Ch 7 H atom, quantum theory, spectroscopy PDF

Ch. 7 – Hydrogen Atom: Spectroscopy, Quantum Theory and Atomic Structure Chemistry The Molecular Nature of Matter and Change Ninth Edition Martin S. Silberberg and Patricia G. Amateis ©McGraw-Hill Education. All rights reserved. Authorized only for instructor use in the classroom. No reproduction or...

Ch. 7 – Hydrogen Atom: Spectroscopy, Quantum Theory and Atomic Structure Chemistry The Molecular Nature of Matter and Change Ninth Edition Martin S. Silberberg and Patricia G. Amateis ©McGraw-Hill Education. All rights reserved. Authorized only for instructor use in the classroom. No reproduction or further distribution permitted without the prior written consent of McGraw-Hill Education. Neon Light Neon gas, enclosed in a glass tube, is subjected to electric power arcing (discharge) between 2 electrodes. Once the arc strikes, Neon emits lights of certain colors only, not the full visible spectrum. The predominant color for Ne is orange-red. The blue lights in the figure are generated by discharge in other gases. What causes Ne and other gases to emit visible light? ©McGraw-Hill Education. The Nature of Light: Its Wave Nature Light: a form of electromagnetic radiation – Composed of perpendicular oscillating waves, one for the electric field and one for the magnetic field An electric field is a region where an electrically charged particle experiences a force. A magnetic field is a region where a magnetized particle experiences a force. All electromagnetic waves move through space (vacuum) at the same constant speed. – 𝑐 = 2.997 924 58 × 108 𝑚 𝑠 ≈ 3.00 × 108 𝑚 𝑠 [3 sig. figs.] – the speed of light in materials (e.g., air) is smaller. © 2017 Pearson Education, Inc. What is an Electromagnetic Wave? Consists of 2 waves: an electric field wave + a magnetic field wave The fields oscillate according to a sine function. The fields are periodic: 1 period = 1 complete oscillation. The fields are perpendicular to each other and in phase, according to the laws of electromagnetism in Physics. © 2017 Pearson Education, Inc. Speed of a Wave An acoustic wave is NOT an electromagnetic wave. Sound can only propagate in a medium (e.g., air), it cannot propagate in vacuum. © 2017 Pearson Education, Inc. Description of Waves The wavelength (l, Greek lambda) is: – The distance on the x axis from one crest to the next. – Alternative definitions: the distance from one trough to the next, or the distance between alternate nodes. – Over this distance, the wave completes 1 cycle. The amplitude is the height of the wave. – The magnitude of the sine function on the y axis: measured from node to crest or node to trough. – The amplitude is a measure of light intensity—the larger the amplitude, the brighter the light. © 2017 Pearson Education, Inc. Description of Waves The frequency (n, Greek nu) is the number of waves that pass a point in a given period of time. – The number of waves = the number of cycles. – Units are hertz (Hz) or cycles/s = s−1 (1 Hz = 1 s−1). – It is the inverse of the time (duration) T required for a wave crest to advance to the position of the following crest, known as the wave period. A wave period is the time the wave needs to advance exactly 1 λ: 1 𝜈= 𝑇 © 2017 Pearson Education, Inc. Amplitude and Wavelength in Classical Electromagnetism In the classical wave theory, the total energy of a wave is proportional to the squared amplitude of the wave (A2) and the frequency. The larger the amplitude, the more force it has. The more frequently the waves strike, the more total force there is. © 2017 Pearson Education, Inc. The Relationship between Wavelength and Frequency The definition of the speed of light based on the new concepts of wavelength, period and frequency: 𝜆 𝑐 = = 𝜆𝜈 𝑇 𝑐 𝑐 𝜈= 𝑜𝑟 𝜆 = 𝜆 𝜈 Example: the classroom red light pointer used to be a He-Ne laser which emits orange-red light at a wavelength of 633 nm. What is the frequency of this color? 