Chapter 28 Atomic Physics

Transcription

1 614 Chapter 28 Atomic Physics GOALS After you have mastered the contents of this chapter, you will be able to achieve the following goals: Definitions Define each of the following terms and use it in an operational definition: atomic number ionization energy Bohr radius Beer's law quantum number Lambert's law energy level optical density spectra absorbance laser Bohr Model State the assumptions and predictions of the Bohr model of the hydrogen atom. Deduce the concept of ionization potential. Pauli Exclusion Principle Use the Pauli exclusion principle and the four quantum numbers; n, l, ml, and ms to account for the periodic chart of the elements. Bohr Model Problems Solve problems using the Bohr equations for hydrogen. Energy Level Diagrams Use energy level diagrams to explain the emission and absorption of radiation by atomic systems. Spectrometry and Lambert's Law Sketch a simple spectrometer, and discuss how it can be used in measuring the concentration of an element in a solution. Laser Explain the principle of laser action. PREREQUISITES Before you begin this chapter you should have achieved the goals in Chapter 4, Forces and Newton's Laws, Chapter 5, Energy, Chapter 21, Electrical Properties of Matter, and Chapter 27, Quantum and Relativistic Physics.

2 615 Chapter 28 Atomic Physics 28.1 Introduction The realm of quantum physics is not an obvious part of your daily experience. Where do you expect to find observable evidence of the quantum model? Although the quantum model was originated to solve the problem of thermal radiation, it was the atomic model of matter that gave rise to the impressive early successes of the quantum model. Do you know the atomic basis for the operation of the common fluorescent light? Emission and absorption spectroscopy play a central role in the chemical analysis of specimens in medical laboratories. The atomic and quantum models have been combined to provide physical explanations for these and other atomic phenomena. In this chapter you will be introduced to a quantum model of the atom and some of its important applications Rutherford's Nuclear Atom Model The early model of the atom proposed by J.J. Thomson in 1904 consisted of a positively charged spherical volume with point particle electrons distributed throughout the volume. (This was referred to as the "plum pudding" model of the atom, with the electrons playing the role of raisins.) This model, with its various symmetrical distributions of electrons attempted to explain the periodicity of chemical properties of elements as well as the electromagnetic spectra of atoms. The empirical equations for the various atomic spectral series could not be derived from the then existing models. The model met with very little success, and in 1911 Lord Rutherford published a paper which showed that the atom was much different than the Thomson model. Rutherford's model was based on the scattering of alpha particles (discovered to be helium nuclei of atomic mass 4) by gold foil(figure 28.1).

3 616 Since the alpha particle is much more massive than the electron, (mα/m e 7360), you should expect very little deflection of the alpha particle by the electrons in the gold. Rutherford found, however, that some of the alpha particles were scattered at very large angles, as though they had collided head on with a small, positively charged particle. Rutherford concluded that the atom must be made up of a small, positively charged nucleus surrounded by the electrons. Rutherford estimated the positive nucleus radius to be about 1/3000 times the atomic radius. (He was about a factor of 3 too large in his estimate.) Rutherford imagined the electrons circling the nucleus, held in circular orbits by the Coulomb force. The atom as a whole has a neutral charge; that is, the positive charge on the nucleus is equal in magnitude to the sum of the electron charges. In his early work Rutherford represented the nuclear charge as Ze, where Z is the atomic number of the element and e is the electron charge. (Z is also equal to the number of electrons in the atom.) Elements are arranged in the periodic table in accordance with the individual atomic number Bohr's Model for the Hydrogen Atom Niels Bohr, a young Danish theoretical physicist, had the insight and courage to propose a model that applied quantum physics concepts to the hydrogen atom in Planck's quantum hypothesis had accounted for blackbody radiation. Einstein had used the quantum concept and the photon to explain the photoelectric effect. Bohr's atomic model was even more remarkable. His model was designed to explain the stability of the atom and to account for the characteristic line spectra of hydrogen. Bohr's model incorporated Ernest Rutherford's nuclear model of the atom. Rutherford's alpha-particle scattering experiments had shown that most of the mass was associated with the positive charge of the atom in a small volume at the center of the atom. In Rutherford's nuclear model, electrons revolved around the positive nucleus. However, there was a serious flaw in this classical physics model. In classical electromagnetism, any accelerating charge becomes a source of electromagnetic radiation. Rutherford's model described the electron as being under constant centripetal acceleration, and classical physics predicted the ultimate spiral of the electron into the nucleus. To solve this dilemma, Bohr presented the following postulates: Postulate 1: Classical dynamical equilibrium of the atom involves stationary states given by the quantum condition on the angular momentum for stable orbits of the electron. L = nh/2π (28.1) where n is the quantum number, n = 1,2,3,..., and L is angular momentum, mvr. These stationary-state orbits are not required to satisfy the radiation laws of classical physics. Postulate 2: Radiation is emitted or absorbed when the system makes transition from one stationary state to another. hƒ = E f -E i (28.2) where E f is the final energy and E i is the initial energy of the states involved when a photon of energy hf is emitted or absorbed. Postulate 1 coupled with classical physics

