(38.9)
The number n is called the angular-momentum quantum number.
Let us now see how to calculate the stationary states and the spectrum of the hydro-gen atom on the basis of these postulates. For the sake of simplicity, we will assume that the electron moves in a circular orbit around the proton, which remains at rest (Fig.
38.11). The centripetal acceleration of the electron is This centripetal accelera-tion is produced by the force of attracaccelera-tion between the electron and the proton, that is, the Coulomb force Thus, the equation of motion for the electron is
(38.10) The orbital angular momentum for a circular orbit is [see Eq. (13.34)].
According to Bohr’s postulate, this orbital angular momentum must be multi-plied by an integer n:
(38.11)
1296 CHAPTER 38 Spectral Lines, Bohr's Theory, and Quantum Mechanics
stationary state
FIGURE 38.11 Electron in circular orbit around a proton.
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It is convenient to rewrite this as
(38.12) where (pronounced “h-bar”) is Planck’s constant h divided by
(38.13)
From Eq. (38.12), we obtain
(38.14)
and when we substitute this into Eq. (38.10), we obtain an equation for the radius of the orbit,
(38.15)
Solving this for the radius, we find
(38.16)
From Eq. (38.16) we see that the radii of the permitted orbits are proportional to The smallest value of n is and this leads to the radius of the smallest permit-ted orbit, which is called the Bohr radius and designapermit-ted by
(38.17)
The radii of the other permitted orbits are multiples of the Bohr radius:
(38.18)
For this gives respectively. Figure 38.12 shows
the permitted circular orbits, drawn to scale.
r 4a0, 9a0, 16a0, …, n 2, 3, 4, …,
r4p0n2U2 mee2 n2a0 a0 4p0U2
mee2 0.529 1010 m 0.0529 nm a0: n 1,
n2. r 4p0n2U2
mee2 mean U
merb21
r 1
4p0
e2 r 2 v n U
mer U h
2p 6.63 1034 J s
2p 1.05 1034 J s 2p:
U
mevr nU
38.4 Bohr’s Theory 1297
Bohr radius
n 1 n 2 n 3 n 4
0 2 4 6 8 10
1010 m Orbits permitted by
Bohr theory are circular, …
…with radii r n2a0 that are square-integer multiples of Bohr radius a0 0.0529 nm.
FIGURE 38.12 The possible Bohr orbits of an electron in the hydrogen atom.
h (h-bar)
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The energy of the electron in one of these orbits is a sum of kinetic and potential energies. The kinetic energy is and the potential energy is the electrostatic
potential energy [see Eq. (25.11)]. Hence
the total energy is
(38.19) Next, we substitute from Eq. (38.14) and then r from Eq. (38.16), and we obtain
(38.20)
(38.21)
or
(38.22)
This energy depends on the quantum number n, and as a reminder of this depend-ence, we label this energy with a subscript n:
(38.23)
According to this equation, the energies of all the stationary states are negative. The stationary state with the least energy, or the most negative energy, is the state with
for which
(38.24) And, by Eq. (28.23), the energies of the other stationary states are fractions of this energy:
(38.25)
Thus, and so on.
Figure 38.13 displays these quantized energies in an energy-level diagram. Each horizontal line represents one of the energies given by Eq. (38.25). According to Bohr’s
E3 (13.6
9) eV,1298 CHAPTER 38 Spectral Lines, Bohr's Theory, and Quantum Mechanics
energies of stationary states
energy-level diagram
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assumptions, the electron emits a photon when it makes a transition, or a quantum jump, from one stationary state to a lower stationary state. Such quantum jumps have been indicated by arrows in Fig. 38.13. The stationary state of lowest energy is called the ground state, the next state is called the first excited state, the next state is called the second excited state, and so on.
Ordinarily, the electron of the hydrogen atom is in the ground state, that is, the circular orbit with radius and energy This is the configuration of least energy, and it is the configuration into which the atom tends to settle when it is left undisturbed. As long as the atom remains in the ground state, it does not emit light.
To bring about the emission of light, we must first kick the electron into one of the excited states, that is, a circular orbit of larger radius and higher energy. We can do this by heating a sample of atoms or by passing an electric current through the sample.
Collisions between the atoms will then disturb the electronic motions and occasion-ally kick an electron into a larger orbit. From there, the electron will spontaneously jump into a smaller orbit, emitting a photon. Note that the quantum jumps indicated by colored arrows in Fig. 38.13 form several series: one series consists of all those jumps (indicated by blue arrows) that end in the ground state, another series consists of all those jumps (indicated by red arrows) that end in the first excited state, etc. These series of jumps give rise to the series of spectral lines: the Lyman series, the Balmer series, etc.
From our formula for the energies of the states of the hydrogen atom we can cal-culate the frequency of the light emitted in a quantum jump from some initial state to a final state, as in the following example.
