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Chapter 35 Bohr Theory of Hydrogen CHAPTER 35 HYDROGEN The hydrogen atom played a special role in the history of physics by providing the key that unlocked the new mechanics that replaced Newtonian mechanics. It started with Johann Balmer's discovery in 1884 of a mathematical formula for the wavelengths of some of the spectral lines emitted by hydrogen. The simplicity of the formula suggested that some understandable mechanisms were producing these lines. The next step was Rutherford's discovery of the atomic nucleus in 1912. After that, one knew the basic structure of atoms—a positive nucleus surrounded by negative electrons. Within a year Neils Bohr had a model of the hydrogen atom that "explained" the spectral lines. Bohr introduced a new concept, the energy level. The electron in hydrogen had certain allowed energy levels, and the sharp spectral lines were emitted when the electron jumped from one energy level to another. To explain the energy levels, Bohr developed a model in which the electron had certain allowed orbits and the jump between energy levels corresponded to the electron moving from one allowed orbit to another. Bohr's allowed orbits followed from Newtonian mechanics and the Coulomb force law, with one small but crucial modification of Newtonian mechanics. The angular momentum of the electron could not vary BOHR THEORY OF continuously, it had to have special values, be quantized in units of Planck's constant divided by 2π , h/2π . In Bohr's theory, the different allowed orbits corresponded to orbits with different allowed values of angular momentum. Again we see Planck's constant appearing at just the point where Newtonian mechanics is breaking down. There is no way one can explain from Newtonian mechanics why the electrons in the hydrogen atom could have only specific quantized values of angular momentum. While Bohr's model of hydrogen represented only a slight modification of Newtonian mechanics, it represented a major philosophical shift. Newtonian mechanics could no longer be considered the basic theory governing the behavior of particles and matter. Something had to replace Newtonian mechanics, but from the time of Bohr's theory in 1913 until 1924, no one knew what the new theory would be. In 1924, a French graduate student, Louis de Broglie, made a crucial suggestion that was the key that led to the new mechanics. This suggestion was quickly followed up by Schrödinger and Heisenberg who developed the new mechanics called quantum mechanics. In this chapter our focus will be on the developments leading to de Broglie's idea. 35-2 Bohr Theory of Hydrogen THE CLASSICAL HYDROGEN ATOM With Rutherford's discovery of the atomic nucleus, it became clear that atoms consisted of a positively charged nucleus surrounded by negatively charged electrons that were held to the nucleus by an electric force. The simplest atom would be hydrogen consisting of one proton and one electron held together by a Coulomb force of magnitude p Fe 2 Fe e Fe = e2 (1) r r (For simplicity we will use CGS units in describing the hydrogen atom. We do not need the engineering units, and we avoid the complicating factor of 1/4πε0 in the electric force formula.) As shown in Equation 1, both the proton and the electron attract each other, but since the proton is 1836 times more massive than the electron, the proton should sit nearly at rest while the electron orbits around it. Thus the hydrogen atom is such a simple system, with known masses and known forces, that it should be a straightforward matter to make detailed predictions about the nature of the atom. We could use the orbit program of Chapter 8, replacing the gravitational force GMm/r 2 by e 2 /r 2 . We would predict that the electron moved in an elliptical orbit about the proton, obeying all of Kepler's laws for orbital motion. There is one important point we would have to take into account in our analysis of the hydrogen atom that we did not have to worry about in our study of satellite motion. The electron is a charged particle, and accelerated charged particles radiate electromagnetic waves. Suppose, for example, that the electron were in a circular orbit moving at an angular velocity ω as shown in Figure (1a). If we were looking at the orbit from the side, as shown in Figure (1b), we would see an electron oscillating up and down with a velocity given by v = v0 sin ωt . In our discussion of radio antennas in Chapter 32, we saw that radio waves could be produced by moving electrons up and down in an antenna wire. If electrons oscillated up and down at a frequency ω , they