35. Show that Equation 41.26 is a solution of Equation 41.24 with energy E512Uv.
36. A one-dimensional harmonic oscillator wave function is
c 5 Axe2bx2
(a) Show that c satisfies Equation 41.24. (b) Find b and the total energy E. (c) Is this wave function for the ground state or for the first excited state?
37. A quantum simple harmonic oscillator consists of an elec- tron bound by a restoring force proportional to its position relative to a certain equilibrium point. The proportional- ity constant is 8.99 N/m. What is the longest wavelength of light that can excite the oscillator?
38. A quantum simple harmonic oscillator consists of a particle of mass m bound by a restoring force proportional to its position relative to a certain equilibrium point. The proportionality constant is k. What is the longest wave- length of light that can excite the oscillator?
(e), what is the energy of the second excited state? (g) What is the energy difference between the ground state and the second excited state? (h) What is the wavelength of a pho- ton that will cause the transition between the ground state and the second excited state?
51. For a quantum particle described by a wave function c(x), the expectation value of a physical quantity f(x) asso- ciated with the particle is defined by
8f1x2 9;3
`
2`
c*f1x2c dx
For a particle in an infinitely deep one-dimensional box extending from x 5 0 to x 5 L, show that
8x295L2 3 2 L2
2n2p2
52. A quantum particle is described by the wave function
c1x2 5 •A cos a2px
L b for 2L
4#x#L 4
0 elsewhere
(a) Determine the normalization constant A. (b) What is the probability that the particle will be found between x 5 0 and x 5 L/8 if its position is measured?
53. A quantum particle of mass m is placed in a one- dimensional box of length L. Assume the box is so small that the particle’s motion is relativistic and K 5 p2/2m is not valid. (a) Derive an expression for the kinetic energy levels of the particle. (b) Assume the particle is an electron in a box of length L 5 1.00 3 10212 m. Find its lowest pos- sible kinetic energy. (c) By what percent is the nonrelativis- tic equation in error? Suggestion: See Equation 39.23.
54. Why is the following situation impossible? A particle is in the ground state of an infinite square well of length L. A light source is adjusted so that the photons of wavelength l are absorbed by the particle as it makes a transition to the first excited state. An identical particle is in the ground state of a finite square well of length L. The light source sends photons of the same wavelength l toward this particle. The photons are not absorbed because the allowed energies of the finite square well are different from those of the infi- nite square well. To cause the photons to be absorbed, you move the light source at a high speed toward the particle in the finite square well. You are able to find a speed at which the Doppler-shifted photons are absorbed as the particle makes a transition to the first excited state.
55. A quantum particle has a wave function
c1x2 5 • Å 2
a e2x/a for x.0
0 for x,0
(a) Find and sketch the probability density. (b) Find the probability that the particle will be at any point where a freely moving particle, with the wave function given by
Equation 41.4, and (b) for a particle in a box, with the wave function given by Equation 41.13.
45. Prove that assuming n 5 0 for a quantum particle in an infinitely deep potential well leads to a violation of the uncertainty principle Dpx Dx$U/2.
46. An electron in an infinitely deep potential well has a ground-state energy of 0.300 eV. (a) Show that the photon emitted in a transition from the n 5 3 state to the n 5 1 state has a wavelength of 517 nm, which makes it green vis- ible light. (b) Find the wavelength and the spectral region for each of the other five transitions that take place among the four lowest energy levels.
47. Calculate the transmission probability for quantum- mechanical tunneling in each of the following cases. (a) An electron with an energy deficit of U 2 E 5 0.010 0 eV is incident on a square barrier of width L 5 0.100 nm. (b) An electron with an energy deficit of 1.00 eV is incident on the same barrier. (c) An alpha particle (mass 6.65 3 10227 kg) with an energy deficit of 1.00 MeV is incident on a square barrier of width 1.00 fm. (d) An 8.00-kg bowling ball with an energy deficit of 1.00 J is incident on a square barrier of width 2.00 cm.
48. A marble rolls back and forth across a shoebox at a con- stant speed of 0.8 m/s. Make an order-of-magnitude esti- mate of the probability of it escaping through the wall of the box by quantum tunneling. State the quantities you take as data and the values you measure or estimate for them.
