12 chapter 1 atoms: the quantum world wavefunctions and energy levels FIGURE 1.14 The Born interpretation of the wavefunction. The probability density (the blue line) is given by the square of the wavefunction and depicted by the density of shading in the band beneath. Note that the probability density is zero at a node. A node is a point where the wavefunction (the orange line) passes through zero, not merely approaches zero. Note the "passes through"; merely becoming zero is not enough to be considered a node. The symbol Vis commonly used to denote potential energy, rather than £P, in this context. Derivatives are reviewed in Appendix IF. FIGURE 1.15 Frwin Schrodinger (1887-1961), thinking about his equation. nothing mysterious about the form of vvavefunctions: they are just mathematical functions such as sin x, a function that varies like a wave, and ea function that decays exponentially toward zero. The German physicist Max Born suggested how the wavefunction should be interpreted physically. According to the Born interpretation of the wavefunction, the probability of finding the particle in a region is proportional to the value of iji2 (Fig. 1.14). To be precise, vJj2 is a probability density, the probability that the particle will be found in a small region divided by the volume of the region. It follows that, to calculate the probability that the particle is in a small region of space, wc multiply i|»2 by the volume of the region. For instance, if ty1 = 0.1 pm--1 at a point, then the probability of finding the particle in a region of volume 2 pm3 located at that point would be (0.1 pm~3) X (2 pm3) = 0.2, or 1 chance in 5. According to the Born interpretation, wherever i|j2 is large, the particle has a high probability density; wherever v|/2 is small, the particle has only a low probability density. Because the square of any number is positive, we don't have to worry about having a negative sign in some regions of space (as a function such as sin x has): probability density is never negative. Wherever i|f, and hence iji2, is zero, there is zero probability density for the particle. A location where i|i passes through zero is called a node of the wavefunction; so we can say that a particle has zero probability density wherever the wavefunction has nodes. The Schrodinger equation is an equation for calculating wavefunctions. We can see what the equation looks like in Fig. 1.15, which shows it emerging from Schrodinger's head. More formally, the equation for a particle of mass m moving in a region where the potential energy is V(x) is A1 cft|i 2m dx2 + V(x)lJj = Eu) The Schrodinger equation is a "differential equation," an equation that relates derivatives of a function (in this case, a second derivative of d2i|r/dx2) to the value of the function at each point. It is impossible to solve exactly except in certain simple cases. However, in this text, we need only the form of some of its solutions, not how those solutions are found. The left-hand side of the Schrodinger equation is commonly written H*\i, where H is called the hamiltonian for the system, then the equation takes the deceptively simple form (8b)* One of the simplest examples of an actual wavefunction is that for a particle of mass m confined between two rigid walls a distance I. apart, the so-called particle in a box (Fig. 1.16), The shapes of the wavefunctions for this one-dimensional system, some of which are shown in the illustration, make sense when we view the particle as a wave. Only certain wavelengths can exist in the box, just as a stretched string can support only certain wavelengths. Think of a i guitar string: because it is tied down at each end, it can support only shapes like the ones shown in Fig. 1.16. The shapes of the wavefunctions are identical with the displacements of a stretched string when it vibrates, and their mathematical form is that of a standing wave: *,(*>= It / mrx\ n = 1.2,.. (9) The integer n labels the wavefunctions and is called a "quantum number." In general, a quantum number is an integer (sometimes, a half-integer) that labels a wavefunction, specifies a state, and can be used to calculate the value of a property of the system. For example, we can use n to find an expression for the energy corresponding to each wavefunction. HOW DO WE DO THAT (LEVEL 1)? | [ The kinetic energy of a particle of mass m is related to its speed, v, by £K = \trnP: We can relate this energy to the wavelength of the particle by noting that the linear momentum is p = mv and then using the de Broglie relation (Eq. 6b): b2 2m 2m\2 The potential energy of the particle is zero everywhere inside the box, and so the total energy, E, is given by the expression for EK alone. At this point, we recognize that only whole-number multiples of half-wavelengths can fit into the box (see Fig 1 16- the waves have one bulge, two bulges, three bulges, and so forth, with each "bulge" a half-wavelength). That is, for a box of length L, L - j'1. V ■ • • - « x J'1' with « = 