4 617 can be used to derive equations for the energy value and the circular orbit radius associated with Bohr's allowed stationary states for the hydrogen atom. The energy of the nth state of the hydrogen atom is derived as follows(figure 28.2). From Newton's and Coulomb's laws force of attraction = centripetal force F c = (1/4πε o )e 2 /r 2 =mv 2 /r From conservation of energy kinetic energy + potential energy =E total (1/2) mv 2 - (1/4πε o )e 2 /r =E total From the force equation we have mv 2 = (1/4πε o )e 2 /r Thus E total = (½) ((1/4πε o )e 2 /r) - ((1/4πε o )e 2 /r) = -(1/2) ((1/4πε o )e 2 /r) The negative sign for E total indicates that electron is bound to the nucleus. The quantum condition of angular momentum is L = mvr = nh/2π where n = 1,2,... and from the force equation we have mv = (1/4πε o ) (me 2 /r) 1/2 and (me 2 r/4πε o ) 1/2 = nh/2π Thus, the radius of the nth electron orbit of hydrogen is given by r n = n 2 [(h/2π) 2 (4πε o /me 2 )] =n 2 a o (28.3)

5 618 where a o = h 2 ε o /πme 2 = x m and is called the Bohr radius of the hydrogen atom. The quantized energies for the hydrogen atom are given by E n = (1/n 2 )(me 4 /8ε o 2 h 2 ) = ev/n 2 (28.4) where n = 1,2,... (Bohr's quantum number), m the electron mass, e, the electron charge, h, Planck's constant, and ε o, the permittivity of free space. (What is the significance of the negative sign for this energy?) EXAMPLE The energy level of the ground state (n = 1) for the hydrogen atom is given by E 1 = ev/1 2. If the hydrogen atom receives this much energy from the external source, the electron will be freed from the atom, and we say the atom is ionized. The ionization energy of hydrogen in its lowest energy or ground state is 13.6 ev. The second postulate of Bohr's model for the hydrogen atom states that if the orbit of the electron remains the same, the atom will neither gain nor lose energy. In order for the atom to either absorb or emit energy, the principle quantum number (n) must change to some other positive integer, say m. The energy change is given by E m - E n = (- 13.6/m 2 ) - (-13.6/n 2 ) = -13.6(1/m 2-1/n 2 ) ev (28.5) On an energy level diagram an arrow pointing vertically upward indicates an increase in the quantum number for the atom. This corresponds to the absorption of energy, say a photon of energy hf. What energy must be absorbed to raise the hydrogen atom from the n = 1 state to the n = 3 state? On an energy level diagram an arrow pointing vertically downward indicates a decrease in the quantum number of the atom. This corresponds to the emission of energy, a photon of light. How much energy is emitted when an n = 5 hydrogen atom goes into the n = 2 state? Transitions are shown for the various energy levels from n = 1 n = 6 in Figure We calculate the absorption and emission energies given by the arrows 1 through 6.