produced radio waves of the same frequency. Thus it is a prediction of Maxwell's equations that the electron in the hydrogen atom should emit electromagnetic radiation, and the frequency of the radiation should be the frequency at which the electron orbits the proton. For an electron in a circular orbit, predicting the motion is quite easy. If an electron is in an orbit of radius r, moving at a speed v, then its acceleration a is directed toward the center of the circle and has a magnitude 2 a = vr (2) Using Equation 1 for the electric force and Equation 2 for the acceleration, and noting that the force is in the same direction as the acceleration, as indicated in Figure (2), Newton's second law gives F = m a e2 = m v2 r r2 (3) One factor of r cancels and we can immediately solve for the electron's speed v to get v 2 = e 2/mr, or velectron = e mr (4) The period of the electron's orbit should be the distance 2πr travelled, divided by the speed v, or 2πr/v seconds per cycle, and the frequency should be the inverse of that, or v/2πr cycles per second. Using Equation 4 for v, we get frequency of e = v = electron in orbit 2πr 2πr mr (5) According to Maxwell's theory, this should also be the frequency of the radiation emitted by the electron. v0 e p a) electron in circular orbit Figure 1 The side view of circular motion is an up and down oscillation. e v = v0sin(ωt) p b) side view of circular orbit 35-3 Electromagnetic radiation carries energy. Thus, to see what effect this has on the electron’s orbit, let us look at the formula for the energy of an orbiting electron. From Equation 3 we can immediately solve for the electron's kinetic energy. The result is 1 mv 2 = e 2 electron kinetic (6) 2 2r energy The electron also has electric potential energy just as an earth satellite had gravitational potential energy. The formula for the gravitational potential energy of a satellite was potential energy = – GMm r of an earth satellite (10-50a) where M and m are the masses of the earth and the satellite respectively. This is the result we used in Chapter 8 to test for conservation of energy (Equations 8-29 and 8-31) and in Chapter 10 where we calculated the potential energy (Equations 10-50a and 10-51). The minus sign indicated that the gravitational force is attractive, that the satellite starts with zero potential energy when r = ∞ and loses potential energy as it falls in toward the earth. We can convert the formula for gravitational potential energy to a formula for electrical potential energy by comparing formulas for the gravitational and electric forces on the two orbiting objects. The forces are Fgravity = GMm ; r2 2 Felectric = e2 r Since both are 1/r 2 forces, we can go from the gravitational to the electric force formula by replacing the v e a Fe p r Figure 2 For a circular orbit, both the acceleration a and the force F point toward the center of the circle. Thus we can equate the magnitudes of F and ma. constant GMm by e2 . Making this same substitution in the potential energy formula gives 2 PE = – re electrical potential energy of the electron in the hydrogen atom (7) Again the potential energy is zero when the particles are infinitely far apart, and the electron loses potential energy as it falls toward the proton. (We used this result in the analysis of the binding energy of the hydrogen molecule ion, explicitly in Equation 18-15.) The formula for the total energy E total of the electron in hydrogen should be the sum of the kinetic energy, Equation 6, and the potential energy, Equation 7. potential E total = kinetic energy + energy 2 = e 2r 2 Etotal = – e 2r 2 – er total energy of electron (8) The significance of the minus (–) sign is that the electron is bound. Energy is required to pull the electron out, to ionize the atom. For an electron to escape, its total energy must be brought up to zero. We are now ready to look at the predictions that follow from Equations 5 and 8. As the electron radiates light it must lose energy and its total energy must become more negative. From Equation 8 we see that for the electron's energy to become more negative, the radius r must become smaller. Then Equation 5 tells us that as the radius becomes smaller, the frequency of the radiation increases. We are lead to the picture of the electron spiraling in toward the proton, radiating even higher frequency light. There is nothing to stop the process until the electron crashes into the proton. It is an unambiguous prediction of Newtonian mechanics and Maxwell's equations that the hydrogen atom is unstable. It should emit a continuously increasing frequency of light until it collapses. 