13.6 eV.
ground state and excited states E line represents one of the allowed energies…
FIGURE 38.13Energy-level diagram for the hydrogen atom. Jumps (blue) that end in the ground state (n = 1) give rise to the Lyman series; jumps (red) that end in the first excited state give rise to the Balmer series; etc.
(n 2)
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wavelength of emitted photon frequency of emitted photon
Calculate the frequency and the wavelength of the light emitted by the electron in a quantum jump from the second excited state to the first excited state
SOLUTION: The energy of the initial state is and the energy of the final state is Hence the electron releases an energy
According to Bohr’s postulates, this energy is radiated as a single photon of fre-quency
and wavelength
This wavelength agrees with wavelength of the first spectral line of the Balmer series (see Table 38.1), except for a round-off error.
More generally, we can calculate the frequency of the light emitted in a quantum jump from some initial state of quantum number to some final state of quantum number nf. The initial energy of the electrons is and the final energy is Thus, the electron releases an energy
(38.26)
This energy is radiated as a single photon of frequency
(38.27)
From the frequency of the light, we can calculate the wavelength. Since the wave-length is inversely proportional to the frequency, Eq. (38.27) implies the following expression for the wavelength of the emitted light:
(38.28)
or, with Eq. (38.26),
(38.29)
Comparison of Eqs. (38.7) and (38.29) yields the following theoretical formula for the Rydberg constant:
1
l mee4
4p(4p0)2U3ca 1 n2f 1
n2ib 1
l Ei Ef
hc Ei Ef 2p U c
l c
f,f ¢E
h Ei Ef
h Ei Ef mee4
2(4p0)2U2 a 1 n2f 1
n2ib Ei Ef,
Ef. Ei
ni lc
f 3.00 108 m/s
4.56 1014 Hz 6.58 107 m 658 nm f ¢E
h 1.89 eV
h 1.89 eV 1.60 1019 J/eV
6.63 1034 J s 4.56 1014 Hz
¢E E3 E2 13.6
9 eV a13.6
4 eVb 1.89 eV E2 (13.6
4) eV. E3 (13.69) eV,(n 2).
(n 3) E X A M P L E 2
1300 CHAPTER 38 Spectral Lines, Bohr's Theory, and Quantum Mechanics
Concepts in Context
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(38.30) Upon insertion of the accurate values of the fundamental constants given in Appendix 6, we obtain
(38.31) This theoretical value of R agrees quite well with the experimental value quoted in Eq. (38.3).1
Suppose that the atoms in a sample of hydrogen gas are ini-tially in the ground state. If we illuminate these atoms with light (from some kind of lamp), what frequencies will the atoms absorb?
SOLUTION: Absorption of light is the reverse of emission. When an electron in an atom absorbs a photon (supplied by the lamp), it jumps from the initial state to a state of higher energy. The energy of the photon must match the energy differ-ence between the states. Thus, the frequencies of the photons that the electrons can absorb when jumping upward from the ground state are exactly those fre-quencies that they emit when jumping downward into the ground state, that is, the frequencies of the Lyman series (see Fig. 38.13).
What is the ionization energy of the hydrogen atom, that is, what energy must we supply to remove the electron from the atom when it is initially in the ground state?
SOLUTION:To remove the electron from the atom, we must lift it into an orbit of infinite radius, that is, an orbit of quantum number The energy required to accomplish this transition from to is
This is the ionization energy.
With his theory Bohr attained the goal of explaining the regularities in the spectrum of hydrogen in terms of the regularities of the structure of the atom. By showing that this structure is based on a simple numerical sequence, he fulfilled the ancient dream of Pythagoras of a universe based on simple numerical ratios, a dream that arose from an analogy with musical instruments. Bohr’s theory tells us how the atom plays its tune.
E1 Eq 13.6 eV a 1 12 1
q2b 13.6 eV n q
n 1
n q.
E X A M P L E 4 E X A M P L E 3
1.097 37 107 m1
R mee4 4p(4p0)2U3c
38.4 Bohr’s Theory 1301
1The small disagreement between the theoretical value of R given in Eq. (38.31) and the experimental value given in Eq. (38.3) is due to the motion of the nucleus of the hydrogen atom, which we have neglected in our calculation. A careful calculation that takes into account the motion of electron and proton about their common center of mass requires the substitution of the so-called reduced mass for the electron mass, and eliminates the disagreement. See also Problem 31.
m memp(me mp)
Concepts in Context
R 9.109 53 1031 kg (1.602 189 1019 C)4
4p(4p 8.854 178 1012 F/m)2 (1.054 589 1034 J s)3 2.997 925 108 m/s
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C h e c k u p 3 8 . 4
QUESTION 1:Why is Bohr’s postulate of stationary states in direct contradiction with classical mechanics and electromagnetism?