6 619 Assume that the energy changes above correspond to the absorption or emission of a photon. Remember that λ (nm) 1240(eV-nm)/(eV). The measured wavelengths of the six photons emitted are: λ 1 = nm λ 2 = nm λ 3 = nm Which of these photons correspond to visible light? λ 4 = nm λ 5 = nm λ 6 = nm Bohr's model was successful because it predicted experimentally verifiable atomic spectra. Bohr's model could be used to derive the previously-known empirical equations for spectral series. Unlike Planck's model, Bohr had no adjustable parameters to use to fit the experimental data! 28.4 Atoms Other than Hydrogen The Bohr model served well as a first approximation for a single electron atom. Bohr himself supported his model as only the first step toward a comprehensive quantum theory. In spite of refinements (such as elliptical orbits), the Bohr model could not account for some of the fine structure, and line intensities of atomic spectra of systems with more than one electron such as helium remained unexplained. On the philosophical level, the Bohr model was a mixture of classical (orbits) and quantum (photons) concepts. The quantum jumps as electrons go between orbits were very mysterious indeed. In 1925 Schrodinger and Heisenberg developed their equivalent versions of modern quantum theory of atoms. Schrodinger's formulation has become known as wave mechanics, reflecting the mathematical form of its equations. In modern quantum theory the electron in the atom is characterized by a probability amplitude (known as a wave function). These wave functions, which are solutions to Schrodinger's quantum wave equation, give stationary states that are characterized by four quantum numbers (n, l, m l, and m s ) instead of the single Bohr quantum number. Like the Bohr model the stationary states have definite energies and are the only allowed states for the electron in the atom. We can assign the following significance to the quantum numbers: n, the principle quantum number, corresponds most closely to Bohr's quantum number. n largely determines the energy of the stationary states; l, the orbital angular momentum quantum number, determines the spatial symmetry of the electron probability distribution in the atom; m l, the magnetic quantum number, determines the orientation of the electron distribution in an external magnetic field; m s, the spin quantum number, determines the intrinsic magnetic property of the electron. The word "spin" is derived from a model of the electron as an electric charge spinning on its axis. These quantum numbers are found to be governed by the following rules: 1. The principle quantum number n must be an integer, n = 1, 2, The orbital angular momentum quantum number l must be an integer; it may take all values up to n - 1, l = 0, 1, 2,..., n The magnetic quantum number m l must be an integer, taking all values from - l to + l, m l - l, - l + 1,... 0, 1,... + l

7 The spin quantum number m s is either +1/2 or -1/2, corresponding to spin up and spin down. Wolfgang Pauli formulated the exclusion principle in The exclusion principle states that no two electrons in a given atom can be in the same quantum state. This means that no two electrons can have identical sets of quantum numbers. We can think of the set of four quantum numbers as a set of unique coordinates for each electron in an atom (n, l, m l, m s ). Consider the following example. The lowest state in an atom corresponds to n = 1, l = 0, m l = 0 with m s being either +1/2 or -1/2. We see that the n = 1 states are (1, 0, 0, +1/2) and (1, 0, 0, -1/2). Thus two electrons exhaust the available n = 1 states consistent with exclusion principle. The two electron atom is helium, an inert gas. Pauli used the exclusion principle to explain the periodic table of elements. The periodicity in the appearance of chemical properties of an atom is primarily associated with electrons in the highest quantum states. These electrons are said to be in outer orbits and more loosely bound than "inner" electrons. When the electrons of an atom fill the allowed states for a given atom for a given n value, the atom is relatively inert (as is helium). Atoms with one electron outside an inert core of electrons are said to be hydrogenlike. Sodium is an example of such an atom. List three other hydrogenlike atoms. The order in which atomic systems fill the allowed quantum states is determined by the rule that lowest energy states are filled first. The situation is complicated by the fact that lower n values have lower energies, but smaller l values have lower energy values than large l values. For example, the n = 4, l = 0 states have low energy than the n = 3, l = 2 states. EXAMPLE Find the element in which atoms have all n = 2 states filled. From our previous example, we know that there are two n = 1 states. Then = 2 states are: (2, 1, 0, +1/2), (2, 1, 0, -1/2), (2, 1, 1, +1/2), (2, 1, 1, -1/2), (2, 1, -1, +1/2), (2, 1, -1, -1/2), (2, 0, 0, +1/2), (2, 0, 0, -1/2). The total number of states for an n = 2 system is thus 10. The atom with 10 electrons is neon, another inert gas. The charge distribution for electron clouds associated with a given l value are called orbitals. Each value of m l for a given l value provides a different orientation for the orbital. The directional and shape properties of atomic orbitals provides a basis for a chemical bonding theory as developed by Heitler, London, Slater, and Pauling. Figure 28.4 shows some examples of electron orbitals.

8 Spectra: The Fingerprints of Nature The basic model is the same for all the elements. Every atom has a series of energy levels that may be occupied by electrons. In its lowest energy state the atom will have all of its electrons in the lowest possible electron states. If an atom is to emit or absorb energy, the electrons must change from one electron state to another, and so the atom is only capable of absorbing or emitting energy of certain values. It is not possible for hydrogen to emit or absorb energy of 12 ev. The other elements can emit or absorb energy only at certain discrete energies. The energy level diagram for sodium is shown in Figure 28.5.

9 622 The energy level diagram for atoms of each element is absolutely unique. It is not duplicated by the atoms of any other element. So the emission and absorption characteristics of a material are indications of the kinds of atoms that are present in that material. The chemical composition of a material can be deduced from an examination of the wavelengths of optical photons (spectra), which are emitted or absorbed by a material. In general, to reduce the interaction effects that can occur between neighboring atoms, these spectral properties of materials are best examined with the material in the gaseous state, so that the atoms of the material are as far apart as feasible Emission Photometry To observe the emission spectra of an atom, energy must be absorbed by the atom so that at least one of its electrons is in an excited energy state. Then as these electrons make a transition to a lower energy level, characteristic photons will be emitted, and these photons can be observed (Figure 28.6).

10 623 In the usual laboratory setting the excitation energy is supplied by a heat source such as a flame. A schematic diagram of a typical flame photometry system is shown in Figure The physics of such a system is now within your grasp. The atomizer is used to spray a fine mist of a solution containing the material to be observed into the flame. The thermal energy of the flame excites some of the electrons in the atoms in the solution. Some of the excited atoms will make transitions to lower energy states with the emission of characteristic photons. The observation of the numbers and energies of the emitted photons gives an indication of the kinds of atoms (elements) that are present in the material as well as the concentration of each element. In order to perform quantitative analyses using a flame photometer, an internal standard element is introduced into the solution in known amount C s. The unknown concentration of the element C x is calculated from the ratio of the intensity of the emission of the unknown concentration to the intensity of the light emitted by the standard: C x /C s =I x /I s (28.6) where I x is the intensity of emission from the unknown concentration at wavelength λ s and I s is the intensity of emission from the known concentration of the internal standard at wavelength λ s.

11 Absorption Photometry Most of the atoms in a flame remain in their lowest energy states; so the sample as a whole is more likely to absorb radiation than to emit it. Therefore, the measurement of the amount of radiation absorbed by a substance in the flame will be an order of magnitude more sensitive than the measurement of flame emission. Once again on the basis of your study of the hydrogen atom, the physics of absorption photometry your study of the hydrogen atom, the physics of absorption photometry can be understood. Just as all elements emit characteristics wavelengths of light when returning from an excited state to a lower energy state, elements will only absorb characteristic wavelengths of light as they are excited from their lowest energy states. The quantitative use of absorption photometry relies upon the use of Beer's law: The change in light intensity is proportional to the product of the concentration of the absorbing atoms and the sample thickness. Atomic absorption photometry may be used for the chemical analysis of blood and urine samples. This technique not only makes use of the inherent sensitivity of the absorption technique, but uses the spectral characteristics of a cold-cathode excitation lamp to predetermine the element whose concentration will be measured. For example, when it is desired to know the amount of sodium in a sample, a sodium lamp is used to irradiate the vapor, and only the absorption of a characteristic sodium line is measured. The presence of other substances in the host materials will not influence the measurement of the sodium concentration. Atomic absorption measurements usually proceed in three steps. First, the sample to be examined is vaporized, usually by a flame, so that most of its constituents are in atomic form. Second, the atomic vapor is irradiated by light characteristic of the element being sought. Finally, the absorption of the characteristic light is related to the concentration of the element. An example of an atomic absorption system used to measure sodium concentration is sketched in Figure 28.8 According to Beer's law the absorbance or optical density is proportional to the concentration of material in the solution (Figure 28.9). The absorbance is given by, A = log 10 I o /I (28.7) where I 0 is the intensity of the light with none of the unknown element present, and I is the intensity with some of the unknown material present. Hence, I is a smaller number than I 0, and the ratio of I 0 to I is a positive number larger than one.

12 625 Therefore, A is always a number greater than zero. In practice, an atomic absorption system is calibrated with a series of known solutions (C s ). The concentrations of unknown (C x ) is determined directly from a graph or ratio equation: C x = A x C s /A s (28.8) where A s is the absorbance of the known solution and A x is the absorbance of the unknown The Colorimeter and the Spectrophotometer As we have noted, the absorption of light takes place when the atoms in the sample absorb photons. In this process the photon energy is used to excite atomic electrons to higher energy levels. Each atomic system has its own unique absorption spectrum which can be used in the chemical analysis of complex systems. For a given wavelength of incident light, the intensity of light that is transmitted through a sample of thickness x is given by Lambert's law which is written as follows: I =I o exp(-µx) (28.9) where I 0 is the incident intensity, µ, the absorption coefficient, x, the thickness of sample. Beer's law states that for low concentrations the absorption coefficient is proportional to the concentration of the absorbing atomic system in the sample. This can be expressed in equation form as: µ = αc where a is a constant depending on the material and the wavelength of light used and C is the concentration of the absorbing system. From Lambert's law we find the following equation for x: x = (1/αC) ln(i o /I) or αc = (1/0.434x) log 10 (I o /I) (28.10) where log 10 (I 0 /I) is called the optical density (OD) of the sample, or the absorbance. The colorimeter is an instrument designed to measure the absorption of light at certain selected wavelengths. The wavelength used is determined by the particular atomic system for which you are looking in the sample. In many colorimeters the output is the optical density of the sample for the wavelength selected. In a simple colorimeter the output may simply be a linear intensity scale. In this case the solvent is first placed in the sample cell and the output reading is the value to be used as I 0. The sample is then placed in the instrument and the value of I is obtained. Consider the following example.

13 626 EXAMPLE The sample cell and pure solvent give a reading of 95 (linear scale), while the sample cell with absorbing atomic system NaCl added gives a reading of 38. a. Find the optical density of the absorbing sample. OD = log 10 (I o /I) = log 10 (95/38) =.39 b. If solvent is added to the sample until the reading is 45, find the ratio of the original concentration of NaCl to the final concentration. αc 1 (0.434) = log 10 (I o /I 1 ) = log 10 (95/38) = 0.39 αc 2 (0.434) = log 10 (I o /I 2 ) = log 10 (95/45) = 0.32 C 1 / C 2 =.39/.32 = 1.23 The spectrophotometer is an instrument that is designed to measure and display the absorption spectra, that is, the transmission intensity versus wavelength, for different samples. Figure shows both a single beam and a double beam spectrophotometer. In the single beam instrument a blank sample cell is used to give an I 0 reading for each wavelength, and the sample gives a corresponding value for I (transmitted intensity). The more sophisticated (and also much more expensive) double beam instrument uses a chopped beam and electronics that determine the ratio of I/I 0 for each wavelength as the instrument scans through a wavelength range Quantum Efficiency in Photobiology Interactions between light and matter seem to involve single photon absorption. This model predicts a direct relationship between the number of incident photons and the photoproducts resulting from the interaction. The concept ofquantum yield or quantum efficiency is defined as

14 627 quantum efficiency photo-reaction rate / photon incidence rate (28.11) The photon incidence rate can be determined from the intensity of the incident radiation as follows: intensity = (energy / photon) x (photons / (unit area x sec)) Then we find photons/sec = intensity/(energy/photon) x area of sample or ΔN / Δt (photons/sec) =I(watts/m 2 )A(m 2 )/hƒ(joules/photon) (28.12) A determination of quantum efficiency is very helpful in studying the relative importance of light and living matter reactions. An example is given in Figure which shows the relative effectiveness of different photons in producing photosynthesis reactions in a sample of Euglena cells A Model for Laser Operation As we pointed out in the chapter on the wave properties of light, the laser is a most significant light source. We can now describe an energy level model that serves as the basis for many laser systems. Atoms of lasting material are excited by an external energy source (usually by a flash tube or radio-frequency excitation). This input energy raises electrons to an excited state. Some of these electrons end up in an excited state with an unusually long lifetime (called a metastable state). This produces a population inversion with more electrons in the metastable state than normal. When an electron does drop back to the ground state, the emitted photon stimulates other excited atoms to decay. The ends of the laser cavity have reflecting coatings so that these photons make several passes down the tube producing an avalanche of photons. This represents a great amplification of the initial photon by emitted radiation. The unique feature of this radiation is that the atoms emit radiation in phase with each other. The high intensity is due to the collective behavior of the atoms, and the monochromatic nature of the radiation is due to the sharpness of the energy level transition involved in the emission process. An energy level diagram of a typical laser system is shown in Figure 28.12

15 628 ENRICHMENT Mathematical Derivation of Lambert's Law We consider a beam of light passing through a light absorbing solution. Imagine a layer of this solution dx thick (perpendicular to the incident beam). If the intensity of the incident beam is I, and I' is the intensity of the transmitted beam, then the fraction of light absorbed is given as (I - I')/I This should be proportional to the number of absorbing systems in the beam in the solution. If C is the concentration, that is the amount of solute per unit volume, then the quantity of absorbers per unit area of solution is amount of solute (absorbers)/unit area = Cdx Thus (I - I')/I C dx or, since I - I' = di and I' is less than I, we have - dl / I = αc dx where α is a proportionality constant. If we let αc =µ, the absorption coefficient, -dl/i = µdx or I = -µx + constant If I 0 is the intensity of incident beam (x = 0), then the constant equals ln I 0, and we have ln I = ln (I o µx) or I =I o exp(-µx) If the solution contains several different absorbing solutes with different concentrations we get the following equation: -dl/i = (α 1 C 1 +α 2 C )dx I = I o exp [-(α 1 C 1 +α 2 C )x] The optical density OD or absorbance is defined as follows: OD = log 10 I o /I x = (α/2.3) C x since ln y = 2.3 log 10 y. For multiple absorbing solutes this becomes OD = i=1 N (x/2.3)α 1 C 1 where x is the thickness of the absorbing solution.

16 629 SUMMARY Use these questions to evaluate how well you have achieved the goals of this chapter. The answers to these questions are given at the end of this summary with the number of the section where you can find related content material. Bohr Model 1. Bohr's quantum postulate stated that one of the following quantities came in discrete units of h/2π: a. linear momentum b. energy c. position d. angular momentum e. mass 2. Bohr's model for the hydrogen atom predicts that the absorption spectra involves a. accelerating electrons b. same wavelengths as emission spectra c. electrons going to higher energy levels d. electrons dropping to lower energy levels e. deaccelerating electrons 3. Which of the following characteristics of hydrogen atoms was most difficult for classical physics to interpret: a. mass b. charge c. line spectra d. ionization e. size Pauli Exclusion Principle 4. Give the possible sets of quantum numbers for n = 2. Bohr Model Problems 5. The series of hydrogen spectral lines corresponding to energy transitions ending on the n = 2 level is called the Balmer series. The shortest wavelength of the Balmer series is a. 122 nm b nm c. 656 nm d nm e. 365 nm

17 The longest wavelength of the Balmer series will be a. 122 nm b nm c. 656 nm d nm e. 365 nm 7. Since room temperature thermal energy corresponds to about 0.03 ev, you would expect hydrogen atoms at room temperature to be a. ionized b. in n = 2 state c. in n = 20 state d. in n = 1 state e. in n = 10 state Energy Level Diagrams 8. Which of the following transitions would give the shortest wavelength emission line: a. n = 1 to n = 2 b. n = 1 to n = 3 c. n = 3 to n = 1 d. n = 4 to n = 2 e. n = 5 to n = 4 9. Which of the following transitions would give the longest wavelength absorption line: a. n = 1 ton = 2 b. n = 2 to n = 3 c. n = 1 to n = 3 d. n = 4 to n = 2 e. n = 5 to n = 4 Spectrometry and Lambert's Law 10. For low concentrations Beer's law predicts that the absorption coefficient of a solution will be linear with a. wavelength b. frequency c. concentration d. energy e. intensity