35-4 Bohr Theory of Hydrogen Energy Levels By 1913, when Neils Bohr was trying to understand the behavior of the electron in hydrogen, it was no surprise that Maxwell's equations did not work at an atomic scale. To explain blackbody radiation and the photoelectric effect, Planck and Einstein were led to the picture that light consists of photons rather than Maxwell's waves of electric and magnetic force. To construct a theory of hydrogen, Bohr knew the following fact. Hydrogen gas at room temperature emits no light. To get radiation, it has to be heated to rather high temperatures. Then you get distinct spectral lines rather than the continuous radiation spectrum expected classically. The visible spectral lines are the H α , H β and H γ lines we saw in the hydrogen spectrum experiment. These and many infra red lines we saw in the spectrum of the hydrogen star, Figure (3328) reproduced below, make up the Balmer series of lines. Something must be going on inside the hydrogen atom to produce these sharp spectral lines. Viewing the light radiated by hydrogen in terms of Einstein's photon picture, we see that the hydrogen atom emits photons with certain precise energies. As an exercise in the last chapter you were asked to calculate, in eV, the energies of the photons in the H α , H β and H γ spectral lines. The answers are E Hα = 1.89 eV E Hβ = 2.55 eV E Hγ = 2.86 eV 3.65 10 –5 H40 H30 3.70 10 H20 (9) –5 3.75 10 H15 H14 H13 H12 –5 wavelength H11 3.80 10 H10 The question is, why does the electron in hydrogen emit only certain energy photons? The answer is Bohr's main contribution to physics. Bohr assumed that the electron had, for some reason, only certain allowed energies in the hydrogen atom. He called these allowed energy levels. When an electron jumped from one energy level to another, it emitted a photon whose energy was equal to the difference in the energy of the two levels. The red 1.89 eV photon, for example, was radiated when the electron fell from one energy level to another level 1.89 eV lower. There was a bottom, lowest energy level below which the electron could not fall. In cold hydrogen, all the electrons were in the bottom energy level and therefore emitted no light. When the hydrogen atom is viewed in terms of Bohr’s energy levels, the whole picture becomes extremely simple. The lowest energy level is at -13.6 eV. This is the total energy of the electron in any cold hydrogen atom. It requires 13.6 eV to ionize hydrogen to rip an electron out. Figure 3 Energy level diagram for the hydrogen atom. All the energy levels are given by the simple formula En = – 13.6/n 2 eV. All Balmer series lines result from jumps down to the n = 2 level. The 3 jumps shown give rise to the three visible hydrogen lines. –5 3.85 10 0 –.544 –.850 –1.51 –3.40 n=5 n=4 n=3 Hα Hβ Hγ n=2 –5 H9 Figure 33-28 Spectrum of a hydrogen star –13.6 n=1 35-5 The first energy level above the bottom is at –3.40 eV which turns out to be (–13.6/4) eV. The next level is at –1.51 eV which is (–13.6/9) eV. All of the energy levels needed to explain every spectral line emitted by hydrogen are given by the formula E n = – 13.62 eV n (10) where n takes on the integer values 1, 2, 3, .... These energy levels are shown in Figure (3). All of the lines in the Balmer result from jumps down to the second energy level. For historical interest, let us see how Balmer's formula for the wavelengths in this series follows from Bohr's formula for the energy levels. For Balmer's formula, the lines we have been calling H α , H β and H γ are H 3 , H 4 , H 5 . An arbitrary line in the series is denoted by H n , where n takes on the values starting from 3 on up. The Balmer formula for the wavelength of the H n line is from Equation 33-6 λ n = 3.65 × 10 – 5cm × Exercise 1 Use Equation 10 to calculate the lowest 5 energy levels and compare your answer with Figure 3. Let us see explicitly how Bohr's energy level diagram explains the spectrum of light emitted by hydrogen. If, for example, an electron fell from the n=3 to the n=2 level, the amount of energy E 3–2 it would lose and therefore the energy it would radiate would be E 3–2 = E 3 – E 2 = – 1.51 eV – ( – 3.40 eV) = 1.89 eV = (11) energy lost in falling from n = 3 to n = 2 level n2 n2 – 4 (33-6) Referring to Bohr's energy level diagram in Figure (3), consider a drop from the nth energy level to the second. The energy lost by the electron is ( E n – E 2 ) which has the value E n – E 2 = 13.62eV – 13.62eV n 2 energy lost by electron going from nth to second level This must be the energy E H n carried out by the photon in the H n spectral line. Thus 1 1 E H n = 13.6 eV – 2 4 n = 13.6 eV n2 – 4 which is the energy of the red photons in the H α line. (12) 4n 2 We now use the formula Exercise 2 Show that the Hβ and Hγ lines correspond to jumps to the n = 2 level from the n = 4 and the n = 5 levels respectively. From Exercise 2 we see that the first three lines in the Balmer series result from the electron falling from the third, fourth and fifth levels down to the second level, as indicated by the arrows in Figure (3). –5 λ = 12.4 × 10 cm ⋅ eV E photon in eV (34-8) relating the photon's energy to its wavelength. Using Equation 12 for the photon energy gives –5 2 λ n = 12.4 × 10 cm ⋅ eV 4n 2 13.6 eV n –4 λ n = 3.65 × 10 – 5cm which is Balmer's formula. n2 n2 – 4 35-6 Bohr Theory of Hydrogen It does not take great intuition to suspect that there are other series of spectral lines beyond the Balmer series. The photons emitted when the electron falls down to the lowest level, down to -13.6 eV as indicated in Figure (4), form what is called the Lyman series. In this series the least energy photon, resulting from a fall from -3.40 eV down to -13.6 eV, has an energy of 10.2 eV, well out in the ultraviolet part of the spectrum. All the other photons in the Lyman series have more energy, and therefore are farther out in the ultraviolet. It is interesting to note that when you heat hydrogen and see a Balmer series photon like H α , H β or H γ , eventually a 10.2 eV Lyman series photon must be emitted before the hydrogen can get back down to its ground state. With telescopes on earth we see many hydrogen stars radiating Balmer series lines. We do not see the Lyman series lines because these ultraviolet photons do not make it down through the earth's atmosphere. But the Lyman series lines are all visible using orbiting telescopes like the Ultraviolet Explorer and the Hubble telescope. Another series, all of whose lines lie in the infra red, is the Paschen series, representing jumps down to the n = 3 energy level at -1.55 eV, as indicated in Figure (5). There are other infra red series, representing jumps down to the n = 4 level, n = 5 level, etc. There are many series, each containing many spectral lines. And all these lines are explained by Bohr's conjecture that the hydrogen atom has certain allowed energy levels, all given by the simple formula En = (– 13.6/n 2) eV . This one simple formula explains a huge amount of experimental data on the spectrum of hydrogen. Exercise 3 Calculate the energies (in eV) and wavelengths of the 5 longest wavelength lines in (a) the Lyman series (b) the Paschen series On a Bohr energy level diagram show the electron jumps corresponding to each line. Exercise 4 0 –.544 –.850 –1.51 n=5 n=4 n=3 –3.40 n=2 In Figure (33-28), repeated 2 pages back, we showed the spectrum of light emitted by a hydrogen star. The lines get closer and closer together as we get to H40 and just beyond. Explain why the lines get closer together and calculate the limiting wavelength. Figure 4 0 The Lyman series consists of all jumps down to the –13.6eV level. (Since this is as far down as the electron can go, this level is called the “ground state”.) Figure 5 The Paschen series consists of all jumps down to the n = 3 level. These are all in the infra red. –13.6 n=1 –.278 –.378 –.544 –.850 n=7 n=6 n=5 n=4 –1.51 n=3 35-7 of hydrogen, we saw that an electron in an orbit of radius r had a total energy E(r) given by THE BOHR MODEL Where do Bohr's energy levels come from? Certainly not from Newtonian mechanics. There is no excuse in Newtonian mechanics for a set of allowed energy levels. But did Newtonian mechanics have to be rejected altogether? Planck was able to explain the blackbody radiation formula by patching up classical physics, by assuming that, for some reason, light was emitted and absorbed in quanta whose energy was proportional to the light's frequency. The reason why Planck's trick worked was understood later, with Einstein's proposal that light actually consisted of particles whose energy was proportional to frequency. Blackbody radiation had to be emitted and absorbed in quanta because light itself was made up of these quanta. total energy of an electron in a circular orbit of radius r 2 E(r) = – e 2r If the electron can have only certain allowed energies E n = –13.6/n 2 eV , then if Equation (8) holds, the electron orbits can have only certain allowed orbits of radius r n given by 2 (13) En = – e 2r n The r n are the radii of the famous Bohr orbits. This leads to the rather peculiar picture that the electron can exist in only certain allowed orbits, and when the electron jumps from one allowed orbit to another, it emits a photon whose energy