QUESTION 2:If an electron in a hydrogen atom makes a transition from some state to a lower state, does its kinetic energy increase or decrease? Its potential energy? Its orbital angular momentum?
QUESTION 3:Suppose that an electron is initially in the state. It then makes three quantum jumps in succession, such that the first jump produces a photon in the Paschen series, the second a photon in the Balmer series, and the third a photon in the Lyman series. In the energy-level diagram of Fig. 38.13, find the arrows that indi-cate these transitions.
QUESTION 4:In the energy-level diagram of Fig. 38.13, what transition would pro-duce the series limit of the Balmer series?
(A) to (B) to (C) to
(D) to (E) to
3 8 . 5 Q U A N T U M M E C H A N I C S ; T H E S C H R O
..D I N G E R E Q U AT I O N
Bohr’s theory is a hybrid. It relies on some basic classical features (orbits) and grafts onto these some quantum features (quantum jumps, quanta of light). In the 1920s the coop-erative efforts of several brilliant physicists—L. de Broglie, E. Schrödinger, W. Heisenberg, M. Born, P. Jordan, P. A. M. Dirac—established that the remaining classical features had to be eradicated from the theory of the atom. Bohr’s semiclassical theory had to be replaced by a new quantum mechanics with an entirely different equa-tion of moequa-tion.
The basis of the new quantum mechanics was laid by the discovery that electrons—
as well as protons, neutrons, and all the other “particles” found in nature—have not only particle properties but also wave properties. When a beam of electrons is made to pass through an extremely narrow slit, the electrons exhibit diffraction. This means that elec-trons are neither classical particles nor classical waves. Elecelec-trons, just like photons, are a new kind of object with a subtle combination of particle and wave properties. Electrons are wavicles. As in the case of photons [see Eq. (37.15)], the wavelength associated with an electron or some other wavicle is inversely proportional to its momentum:
(38.32)
This formula was postulated by de Broglie, and it is called the de Broglie wavelength.
This wavelength is quite small, even for electrons of the lowest energies attainable in experiments with beams of electrons.
What is the de Broglie wavelength of an electron of kinetic energy 1.0 eV, which is about the lowest energy that can be attained in experiments with beams of electrons?
E X A M P L E 5
l h p
n 2 n 11
n 2 n q
n 1 n q
n 2 n 3
n 1 n 2
n 5
✔
1302 CHAPTER 38 Spectral Lines, Bohr's Theory, and Quantum Mechanics
de Broglie wavelength
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SOLUTION: In joules, a kinetic energy of 1.0 eV is This kinetic energy is related to the momentum as follows:
Hence
According to the de Broglie equation, the wavelength of the electron is then
The wave properties of electrons were confirmed experimentally by C. J. Davisson and L. Germer, who observed interference effects with beams of electrons scattered by a crystal. In these experiments the crystal acts as a grating for electron waves, in much the same way as it acts as a grating for X rays in X-ray interference experiments (see the discussion of X-ray interference and the production of Laue spots in Section 37.5). Davisson and Germer found that constructive interference of the electron waves scattered by the rows of atoms in the crystal produced strong interference maxima in selected directions, and they were able to confirm that these directions agreed with calculations based on the de Broglie formula (38.32).
In the new quantum mechanics, or wave mechanics, the motion of an electron is described by a wave equation, the Schrödinger equation. This Schrödinger equation plays the same role for electrons as the wave equation derived from the Maxwell equa-tions plays for photons [see Eq. (33.33)]. As in the case of photons, the behavior of an electron is governed by a probabilistic law. The electron is represented by an elec-tron wavefunction (Greek letter psi) calculated from the Schrödinger equation, and the intensity of this electron wave at some point determines the probability that there is an electron particle at that point [compare Eq. (37.24)]:
(38.33)
Furthermore, as a consequence of their wave properties, electrons obey the Heisenberg uncertainty relations for position and momentum [see Eq. (37.30)],
(38.34) These quantum uncertainties are of crucial importance for the behavior of an elec-tron inside an atom. For such an elecelec-tron, the uncertainty in the position is very large—
about as large as the size of the atom. This implies that the electron follows no definite orbit. It is therefore not surprising that the Bohr theory should have failed in all attempts at calculating the electron motions in the helium atom and in other atoms with several electrons; what is surprising is that this theory should have succeeded as well as it did in the case of the hydrogen atom.
¢y ¢py U 2
[probability for an electron at a point] r c2 c
1.2 109 m 1.2 nm l h
p 6.63 1034 J s 5.4 1025 kg m/s
5.4 1025 kg m/s
p 22meE 22 9.11 1031 kg 1.6 1019 J E p2
2me
1.6 1019 J.