18 If the optical density of a sample is 1.0, then the ratio of transmitted to incident radiation is a. 0.1 b. 10 c. 100 d e The spectrophotometer is an instrument designed to measure and display a. frequency vs. wavelength b. transmission intensity vs. wavelength c. intensity vs. thickness d. intensity vs. concentration e. absorption coefficient vs. concentration Answers 1. d (Section 28.3) 2. b, c (Section 28.3) 3. c (Section 28.3) 4. l = 1, m l = -1, 0, + 1, m S = ±1/2; l = 0, m l = 0, m S = ±1/2; total of 8 (Section 28.4) 5. e (Section 28.3) 6. c (Section 28.3) 7. d (Section 28.3) 8. c (Section 28.3) 9. e (Section 28.3) 10. c (Section 28.8) 11. a (Section 28.8) 12. b (Section 28.8) ALGORITHMIC PROBLEMS Listed below are the important equations from this chapter. The problems following the equations will help you learn to translate words into equations and to solve singleconcept problems. Equations L = nh/2π (28.1) hƒ = E f -E i (28.2) r n = n 2 [(h/2π) 2 (4πε o /me 2 )] =n 2 a o where a o = x m (28.3) E n = (1/n 2 )(me 4 /8ε 2 o h 2 ) = ev/n 2 (28.4) E m - E n = (- 13.6/m 2 ) - (-13.6/n 2 ) = -13.6(1/m 2-1/n 2 ) ev (28.5) A = OD = log 10 I o /I (28.7) C x = A x C s /A s (28.8) I =I o exp(-µx) (28.9)

19 632 Problems 1. What would be the velocity of an electron in an orbit for n = 1 according to the Bohr model? The radius r 1 =.053 nm. How does it compare with the velocity in an orbit for n = 2? 2. If radiation is emitted by an atom in going from an energy level of ev (3.04 x J) to an energy level of ev (5.44 x J), what is a. the frequency of the radiation b. its wavelength 3. What is the absorbance or optical density of a sample in which the ratio of I 0 /I is 3? 4. What is the ratio of I/I 0 for a specimen 0.5 cm thick of a material which has an absorption coefficient of 2 per centimeter? Answers x 10 6 m/sec; v 1 /v 2 = 2 2. a x Hz; b. 829 nm EXERCISES These exercises are designed to help you apply the ideas of a section to physical situations. When appropriate the numerical answer is given in brackets at the end of each exercise. Section Suppose you could observe the hydrogen atom through a magnifying system such that the hydrogen nucleus appeared to be the size of a baseball (r 4 cm). What would the apparent radius of the n = 1 electron orbit? Assume the radius of a proton = 1.5 x m. [1.42 x 10 4 cm] 2. Assume you want to find an electron orbit that is as far away from the n = 1 electron orbit as is the nucleus. What is the n value for such a fictitious orbit? [SQR RT(2)] 3. It has been suggested that the volume of an atom is mostly empty space. What percentage of the volume of an n = 1 hydrogen atom is occupied by matter? Radius of proton = 1.5 x m [Volume of atom about 4 x times as large as volume of proton] 4. What is the velocity of the electron in a hydrogen atom for n = 2? The average time in the excited state is usually about 10-8 sec. How many revolutions does this electron make during this time? [v = 0.6 x 10 6 m/sec; 4.73 x 10 6 revolutions]

20 Compute the energy associated with a hydrogen atom for n = 1, 2, 3, and 4. Draw an energy level diagram. [n = 1, ev; n = 2, -3.4 ev; n = 3, ev; n = 4, ev] 6. A large number of hydrogen atoms are excited to state n = 4. What transitions and spectra lines are then possible? [4 3, 4 2, 4 1, 3 2, 3 1, 2 1] Section Find the n and l numbers for the final electron ground state of krypton (Z = 36). [n = 4, l = 1.] Section An atomic absorption instrument is being used to measure the concentration of sodium in an aqueous solution. The following calibration measurements are taken: Solution Readout Current (ma) Distilled water ppm Na in H2 O ppm Na in water ppm Na in water ppm Na in water ppm Na in water 0.09 What are the concentrations of sodium in unknown solutions whose readout currents are given by: unknown solution 1, 200 ma; unknown solution 2, 40.0 ma; unknown solution 3, 8.00 ma. Na in 1 = ppm Na in 2 = ppm Na in 3 = ppm [4, 19, 35] 9. The absorbance of a 10-ml sample is found to be 3. The sample is accidently diluted. The lab technician (who had completed this physics course) ran another absorbance spectrum and found the new absorbance to be 2.7. a. Find the ratio of sample concentrations and volume of the diluted sample. b. Find the ratio of transmitted intensities for these two measurements. [a. C 2 /C 1 = 0.9,V 2 = 11.1 ml; b. I 2 /I 1 = 2] 10. A blood sugar study shows the absorbance of a morning sample to be 2 and that of an evening sample to be 1.8. Find the ratio of sugar concentration in the two samples. [C E = 0.9C M ]

21 634 PROBLEMS Each of the following problems may involve more than one physical concept. The answer is given in parentheses at the end of the problem. 11. Find the atom that completes the third period in the periodic table (all states up to n = 3, l = 1 states filled). [Argon] 12. A helium He+ ion is similar to a hydrogen atom. Compare the radii of orbits for n = 1 for He+ ion and the hydrogen atom. (Use reduced the mass correction from problem 14.) [r He = 1/2 r H ] 13. Assume the sodium atom is hydrogen type (a nucleus with +11e charge surrounded by the 10 inner electrons) and that it has an ionization potential of 5.12 ev. What is the radius of the eleventh electron orbit? Compare this with the size of the Na atom. [r 11 = 14 nm] 14. One of the first corrections that was made in the simple Bohr model was the correction for the relative motion of the proton and the electron. This correction involves the relative motion of the proton and electron about the center of mass of the system. The results show that this correction calls for the use of the reduced mass in place of the electron mass. The reduced mass m is given by µ = m e M p /m e + M p where m e = electron mass, M p = proton mass = 1840 m e. Use this correction to compare the Balmer series limit wavelength (n = n = 2) of deuterium with the corresponding hydrogen line. [Differs by about 30 parts in 109,707] 15. Imagine a new kind of atom in which the electron is attached to the nucleus of the hydrogen atom by a quantum mechanical spring. Hence the force of attraction is given by F = -kr where k has a value of 32.4π 2 N/m. (This atom is not the usual Bohr atom where the force of attraction is given by Coulomb's law!) a. Compute the value of r in terms of m, k, and v for an electron traveling in a circular orbit of radius r with a velocity v in this new kind of atom. b. Use Bohr's postulate to quantize the angular momentum of the electron, and express the radius of the electron orbit in this new atom as a function of n, the principle quantum number where n = 1,2,3,... c. Since you know that for a force of the form kr, the potential energy is given by (½)kr 2, calculate the total energy of the electron in terms of n. d. Sketch the energy level diagram for the new atom. How does it differ from the energy level diagram for the Bohr model of the hydrogen atom? [a. r = (SQR RT(m/k))v; b. r = (n(h/2π)sqr RT(1/mk)) 1/2 = [n(h/2π) 1/mk] 1/2 ; c. E = - Eo +n(h/2π) SQR RT(k/m) = -Eo + (n(h/2π)/2π)sqr RT(k/m)]

22 Light from an ultraviolet light source consists of all wavelengths between 100 nm and 200 nm. If this light is incident upon a cell containing hydrogen gas, discuss quantitatively the physical phenomena you expect to result from this interaction. 17. Beer's law may be stated in the following form: I transmitted = I incident exp(-kx) Where x is length of path in absorbing medium and k is the absorption coefficient. What are the units of k? If I transmitted is 75 percent of I incident for a given thickness x, what thickness is required to reduce I transmitted to 50 percent? [x 1 = x + (1/k) ln(3/2)]