1,2,... Therefore, the allowed wavelengths are \ = 2LM, with n = 1, 2,... pThen this expression for X is inserted into the expression for the energy, we obtain E = — " 8mL2 HOW DO WE DO THAT (LEVEL 2)? I 1 Th,° more sophisticated—and more general—way of finding the energy levels of a particle in a box is to solve the Schrodinger equation. First, we note that the potential energy of the particle is zero everywhere inside the box; so V(x) = 0, and the equation that we have to solve is _*Lf* = £l|I 2m dx2 I This equation has the solutions {x) would be 0 everywhere, which would mean that the particle would be nowhere. Therefore, to guarantee that i|)(L) = 0, we use the relation sin wtt = 0 and set kL = nit, with n = 1,2,.... We now have ty{x) = A sin- L n = 1,2, To find the one remaining constant, A, we use the fact that the probability of finding the particle in a region of length dx at x is \\i(x)2dx, and therefore the total probability of finding the particle between x = 0 and x = L is the sum (integral) of these probabilities and must be equal to 1 (the particle must be somewhere in the box). Therefore, ty{x)2dx = A2 ■ 2 n7TX J i sin —j— ax1 The integral evaluates to L/2; so A = (2/L)1'2 and the final form of the wavefunction is that given in Eq. 9. Moreover, because k is limited to the values mr/L, it follows that the energy of the particle is limited to E = [n-n/L)2h2 _ n~h2 SmL2 exactly as we found in level 1, We see that the quantization of energy (see following text) stems directly from the imposition of "boundary conditions" on the wavefunction, conditions that 41 must obey in order for it to be acceptable. In this example, the boundary conditions are that \\i must be zero at each wall of the box. The origin of quantization from the imposition of boundary conditions on the wavefunction is a general result for any system. We have established that the allowed energies of a particle of mass m in a onc-dimensional box of length L arc n2h2 SmL2 (10)* A striking implication of this result is that the energy of the particle is quantized, meaning that it is restricted to a series of discrete values, called energy levels. According to classical mechanics, an object can have any total energy—high, low, and anything in between. For example, a particle in a box could bounce from wall to wall with any speed and hence any kinetic energy. According to quantum mechanics, however, energy is quanrized: only certain wavelengths fit into the box. Quantization is a little like pouring water into a bucket. Water seems to be a continuous fluid that can be transferred in any amount, no matter how small. However, the smallest amount of water that we can transfer is one H20 molecule. Energy seems to be unrestricted on the macroscopic scale; but, on the subatomic scale, we can transfer only discrete amounts to the particle in a box and it can lose energy only in discrete amounts corresponding to the differences between the energy levels. As we saw in the derivation of Eq. 10, the existence of quantization stems from the boundary conditions on the wavefunction, the constraints that the wavefunction must satisfy at different points of space (such as fitting into a container correctly). We can use F.q. 10 to calculate the energy separation between two neighboring levels with quantum numbers n and n + 1: [n + l)2b2 n2b2 _ [In + ^)h1 SmL SmL HmJj (11) We see that, as L or m increases, the separation between neighboring energy levels decreases (Fig. 1.17). That is why no one noticed that energy is quantized until they investigated very small systems such as an electron in a hydrogen atom: the separation between levels is so small for ordinary particles in ordinary-sized vessels that it is completely undetectable. We can, in fact, ignore the quantization of the motion of the atoms of a gas in a typical flask. jpCAMPLE 1.5 Calculating the energies of a particle in a box Treat a hydrogen atom as a one-dimensional box of length 150. pm (the approximate diameter of the atom) containing an electron and predict the wavelength of the radiation emitted when the electron falls to the lowest energy level from the next higher energy level. STRATEGY The lowest energy level has n = 1, and so we can use Eq. 11 with n = 1 and rn = me> the mass of the electron. The energy difference is carried away as a photon of radiation; so we set the energy difference equal to bv and express v in terms of the corresponding wavelength by using Eq. 1 (X = civ). As usual, it is good practice to go as far as possible symbolically and then to insert numerical values at the last possible stage. SOLUTION According to Eq. 11, with n = 1: [(2 x 1) + l]h2 _ 3b2 SmPL2 ~ SmX2 Et It follows that, from bv = 3b2 8meL2 and, from Xv = c, Finally, insert the data: 8 X (9.109 X 10 X - 3b ' 8mcL2 c 8mEcL2 v ~ 3b X (2.998 X 10s m-s"1) X (1.50 X 10" = 2.47 X 10 _8 kg X (nvs 1) X m2 3 X (6.626 X 10"34Js) (kg-m 2.