8𝑚 3.00 × 10 𝑐 𝑠 14 𝐻𝑧 = 474 𝑇𝐻𝑧 𝜈= = = 4.74 × 10 1 × 10−9 𝑚 𝜆 633 𝑛𝑚 × 1 𝑛𝑚 © 2017 Pearson Education, Inc. Sample Problem 7.1 – Problem and Plan Interconverting Wavelength and Frequency PROBLEM: A dental hygienist uses xrays (λ = 0.50 Å) to take a series of dental radiographs while the patient listens to a radio station (λ = 325 cm)and looks out the window at the blue sky (λ = 473 nm). What is the frequency (in s−1) of the electromagnetic radiation from each source? (Assume that the radiation travels at the speed of light, 3.00×108 m/s.) PLAN: Use the equation c = nl to convert wavelength to frequency. Wavelengths need to be in meters because c has units of m/s. ©McGraw-Hill Education. Sample Problem 7.1 – Solution: 10−10 m For the x-rays: l = 0.50 Å x = 5.00 x 10−11 m 1Å 1 Å Ångström = 10−10 𝑚 (1 𝑎𝑡𝑜𝑚𝑖𝑐 𝑑𝑖𝑎𝑚𝑒𝑡𝑒𝑟) 8 𝑚 3.00 x 10 c 𝑠 = 6.00 x 𝟏𝟎𝟏𝟖 s−1 n= = l 5.00 x 10−11 m For the radio signal: 𝑚 3.00 x 10 c 𝑠 = 9.23 x 𝟏𝟎𝟕 𝒔−𝟏 = 𝟗𝟐. 𝟑 𝑴𝑯𝒛 (𝑭𝑴) n= = l 325 cm x 10−2 m 1 cm 8 For the blue sky: 8 𝑚 3.00 x 10 c 𝑠 = 6.34 x 𝟏𝟎𝟏𝟒 𝒔−𝟏 = 𝟔𝟑𝟒 𝑻𝑯𝒛 n= = l 473 nm x 10−9 m 1 nm ©McGraw-Hill Education. White Light: sum of all visible colors (Newton) © 2017 Pearson Education, Inc. Color The color of light is determined by its wavelength or frequency. White light is a mixture of all the colors of visible light. – A spectrum is a collection of multiple colors traveling together – Red Orange Yellow Green Blue (Cyan) Indigo (Navy Blue) Violet (Purple) When an object absorbs some of the wavelengths of white light and reflects others, it appears colored; the observed color is predominantly the colors reflected. Why is the sweater red? The dye either absorbs all colors except for red, or absorbs only cyan, leaving red uncompensated in the white light (red and cyan are complementary colors). © 2017 Pearson Education, Inc. Complementary Colors (Rosetta of VIS spectrum) © 2017 Pearson Education, Inc. Electromagnetic Spectrum © 2017 Pearson Education, Inc. The Electromagnetic Spectrum Visible light comprises only a small fraction of all the wavelengths of light, called the electromagnetic spectrum. Shorter wavelength (high-frequency) light has higher energy. – Radio wave light has the lowest energy. – Gamma ray light has the highest energy. High-energy electromagnetic radiation can potentially damage biological molecules: UV, X, γ – Ionizing radiation © 2017 Pearson Education, Inc. Thermal Imaging Using Infrared Light © 2017 Pearson Education, Inc. Interference The interaction between overlapping waves is called interference. https://www.youtube.com/watch?v=v_uBaBuarEM © 2017 Pearson Education, Inc. Interference © 2017 Pearson Education, Inc. Diffraction of Waves When traveling waves encounter an obstacle or opening in a barrier that is about the same size as the wavelength, they bend around it; this is called diffraction. The diffraction of light through two slits separated by a distance comparable to the wavelength results in an interference pattern of the diffracted waves. © 2017 Pearson Education, Inc. Refraction and Dispersion Light of a given frequency travels at different speeds through various transparent media. When changing media, both the speed of light and the wavelength change, but the frequency stays constant. However, because light is a wave, the trajectory also changes. This is known as refraction. Dispersion: different angles of refraction for different frequencies, at the same angle of incidence. An optical prism is able to disperse the white light into monochromatic components because of the dispersion phenomenon. ©McGraw-Hill Education. Blackbody Radiation Observation: A solid object emits a continuous spectrum of visible light when it is heated to about 1000 K or higher, not just particular wavelengths. This is called blackbody radiation. In blackbody radiation, the color (and the intensity ) of the light changes as the temperature changes. Color is related to wavelength and frequency, while temperature is related to energy. https://www.youtube.com/watch?v=h5jOAw57OXM&t=352s ©McGraw-Hill Education. Blackbody Radiator Blackbody Radiator and credit: https://phys.libretexts.org/Bookshelves/University_Physics/Book% 3A_University_Physics_(OpenStax)/University_Physics_III__Optics_and_Modern_Physics_(OpenStax)/06%3A_Photons_and_ Matter_Waves/6.02%3A_Blackbody_Radiation ©McGraw-Hill Education. The Quantum Theory of Electromagnetic Energy (Max Planck, 1900) Any object (including atoms) can emit or absorb only certain quantities of energy. Energy emitted or absorbed by matter is quantized; it occurs in fixed quantities, rather than being continuous. Each fixed quantity of energy emitted or absorbed is called a quantum of light (Albert Einstein – 