energy electrons collide with a target atom, causing inner-shell electrons to be ejected. When an electron from a higher energy level falls into the vacancy left by the ejected electron, it emits energy in the form of X-rays. The Kα and Kβ X-rays correspond to transitions from the L shell to the K shell and from the M shell to the K shell, respectively, each with specific energy differences.
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Wavelength Calculation

The wavelength of emitted X-rays can be calculated using the energy of the emitted photon, which is related to the energy difference between the two electron shells involved in the transition. The relationship is given by the equation E = hc/λ, where E is the energy of the photon, h is Planck's constant, c is the speed of light, and λ is the wavelength. By rearranging this equation, one can determine the wavelength from the energy difference.
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