„-2 } X s = 2.47 X 10 Note how the complicated collection of units is treated: arriving at the correct units for the answer is a sign that you have set up the equation correctly. This wavelength corresponds to 24.7 nm. The experimental value for an electron in a hydrogen atom is 122 nm. Although there is a big discrepancy, the fact that the predicted wavelength has nearly the same order of magnitude as the actual value suggests that a quantum theory of the atom, based on a more realistic three-dimensional model, should give good agreement. Self-Test 1.7A Use the same model for helium but suppose that the box is 100. pm long, because the atom is smaller. Estimate the wavelength of the same transition. [Answer: 11.0 nm] Self-Test 1.7B Use the same model for hydrogen and estimate the wavelength for the transition from the n = 3 energy level to the n = 2 level. Another surprising implication of Eq. 10 is that a particle in a container cannot have zero energy. Because the lowest value of ft is .1 (corresponding to a wave of one-half wavelength fitting into the box), the lowest energy is £1 — b2i%mhl. This lowest possible energy is called the zero-point energy. The existence of a zero-point energy means that, according to quantum mechanics, a particle can never be perfectly still when it is confined between two walls: it must always possess an energy—in this case, a kinetic energy—of at least b2/SmL2. This result is consistent with the uncertainty principle. When a particle is confined between two walls, the uncertainty in its position cannot be larger than the distance between the two walls. Because the position is not completely uncertain, the linear momentum must be uncertain, too, and so we cannot say that the particle is completely still. The particle must therefore have some kinetic energy. The zero-point energy is a purely quantum mechanical phenomenon and is very small for macroscopic systems. For example, a billiard ball on a pool table has a completely negligible zero-point energy of about lO-67 J. chapter 1 atoms: the quantum world FIGURE 1.18 The two lowest energy wavcfunctions {\\s, orange] for a particle in a box and the corresponding probability densities (i[i2, blue). The probability densities are also shown by the density of shading of the hands beneath each wavefunction. Finally, the shapes of the wavcfunctions of a particle in a box also reveal some interesting information. Let's look at the two lowest energy wavefunctions, corresponding to n = 1 and n — 2. Figure 1.18 shows, by the density of shading, the likelihood of finding a particle: we see that when a particle is described by the wavefunction (and has energy /?2/8mL2), then it is most likely to be found in the center of the box. Conversely, if the particle is described by the wavefunction ij;2 (and has energy b^/lmL2), then it is most likely to be found in regions between the center and the walls and is unlikely to be found in the middle of the box. Remember that the wavefunction itself does not have any direct physical significance: we have to take the square of i|' before we can interpret it in terms of the probability of finding a particle somewhere. EXAMPLE 1.6 Plotting the wavcfunctions for a particle in a box (a) Using the Living Graphs on the "Web site for this book, plot the particle-in-a-box wavefunction for n = 2 and L = 1 m. (b) How many nodes does the wavefunction have? Where do these nodes occur? (c) Repeat parts (a) and (b) for n = 3. (d) What general conclusion can you draw about the relation between n and the number of nodes present in a wavefunction? (e) Convert the n = 2 plot into a probability density distribution: at what values of x is the particle most likely to be found? (f) Repeat part (e) for n = 3. STRATEGY The wavefunctions for a particle in a box are given by Eq. 9, Use the Living Graphs "Particle-in-a-Box Wavefunctions" on the Web site for this book to plot them, (a) To plot the wavefunction, enter the parameters L = 1 m and n = 2, set the range for the entire box (x — 0 Til to 1 m), check the f(x) box, and hit New Plot. (b) The nodes are found where the curve passes through zero. To find the value of # at a point, click on the plot at that point, (c) Create a new plot as before, but set n = 3. (d) Look for the patterns in the number of nodes and the value of the quantum number n. (e) Repeat part (a) with the f2{x) box checked and look for the maxima in the | probability density, (f) Repeat part (e) for n = 3. SOLUTION (b) The plot for n = 2 has one node at x = 0.500 m. (c) The plot for n = 3 has two nodes at x ~ 0.333 m and 0.667 m. (d) The number of nodes is equal to n — 1. (e) Maxima occur at x = 0.25 m and 0.75 m. ff) Maxima occur at x = 0.17 m, 0.50 m, and 0.83 m. Self-Test 1.8A Verify the conclusion in part (d) of Example 1.6 by plotting the i wavefunction for n = 4 and determining the number of nodes. [Answer: Three nodesj Self-Test 1.8B Verify the conclusion in part (d) of Example l.f wavefunction for n = 5 and determining the number of nodes. by plotting the The probability density for a particle at a location is proportional to the square of the wavefunction at that point; the wavefunction is found by solving the Schrodinger equation for the particle. When the equation is solved subject to the appropriate boundary conditions, it is found that the particle can possess only certain discrete energies. Ultraviolet radiation atomic spectra and energy levels FIGURE 1.19 (a) The visible spectrum, (b) The complete spectrum of atomic hydrogen. The spectral lines have been assigned to various groups called series, two of which are shown with their names. 