1905). The name “quantum of light” was later changed to photon, as proposed by G.N. Lewis in 1926. ©McGraw-Hill Education. Max Planck, 1858-1947 Nobel Prize in Physics, 1918 Max Planck (1900): Energy of light 𝑬 = 𝒉𝝂 The energy of a photon of light is directly proportional to its frequency. – Inversely proportional to its wavelength – The proportionality constant is called Planck’s Constant, (h) and has the value 6.626 × 10−34 J ∙ s. Calculate the energy of a photon of 633 nm: 8𝑚 3.00 × 10 𝑠 𝑐 −34 −19 𝐽 𝐸 = ℎ = 6.626 × 10 𝐽𝑠 = 3.14 × 10 10−9 𝑚 𝜆 633 𝑛𝑚 1 𝑛𝑚 © 2017 Pearson Education, Inc. Max Planck (1900): Blackbody radiation The continuous emission spectrum of blackbody radiation cannot be explained by the laws of classical physics. To explain the Blackbody radiation, Max Plank in 1900 introduced two new equations: – Classical Entropy (Clausius) based on Boltzmann’s theories must have the following statistical formula: 𝑆 = 𝑘𝐵 ln 𝑊. This equation was not proposed or derived by Boltzmann as stated in most textbooks, but he came close to it in 1896. This is still based on classical physics. Plank showed that the proportionality factor kB is a fundamental constant, and he was able to calculate it with high accuracy from the blackbody 𝐽 radiation curves: 𝑘𝐵 = 1.38 × 10−23. 𝐾 – Departure from classical physics: not all energies are emitted or absorbed by molecules of a glowing blackbody radiator, only certain discrete energies: 𝜀 = 𝑛ℎ𝜈, where n is an integer and h is another fundamental constant: ℎ = 6.62 × 10−34 𝐽 𝑠. © 2017 Pearson Education, Inc. Max Planck (1900): Blackbody radiation Plank proposed the following equation for the emission curve of a blackbody radiator. The energy density is: 2ℎ𝜈 3 𝐼 𝜈, 𝑇 = 2 𝑐 1 ℎ𝜈 𝑒 𝑘𝐵 𝑇 −1 This equation was able to explain all known facts about the blackbody radiation curve and allowed Planck to calculate the values for h and kB from available experimental curves. © 2017 Pearson Education, Inc. Sample Problem 7.2 – Problem, Plan and Solution Interconverting Energy, Wavelength and Frequency PROBLEM: A student uses a microwave oven to heat a meal. The wavelength of the radiation is 1.20 cm. What is the energy of one photon of this microwave radiation? PLAN: We know l in cm, so we convert to m and find the frequency using the speed of light. We then find the energy of one photon using E = hn. SOLUTION: −34 8 𝑚 (6.626 x 10 J∙s)(3.00 x 10 ) hc 𝑠 E = hn= = = 1.66 x 𝟏𝟎−𝟐𝟑 J −2 10 m l (1.20 cm) 1 cm ©McGraw-Hill Education. The Photoelectric Effect (1901-P. Lenard, 1914R. Millikan) Figure 7.7 ©McGraw-Hill Education. It was observed that all metals emit electrons when a light of a minimum frequency shines on the surface of metal. – This is called the photoelectric effect. Classic wave theory attributed this effect to the light energy being transferred to the electron. According to the classical theory, if the wavelength of light is made shorter, or the light wave’s intensity made brighter, more electrons should be ejected. – Remember that the energy of a wave is directly proportional to its amplitude and its frequency. – This idea predicts if a dim light (low amplitude) or light of any frequency were used there would be a lag time before electrons were emitted (to give the electrons time to absorb enough energy). The Photoelectric Effect: The Problem Experimental observations indicate the following (R. Millikan – 1914, predicted by A. Einstein – 1905): – A minimum frequency was needed before electrons would be emitted, regardless of the intensity, called the threshold frequency. – Above-threshold frequency light from a dim source caused electron emission without any lag time. © 2017 Pearson Education, Inc. Einstein’s Explanation (1905) Einstein proposed that the light energy was delivered to the atoms in packets, called quanta (later called photons). One photon at the threshold frequency gives the electron just enough energy for it to escape the atom. – Binding energy, f (or surface work function) When irradiated with a shorter wavelength photon, the electron absorbs more energy than is necessary to escape. This excess energy becomes kinetic energy of the ejected electron: Kinetic Energy = Ephoton – Ebinding KE = hn − f © 2017 Pearson Education, Inc. Class Question Suppose a metal will eject electrons from its surface when struck by yellow light. What will happen if the surface is struck with ultraviolet light? a. No electrons would be ejected. b. Electrons would be ejected, and they would have the same kinetic energy as those ejected by yellow light. c. Electrons would be ejected, and they would have greater kinetic energy than those ejected by yellow light. d. Electrons would be ejected, and they would have lower kinetic energy than those ejected by yellow light. © 2017 Pearson Education, Inc. 7.2 Atomic Spectra When individual atoms or molecules absorb energy, that energy is often released quickly as light energy (photons). – Fireworks, neon lights, etc. When that emitted light is passed through a prism or a diffraction grating, a pattern of particular wavelengths of light is seen that is unique to that type of atom or molecule; the pattern is called an emission spectrum. – Non-continuous: only specific wavelengths are present – Can be used to identify the material – Flame tests © 2017 Pearson Education, Inc. Exciting Gas Atoms to Emit Light CHEM 123 experiment © 2017 Pearson Education, Inc. Light is emitted when gas atoms are excited via external energy (e.g., electricity or flame). Each element emits a characteristic color of light. What you see here to the left is only the dominant color in the emission spectrum, overlapping a white background, which is the result of the combination of all other lines. Emission Spectra: “lines” are generated by a slit in the monochromator of spectrograph To see the emission lines, the white light must be dispersed, by a prism or a diffraction grating. A diffraction grating works better. © 2017 Pearson Education, Inc. Examples of Spectra © 2017 Pearson Education, Inc. Identifying Elements with Flame Tests Na © 2017 Pearson Education, Inc. K Li Ba Emission versus Absorption Spectra Emission spectrum Absorption spectrum Spectra of Mercury © 2017 Pearson Education, Inc. The Line Spectrum of Hydrogen Figure 7.8 ©McGraw-Hill Education. Spectral Lines of Atomic Hydrogen Vacuum Ultraviolet (VUV) series (Lyman): around 100 nm Visible (VIS-UV) series (Balmer): 300 – 700 nm Near-Infrared (NIR) series (Paaschen): 800 – 1900 nm Figure 7.9 ©McGraw-Hill Education. Rydberg’s Analysis of Hydrogen Spectrum Rydberg analyzed the spectrum of atomic hydrogen and found that it could be described with an equation that involved an inverse square of integers (1888): 1 1 1 =𝑅 − 2 2 𝜆 𝑚 𝑛 𝑤ℎ𝑒𝑟𝑒 𝑚, 𝑛 𝑎𝑟𝑒 𝑖𝑛𝑡𝑒𝑔𝑒𝑟𝑠 𝑅 = 1.097 × 107 𝑚−1 𝑖𝑠 𝑡ℎ𝑒 𝑅𝑦𝑑𝑏𝑒𝑟𝑔 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 It is an empirical equation. Johannes Rydberg Physicists could not come up with a Swedish physicist 1854 - 1919 theory to explain this formula until the early 20th century. © 2017 Pearson Education, Inc. Rydberg’s Analysis of Hydrogen Spectrum 1 1 1 Rydberg equation: = R 2 − 2 n1 n2 l where R = 1.096776×107 m−1 and n1 < n2 1) VUV (Lyman) series: n1 = 1, n2 = 2, 3, …, ∞ 2) VIS (Balmer) series: n1 = 2 and n2 = 3, 4,..., ∞ 3) NIR (Paaschen) series: n1 = 3 and n2 = 4, 5, …, ∞ © 2017 Pearson Education, Inc. Atomic Spectroscopy Explained (Niels Bohr, 1913) Each wavelength in the discrete spectrum of an atom corresponds to an electron transition between levels. When an electron is excited, it transitions from a lower energy level to a higher energy level by absorbing energy. The source of energy can be manifold: – A photon: the photon energy must exactly match the Niels Bohr, 1885-1962 energy difference between levels, else the excitation Danish physicist Nobel prize 1922 does not happen and the photon is not absorbed. – Electric discharge in a gas – Internal rearrangement of electrons within the atom or molecule. When an electron relaxes, it transitions from a higher energy level to a lower energy level. When an electron relaxes, a photon of light is released whose energy equals the energy difference between the levels. © 2017 Pearson Education, Inc. Electron Transitions To transition to a higher energy state, the electron must gain the correct amount of energy corresponding exactly to the difference in energy between the final and initial levels (or quantum states). Electrons in high energy levels are unstable and tend to lose energy and transition to lower energy levels. Each line in the emission spectrum corresponds to the difference in energy between two energy levels. © 2017 Pearson Education, Inc. Predicting the Spectrum of Hydrogen The wavelengths of lines in the emission spectrum of hydrogen and oneelectron ions (He+, Li2+ etc.) can be predicted by calculating the difference in energy between any two levels. The energy of a photon released is equal to the difference in energy between the two levels between which the electron is jumping. It can be calculated by subtracting the energy of the initial state from the energy of the final state: DEelectron = Efinal state − Einitial state Eemitted photon = −DEelectron 𝑍2 𝐸𝑒 − = −2.18 × 𝐽 2 𝑛 1 1 −18 = −∆𝐸𝑒 − = +2.18 × 10 2− 2 𝑛𝑖 𝑛𝑓 10−18 𝐸ℎ𝜈 1 1 1 𝑅𝑦𝑑𝑏𝑒𝑟𝑔: = 𝑅𝐻 2 − 2 𝜆 𝑛𝑖 𝑛𝑓 © 2017 Pearson Education, Inc. → 𝐸ℎ𝜈 ℎ𝑐 1 1 = = ℎ𝑐𝑅𝐻 2 − 2 𝜆 𝑛𝑖 𝑛𝑓 Energy Levels in Hydrogen © 2017 Pearson Education, Inc. Hydrogen Spectrum Lines Figure 7.11 ©McGraw-Hill Education. A Quantum “Staircase” Figure 7.10 ©McGraw-Hill Education. An Analogy for the Energy of a System DE = Efinal – Einitial = ©McGraw-Hill Education. –2.18x𝟏𝟎−𝟏𝟖 J 𝟏 𝒏𝟐𝒇𝒊𝒏𝒂𝒍 − 𝟏 𝒏𝟐𝒊𝒏𝒊𝒕𝒊𝒂𝒍 Sample Problem 7.3 – Problem and Plan Determining DE and l of an Electron Transition PROBLEM: A hydrogen atom absorbs a photon of VUV light (see Figure 7.11) and its electron enters the n = 4 energy level. Calculate (a) the change in energy of the atom and (b) the wavelength (in nm) of the absorbed photon. PLAN: (a) The H atom absorbs energy, so Efinal > Einitial. We are given nfinal = 4, and Figure 7.11 shows that ninitial = 1 because a VUV photon is absorbed. We apply Equation 7.5 to find DE. (b) Once we know DE, we find frequency and wavelength. Then we convert from meters to nanometers. ©McGraw-Hill Education. Sample Problem 7.3 - Solution SOLUTION: (a) DE = –2.18x10−18 1 1 J 2 − 2 = 2.04 x 𝟏𝟎−𝟏𝟖 J 4 1 (6.626x10−34 J∙s)(3.00x108 m/s) (b) l = = = 9.74 x 𝟏𝟎−𝟖 m 8 DE 2.04x10 J ℎ𝑐 9.74x10−8 ©McGraw-Hill Education. 1 nm m x −9 = 97.4 nm 10 m 7.3 The Wave-Particle Duality of Matter and Energy Matter and Energy are interconvertible (A. Einstein, theory of special relativity, 1905). E = mc2 All matter exhibits properties of both particles and waves. Electrons have wave-like motion and therefore have only certain allowable frequencies and energies. Matter behaves as though it moves in a wave, and the de Broglie wavelength for any particle is given by (L. de Broglie, 1924, PhD thesis): h l= mu m = mass ©McGraw-Hill Education. u = speed in m/s French physicist Nobel prize in Physics, 1929 Electron Diffraction and Interference Proof that the electron had wave nature came in 1927 (Davisson and Germer) with the demonstration that a beam of electrons would produce a diffraction pattern the same as waves do. Electrons also present an interference pattern, demonstrating they behave like waves. Fig.7.13.b. Electron diffraction on polycrystalline Al. © 2017 Pearson Education, Inc. Low Energy Electron Diffraction on a single crystal Al(111) surface. Values of the de Broglie Wavelengths Equations for photons and particles are NOT interchangeable. ©McGraw-Hill Education. Sample Problem 7.4 – Problem, Plan and Solution Calculating the de Broglie Wavelength of an Electron PROBLEM: Find the deBroglie wavelength of an electron with a speed of 1.00x106 m/s (electron mass = 9.11x10-31 kg; h = 6.626x10-34 kgm2/s). PLAN: We know the speed and mass of the electron, so we substitute these into Equation 7.6 to find l. SOLUTION: 2 𝑚 −34 6.626x10 kg∙ 2 𝑠 h 𝑠 −𝟏𝟎 m =7.27 Å λ= = = 7.27 x 𝟏𝟎 mu 9.11x10−31 kg x 1.00x106 𝑚 𝑠 1 Å Ångström = 10−10 𝑚 (1 𝑎𝑡𝑜𝑚𝑖𝑐 𝑑𝑖𝑎𝑚𝑒𝑡𝑒𝑟) ©McGraw-Hill Education. From Classical Theory to Quantum Theory Figure 7.15 ©McGraw-Hill Education. Heisenberg’s Uncertainty Principle (1927) In classical Physics, a moving particle has a definite position at any time; a wave is spread out in space. If an electron is both a particle and a wave, can we determine the electron position? Heisenberg’s Uncertainty Principle states that it is not possible to know both the position and momentum (𝑝 = 𝑚 ∙ 𝑢) of a moving particle at the same time: ℎ ∆𝑥 ∙ 𝑚∆𝑢 ≥ 4𝜋 x = position u = speed The more accurately we know the speed, the less accurately we know the position, and vice versa. The best we can do is to describe the probability an electron will be found in a particular region using statistical functions. ©McGraw-Hill Education. Trajectory versus Probability © 2017 Pearson Education, Inc. Sample Problem 7.5 – Problem, Plan and Solution Applying the Uncertainty Principle PROBLEM: An electron moving near an atomic nucleus has a speed 6x106 m/s ± 1%. What is the uncertainty in its position (Dx)? PLAN: The uncertainty in the speed (Du) is given as ±1% (0.01) of 6x106 m/s. We multiply u by 0.01 and substitute this value into Equation 7.6 to solve for Δx. SOLUTION: Du = (0.01)(6x106 m/s) = 6x104 m/s h Dx∙mDu ≥ 4p h 6.626x10−34 kg∙m2/s −𝟗 m Dx ≥ ≥ ≥ 1 x 𝟏𝟎 4pmDu 4p (9.11x10−31 kg)(6x104 m/s) ©McGraw-Hill Education. 1926: Erwin Schrödinger and the Quantum Mechanics equation for the quantum state Rather than solving Newton’s equations of motion for a particle, an appropriate wave equation which incorporates the momentum of a particle needs to be solved. Schrödinger was the first to formulate such an equation successfully. A quantum mechanical system is characterized by a wave function. A wave function is a complete description of the system: any measurable property can be obtained if the wave function is known. Austrian physicist Nobel prize, 1933 Time-independent non-relativistic Schrödinger 1st Equation: for systems that do not evolve in time (e.g., a “bound” particle in a nonevolving potential well) ℎ2 𝜕2 𝜕2 𝜕2 − 2 + + + 𝑉 𝑥, 𝑦, 𝑧 Ψ 𝑥, 𝑦, 𝑧 = 𝐸Ψ(𝑥, 𝑦, 𝑧) 8𝜋 𝑚 𝜕𝑥 2 𝜕𝑦 2 𝜕𝑧 2 where: Ψ 𝑥, 𝑦, 𝑧 - the wave function, V(x,y,z) - the analytical expression of the potential energy E is the scalar value of the total energy of the quantum system 1st great success: The solutions to this equation describe the Hydrogen atom. Challenge: the wave function solutions of this differential equation are complex (contain the imaginary part 𝑖 = −1). 7.4 Schrödinger Equation Schrödinger’s equation allows us to calculate the probability of finding an electron with a particular amount of energy at a particular location in the atom. In this equation, Ψ is an electron wavefunction inside the atom, and E is the total energy (sum of potential and kinetic energies) of that electron corresponding to a particular wavefunction. H is the Hamiltonian operator: 𝐻 = ℎ2 𝜕2 − 2 8𝜋 𝑚 𝜕𝑥 2 + 𝜕2 𝜕𝑦 2 + 𝜕2 𝜕𝑧 2 + 𝑉 𝑥, 𝑦, 𝑧 A wavefunction is a mathematical function associated with the de Broglie wave of the electron in the electric field of a proton (this is what a Hydrogen atom is). Solutions to Schrödinger’s equation produce many wavefunctions, Y and many energies, E. Interestingly, the many possible solutions to the Schrödinger’s equation differ among themselves by certain integer exponents and terms: n, l and m. © 2017 Pearson Education, Inc. Schrödinger equation solutions for the Hydrogen atom 1. Total energies E of the Hydrogen atom (proton + electron) are quantized and depend on a single quantum number (positive integer), n: 𝜇𝑒 4 1 1 1 1 𝐸 = − 2 2 ∙ 2 𝑤ℎ𝑒𝑟𝑒 = + 𝑖𝑠 𝑡ℎ𝑒 𝑅𝑒𝑑𝑢𝑐𝑒𝑑 𝑀𝑎𝑠𝑠 𝑜𝑓 𝐻 𝑛 𝜇 𝑚 𝑚 8𝜀𝑜 ℎ 𝑒 𝑝 Principal quantum number n is the shell or energy level of the electron. In the ground state n=1. 2. Eigenfunctions Y, solutions to the Schrödinger equation for the H atom, depend on 3 quantum numbers n, l, and ml, and may be complex: Ψ𝑛,𝑙,𝑚 𝑟, 𝜃, 𝜑 = 2 𝑛𝑎𝑜 3 𝑛 − 𝑙 − 1 ! −𝑛𝑎𝑟 2𝑟 𝑜 𝑒 2𝑛 𝑛 + 𝑙 ! 𝑛𝑎𝑜 3 2 𝑙 𝐿2𝑙+1 𝑛−𝑙−1 2 𝑌𝑙𝑚 𝜃, 𝜑 𝑛𝑎𝑜 1 1 𝑟 𝑟 2 −3𝑎𝑟 Ψ31±1 𝑟, 𝜃, 𝜑 = 6 − 2 𝑒 𝑜 sin 𝜃 𝑒 ±𝑖𝜑 𝑎𝑜 𝑎𝑜 81 𝜋 𝑎𝑜 2 𝜀𝑜 ℎ 𝑎𝑜 = = 0.529 Å 𝑖𝑠 𝑐𝑎𝑙𝑙𝑒𝑑 𝑡ℎ𝑒 𝐵𝑜ℎ𝑟 𝑟𝑎𝑑𝑖𝑢𝑠 𝜋𝜇𝑒 2 n, l, m – Quantum numbers governing the solutions The n, l and m integers are called quantum numbers: – Principal quantum number, n – Angular momentum quantum number, l – Magnetic quantum number of the angular momentum, ml The total energy, E, solutions in the Schrödinger’s equation depend only on the n quantum number. The wavefunctions depend on all 3 quantum numbers. © 2017 Pearson Education, Inc. Principal Quantum Number, n Characterizes the energy of the electron in all wavefunctions with the same value of n. All these wavefunctions are said to belong to the same energy level, same electron shell, or simply same level. 