49. An atom in an excited state 1.80 eV above the ground state remains in that excited state 2.00 ms before moving to the ground state. Find (a) the frequency and (b) the wave- length of the emitted photon. (c) Find the approximate uncertainty in energy of the photon.
50. An electron is confined to move in the xy plane in a rectangle whose dimensions are Lx and Ly. That is, the electron is trapped in a two-dimensional potential well having lengths of Lx and Ly. In this situation, the allowed energies of the electron depend on two quantum numbers nx and ny and are given by
E5 h2 8meanx
2
Lx21ny2
Ly2b
Using this information, we wish to find the wavelength of a photon needed to excite the electron from the ground state to the second excited state, assuming Lx 5 Ly 5 L.
(a) Using the assumption on the lengths, write an expres- sion for the allowed energies of the electron in terms of the quantum numbers nx and ny. (b) What values of nx and ny
correspond to the ground state? (c) Find the energy of the ground state. (d) What are the possible values of nx and ny
for the first excited state, that is, the next-highest state in terms of energy? (e) What are the possible values of nx and ny for the second excited state? (f) Using the values in part
| Problems 1249
where k1 5 2p/l1 and k2 5 2p/l2 are the wave numbers for the incident and transmitted particles, respectively. Pro- ceed as follows. Show that the wave function c1 5 Aeik1x 1 Be2ik1x satisfies the Schrửdinger equation in region 1, for x , 0. Here Aeik1x represents the incident beam and Be2ik1x represents the reflected particles. Show that c2 5 Ceik2x satisfies the Schrửdinger equation in region 2, for x . 0.
Impose the boundary conditions c15 c2 and dc1/dx 5 dc2/dx, at x 5 0, to find the relationship between B and A.
Then evaluate R 5 B2/A2. A particle that has kinetic energy E 5 7.00 eV is incident from a region where the potential energy is zero onto one where U 5 5.00 eV. Find (b) its probability of being reflected and (c) its probability of being transmitted.
x , 0. (c) Show that c is normalized and then (d) find the probability of finding the particle between x 5 0 and x 5 a.
56. A two-slit electron diffraction experiment is done with slits of unequal widths. When only slit 1 is open, the number of electrons reaching the screen per second is 25.0 times the number of electrons reaching the screen per second when only slit 2 is open. When both slits are open, an interfer- ence pattern results in which the destructive interference is not complete. Find the ratio of the probability of an electron arriving at an interference maximum to the prob- ability of an electron arriving at an adjacent interference minimum. Suggestion: Use the superposition principle.
57. The normalized wave functions for the ground state, c0(x), and the first excited state, c1(x), of a quantum har- monic oscillator are
c01x25 aa
pb1/4 e2ax2/2 c11x2 5a4a3
p b1/4 xe2ax2/2 where a5mv/U. A mixed state, c01(x), is constructed from these states:
c011x2 5 1
!23c01x2 1 c11x2 4
The symbol 8q9s denotes the expectation value of the quan- tity q for the state cs(x). Calculate the expectation values (a) 8x90, (b) 8x91, and (c) 8x901.
Challenge Problems
58. An electron is represented by the time-independent wave function
c1x25 eAe2ax for x.0 Ae1ax for x,0
(a) Sketch the wave function as a function of x. (b) Sketch the probability density representing the likelihood that the electron is found between x and x 1 dx. (c) Only an infinite value of potential energy could produce the discontinuity in the derivative of the wave function at x 5 0. Aside from this feature, argue that c(x) can be a physically reasonable wave function. (d) Normalize the wave function. (e) Deter- mine the probability of finding the electron somewhere in the range
2 1
2a # x # 1 2a
59. Particles incident from the left in Figure P41.59 are con- fronted with a step in potential energy. The step has a height U at x 5 0. The particles have energy E . U. Classi- cally, all the particles would continue moving forward with reduced speed. According to quantum mechanics, how- ever, a fraction of the particles are reflected at the step.
(a) Prove that the reflection coefficient R for this case is
R5 1k12k222 1k11k222
Incoming particle
U U 0
E
Figure P41.59
60. Consider a “crystal” consisting of two fixed ions of charge 1e and two electrons as shown in Figure P41.60.
(a) Taking into account all the pairs of interactions, find the potential energy of the system as a function of d. (b) Assum- ing the electrons to be restricted to a one-dimensional box of length 3d, find the minimum kinetic energy of the two electrons. (c) Find the value of d for which the total energy is a minimum. (d) State how this value of d compares with the spacing of atoms in lithium, which has a density of 0.530 g/cm3 and a molar mass of 6.94 g/mol.
d d d
Figure P41.60
61. An electron is trapped in a quantum dot. The quantum dot may be modeled as a one-dimensional, rigid-walled box of length 1.00 nm. (a) Taking x 5 0 as the left side of the box, calculate the probability of finding the electron between x1 5 0.150 nm and x2 5 0.350 nm for the n 5 1 state. (b) Repeat part (a) for the n 5 2 state. Calculate the energies in electron volts of (c) the n 5 1 state and (d) the n 5 2 state.
62. (a) Find the normalization constant A for a wave func- tion made up of the two lowest states of a quantum particle in a box extending from x 5 0 to x 5 L:
c1x2 5Acsin apx
L b14 sin a2px L b d
is a solution to the simple harmonic oscillator problem.
(a) Find the energy of this state. (b) At what position are you least likely to find the particle? (c) At what positions are you most likely to find the particle? (d) Determine the value of B required to normalize the wave function.
(e) What If? Determine the classical probability of finding the particle in an interval of small length d centered at the position x521U/mv21/2. (f) What is the actual probability of finding the particle in this interval?
(b) A particle is described in the space 2a # x # a by the wave function
c1x2 5A cos apx
2ab1B sin apx a b
Determine the relationship between the values of A and B required for normalization.
63. The wave function
c1x2 5Bxe21mv/2U2x2
1251 42.1 Atomic Spectra of Gases
42.2 Early Models of the Atom
42.3 Bohr’s Model of the Hydrogen Atom 42.4 The Quantum Model of the Hydrogen Atom 42.5 The Wave Functions for Hydrogen
42.6 Physical Interpretation of the Quantum Numbers
42.7 The Exclusion Principle and the Periodic Table 42.8 More on Atomic Spectra: Visible and X-Ray 42.9 Spontaneous and Stimulated Transitions 42.10 Lasers
In Chapter 41, we introduced some basic concepts and techniques used in quantum mechanics along with their applications to vari- ous one-dimensional systems. In this chapter, we apply quantum mechanics to atomic sys- tems. A large portion of the chapter is focused on the application of quantum mechanics to the study of the hydrogen atom. Understand- ing the hydrogen atom, the simplest atomic system, is important for several reasons:
• The hydrogen atom is the only atomic sys- tem that can be solved exactly.
• Much of what was learned in the 20th
century about the hydrogen atom, with its single electron, can be extended to such single-electron ions as He1 and Li21.
• The hydrogen atom is an ideal system for performing precise tests of theory against experiment and for improving our overall understanding of atomic structure.
• The quantum numbers that are used to characterize the allowed states of hydro- gen can also be used to investigate more complex atoms, and such a description
This street in the Ginza district in Tokyo displays many signs formed from neon lamps of varying bright colors. The light from these lamps has its origin in transitions between quantized energy states in the atoms contained in the lamps. In this chapter, we investigate those transitions. (© Ken Straiton/Corbis)
chapter 42
Atomic Physics
enables us to understand the periodic table of the elements. This understanding is one of the greatest triumphs of quantum mechanics.
• The basic ideas about atomic structure must be well understood before we attempt to deal with the complexities of molecular structures and the electronic structure of solids.
The full mathematical solution of the Schrửdinger equation applied to the hydrogen atom gives a complete and beautiful description of the atom’s properties. Because the mathematical procedures involved are beyond the scope of this text, however, many details are omitted. The solutions for some states of hydrogen are discussed, together with the quantum numbers used to characterize various allowed states. We also discuss the physical significance of the quantum numbers and the effect of a magnetic field on certain quantum states.
A new physical idea, the exclusion principle, is presented in this chapter. This principle is extremely important for understanding the properties of multielectron atoms and the arrangement of elements in the periodic table.
Finally, we apply our knowledge of atomic structure to describe the mechanisms involved in the production of x-rays and in the operation of a laser.
42.1 Atomic Spectra of Gases
As pointed out in Section 40.1, all objects emit thermal radiation characterized by a continuous distribution of wavelengths. In sharp contrast to this continuous- distribution spectrum is the discrete line spectrum observed when a low-pressure gas undergoes an electric discharge. (Electric discharge occurs when the gas is sub- ject to a potential difference that creates an electric field greater than the dielectric strength of the gas.) Observation and analysis of these spectral lines is called emis- sion spectroscopy.
When the light from a gas discharge is examined using a spectrometer (see Active Fig. 38.15), it is found to consist of a few bright lines of color on a generally dark background. This discrete line spectrum contrasts sharply with the continu- ous rainbow of colors seen when a glowing solid is viewed through the same instru- ment. Figure 42.1a shows that the wavelengths contained in a given line spectrum are characteristic of the element emitting the light. The simplest line spectrum is that for atomic hydrogen, and we describe this spectrum in detail. Because no two elements have the same line spectrum, this phenomenon represents a practical and sensitive technique for identifying the elements present in unknown samples.
Another form of spectroscopy very useful in analyzing substances is absorption spectroscopy. An absorption spectrum is obtained by passing white light from a continuous source through a gas or a dilute solution of the element being ana- lyzed. The absorption spectrum consists of a series of dark lines superimposed on the continuous spectrum of the light source as shown in Figure 42.1b for atomic hydrogen.
The absorption spectrum of an element has many practical applications. For example, the continuous spectrum of radiation emitted by the Sun must pass through the cooler gases of the solar atmosphere. The various absorption lines observed in the solar spectrum have been used to identify elements in the solar atmosphere. In early studies of the solar spectrum, experimenters found some lines Pitfall Prevention 42.1
Why Lines?
The phrase “spectral lines” is often used when discussing the radiation from atoms. Lines are seen because the light passes through a long and very narrow slit before being sepa- rated by wavelength. You will see many references to these “lines” in both physics and chemistry.
42.1 | Atomic Spectra of Gases 1253
that did not correspond to any known element. A new element had been discov- ered! The new element was named helium, after the Greek word for Sun, helios.
Helium was subsequently isolated from subterranean gas on the Earth.
Using this technique, scientists have examined the light from stars other than our Sun and have never detected elements other than those present on the Earth.
Absorption spectroscopy has also been useful in analyzing heavy-metal contamina- tion of the food chain. For example, the first determination of high levels of mer- cury in tuna was made with the use of atomic absorption spectroscopy.
The discrete emissions of light from gas discharges are used in “neon” signs such as those in the opening photograph of this chapter. Neon, the first gas used in these types of signs and the gas after which these signs are named, emits strongly in the red region. As a result, a glass tube filled with neon gas emits bright red light when an applied voltage causes a continuous discharge. Early signs used different gases to provide different colors, although the brightness of these signs was gener- ally very low. Many present-day “neon” signs contain mercury vapor, which emits strongly in the ultraviolet range of the electromagnetic spectrum. The inside of a present-day sign’s glass tube is coated with a material that emits a particular color when it absorbs ultraviolet radiation from the mercury. The color of the light from the tube results from the particular material chosen. A household fluorescent light operates in the same manner, with a white-emitting material coating the inside of the glass tube.
From 1860 to 1885, scientists accumulated a great deal of data on atomic emis- sions using spectroscopic measurements. In 1885, a Swiss schoolteacher, Johann Jacob Balmer (1825–1898), found an empirical equation that correctly predicted the wavelengths of four visible emission lines of hydrogen: Ha (red), Hb (blue- green), Hg (blue-violet), and Hd (violet). Figure 42.2 shows these and other lines (in the ultraviolet) in the emission spectrum of hydrogen. The four visible lines occur at the wavelengths 656.3 nm, 486.1 nm, 434.1 nm, and 410.2 nm. The com- plete set of lines is called the Balmer series. The wavelengths of these lines can be described by the following equation, which is a modification made by Johannes Rydberg (1854–1919) of Balmer’s original equation:
1
l5RHa1 22 2 1
n2b n53, 4, 5,c (42.1)
where RH is a constant now called the Rydberg constant with a value of 1.097 373 2 3 107 m21. The integer values of n from 3 to 6 give the four visible lines from 656.3 nm (red) down to 410.2 nm (violet). Equation 42.1 also describes the ultraviolet spectral lines in the Balmer series if n is carried out beyond n 5 6. The series limit is the shortest wavelength in the series and corresponds to n S `, with
Balmer series W
400 500 600 700
H
400 500 600 700
Ne Hg H
l(nm)
a
b
Figure 42.1 (a) Emission line spectra for hydrogen, mercury, and neon. (b) The absorption spectrum for hydrogen. Notice that the dark absorption lines occur at the same wavelengths as the hydrogen emis- sion lines in (a). (K. W. Whitten, R. E. Davis, M. L. Peck, and G. G.
Stanley, General Chemistry, 7th ed., Belmont, CA, Brooks/Cole, 2004.)
410.2 434.1 Ultraviolet
486.1 656.3 364.6
(nm) l
The lines shown in color are in the visible range of wavelengths.
This line is the shortest wavelength line and is in the ultraviolet region of the electromagnetic spectrum.
Figure 42.2 The Balmer series of spectral lines for atomic hydrogen, with several lines marked with the wavelength in nanometers. (The horizontal wavelength axis is not to scale.)
a wavelength of 364.6 nm as in Figure 42.2. The measured spectral lines agree with the empirical equation, Equation 42.1, to within 0.1%.
Other lines in the spectrum of hydrogen were found following Balmer’s discov- ery. These spectra are called the Lyman, Paschen, and Brackett series after their discoverers. The wavelengths of the lines in these series can be calculated through the use of the following empirical equations:
1
l5RHa12 1
n2b n52, 3, 4,c (42.2)
1
l 5RHa1 322 1
n2b n54, 5, 6,c (42.3)
1
l 5RHa1 422 1
n2b n55, 6, 7,c (42.4)
No theoretical basis existed for these equations; they simply worked. The same con- stant RH appears in each equation, and all equations involve small integers. In Sec- tion 42.3, we shall discuss the remarkable achievement of a theory for the hydrogen atom that provided an explanation for these equations.
42.2 Early Models of the Atom
The model of the atom in the days of Newton was a tiny, hard, indestructible sphere. Although this model provided a good basis for the kinetic theory of gases (Chapter 21), new models had to be devised when experiments revealed the electri- cal nature of atoms. In 1897, J. J. Thomson established the charge-to-mass ratio for electrons. (See Fig. 29.15 in Section 29.3.) The following year, he suggested a model that describes the atom as a region in which positive charge is spread out in space with electrons embedded throughout the region, much like the seeds in a water- melon or raisins in thick pudding (Fig. 42.3). The atom as a whole would then be electrically neutral.
In 1911, Ernest Rutherford (1871–1937) and his students Hans Geiger and Ernest Marsden performed a critical experiment that showed that Thomson’s model could not be correct. In this experiment, a beam of positively charged alpha particles (helium nuclei) was projected into a thin metallic foil such as the target in Fig- ure 42.4a. Most of the particles passed through the foil as if it were empty space, but some of the results of the experiment were astounding. Many of the particles deflected from their original direction of travel were scattered through large angles.
Some particles were even deflected backward, completely reversing their direction of travel! When Geiger informed Rutherford that some alpha particles were scat- tered backward, Rutherford wrote, “It was quite the most incredible event that has Lyman series X
Paschen series X
Brackett series X
Joseph John Thomson English physicist (1856–1940) The recipient of a Nobel Prize in Physics in 1906, Thomson is usually considered the discoverer of the electron. He opened up the field of subatomic particle physics with his extensive work on the deflection of cathode rays (electrons) in an electric field.
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The electrons are small negative charges at various locations within the atom.
The positive charge of the atom is distributed continuously in a spherical volume.
Figure 42.3 Thomson’s model of the atom.
Scintillation screens Lead
screens Source of
alpha particles
a b
Target
Figure 42.4 (a) Rutherford’s technique for observing the scattering of alpha particles from a thin foil target. The source is a naturally occurring radioactive substance, such as radium. (b) Rutherford’s planetary model of the atom.