1.6 Atomic Spectra and Energy Levels Compelling evidence for the validity of quantum mechanics came from its ability to explain atomic spectra. When an electric current is passed through a low-pressure sample of hydrogen gas, the sample emits light. The electric current, which is like a storm of electrons, breaks up the H2 molecules and excites the free hydrogen atoms to higher energies. These excited atoms quickly discard their excess energy by giving off electromagnetic radiation; then they combine to form H2 molecules again. When white light is passed through a prism, a continuous spectrum of light results (Fig. 1.19a). However, when the light emitted by excited hydrogen atoms is passed through a prism, the radiation is found to consist of a number of components, or spectral lines (Fig. 1.19b). The brightest line (at 656 rim) is red, and the excited atoms in the gas glow with this red light. Excited hydrogen atoms also eipit ultraviolet and infrared radiation, which are invisible to the eye but can be detected electronically and photographically. The series of discrete lines making up the spectrum of atomic hydrogen was highly puzzling to early spectroscopists. They wondered how an atom could emit only particular frequencies of electromagnetic radiation and not all possible frequencies. The answer must be that an atom can lose energy only in certain discrete amounts. This answer, in turn, suggests that the electron in the atom can exist only in a series of discrete energy levels, just like a particle in a box. When an electron undergoes a transition, a change of state, in which its energy changes from a higher energy level to a lower one, the difference in energy, AE = Euppcr — -^lowcn 1S carried away as a photon. Because the energy of a photon is hv, where h is Planck's constant, the frequency v, of an individual line in a spectrum, is related to the energy difference between two energy levels (Fig. 1.20): hv = K (12)! This relation is called the Bohr frequency condition. Each spectral line arises from a specific transition. By analyzing the appearance of the spectrum, we can build a picture of the ladder of energy levels for the atom, which is called an energy-level diagram (Fig. 1.21). FIGURE 1.21 The spectrum of atomic hydrogen (reproduced on the right) tells us the arrangement of the energy levels of the atom because each line results from Lhc transition of elections between two specific states. The frequency of the radiation emitted in a transition is proportional to the energy ■difference between the two energy levels. The zero of energy corresponds to the completely separated proton and electron. The numbers on the right identify the energy levels: they are examples of quantum numbers. FIGURE 1.20 When an atom undergoes a transition from a state of higher energy to one of lower energy it loses energy that is carried away as a photon. The greater the energy loss (A compared with B), the higher the frequency (and the shorter the wavelength) of the radiation emitted. The components of different frequency or wavelength are called lines because, in the early spectroscopic experiments, the radiation from the sample was passed through a slit and then through a prism; the image of the slit was then focused on a photographic plate, where it appeared as a line. Spectrum 18 CHAPTER 1 ATOMS: THE QUANTUM WORLD 1.7 THE PRINCIPAL QUANTUM NUMBER 0 Lyman series i FIGURE 1.22 The permitted energy levels of a hydrogen atom as calculated from Eq. 14. The levels are labeled with the quantum number n, which ranges from I (for the lowest state) to infinity (for the separated proton and electron). FIGURE 1.23 When white light shines through an atomic vapor, radiation is absorbed at frequencies that correspond to excitation energies of the atoms. I lere is a small section of the spectrum of the Sun, in which atoms in its outer layers absorb the radiation from the incandescence below. Many of the lines have been ascribed to hydrogen, showing that hydrogen is present in the cooler outer layers of the sun. The first person to identify a pattern in the lines of the visible region of the spectrum was Joseph Balmer, a Swiss schoolteacher. In 1885, he noticed that the frequencies of all the lines then known could be generated by the expression 1 1 n = 3,4, (13) As experimental techniques advanced, more lines were discovered, and the Swedish spectroscopist Johann Rydberg noticed that all of them could be predicted from the expression %.