38.5 Quantum Mechanics; the Schrödinger Equation 1303
Schrödinger equation
probability interpretation of wavefunction
LOUIS VICTOR, PRINCE DE BROGLIE (DE BROY) (1885–1962) French theoretical physicist. He discovered the de Broglie wavelength by reasoning that if waves have particle properties, then maybe particles have wave properties. For his discovery of the wave properties of matter he was awarded the Nobel Prize in 1929, after the existence of these wave properties was confirmed experi-mentally.
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Consider an electron in the ground state of hydrogen. Show that a well-defined orbit is inconsistent with the Heisenberg uncertainty relations.
SOLUTION: If the electron is to follow a well-defined orbit, the uncertainty in its momentum (in any direction) must be much smaller than the magnitude of the momentum. According to Eq. (38.14), the speed of the electron in the smallest circular orbit is and the magnitude of the momentum is For a well-defined orbit, the uncertainty of the momentum must be much smaller than this magnitude of the momentum:
(38.35) Furthermore, for a well-defined orbit, we must require that the uncertainty in the position must be much smaller than the size of the orbit:
Vr (38.36)
In view of the inequalities (38.35) and (38.36), the product y pymust be smaller than :
V (for a well-defined orbit) (38.37) This is inconsistent with the Heisenberg uncertainty relation (38.34).
In wave mechanics, the quantization of the energy in the hydrogen atom and other atoms is an automatic consequence of the wave properties of the electron. The attractive electric force of the nucleus confines the electron wave to some region near the nucleus and causes the wave to reflect back and forth across the region, forming a standing wave. The different stationary states of the atom correspond to different standing-wave modes. As in the case of standing waves on a string, the standing electron waves in the atom have a discrete set of eigenfrequencies. For electrons and other particles, just as for photons, the energy is related to the frequency by
(38.38) Thus a discrete set of frequencies for electron standing waves corresponds to a dis-crete set of energies. The ground state of the atom is analogous to the fundamental mode of the string, the first excited state is analogous to the first overtone, and so on.
However, whereas the determination of the eigenfrequencies of standing waves on a string is a quite trivial mathematical exercise, the determination of the eigenfrequen-cies of the electron waves in an atom is a formidable mathematical problem, which requires an investigation of the solutions of the Schrödinger wave equation.
Although we will not deal here with the mathematical complexities of the Schrödinger wave equation in three dimensions, we can gain some insight into how elec-tron waves determine the discrete energies in the hydrogen atom by means of the fol-lowing simple calculation. Let us assume that the electron travels around the nucleus along an orbit of radius r, but instead of thinking of the electron as a particle, as in the Bohr theory, let us think of it as a wave. Figure 38.14 shows a “snapshot” of such an electron wave at one instant of time. If the wave is to have a well-defined amplitude at all points, it must repeat whenever we go once around the circumference—if it did
E hf U
¢y¢py U
¢y
¢py V U r p mev U
(nr. 1) v Umer,E X A M P L E 6
1304 CHAPTER 38 Spectral Lines, Bohr's Theory, and Quantum Mechanics
energy of stationary state in terms of frequency ERWIN SCHRÖDINGER(1887–1961) Austrian theoretical physicist. Another of the founders of the new quantum mechanics, he received the Nobel Prize in 1933 for the dis-covery of his wave equation.
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not, then the wave amplitude would have two different values at a single point, which makes no sense. Hence, the wave is subject to the condition that some integer number of wavelengths must fit around the circumference:
(38.39) Although this equation looks like the condition for a standing wave on a string of length we are here dealing with a traveling wave, for which the entire wave pat-tern in Fig. 38.14 rotates rigidly around the center. With the de Broglie relation
Eq. (38.39) becomes
or
(38.40) Since rp is the orbital angular momentum, this equation coincides with Bohr’s quan-tization condition for the angular momentum, Eq. (38.9). Thus, the wave picture of the electron implies the quantization of the angular momentum and, therefore, the quantiza-tion of the energy. But we must not take this calculaquantiza-tion too seriously—its legitimacy is questionable, since it relies in part on the wave picture and in part on the particle pic-ture. Furthermore, analysis of the Schrödinger equation shows that it is not enough to consider the behavior of the wave around the circumference; we must also consider the behavior of the wave outward along each radius. Thus, this simple calculation pro-vides no more than a crude qualitative sketch of the role of the wave properties of elec-trons in the atom.
There are some applications where a simple one-dimensional calculation gives accurate and meaningful results, such as electron reflection from a barrier, electron transmission through a barrier (known as “tunneling”; see Physics in Practice:
There are some applications where a simple one-dimensional calculation gives accurate and meaningful results, such as electron reflection from a barrier, electron transmission through a barrier (known as “tunneling”; see Physics in Practice: