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5.2Important LinksDigital Support Online quizzes, study notes, and morewww.edvantagescience.comOrder your own copy of AP Chemistry 1…

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5.2Important LinksDigital Support Online quizzes, study notes, and morewww.edvantagescience.comOrder your own copy of AP Chemistry 1 www.edvantageinteractive.com5.2 Quantum Theory and the Bohr Model of the Atom Warm Up 1. (a) List three properties of waves such as water waves or visible light. (b) What would you expect to happen when two water waves moving in opposite directions met each other? 2. (a) List three properties of solid objects or particles such as the marbles in a bag. (b) What would you expect to happen when two marbles moving in opposite directions met each other? 3. Are any of your answers to question 1 identical to question 2? (Stay tuned…) A serious challenge to Rutherford’s atomic model arose almost immediately. A very secure predictionWaves Behaving like of the physics available at the end of the 1800s was that accelerating charges should radiate energy. Particles? Well Hit Me Because orbiting electrons are accelerating charges, electrons in atoms should lose energy. That prediction was catastrophic for Rutherford’s model. It meant that all atoms, and so also all matter, with a Planck! should collapse in a fraction of a second as their electrons lost energy and spiraled into the nucleus! Obviously, a significant piece of the atomic puzzle was missing and even Rutherford himself was ready to abandon his view of the atom. Yet, his conclusions and his nuclear model were correct. The real problem was that the physics of the day needed to be re-written to explain the behavior of electrons in atoms. To begin to understand how the solution came about, we must consider the work of German physicist Max Planck. In 1900, this conservative professor began nothing short of a revolution in physics. He proposed that energy, long considered to be strictly a wave phenomenon, could be shown to behave like particles in the form of very tiny, discreet energy packets or bundles he called quanta (plural for quantum). Planck called this the quantum theory and arrived at his conclusions (reluctantly) by studying the energy radiated from heated solids. Planck developed the following equation for the energy associated with each packet or quantum: E = hν where E = energy, ν = frequency, h = a very tiny proportionality constant (h = 6.626 × 10–34J.s) called Planck’s constant. According to Planck, energy could only be absorbed or emitted in whole numbers of quanta, that is, one quantum of energy (E = hν), two quanta (E = 2hν), three quanta (E = 3hν) and so on, but nowhere in between. Think of each energy quantum as a glass marble in a bag of identical marbles (Figure 5.2.1). In the same way that you could only ever add or remove a specific amount of glass from the bag in the form of a whole number of marbles, so too could amounts of energy only be absorbed or emitted in the form of whole numbers of quanta. At the end of the 1800s, the behavior of waves and the behavior of particles were seen as very different and mutually exclusive. Waves were disturbances that moved through space, could pass through and interfere with each other, and could have any Figure 5.2.1 Quanta of energy are like value within a range. Particles were objects with definite boundaries that bounced off marbles in bag. The marbles can only be each other when they collided and could only exist in certain whole-number quantities. A removed as whole units.© Edvantage Interactive 2017Chapter 5 A Closer Look at Matter 273firm experimental and mathematical foundation supported the idea that waves were fundamentally different from particles. To now suggest that waves could behave like particles was almost sacrilegious! Planck himself wrote about his work: “By nature, I am peacefully inclined and reject all doubtful adventures. …However, a theoretical interpretation had to be found at any cost….I was ready to sacrifice every one of my previous convictions about physical laws.” Planck’s theory wasn’t taken very seriously at first. But in 1905, a 26-year-old clerk in a Swiss patent office named Albert Einstein wrote five papers that changed the scientific world forever. One of those papers used the quantum theory to explain a phenomenon involving light called the “photoelectric effect” that had baffled physicists until then. According to Einstein, the only way to make sense of the photoelectric effect was to consider light as being composed of tiny discreet packets of energy. (These were later called “photons” by American chemist Gilbert Lewis.) Einstein’s paper was the first practical application of the quantum theory and as a result, the theory soon began to gain widespread acceptance. In 1921, Einstein was awarded the Nobel Prize in physics for his explanation of the photoelectric effect.Quick Check 1. Briefly state what it means for something to be a “quantized.” 