is equal to the difference in energy between the two orbits. This model is indicated schematically in Figure (6). By 1913 it had become respectable, frustrating perhaps, but respectable to modify classical physics in order to explain atomic phenomena. The hope was that a deeper theory would come along and naturally explain the modifications. Exercise 5 What kind of a theory do we construct to explain the allowed energy levels in hydrogen? In the classical picture we have a miniature solar system with the proton at the center and the electron in orbit. This can be simplified by restricting the discussion to circular orbits. From our earlier work with the classical model From Equation 13 and the fact that E1 = – 13.6 eV , calculate the radius of the first Bohr orbit r1 . [Hint: first convert eV to ergs.] This is known as the Bohr radius and is in fact a good measure of the actual radius of a cold hydrogen atom. [The answer is –8 ° r1 = .529 × 10 cm = .529A .] Then show that rn = n2 r1 . Figure 6 Lyma ns e r2 r3 Paschen series r1 eries er s lm Ba s rie The Bohr orbits are determined by equating the allowed energy E n = – 13.6 n 2 to the energy E n = – e2 2rn for an electron in an orbit of radius rn. The Lyman series represents all jumps down to the smallest orbit, the Balmer series to the second orbit, the Paschen series to the third orbit, etc. (The radii in this diagram are not to scale, the radii r n increase in size as n 2, as you can easily show by equating the two values for E n.) (8 repeated) 35-8 Bohr Theory of Hydrogen Angular Momentum in the Bohr Model Nothing in Newtonian mechanics gives the slightest hint as to why the electron in hydrogen should have only certain allowed orbits. In the classical picture there is nothing special about these particular radii. But ever since the time of Max Planck, there was a special unit of angular momentum, the amount given by Planck's constant h. Since Planck's constant keeps appearing whenever Newtonian mechanics fails, and since Planck's constant has the dimensions of angular momentum, perhaps there was something special about the electron's angular momentum when it was in one of the allowed orbits. The next step is to express r in terms of the angular momentum L. Squaring Equation 13 gives L 2 = e 2 mr or 2 r = L2 e m Finally we can eliminate the variable r in favor of the angular momentum L in our formula for the electron's total energy. We get total energy – e2 E = of the electron 2r = We can check this idea by re expressing the electron's total energy not in terms of the orbital radius r, but in terms of its angular momentum L. We first need the formula for the electron's angular momentum when in a circular orbit of radius r. Back in Equation 4, we found that the speed v of the electron was given by v = e mr (4 repeated) Multiplying this through by m gives us the electron's linear momentum mv mv = me = e mr m r (14) L = mv r = e = e mr – e2 2 L 2 e2 m 2 2 = –e e m 2 L2 4 = – e 2m 2L (17) In the formula – e 4m/2L 2 for the electron's energy, only the angular momentum L changes from one orbit to another. If the energy of the nth orbit is E n , then there must be a corresponding value L n for the angular momentum of the orbit. Thus we should write 4 En = – e m 2L2n The electron's angular momentum about the center of the circle is its linear momentum mv times the lever arm r, as indicated in the sketch of Figure (7). The result is m r r (16) (18) v L = mvr m r (15) where we used Equation 14 for mv. Figure 7 Angular momentum of a particle moving in a circle of radius r. 35-9 At this point, Bohr had the clue as to how to modify Newtonian mechanics in order to get his allowed energy levels. Suppose that angular momentum is quantized in units of some quantity we will call L0 . In the smallest orbit, suppose it has one unit, i.e., L1 = 1 × L0 . In the second orbit assume it has twice as much angular momentum, L2 = 2 L0 . In the nth orbit it would have n units Ln = nL0 quantization of angular momentum (19) Substituting Equation 19 into Equation 18 gives 4 1 En = – e m 2 n2 2L 0 (20) as the total energy of an electron with n units of angular momentum. Comparing Equation 20 with Bohr's energy level formula E n = –13.6 eV 12 n (10 repeated) we see that we can explain the energy levels by assuming that the electron in the nth energy level has n units of quantized angular momentum L0 . We can also evaluate the size of L0 by equating the constant factors in Equations 10 and 20. We get e4m = 13.6 eV 2L20 (21) Converting 13.6 eV to ergs, and solving for L0 gives e4m = 13.6 eV × 1.6 × 10 – 12 ergs eV 2L20 With e = 4.8 × 10 – 10esu and m = .911 × 10 – 27gm in CGS units, we get gm cm 2 L 0 = 1.05 × 10 – 27 sec (22) which turns out to be Planck's constant divided by 2π . 