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Ch.7 - Periodic Properties of the Elements

Chapter 7, Problem 111d

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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Hi everyone. This problem reads ultraviolet photoelectron spectroscopy. Pes a method based on the photo electric effect is used to measure ionization energies, electrons are emitted by illuminating monochromatic light onto a sample. Then the kinetic energy of the emitted electrons is measured. The ionization energy is the difference between the energy of the photons and the kinetic energy of the electrons assume that a PS experiment is conducted in which xenon gas is illuminated with 100 and 2.3 nanometers identify the halogen whose initial ionization energy is closest to xenon and were given the ionization energies of a couple of halogen. Alright, so our goal here is to identify the halogen which or whose initial ionization energy is closest to xenon. Alright, so in this problem we're dealing with wavelengths, we're dealing with photons photo electrons and so what we're going to need to do here is think about the relationship between the energy of a photon and the wavelength of light. Okay, and there is an inverse relationship that is described by the following equation. Okay, and that equation is energy is equal to Planck's constant, times the speed of light over wavelength. And what this equation is going to tell us is energy in jewels per photon. Alright, that's the first thing we're going to need to do is calculate the energy and we'll take a look at what we have based off of what's given in the problem. All right. So h represents plank's constant. And this is a value We should know which is 6.626 times 10 to the negative 34 jewels times seconds C is the speed of light which is also another value we should know which is 3.0 times 10 to the eighth meters per second. And lambda represents way wavelength. And then the problem we're told that the wavelength is 102.3 nanometers. But something that's important is we need to make sure our units match and they're consistent. So if we look at our speed of light, the unit is in meters. And so that means we're going to need to convert our wavelength from nanometers two m. So let's go ahead and do that In one nanometer, there is 10 to the -9 m. Okay. And so we'll see here that are nanometers cancel and we're left with meters. And so once we do that we get 1.23 times 10 to the negative seven m is our wavelength converted. So now that we have all the values, let's go ahead and calculate our energy. Okay, so once we plug everything in we'll go ahead and solve so we know what is our energy and jewels per photon. Okay. Mhm. Alright, so now we have everything we need and so we get our energy is 1.9431 times 10 to the negative 18. Let's clear that up a little bit times 10 to the negative jules per photon. So in one photon this is our energy. Okay, so now we want to identify the halogen, whose initial ionization energy is closest to zenon. And as we can see here our units for ionization energy are in kila jewels per mole. So what that means here is we want to go from jewels or let's write it in a different color. We want to go from jewels per photon, two kila jules Permal. This is our goal. So let's go ahead and start off with what we just calculated. So we have 1.9431 times to the negative 18 jewel per photon. So first we can go we know we need to be in kill a jewel. So we can go from jewels to kill a jules first. Alright, so in one kill a jewel There is 1000 jewels and we'll make sure our units cancel. So our jewels cancel. And now we're in killer jewels. Next we want to go from photon two moles. And remember all the God rose number here and one mole. There is 6. There's 6.022 times 10 to the 23 photons. Okay and again that is over. God rose number. So our photons cancel. And now we're left with killer joules per mole which is what we wanted. All right, so once we do this math, we get 1,170. Kill it jules per mole. And the question is asking us to identify the halogen whose initial ionization energy is closest to xenon. Okay, so looking at the answer choices given which one is closest to 1170.1. Okay. The closest is going to be bro ming. Alright, so browning at 1140 kila jewels per mole is going to be the correct answer. Alright, so that is it for this problem. I hope this was helpful.
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 (d) Use the result from part (b) to predict the X-ray wavelength emitted by iron.

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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 (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. (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
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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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?

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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?

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