SECTION 3.2 The Nature of Light PDF

# SECTION 3.2 The Nature of Light ## Worked Example 3.3 ### Worked Example 3.3 One type of laser used in the treatment of vascular skin lesions is a neodymium-doped yttrium aluminum garnet, or Nd:YAG, laser. The wavelength commonly used in these treatments is 532 nm. What is the frequency of this...

# SECTION 3.2 The Nature of Light ## Worked Example 3.3 ### Worked Example 3.3 One type of laser used in the treatment of vascular skin lesions is a neodymium-doped yttrium aluminum garnet, or Nd:YAG, laser. The wavelength commonly used in these treatments is 532 nm. What is the frequency of this radiation? #### Strategy We must convert the wavelength to meters and solve for frequency using Equation 3.3 (c = λν). #### Setup Rearranging Equation 3.3 to solve for frequency gives ν = c/λ. #### Solution 532 nm x 1 nm /1 x 10^-9 m = 5.32 x 10^-7 m The speed of light, c, is 3.00 x 10⁸ m/s. ν = 3.00 x 10⁸ m/s /5.32 x 10^-7 m = 5.64 x 10¹⁴ s^-1 #### Think About It Make sure your units cancel properly. A common error in this type of problem is neglecting to convert wavelength to meters. ## Practice Problems - **ATTEMPT** What is the wavelength (in meters) of an electromagnetic wave whose frequency is 1.61 x 10¹² s^-1? - **BUILD** What is the frequency (in reciprocal seconds) of electromagnetic radiation with a wavelength of 1.03 cm? - **CONCEPTUALIZE** Which of the following sets of waves best represents the relative wavelengths/frequencies of visible light of the colors shown? - (i) _wavy line_ - (ii) _wavy line_ - (iii) _wavy line_ - (iv) _wavy line_ - (v) _wavy line_ ## Section 3.2 Review - **3.2.1** Calculate the wavelength (in nanometers) of light with frequency 3.45 x 10¹⁴ s^-1. - (a) 1.15 x 10⁶ nm - (b) 1.04 x 10²³ nm - (c) 8.70 x 10² nm - (d) 115 nm - (e) 9.66 x 10^-24 nm - **3.2.2** Calculate the frequency of light with wavelength 126 nm. - (a) 2.38 x 10¹⁵ s^-1 - (b) 4.20 x 10^-16 s^-1 - (c) 37.8 s^-1 - (d) 2.65 x 10^-2 s^-1 - (e) 3.51 x 10¹⁹ s^-1 # PTER 3 Quantum Theory and the Electronic Structure of Atoms - **3.2.3** Of the waves pictured, which has the greatest frequency, which has the greatest wavelength, and which has the greatest amplitude? - (i) _ wavy line_ - (ii) _ wavy line_ - (iii) _ wavy line_ - (a) i, ii, iii - (b) i, iii, ii - (c) ii, i, ii - (d) ii, i, iii - (e) ii, iii, ii - **3.2.4** When traveling through a translucent medium, such as glass, light moves more slowly than it does when traveling through a vacuum. Red light with a wavelength of 684 nm travels through Pyrex glass with a frequency of 2.92 x 10¹⁴ s^-1. Calculate the speed of this light. - (a) 3.00 x 10⁸ m/s - (b) 2.00 x 10⁸ m/s - (c) 2.92 x 10⁶ m/s - (d) 4.23 x 10⁷ m/s - (e) 2.23 x 10⁸ m/s ## 3.3 Quantum Theory Early attempts by nineteenth-century physicists to figure out the structure of the atom met with only limited success. This was largely because they were attempting to understand the inner workings of atoms using the laws of classical physics, which describe the behavior of macroscopic objects. It took a long time to realize-and an even longer time to accept-that the behavior of subatomic particles is not governed by the same physical laws as larger objects. ### Quantization of Energy When a solid is heated, it emits electromagnetic radiation, known as blackbody radiation, over a wide range of wavelengths. The red glow of the element of an electric stove and the bright white light of a tungsten lightbulb are examples of blackbody radiation. Measurements taken in the latter part of the nineteenth century showed that the amount of energy given off by an object at a certain temperature depends on the wavelength of the emitted radiation. Attempts to account for this dependence in terms of established wave theory and thermodynamic laws were only partially successful. One theory was able to explain short-wavelength dependence but failed to account for the longer wavelengths. Another theory accounted for the longer wavelengths but failed for short wavelengths. With no one theory that could explain both observations, it seemed that something fundamental was missing from the laws of classical physics. In 1900, German physicist Max Planck provided the solution and launched a new era in physics with an idea that departed drastically from accepted concepts. Classical physics assumed that radiant energy was continuous; that is, it could be emitted or absorbed in any amount. Based on data from blackbody radiation experiments, Planck proposed that radiant energy could be emitted or absorbed only in discrete quantities, like small packages or bundles. Planck gave the name *quantum* to the smallest quantity of energy that can be emitted (or absorbed) in - **Practice Problem CONCEPTUALIZE** A blue billiard ball with a mass of 165 g rests in a shallow well on an otherwise flat surface. When a red billiard ball with the same mass moving at any velocity less than 1.20 m/s strikes the blue ball, the blue ball does not move (i). When the red ball strikes the blue ball moving at exactly 1.20 m/s, the blue ball is just barely dislodged from the well (ii). What will be the velocity (iii) of the blue ball when it is struck by the red ball moving at 1.75 m/s? - (i) _circle_ - (ii) _circle_ - (iii) _circle_ Einstein's theory of light posed a dilemma for scientists. On the one hand, it explains the photoelectric effect. On the other hand, the particle theory of light is inconsistent with the known wavelike properties of light. The only way to resolve the dilemma is to accept the idea that light possesses properties characteristic of both particles and waves. Depending on the experiment, light behaves either as a wave or as a stream of particles. This concept was totally alien to the way physicists had thought about matter and radiation, and it took a long time for them to accept it. We show in Section 3.5 that possessing properties of both particles and waves is not unique to light but ultimately is characteristic of all matter, including electrons. ## Section 3.3 Review - **3.3.1** Calculate the energy per photon of light with wavelength 650 nm. - (a) 1.29 x 10^-31 J - (b) 4.31 x 10^-40 J - (c) 1.02 x 10^-27 J - (d) 1.44 x 10^-48 J - (e) 3.06 x 10^-19 J - **3.3.2** Calculate the wavelength (in centimeters) of light that has energy 1.32 x 10^-23 J/photon. - (a) 5.02 x 10^-9 cm - (b) 6.64 x 10³ cm - (c) 2.92 x 10^-63 cm - (d) 1.51 cm - (e) 66.4 cm - **3.3.3** Calculate the maximum kinetic energy of an electron ejected by a photon of the light in 3.3.1 from a metal with a work function of 1.56 eV. - (a) 5.61 x 10^-20 J - (b) 684 J - (c) 2.50 x 10^-19 J - (d) 5.56 x 10^-19 J - (e) 6.50 x 10^-7 J - **3.3.4** A clean metal surface is irradiated with light of three different wavelengths: λ1, λ2, and λ3. The kinetic energies of the ejected electrons are as follows: λ1: 2.9 x 10^-20 J; λ2: approximately zero; λ3: 4.2 x 10^-19 J. Arrange the light in order of increasing wavelength. - (a) λ1 < λ2 < λ3 - (b) λ2 < λ1 < λ3 - (c) λ3 < λ2 < λ1 - (d) λ3 < λ1 < λ2 - (e) λ2 < λ3 < λ1 - **3.3.5** Shown here are waves of electromagnetic radiation of two different frequencies and two different amplitudes. Assume that intensity of radiation (photons per second) is directly proportional to amplitude. Which of the waves is made up of photons of greater energy? Which wave delivers more photons during a given period of time? Which wave delivers more total energy during a given time period? - (i) _ wavy line_ - (ii) _ wavy line_ - (a) i, i, i - (b) i, ii, i - (c) ii, i, ii - (d) ii, ii, i - (e) ii, ii, ii # 3.4 BOHR'S THEORY OF THE HYDROGEN ATOM In addition to explaining the photoelectric effect, Planck's quantum theory and Einstein's ideas made it possible for scientists to unravel another nineteenth-century mystery in physics: atomic line spectra. In the seventeenth century, Newton had shown that sunlight is composed of various color components that can be recombined to produce white light. Since that time, chemists and physicists have studied the characteristics of such emission spectra. The emission spectrum of a substance can be seen by energizing a sample of material with either thermal energy or some other form of energy (such as a high-voltage electrical discharge if the substance is a gas). A "red-hot" or "white-hot" iron bar freshly removed from a fire produces a characteristic glow. The glow is the visible portion of its emission spectrum. The heat given off by the same iron bar is another portion of its emission spectrum-the infrared region. A feature common to the emission spectrum of the Sun and that of a heated solid is that both are continuous; that is, all wavelengths of visible light are present in each spectrum (Figure 3.6). - **Figure 3.6** The visible white light emitted by (a) the Sun and (b) a white-hot iron bar. In each case, the white light is the combination of all visible wavelengths (see Figure 3.1). - (a) _Image not included._ - (b) _Image not included._ # Figure 3.9 ## Emission spectrum of hydrogen. - _Image not included._ - The transitions shown are as follows: - n = 3 to n = 2 - n = 4 to n = 2 - n = 5 to n = 2 - n = 6 to n = 2 - Electrons in H atoms may be promoted to excited states other than 3, 4, 5, or 6; and electrons in excited states may