6.63 × 10 – 27gm cm/sec L0 = h = 2π 2π gm cm = 1.05 × 10 – 27 sec This quantity, Planck's constant divided by 2π , appears so often in physics and chemistry that it is given the special name “h bar” and is written h "h bar " h ≡ h (23) 2π Using h for L0 in the formula for E n , we get Bohr's formula 4 E n = – e m2 12 2h n (24) where e 4m/2h 2, expressed in electron volts, is 13.6 eV. This quantity is known as the Rydberg constant. [Remember that we are using CGS units, where e is in esu, m in grams, and h is erg-sec.] Exercise 6 Use Equation 21 to evaluate L0 . Exercise 7 What is the formula for the first Bohr radius in terms of the electron mass m, charge e, and Planck's constant h. Evaluate your result and show that ° . (Answer: r = h2/e2m .) r1 = .51 × 10– 8cm = .51A 1 Exercise 8 Starting from Newtonian mechanics and the Coulomb force law F = e2/r2 , write out a clear and concise derivation of the formula 4 En = – e 2m 12 2h n Explain the crucial steps of the derivation. A day or so later, on an empty piece of paper and a clean desk, see if you can repeat the derivation without looking at notes. When you can, you have a secure knowledge of the Bohr theory. 35-10 Bohr Theory of Hydrogen Exercise 9 An ionized helium atom consists of a single electron orbiting a nucleus containing two protons as shown in Figure (8). Thus the Coulomb force on the electron has a magnitude Fe = 2 e 2e = 2e2 2 r r –e Figure 8 Ionized helium has a nucleus with two protons, surrounded by one electron. 2e a) Using Newtonian mechanics, calculate the total energy of the electron. (Your answer should be – e2/r . Note that the r is not squared.) b) Express this energy in terms of the electron's angular momentum L. (First calculate L in terms of r, solve for r, and substitute as we did in going from Equations 16 to 17.) c) Find the formula for the energy levels of the electron in ionized helium, assuming that the electron's angular momentum is quantized in units of h. d) Figure out whether ionized helium emits any visible spectral lines (lines with photon energies between 1.8 eV and 3.1 eV.) How many visible lines are there and what are their wavelengths?) Exercise 10 You can handle all single electron atoms in one calculation by assuming that there are z protons in the nucleus. (z = 1 for hydrogen, z = 2 for ionized helium, z = 3 for doubly ionized lithium, etc.) Repeat parts a), b), and c) of Exercise 9 for a single electron atom with z protons in the nucleus. (There is no simple formula for multi electron atoms because of the repulsive force between the electrons.) DE BROGLIE'S HYPOTHESIS Despite its spectacular success describing the spectra of hydrogen and other one-electron atoms, Bohr's theory represented more of a problem than a solution. It worked only for one electron atoms, and it pointed to an explicit failure of Newtonian mechanics. The idea of correcting Newtonian mechanics by requiring the angular momentum of the electron be quantized in units of h , while successful, represented a bandaid treatment. It simply covered a deeper wound in the theory. For two centuries Newtonian mechanics had represented a complete, consistent scheme, applicable without exception. Special relativity did not harm the integrity of Newtonian mechanics—relativistic Newtonian mechanics is a consistent theory compatible with the principle of relativity. Even general relativity, with its concepts of curved space, left Newtonian mechanics intact, and consistent, in a slightly altered form. The framework of Newtonian mechanics could not be altered to include the concept of quantized angular momentum. Bohr, Sommerfield, and others tried during the decade following the introduction of Bohr's model, but there was little success. In Paris, in 1923, a graduate student Louis de Broglie, had an idea. He noted that light had a wave nature, seen in the 2-slit experiment and Maxwell's theory, and a particle nature seen in Einstein's explanation of the photoelectric effect. Physicists could not explain how light could behave as a particle in some experiments, and a wave in others. This problem seemed so incongruous that it was put on the back burner, more or less ignored for nearly 20 years. De Broglie's idea was that, if light can have both a particle and a wave nature, perhaps electrons can too! Perhaps the quantization of the angular momentum of an electron in the hydrogen atom was due to the wave nature of the electron. The main question de Broglie had to answer was how do you determine the wavelength of an electron wave? 35-11 An analogy with photons might help. There is, however, a significant difference between electrons and photons. Electrons have a rest mass energy and photons do not, thus there can be no direct analogy between the total energies of the two particles. But both particles have mass and carry linear momentum, and the amount of momentum can vary from zero on up for both particles. Thus photons and electrons could have similar formulas for linear momentum. But if the circumference of the circle were an exact integral number of wavelengths as illustrated in Figure (10), there would be no cancellation. This would therefore be one of Bohr's allowed orbits shown in Figure (6). Back in Equation 34-13 we saw that the linear momentum p of a photon was related to its wavelength λ by the simple equation Using the de Broglie formula λ = h/p for the electron wavelength, we get λ = h p de Broglie wavelength (34-13) De Broglie assumed that this same relationship also applied to electrons. An electron with a linear momentum p would have a wavelength λ = h/p . This is now called the de Broglie wavelength. This relationship applies not only to photons and electrons, but as far as we know, to all particles! With a formula for the electron wavelength, de Broglie was able to construct a simple model explaining the quantization of angular momentum in the hydrogen atom. In de Broglie's model, one pictures an electron wave chasing itself around a circle in the hydrogen atom. If the circumference of the circle, 2πr did not have an exact integral number of wavelengths, then the wave, after going around many times, would eventually cancel itself out as illustrated in Figure (9). Suppose (n) wavelengths fit around a particular circle of radius r n . Then we have (25) nλ = 2πr n n hp = 2πr n (26) Multiplying both sides by p and dividing through by 2π gives n h = pr n (27) 2π Now h/2π is just h , and pr n is the angular momentum L n (momentum times lever arm) of the electron. Thus Equation 27 gives (28) nh = pr n = L n Equation 28 tells us that for the allowed orbits, the orbits in which the electron wave does not cancel, the angular momentum comes in integer amounts of the angular momentum h . The quantization of angular momentum is thus due to the wave nature of the electron, a concept completely foreign to Newtonian mechanics. r Figure 9 Figure 10 Figure 10a--Movie De Broglie picture of an electron wave cancelling itself out. If the circumference of the orbit is an integer number of wavelengths, the electron wave will go around without any cancellation. The standing waves on a circular metal band nicely illustrate de Broglie’s waves 35-12 Bohr Theory of Hydrogen When a graduate student does a thesis project, typically the student does a lot of work under the supervision of a thesis advisor, and comes up with some new, hopefully verifiable, results. What do you do with a student that comes up with a strange idea, completely unverified, that can be explained in a few pages of algebra? Einstein happened to be passing through Paris in the summer of 1924 and was asked if de Broglie's thesis should be accepted. Although doubtful himself about a wave nature of the electron, Einstein recommended that the thesis be accepted, for de Broglie just might be right. In 1925, two physicists at Bell Telephone Laboratories, C. J. Davisson and L. H. Germer were studying the surface of nickel by scattering electrons from the surface. The point of the research was to learn more about metal surfaces in order to improve the quality of switches used in telephone communication. After heating the metal target to remove an oxide layer that accumulated following a break in the vacuum line, they discovered that the electrons scattered differently. The metal had crystallized during the heating, and the peculiar scattering had occurred as a result of the crystallization. Davisson and Germer then prepared a target consisting of a single crystal, and studied the peculiar scattering phenomena extensively. Their apparatus is illustrated schematically in Figure (11), and their experimental results are shown in Figure (12). For their experiment, there was a marked peak in the scattering when the detector was located at an angle of 50° from the incident beam. Davisson presented these results at a meeting in London in the summer of 1927. At that time there was a considerable discussion about de Broglie's hypothesis that electrons have a wave nature. Hearing of this idea, Davisson recognized the reason for the scattering peak. The atoms of the crystal were diffracting electron waves. The enhanced