2. Give three common examples of things considered to be “quantized.” 3. According to Planck, could an amount of energy equal to 2.5hν be absorbed or emitted by an object? Explain. The Bohr Model One of the scientists who paid particular attention to the work of Planck and Einstein was a young Danish physicist named Niels Bohr. As a young grad student, Bohr had met Ernest Rutherford at the Cavendish Laboratory in Cambridge and then worked with Rutherford at the University of Manchester. Bohr believed in the nuclear atomic model and started to see a way to “save” it using the quantum theory. If energy was indeed quantized and so could be seen to have only certain values and not others, perhaps the energies associated with electrons orbiting the nucleus had similar restrictions. Bohr was working on this idea at Rutherford’s laboratory in Manchester in 1912. In the middle of his experiments and calculations, he returned to Copenhagen to get married. He was so excited about his work that he managed to convince his new bride to cancel their honeymoon and return to England with him. Soon thereafter, Niels Bohr completed one of the most brilliant papers on atomic structure ever written. In doing so, he managed to rescue Rutherford’s nuclear model of the atom. Scientific theories are carefully constructed on a firm foundation of data gathered from a multitude of meticulously documented, rigorously controlled, and perpetually repeatable experiments. Bohr’s theory was no exception. In the previous section, we discussed the conduction of charges through gases in glass discharge tubes. Those experiments in the later years of the 1800s not only led to Thomson’s discovery of the electron, but also provided Bohr with valuable data for his ideas about the nature of those electrons in atoms.274 Chapter 5 A Closer Look at Matter© Edvantage Interactive 2017Quantum Theory Rescues the Nuclear ModelBohr knew that when high voltage was applied across the electrodes of a sealed glass tube containing a gas such as hydrogen, the gas was heated and emitted light. As part of his investigations, Bohr looked at this light through a spectroscope. A spectroscope is a device similar to a prism, which separates the light into its component wavelengths. When Bohr viewed the light from the heated hydrogen through the spectroscope, he saw only a series of colored lines against a black background, rather than a continuous rainbow of color. For hydrogen, the same pattern of four colored lines was always seen: a red, blue-green, blue, and violet line (Figure 5.2.2). For each gaseous element used, a bright-line pattern unique to that element called a bright-line spectrum always appeared (much like a bar code on a modern grocery item). The phenomenon had mystified scientists. None could explain why only certain colors of light, each corresponding to specific wavelength, frequency, and energy, were emitted by the heated gases — until Bohr. Bohr realized that by applying quantum principles to the behavior of hydrogen’s lone electron, he could not only account for the existence of the bright-line spectrum, but also save Rutherford’s nuclear model of the atom.410.1433.9486.0Wavelength (nm)656.1Figure 5.2.2 Hydrogen’s bright-line spectrumBohr’s Postulates Bohr’s postulates for hydrogen are summarized below: 1. The hydrogen atom had only certain allowed energy levels or stationary states. Each of these states corresponded to a circular electron orbit of a fixed size. The larger the allowed orbit, the greater the energy associated with it. No other orbits existed in the atom. Each allowed energy state was given an integer number “n” that Bohr called a quantum number with allowed values that ranged from 1 to ∞ (i.e., n could equal 1, 2, 3…etc). The lowest energy (smallest) orbit corresponded to the lowest allowed energy state called the ground state and was designated as n = 1. The larger orbits of greater energy were designated as n = 2, n = 3, n = 4, etc. and were said to be excited states. 