return to a state other than n = 2. However, the transitions shown are the ones that give rise to the visible lines in the hydrogen emission spectrum. - Hydrogen atoms in the ground state are excited by the addition of energy. Electrons that have been promoted to higher energy levels (n > 1) return to lower energy levels and emit the excess energy as electromagnetic radiation. ### What's the point? Each line in the visible emission spectrum of hydrogen is the result of an electronic transition from a higher excited state (n = 3, 4, 5, or 6) to a lower excited state (n = 2). The energy gap between the initial and final states determines the wavelength of the light emitted. # CHAPTER 4 Periodic Trends of the Elements ## SECTION 4.3 Electron Configurations and the Modern Periodic Table - **Think About It** Review Figure 4.2 to make sure you have filled the orbitals in the correct order and the sum of electrons is 13. - **Practice Problem ATTEMPT** Write the electron configuration and give the orbital diagram of a rubidium (Rb) atom (Z = 37). - **Practice Problem BUILD** Write the electron configuration and give the orbital diagram of a bromine (Br) atom (Z = 35). - **Practice Problem CONCEPTUALIZE** Imagine an alternate universe in which the allowed values of the magnetic quantum number, ml, can have values of -(l + 1) ... 0 ... +(l + 1). In this alternate universe, what would be the maximum number of electrons that could have the principal quantum number 3 in a given atom? ### Thinking Outside the Box How do we know that ns orbitals fill before (n - 1)d orbitals? As shown in Figures 4.2 and 4.4, the 4s subshell is filled before the 3d subshell in a many-electron atom. Thus, the electron configuration of potassium (Z = 19) is 1s²2s²2p⁶3s²3p⁶4s¹. The placement of the outermost electron in the 4s orbital (rather than in the 3d orbital) of potassium is strongly supported by experimental evidence. The physical and chemical properties of potassium are very similar to those of lithium and sodium, the first two members of Group 1 (alkali metals). In both lithium and sodium, the outermost electron is in an s orbital (there is no doubt that their outermost electrons occupy s orbitals because there is no 1d or 2d subshell). Based on its similarities to the other alkali metals, we expect potassium to have an analogous electron configuration; that is, we expect the last electron in potassium to occupy the 4s orbital rather than a 3d orbital. ## Section 4.2 Review ### Electron Configurations - **4.2.1** Which of the following electron configurations correctly represents the Na atom? - (a) 1s²2s²2p⁶ - (b) 1s²2s²2p⁶3s¹ - (c) 1s²2s²2p⁶3s²3p⁴s² - (d) 1s²2s²2p⁶ - (e) 1s²2s²2p⁶3s²3p⁶ - **4.2.2** What element is represented by the following electron configuration? - (a) Br - (b) As - (c) S - (d) Se - (e) Te 1s²2s²2p⁶3s²3p⁴s²3d¹⁰4p³ - **4.2.3** Which orbital diagram is correct for the ground-state S atom? - (a) 1s² 2s² 2p⁶ 3s² 3p⁶ - (b) 1s² 2s² 2p⁶ 3s² 3p⁶ - (c) 1s² 2s² 2p⁶ 3s² 3p⁶ - (d) 1s² 2s² 2p⁶ 3s² 3p⁶ - (e) 1s² 2s² 2p⁶ 3s² 3p⁶ ## 4.3 Electron Configurations and the Modern Periodic Table Thus far, the electron configurations we have written have all been of manageable length. However, for elements with high atomic numbers, they can be quite long. Fortunately, the process of representing long electron configurations can be simplified by using a noble gas core. A noble gas core shows in brackets the completed-shell electron configuration of the noble gas element that most recently precedes the element in question, followed by the electron configuration in the outermost occupied subshells. The electrons that make up the noble gas core are known as the core electrons. The electrons in the outermost occupied subshells are known as the *valence electrons*. As we saw in Section 4.2, the 4s subshell is filled before the 3d subshell in a many-electron atom (see Figure 4.2). Thus, the electron configuration of calcium (Z = 20) is 1s²2s²2p⁶3s²3p⁶4s². Because 1s²2s²2p⁶3s²3p⁶ is the electron configuration of argon, we can simplify the electron configuration of calcium by writing [Ar]4s², where [Ar] denotes the "argon core." Figure 4.5 gives the ground-state valence-electron configurations of elements from H (Z = 1) through element 118. There is a distinct pattern to the electron configurations of the elements in a particular group. See, for example, the electron configurations of Groups 1 and 2 in Table 4.1. Each member of Group 1 has a noble gas core plus one additional electron, giving each alkali metal the general electron configuration of [noble gas]ns¹. Similarly, the Group 2 alkaline earth metals have a noble gas core and a valence-electron configuration of ns².&#x20; Valence electrons determine how atoms interact with one another. Having the same valence-electron configuration is what causes the elements in a group, such as - **4.3.1** Which of the following electron configurations correctly represents the Ag atom? - (a) [Kr]5s²4d⁹ - (b) [Kr]5s²4d¹⁰ - (c) [Kr]5s¹4d¹⁰ - (d) [Xe]5s²4d⁹ - (e) [Xe]5s¹4d¹⁰ - **4.3.2** What element is represented by the following electron configuration: [Xe]5s²4d¹⁰5p³? - (a) Tc - (b) Br - (c) I - (d) Te - (e) None. This is not a correct electron configuration. - **4.3.3** Which of the following is a p-block element? (Select all that apply.) - (a) Sb - (b) Au - (c) Ca - (d) Zn - (e) U - **4.3.4** What is the correct electron configuration for a chlorine (Cl) atom in the ground state? - (a) 1s²2s²2p⁶3s²3p⁴s²4p³ - (b) 1s²2s²2p⁶3s²3p⁵ - (c) 1s²2s²2p⁶3s²3p⁶ - (d) 1s²2s²2p⁶3p⁵ - (e) 1s²2s²2p⁶3s²3p⁵3d¹⁰ - **4.3.5** What element is represented by the ground-state electron configuration 1s²2s²2p⁶3s²3p⁶4s²3d¹⁰4p⁵5s²? - (a) Pb - (b) Ge - (c) In - (d) Sr - (e) Sb - **4.3.6** Which of the following is a d-block element? (Select all that apply.) - (a) Pb - (b) C - (c) Sr - (d) Xe - (e) Fe ## 4.4 EFFECTIVE NUCLEAR CHARGE As we have seen, the electron configurations of the elements show a periodic variation with increasing atomic number. In this and the next few sections, we will examine how electron configuration explains the periodic variation of physical and chemical properties of the elements. We begin by introducing the concept of effective nuclear charge. Nuclear charge (Z) is simply the number of protons in the nucleus of an atom. Effective nuclear charge (Zeff) is the actual magnitude of positive charge that is "experienced" by an electron in the atom. The only atom in which the nuclear charge and effective nuclear charge are the same is hydrogen, which has only one electron. In all other atoms, the electrons are simultaneously attracted to the nucleus and repelled by one another. This results in a phenomenon known as *shielding*. An electron in a many-electron atom is partially shielded from the positive charge of the nucleus by the other electrons in the atom. One way to illustrate how electrons in an atom shield one another is to consider the amounts of energy required to remove the two electrons from a helium atom, shown in Figure 4.8. Experiments show that it takes 3.94 x 10^-18 J to remove the first electron but 8.72 x 10^-18 J to remove the second one. There is no shielding once the first electron is removed, so the second electron feels the full effect of the +2 nuclear charge and is more difficult to remove. Although all the electrons in an atom shield one another to some extent, those that are most effective at shielding are the core electrons. As a result, the value of Zeff increases steadily from left to right across a period of the periodic table because the number of core electrons remains the same (only the number of protons, Z, and the number of valence electrons increases).&#x20; - **Figure 4.8** Removal of the first electron in He requires less energy than removal of the second electron because of shielding. - _Image not included._ ## Section 4.4 Review ### Electron Configurations and the Modern Periodic Table - **4.4.1** Which of the following is not a correct definition of Z_eff? (Select all that apply.) - (a) The effective nuclear charge experienced by an electron in a many-electron atom. - (b) The total positive charge experienced by a given electron in a many-electron atom. - (c) The charge that an electron actually experiences in a many-electron atom. - (d) The nuclear charge minus the shielding effect of the other electrons. - (e) The actual nuclear charge of the nucleus, Z, minus the core electrons. - **4.4.2** What is the correct electron configuration for a chlorine (Cl) atom given that the Zeff experienced by its valence electrons is 6? - (a) 1s²2s²2p⁶3s²3p⁴s²4p³ - (b) 1s²2s²2p⁶3s²3p⁵ - (c) 1s²2s²2p⁶3s²3p⁶ - (d) 1s²2s²2p⁶3p⁵ - (e) 1s²2s²2p⁶3s²3p⁵3d¹⁰ - **4.4.3** What element is represented by the ground-state electron configuration 1s²2s²2p⁶3s²3p⁶4s²3d¹⁰4p⁵5s²? - (a) Pb - (b) Ge - (c) In - (d) Sr - (e) Sb - **4.4.4** Which of the following is not a good explanation for the trend of Zeff values in the periodic table? - (a) Increasing Z_eff results in decreasing atomic radius. - (b) Increasing Z_eff results in increasing ionization energy. - (c) Increasing Z_eff results in decreasing electron affinity. - (d) Increasing Z_eff results in decreasing shielding. - (e) Increasing Z_eff results in increasing electronegativity.