𝐸𝑛 = −2.18 × 10−18 1 ∙ 2 𝐽 𝑛 n can be any integer  1. The larger the value of n, the more energy the level has. Energies are defined as being negative. – An electron would have E = 0 when it escapes the atom (ionization). As n gets larger, the difference in energy between levels gets smaller. The larger the value of n, the larger the size of the orbital (see later). © 2017 Pearson Education, Inc. Standing (Stationary) Waves Waves confined by hard energy boundaries. Waves traveling in opposite directions, appearing to be standing still. Can only adopt certain frequencies (wavelengths) called harmonics. Are caused by reflection of the wave at the hard boundary, combined with constructive interference of the wave with itself. A standing wave consists of nodes (amplitude equals 0) and antinodes (loops or ventricles). At the hard boundaries, the wave amplitude is always 0 (node) and the wave changes phase. Constructive interference occurs when the length of the confinement exactly equals an integer number of half-wavelengths: 𝜆 𝐿=𝑛 2 An orbital is a standing de Broglie wave of an electron in the electric field of atomic nuclei and other bound electrons. ©McGraw-Hill Education. Interpretation of the Wavefunction: Orbital A plot of Y 2 versus distance originating in the atomic nucleus and going along the radius of the sphere represents an orbital. An orbital is a probability distribution map of a region where the electron is likely to be found in a given nlm state described by the associated wavefunction. y 2 is the probability density. – The probability of finding an electron at a particular point in space – Decreases as you move away from the nucleus Nodes in the functions are where the probability drops to 0. © 2017 Pearson Education, Inc. Electron Probability Contours and Probability Densities Figure 7.19 ©McGraw-Hill Education. Radial Distribution Function The radial distribution function represents the total probability of finding an electron within a thin spherical shell at a distance r from the nucleus. The probability at a point decreases with increasing distance from the nucleus, but the volume of the spherical shell increases. The net result is a plot that indicates the most probable distance of the electron in a 1s orbital of H is 52.9 pm. © 2017 Pearson Education, Inc. Angular Momentum Quantum Number, l The angular momentum quantum number determines the shape of the orbital. It also determines the sublevel within a level. l can have integer values from 0 to (n – 1). Each value of l is called by a particular letter that designates the shape of the orbital. The s, p, d, f notation comes from atomic spectroscopy and stands for the 1st letter of the words sharp, principal, diffuse and fine. Emission spectra of atoms consist of lines with either one of these characteristics. © 2017 Pearson Education, Inc. Magnetic Quantum Number, ml The magnetic quantum number is an integer that specifies the orientation of the orbital: – The direction in space the orbital is aligned relative to the other orbitals – The orientation with respect to an external magnetic field: 1) along the magnetic field lines (z direction), 2) perpendicular to the magnetic field lines (either the x or the y direction) Values are integers from −l to +l – Including zero – Gives the number of orbitals of a particular shape When l = 2, the values of ml are −2, −1, 0, +1, +2, which means there are 5 orbitals with l = 2. © 2017 Pearson Education, Inc. Energy Levels and Sublevels Summarizing: – the number of sublevels within a level is n. – the number of orbitals within a sublevel is 2l + 1. – the number of orbitals in a level is n2. © 2017 Pearson Education, Inc. Sample Problem 7.6 – Problem, Plan and Solution Determining Quantum Numbers for an Energy Level PROBLEM: What values of the angular momentum (l) and magnetic (ml) quantum numbers are allowed for a principal quantum number (n) of 3? How many orbitals are there in this energy level? PLAN: Values of l are determined from the value for n, since l can take values from 0 to (n – 1). The values of ml then follow from the values of l. SOLUTION: For n = 3, allowed values of l are = 0, 1, and 2 For l = 0, ml = 0; For l = 1, ml = –1, 0, or +1; For l = 2, ml = –2, –1, 0, +1, or +2 There are 9 ml values and therefore 9 orbitals with n = 3. ©McGraw-Hill Education. Principal Energy Levels n 1 2 3 4 ©McGraw-Hill Education. Number of Orbitals in Energy Level (n2) Sublevel Designation ml Number of Orbitals in Sublevel (2l + 1) 1s 0 1 1 0 1 2s 2p 0 −1, 0, +1 1 3 4 0 1 2 3s 3p 3d 0 −1, 0, +1 −2, −1, 0, +1, +2 1 3 5 9 1 3 5 7 16 l 0 0 1 2 3 4s 4p 4d 4f 0 −1, 0, +1 −2,−1, 0, +1, +2 −3,−2,−1, 0, +1, +2, +3 Sample Problem 7.7 – Problem, Plan and Solution Determining Sublevel Names and Orbital