2. As long as an electron moved in an allowed orbit or stationary state, the electron (and therefore the atom) did not radiate or absorb energy. 3. The electron could only move from one allowed orbit to another if it absorbed or emitted an amount of energy exactly equal to the energy difference between the two orbits, ∆E. This meant that the hydrogen atom could only change from one stationary energy state to another.Bohr’s Postulates — Another Look© Edvantage Interactive 2017Postulate 1 employed Planck’s theory by quantizing the energies allowed for the hydrogen atom (and thus the electron). Because only certain-sized orbits were allowed, the atom was restricted to existing in only certain energy states and not others. Think of the electron as a ball on a staircase. Just as the ball can only rest on any particular stair and so have only certain amounts of potential energy and not others, so too is the electron restricted to only specific energies.Chapter 5 A Closer Look at Matter 275Postulate 2 meant that the nuclear model of the atom proposed by Rutherford would not collapse as predicted. Although the postulate violated the laws of classical physics, Bohr insisted that it must be true, even though he didn’t know why. Postulates 1 and 3 explained the origin and nature of hydrogen’s bright-line spectrum. An atomic spectrum could not be continuous (i.e., a complete rainbow of colors) because an atom’s energy states could only be certain values and not others. When a sample of hydrogen gas is heated in a discharge tube, the electrons in the hydrogen atoms n=3 absorb sufficient amounts of energy to “jump” to larger orbits. (In any one hydrogen atom, only one electron is involved, but in a sample of the gas, the electrons of many hydrogen atoms are undergoing many transitions). Once in a higher energy ΔE=hv n=2 orbit, any electron could then return to a lower energy orbit by emitting a specific amount of energy corresponding exactly to the difference between the higher n=1 and lower energy orbits (Figure 5.2.3). If the frequency of that emitted energy corresponds to any part of the visible spectrum, then a bright line of that specific color would be seen. Four of hydrogen’s electron transitions emitted energy in the visible spectrum. If an excited electron emits energy and drops to n = 2 from a higher energy orbit, the wavelength of the emitted energy corresponds to a particular color of visible light. If an electron drops from n = 3 to n = 2, the energy difference between the two orbits (and therefore the energy emitted) corresponds to that of red light. Figure 5.2.3 When the electron moves to an inner Hence the red line appears in the emission spectrum. The blue-green line results energy level, it emits energy. from an electron transition from n = 4 to n = 2, the blue line from an electron transition from n = 5 to n = 2, and the violet line from an electron transition from n = 6 to n = 2. This series of four bright lines in the visible spectrum is called the Balmer series, named after the Swiss schoolteacher who first derived a mathematical relationship between the lines in hydrogen’s visible emission spectrum (Figure 5.2.4). Bohr’s model of the hydrogen atom was successful in explaining the mystery of bright line spectra. His calculations and predictions worked for hydrogen and he even calculated the radius of the orbit for hydrogen’s electron in its ground state. In 1922, Niels Bohr was awarded the Nobel Prize in physics. λ(nm) 400500600700500600700H Hg Ne (a) H λ(nm) 400 (b)Figure 5.2.4 (a) The emission spectra of hydrogen, mercury, and neon; (b) The absorption spectrum ofhydrogenThe Emission Spectrum of Hydrogen — Two Views The diagram on the left in Figure 5.2.5 shows the circular orbits Bohr envisioned for the hydrogen electron and the transitions associated with the Lyman, Balmer, and Paschen emission spectra. The diagram on the right shows that the energy differences between various stationary states (n) decrease as the energies of those states increase. The arrows pointing down represent electrons falling to lower energy states and thus emitting energy. Electrons absorbing energy could be indicated by arrows pointing up and would represent absorption spectra. Any electron transitions from an excited state down to n = 1 result in the emission of energy in the ultraviolet region of the electromagnetic spectrum (the Lyman series). Any transitions from excited276 Chapter 5 A Closer Look at Matter© Edvantage Interactive 2017states down to the third and fourth orbits result in energies in the infrared region being emitted (the Paschen and Brackett series respectively).949n=4m 656 n 486 nm434 nmn=1 187n=2Hydrogen atom electron transitions n=7 n=6 n=5122 nm 10 3n m 97nmm 5nLyman series (ultraviolet)nm410 nm 5 nmBrackett series E(n) to E(n=4)n=3 Balmer series (visible)Paschen series E(n) to E(n=3)n=2Balmer series E(n) to E(n=2)1282 nmn=310 94n=4 n=5nmn=6Paschen series (infrared)n=1Lyman series E(n) to E(n=1)Figure 5.2.5 The Lyman, Balmer, and Paschen emission series for hydrogenQuick Check 1. Describe the appearance of hydrogen’s “bright-line” spectrum. 2. Briefly indicate how electrons generate each visible line in hydrogen’s emission spectrum. 3. Which electron transitions in the emission spectrum generate lines in the UV region of the electromagnetic spectrum? Using Some Simple Equations © Edvantage Interactive 2017In 1885, a Swiss schoolteacher named Johann Balmer found an equation that was able to determine the wavelengths of the lines in the visible portion of hydrogen’s emission spectrum. Three years later, the Swedish physicist Johannes Rydberg derived an equation that could be used to calculate the wavelengths of all of hydrogen’s spectral lines. This is worth mentioning because if a mathematical relationship exists for a natural phenomenon, it usually means there’s a theoretical foundation waiting to be discovered. Both Balmer’s and Rydberg’s equations were based on data, rather than theory. Neither equation had any physical meaning in terms of atomic structure, but both worked. The physical meaning was supplied by Niels Bohr. Bohr derived an equation for the energy associated with the electron in each allowed orbit. He correctly predicted the visible, ultraviolet, and infrared spectral lines in hydrogen’s emission spectrum by calculating the energy differences between those stationary states. Although a rigorous mathematical treatment of this material is not intended here, we will present two equations that are quite straightforward. The first equation gives the change in energy ∆EChapter 5 A Closer Look at Matter 277(energy of photon released) when an electron initially in a higher energy orbit (with a higher quantum number nh) drops to a lower energy orbit (and so with a lower quantum number n l).()∆E = b 1/nl2 – 1/nh2where b is a constant with value of 2.18 × 10–18 J. The second equation arises from the fact that, according to Planck’s equation, the energy of this photon, ∆E = hν. Because ν = c/λ, we can replace ∆E with hc/λ . Now dividing both sides of the equation by hc yields:()1/λ = b/hc 1/nl2 – 1/nh2This equation allows us to solve for the wavelength λ of the spectral line we would observe when the electron lost energy as it made the above transition. The combination of the constants, b/hc, is itself a constant. Remembering that Planck’s constant, h = 6.626 × 10–34J.s and that the speed of light, c = 3.00 × 108m/s, we can combine these three constants to give b/hc = 1.097 30 × 107 m–1. Let’s use this to calculate the wavelength of the spectral line we would see when hydrogen’s electron made the transition from n = 4 down to n = 2.Sample Problem — Calculating the Wavelength of Emission Spectral Lines Calculate the wavelength λ (in nm) of the spectral line seen in hydrogen’s emission spectrum when hydrogen’s electron falls from the fourth allowed orbit (n = 4) to the second allowed orbit (n = 2).What to Think aboutHow to Do It1. Consider the equation: 1/λ = (1.097 30 x 107 m–1) 1/nl2 – 1/nh2 The values of nl and nh are given in the question: n = 4 and n = 2.1/λ ()2. Convert this to nanometers. This value corresponds exactly to the green line seen in hydrogen’s emission spectrum.= (1.097 30 × 107 m–1) (1/nl2 – 1/nh2) = (1.097 30 × 107 m–1) (½2 – ¼2) = (1.097 30 × 107 m–1) (0.1875) = 2.0574 × 106 m–1 1 λ= = 4.8604 × 10–7 m 2.0574 × 106 m–1 9 4.860 × 10–7 m × 1.0 × 10 nm = 486.04 nm mPractice Problem()1. Use equation ∆E = b 1/nl2 – 1/nh2 and the value for b given above to calculate the energy released when an excited hydrogen electron drops from the fourth allowed orbit (n = 4) to the second allowed orbit (n = 2).278 Chapter 5 A Closer Look at Matter© Edvantage Interactive 2017

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