scattering at 50° was a diffraction peak, a maximum similar to the reflected maxima we saw back in Figure (33-19) when light goes through a diffraction grating. Davisson had the experimental evidence that de Broglie's idea about electron waves was correct after all. electron gun detector θ= Reflected maximum 50 ° θ electron beam transmitted maximum Figure 33-19 nickel crystal Figure 11 Scattering electrons from the surface of a nickel crystal. Laser beam impinging on a diffraction grating. Figure 12 Plot of intensity vs. angle for electrons scattered by a nickel crystal, as measured by Davisson and Germer. The peak in intensity at 50° was a diffraction peak like the ones produced by diffraction gratings. (The intensity is proportional to the distance out from the origin.) 35-13 Index Symbols 13.6 eV, hydrogen spectrum 35-4 A Allowed orbits, Bohr theory 35-1 Angular momentum Bohr model 35-1, 35-8 Planck's constant 35-8 Atoms Classical hydrogen atom 35-2 B Balmer series Energy level diagram for 35-6 Formula from Bohr theory 35-5 Hydrogen spectrum 35-4 Bell Telephone Lab, electron waves 35-12 Bohr model Allowed orbits 35-1 Angular momentum 35-1, 35-8 Chapter on 35-1 De Broglie explanation 35-1 Derivation of 35-8 Energy levels 35-4 Planck's constant 35-1, 35-8 Quantum mechanics 35-1 Rydberg constant 35-9 Bohr orbits, radii of 35-7 C CGS units Classical hydrogen atom 35-2 Circular orbit, classical hydrogen atom 35-2 Classical hydrogen atom 35-2 Coulomb's law Classical hydrogen atom 35-2 D Davisson & Germer, electron waves 35-12 De Broglie Electron waves 35-11 Formula for momentum 35-11 Hypothesis 35-10 Key to quantum mechanics 35-1 Wavelength, formula for 35-11 Waves, movie of standing wave model 35-11 E Electromagnetic radiation Energy radiated by classical H atom 35-3 Electron In classical hydrogen atom 35-2 Electron scattering First experiment on wave nature 35-12 Electron waves Davisson & Germer experiment 35-12 De Broglie picture 35-11 Scattering of 35-12 Energy Electric potential energy In classical hydrogen atom 35-3 Energy level 35-1 Kinetic energy Bohr model of hydrogen 35-3 Classical hydrogen atom 35-3 Total energy Classical H atom 35-3 Energy level diagram Balmer series 35-6 Bohr theory 35-4 Lyman series 35-6 Paschen series 35-6 F Force Electric force Classical hydrogen atom 35-2 H h bar, Planck's constant 35-9 Hydrogen atom Bohr theory 35-1 Classical 35-2 Hydrogen atom, classical Failure of Newtonian mechanics 35-3 Hydrogen spectrum Balmer series 35-4 Lyman series 35-6 Of star 35-4 Paschen series 35-6 I Infrared light Paschen series, hydrogen spectra 35-6 K Kinetic energy Bohr model of hydrogen 35-3 Classical hydrogen atom 35-3 L Light Hydrogen spectrum Balmer formula 35-5 Spectral lines, hydrogen Bohr theory 35-4 Lyman series, energy level diagram 35-6 35-14 Bohr Theory of Hydrogen M Maxwell's equations Failure of In classical hydrogen atom 35-2 Mechanics Newtonian Classical H atom 35-3 Momentum De Broglie formula for momentum 35-11 Movie Standing De Broglie like waves 35-11 N Newtonian mechanics Classical H atom 35-3 Failure of In the classical hydrogen atom 35-3 Nucleus Discovery of, Rutherford 35-1 O Orbits Bohr, radii of 35-7 Classical hydrogen atom 35-2 P Particle-wave nature De Broglie picture 35-10 Of electrons Davisson and Germer experiment 35-12 De Broglie picture 35-10 Paschen series Energy level diagram 35-6 Hydrogen spectra 35-6 Photon Hydrogen spectrum 35-5 Planck's constant Angular momentum, Bohr model 35-8 Bohr theory 35-1 In de Broglie wavelength formula 35-11 Potential energy Electric potential energy In classical hydrogen atom 35-3 Q Quantized angular momentum In Bohr theory 35-9 In de Broglie's hypothesis 35-10 Quantum mechanics Bohr theory of hydrogen 35-1 R Radiation Radiated energy and the classical H atom 35-3 Radio waves Predicted from the classical hydrogen atom 35-2 Rutherford and the nucleus 35-1 Rydberg constant, in Bohr theory 35-9 S Satellite motion Classical hydrogen atom 35-2 Scattering of waves Davisson-Germer experiment 35-12 Spectral lines Hydrogen Bohr theory of 35-4 Spectrum Hydrogen Bohr theory of 35-4 Lyman series, ultraviolet 35-6 Paschen series, infrared 35-6 Hydrogen star 35-4 Standing waves De Broglie waves Movie 35-11 Star Hydrogen spectrum of 35-4 T Total energy Classical hydrogen atom 35-3 U Unit of angular momentum In Bohr theory 35-9 W Wave De Broglie, standing wave movie 35-11 Electron waves, de Broglie picture 35-11 Wavelength De Broglie 35-11 X x-Ch35 Exercise 1 Exercise 2 Exercise 3 Exercise 4 Exercise 5 Exercise 6 Exercise 7 Exercise 8 Exercise 9 Exercise 10 35-5 35-5 35-6 35-6 35-7 35-9 35-9 35-9 35-10 35-10