Quantum Numbers PROBLEM: Give the name, magnetic quantum numbers, and number of orbitals for each sublevel with the following quantum numbers: (a) n = 3, l = 2 (b) n = 2, l = 0 (c) n = 5, l = 1 (d) n = 4, l = 3 PLAN: Combine the n value and l designation to name the sublevel. Knowing l, we can find ml and the number of orbitals. SOLUTION: n (a) 3 (b) 2 (c) 5 (d) 4 ©McGraw-Hill Education. l 2 0 1 3 Sublevel Name Possible ml Values No. of Orbitals 3d 2s 5p 4f −2, −1, 0, +1, +2 0 −1, 0, +1 −3, −2, −1, 0, +1, +2, +3 5 1 3 7 Sample Problem 7.8 – Problem and Solution Identifying Incorrect Quantum Numbers PROBLEM: What is wrong with each of the following quantum numbers designations and/or sublevel names? n l ml Name (a) 1 (b) 4 (c) 3 1 3 1 0 +1 −2 1p 4d 3p SOLUTION: (a) A sublevel with n = 1 can only have l = 0, not l = 1. The only possible sublevel name is 1s. (b) A sublevel with l = 3 is an f sublevel, not a d sublevel. The name should be 4f. (c) A sublevel with l = 1 can only have ml values of –1, 0, or +1, not –2. ©McGraw-Hill Education. Shapes of Atomic Orbitals: l = 0, the s Orbital Each energy level has one s orbital. Lowest energy orbital in a principal energy state Spherical shape and symmetry Number of spherical nodes = (n – 1): – 1s: no node – 2s: 1 spherical node – 3s: 2 spherical nodes © 2017 Pearson Education, Inc. Internal Nodes in the 1s, 2s and 3s orbitals The higher the n: The higher the size of the orbital; It has more internal nodes; Number of internal nodes for any orbital shape is equal to: n–l–1 Because the internal nodes of s orbitals are always spherical, and because l = 0 for an s orbital, the number of spherical nodes equals n – 1. ©McGraw-Hill Education. l = 1, p orbitals: Angular Shapes Each electron energy level n  2 has three p orbitals forming one p sublevel. Each of the three orbitals points along a different axis, distinguished by the magnetic quantum numbers: – px, py, pz – ml = −1, 0, +1 Shape: Two lobes © 2017 Pearson Education, Inc. Angular nodes in 2p orbitals Courtesy of: https://chem.libretexts.org/LibreTexts/University_of_California_Davis/UCD_Chem_107B%3A_Physic al_Chemistry_for_Life_Scientists/Chapters/4%3A_Quantum_Theory/4.10%3A_The_Schr%C3%B6d inger_Wave_Equation_for_the_Hydrogen_Atom © 2017 Pearson Education, Inc. l = 2, d Orbitals: Angular Shapes Each electron energy level n  3 has five d orbitals. – ml = −2, − 1, 0, +1, +2 Four of the five orbitals are aligned in a different plane. – The fifth is aligned with the z axis, dz2. – dxy, dyz, dxz, dx2 – y2 – distinguished by the magnetic quantum numbers: ml = −2, − 1, 0, +1, +2 Shapes: 4 have four lobes, 1 has two lobes and a toroid © 2017 Pearson Education, Inc. Angular nodes in the 3d orbitals © 2017 Pearson Education, Inc. l = 3, f Orbitals: Angular Shapes Each electron energy level n  4 has seven f orbitals. – ml = −3, −2, −1, 0, +1, +2, +3 The fourth-lowest energy orbitals in a principal energy state Mainly eight-lobed – Some two-lobed with two toroids Planar nodes © 2017 Pearson Education, Inc. Angular nodes in the 4f orbitals © 2017 Pearson Education, Inc. Table of Atomic Orbitals http://www.orbitals.com/orb/index.html © 2017 Pearson Education, Inc. Why Are Atoms Spherical? © 2017 Pearson Education, Inc. The Phase of an Orbital Orbitals are determined from mathematical wavefunctions. A wave function can have positive or negative values. – Nodes are where the wave function = 0 The sign of the wave function is called its phase. When orbitals interact, their wave functions may be in phase (same sign) or out of phase (opposite signs). – This is important in chemical bonding, as will be examined in a later chapter. © 2017 Pearson Education, Inc. Phases © 2017 Pearson Education, Inc. Describing an Orbital Each set of n, l, and ml describes one orbital. Orbitals with the same value of n are in the same principal energy level. – Also called the principal shell Orbitals with the same values of n and l are said to be in the same sublevel. – Also called a subshell © 2017 Pearson Education, Inc. Energy of Sublevels in a Hydrogen atom: a special case Orbital Energy Diagrams In a Hydrogen atom, ALL sublevels within the same level have the same energy. In all other atoms, the sublevels have slightly different energies due to electronelectron repulsions. © 2017 Pearson Education, Inc.