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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Step 1: First, we need to calculate the energy of the ultraviolet light used in the PES experiment. The energy of a photon can be calculated using the equation E = hc/λ, where h is Planck's constant (6.626 x 10^-34 J.s), c is the speed of light (3.00 x 10^8 m/s), and λ is the wavelength of the light in meters. Convert the wavelength from nm to m by multiplying by 1 x 10^-9.

Step 2: The energy calculated in step 1 is the energy of the photons used in the PES experiment. This energy is used to remove electrons from the mercury atoms, so it corresponds to the ionization energy of mercury.

Step 3: Now, we need to compare this ionization energy with the first ionization energies of the halogen elements. The first ionization energies of the halogens can be found in Figure 7.10 or in a table of ionization energies.

Step 4: Compare the ionization energy of mercury calculated in step 2 with the first ionization energies of the halogens. Look for the halogen whose first ionization energy is closest to that of mercury.

Step 5: The halogen with the first ionization energy closest to that of mercury is the answer to the problem. Remember that ionization energy is the energy required to remove an electron from an atom, so a similar ionization energy means that it requires a similar amount of energy to remove an electron from the mercury atom and the halogen atom.

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Key Concepts

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

Photoelectric Effect

The photoelectric effect is a phenomenon where electrons are emitted from a material when it is exposed to light of sufficient energy. This effect is foundational for techniques like photoelectron spectroscopy (PES), as it demonstrates the interaction between light and matter, allowing for the measurement of the energy required to remove electrons from atoms.

Ionization Energy

Ionization energy is the amount of energy required to remove an electron from an atom or ion in its gaseous state. It is a critical property of elements that influences their chemical behavior and reactivity. In the context of PES, ionization energy can be determined by analyzing the kinetic energy of emitted electrons when a sample is irradiated with light.

Ultraviolet Photoelectron Spectroscopy (PES)

Ultraviolet photoelectron spectroscopy (PES) is an analytical technique used to study the electronic structure of atoms and molecules. By directing monochromatic ultraviolet light onto a sample, PES measures the kinetic energy of emitted electrons, allowing researchers to calculate ionization energies and gain insights into the energy levels of electrons in different elements.

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Photoelectric Effect

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?

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. (b) Write an equation that shows the process corresponding to the first ionization energy of Hg.

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?

Textbook Question

Mercury in the environment can exist in oxidation states 0,+1, and +2. One major question in environmental chemistryresearch is how to best measure the oxidation state of mercuryin natural systems; this is made more complicated by thefact that mercury can be reduced or oxidized on surfaces differentlythan it would be if it were free in solution. XPS, X-rayphotoelectron spectroscopy, is a technique related to PES (seeExercise 7.111), but instead of using ultraviolet light to eject valenceelectrons, X rays are used to eject core electrons. The energiesof the core electrons are different for different oxidationstates of the element. In one set of experiments, researchersexamined mercury contamination of minerals in water. Theymeasured the XPS signals that corresponded to electrons ejectedfrom mercury's 4f orbitals at 105 eV, from an X-ray sourcethat provided 1253.6 eV of energy 11 ev = 1.602 * 10-19J2.The oxygen on the mineral surface gave emitted electron energiesat 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 ofthe 4f electrons in mercury and the 1s electrons in oxygenfrom these data to the first ionization energies of mercuryand oxygen from the data in this chapter.

Textbook Question

When magnesium metal is burned in air (Figure 3.6), two products are produced. One is magnesium oxide, MgO. The other is the product of the reaction of Mg with molecular nitrogen, magnesium nitride. When water is added to magnesium nitride, it reacts to form magnesium oxide and ammonia gas. (c) In an experiment, a piece of magnesium ribbon is burned in air in a crucible. The mass of the mixture of MgO and magnesium nitride after burning is 0.470 g. Water is added to the crucible, further reaction occurs, and the crucible is heated to dryness until the final product is 0.486 g of MgO. What was the mass percentage of magnesium nitride in the mixture obtained after the initial burning?

Textbook Question

When magnesium metal is burned in air (Figure 3.6), two products are produced. One is magnesium oxide, MgO. The other is the product of the reaction of Mg with molecular nitrogen, magnesium nitride. When water is added to magnesium nitride, it reacts to form magnesium oxide and ammonia gas. (d) Magnesium nitride can also be formed by reaction of the metal with ammonia at high temperature. Write a balanced equation for this reaction. If a 6.3-g Mg ribbon reacts with 2.57 g NH31g2 and the reaction goes to completion, which component is the limiting reactant? What mass of H21g2 is formed in the reaction?