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Ch.7 - Periodic Properties of the Elements
Chapter 7, Problem 109e

Moseley established the concept of atomic number by studying X rays emitted by the elements. The X rays emitted by some of the elements have the following wavelengths: Element Wavelength (pm) Ne 1461 Ca 335.8 Zn 143.5 Zr 78.6 Sn 49.1 (e) A particular element emits X rays with a wavelength of 98.0 pm. What element do you think it is?

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1. Moseley's law states that the frequency of the characteristic X-rays emitted by an element is proportional to the square of the atomic number (Z) of the element. The relationship is given by the equation: $\nu = a(Z - b)^2$, where $\nu$ is the frequency, $Z$ is the atomic number, $a$ and $b$ are constants.
2. The frequency and wavelength of light are inversely proportional. This relationship is given by the equation: $\nu = c/\lambda$, where $\nu$ is the frequency, $c$ is the speed of light, and $\lambda$ is the wavelength.
3. From the given data, we can see that as the atomic number increases, the wavelength of the emitted X-rays decreases. This is consistent with Moseley's law.
4. Given that the wavelength of the X-rays emitted by the unknown element is 98.0 pm, we can compare this value with the wavelengths of the known elements. The element with the wavelength closest to 98.0 pm is likely to be the unknown element.
5. Looking at the given data, the element Zr has a wavelength of 78.6 pm and the element Sn has a wavelength of 49.1 pm. The wavelength of the unknown element (98.0 pm) falls between these two values, so the unknown element is likely to be one that has an atomic number between those of Zr and Sn.

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Key Concepts

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

Atomic Number

The atomic number of an element is defined as the number of protons in its nucleus, which determines the element's identity and its position in the periodic table. Moseley's work demonstrated that the atomic number is a more fundamental property than atomic mass, as it correlates with the unique X-ray emission patterns of each element.
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Atom Structure

X-ray Emission

X-ray emission occurs when electrons transition between energy levels in an atom, resulting in the release of energy in the form of X-rays. The wavelength of these X-rays is characteristic of the element, allowing scientists to identify elements based on their unique X-ray spectra.
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Characteristics of Gamma Rays

Wavelength and Energy Relationship

The wavelength of electromagnetic radiation is inversely related to its energy, as described by the equation E = hc/λ, where E is energy, h is Planck's constant, c is the speed of light, and λ is the wavelength. Shorter wavelengths correspond to higher energy, which is crucial for understanding the X-ray emissions of elements and identifying them based on their emitted wavelengths.
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Frequency-Wavelength Relationship
Related Practice
Textbook Question

Moseley established the concept of atomic number by studying X rays emitted by the elements. The X rays emitted by some of the elements have the following wavelengths: Element Wavelength (pm) Ne 1461 Ca 335.8 Zn 143.5 Zr 78.6 Sn 49.1 (e) A particular element emits X rays with a wavelength of 98.0 pm. What element do you think it is?

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Textbook Question

Moseley established the concept of atomic number by studying X rays emitted by the elements. The X rays emitted by some of the elements have the following wavelengths: Element Wavelength (pm) Ne 1461 Ca 335.8 Zn 143.5 Zr 78.6 Sn 49.1 (a) Calculate the frequency, n, of the X rays emitted by each of the elements, in Hz.

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Textbook Question

Moseley established the concept of atomic number by studying X rays emitted by the elements. The X rays emitted by some of the elements have the following wavelengths: Element Wavelength (pm) Ne 1461 Ca 335.8 Zn 143.5 Zr 78.6 Sn 49.1 (d) Use the result from part (b) to predict the X-ray wavelength emitted by iron.

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Textbook Question

One way to measure ionization energies is ultraviolet photoelectron spectroscopy (PES), a technique based on the photoelectric effect. (Section 6.2) In PES, monochromatic light is directed onto a sample, causing electrons to be emitted. The kinetic energy of the emitted electrons is measured. The difference between the energy of the photons and the kinetic energy of the electrons corresponds to the energy needed to remove the electrons (that is, the ionization energy). Suppose that a PES experiment is performed in which mercury vapor is irradiated with ultraviolet light of wavelength 58.4 nm. (c) The kinetic energy of the emitted electrons is measured to be 1.72 * 10-18 J. What is the first ionization energy of Hg, in kJ>mol?

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Textbook Question

One way to measure ionization energies is ultraviolet photoelectron spectroscopy (PES), a technique based on the photoelectric effect. (Section 6.2) In PES, monochromatic light is directed onto a sample, causing electrons to be emitted. The kinetic energy of the emitted electrons is measured. The difference between the energy of the photons and the kinetic energy of the electrons corresponds to the energy needed to remove the electrons (that is, the ionization energy). Suppose that a PES experiment is performed in which mercury vapor is irradiated with ultraviolet light of wavelength 58.4 nm. (d) Using Figure 7.10, determine which of the halogen elements has a first ionization energy closest to that of mercury.

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Textbook Question
Mercury in the environment can exist in oxidation states 0, +1, and +2. One major question in environmental chemistry research is how to best measure the oxidation state of mercury in natural systems; this is made more complicated by the fact that mercury can be reduced or oxidized on surfaces differently than it would be if it were free in solution. XPS, X-ray photoelectron spectroscopy, is a technique related to PES (see Exercise 7.111), but instead of using ultraviolet light to eject valence electrons, X rays are used to eject core electrons. The energies of the core electrons are different for different oxidation states of the element. In one set of experiments, researchers examined mercury contamination of minerals in water. They measured the XPS signals that corresponded to electrons ejected from mercury's 4f orbitals at 105 eV, from an X-ray source that provided 1253.6 eV of energy 11 ev = 1.602 * 10-19J2. The oxygen on the mineral surface gave emitted electron energies at 531 eV, corresponding to the 1s orbital of oxygen. Overall the researchers concluded that oxidation states were +2 for Hg and -2 for O. (b) Compare the energies of the 4f electrons in mercury and the 1s electrons in oxygen from these data to the first ionization energies of mercury and oxygen from the data in this chapter.
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