[ —--7 I »1 = 1,2,... «2 = «1 + + 2,. . . (14)* Here i^is an empirical (experimentally determined) constant now known as the Rydberg constant; its value is 3.29 X 10" Hz. This empirical formula for the lines, together with the Bohr frequency condition, strongly suggests that the energy levels themselves are proportional to SJn1. The Balmer series is the set of lines with Bi = 2 (and n2 = 3, 4, . . . ). The Lyman series, a set of lines in the ultraviolet region of the spectrum, has «j = 1 (and n2 = 2, 3, . . . ) (Fig. 1.22). Another consequence of quantization is that an atom can absorb radiation only of certain frequencies. The energy of the incoming photon excites electrons from one quantum level to another. If we pass light through a vapor composed of the atoms of an element, we see its absorption spectrum, a series of dark lines on an otherwise continuous spectrum (Fig 1.23). The absorption lines have the same frequencies as those of the lines in the emission spectrum. Absorption spectra are used by astronomers to identify elements in the outer layers of stars. EXAMPLE 1.7 Identifying a line in the hydrogen spectrum Calculate the wavelength of the radiation emitted by a hydrogen atom when an eleo tron makes a transition between the levels with n2 — 3 and n1 = 2. Identify in Fig. 1.19b the spectral line produced by this transition. STRATEGY The frequency of a line arising from a transition is given by Eq. 14. Convert frequency into wavelength by using Eq. 1. The wavelength should match one of the lines in the Balmer series in Fig. 1.19b. SOLUTION For the transition from a level with n2 = 3 to a level with nx = 2, from Eq. 14 with nl = 2 and k2 = 3, 5 7%. J___1 21 ~ J2J 36' From \v ~ c, _ c _ c 36c v ~ (5/36)3?. ~ 5% Now substitute the values of c and '](: 36 X (2.998 X 108 nvs-') 5 X (3.29 X 1015s_1) 36 X 2.998 X 10" 5 X 3.29 X 10" ' ;.57 X 10" This wavelength, 657 nm, corresponds to the red line in the Balmer series of lines in the spectrum. Light of this wavelength is absorbed when the electron is promoted from the n = 2 level to the n = 3 level and is given off when the electron falls back to the « = 2 level. Ultravi. X-rays frays SELF-Test 1.9A Repeat the calculation for the transition from the state with n = 4 to that with n = 2 and identify the spectral line in Fig. 1.19b. [Answer: 486 nm; blue linej SELF-TnST 1.9B Repeat the calculation for the transition from the state with n = 5 to that with n = 2 and identify the spectral line in Fig. 1.19b, The observation of discrete spectral lines suggests that an electron in an atom can have only certain energies. Transitions between these energy levels generate or absorb photons in accord with the Bohr frequency condition. MODELS OF ATOMS Our next task is to combine the wavelike properties of electrons with the nuclear model of the atom and explain the strange ladder of energy levels observed experimentally in a hydrogen atom. We proceed in two steps. First, wc develop the quantum mechanical model of a hydrogen atom. Then wc use that model to account for the structures of atoms with more than one electron and see how a few simple ideas account not only for atomic spectra but also for the structure of the periodic table. 1.7 The Principal Quantum Number An electron in an atom is like a particle in a box, in the sense that it is confined within the atom by the pull of the nucleus. We can therefore expect the electron's wavefunctions to obey certain boundary conditions, like the constraints we encountered when fitting a wave between the walls of a container. As wc saw for a particle in a box, these constraints result in the quantization of energy and the existence of discrete energy levels. "We find the actual energy levels of an electron in a hydrogen atom by setting up and solving the appropriate Schrodinger equation. To set up the equation, we insert the appropriate expression for the potential energy, V, in Eq. 8. For an electron in a hydrogen atom, the Coulomb potential energy of the electron of charge —e (Section A) varies with distance, i; from the nucleus of charge +e as follows (as remarked earlier, V is commonly used rather than EP in this context): -e){+e) 4TfE,/ V(r) (IS) Solving the Schrodinger equation for a particle with this potential energy is difficult, but Schrodinger himself achieved it in 1927. He found that the allowed energy levels are n = 1,2, , (16a)" These energy levels have exactly the form suggested by the wavelengths of the lines measured spectroscopically, but now we have an expression for 'X'm terms of more fundamental constants. When the fundamental constants are inserted into the expression for the value obtained is 3.29 X 1015 Hz, the same as the experimental value of the Rydberg constant. This agreement is a triumph for Schrodinger's theory and for quantum mechanics. A very similar expression applies to other one-electron ions, such as He' and even C> + , with atomic number Z: (16b) Wavelength, X (nm The greater the value of Z, the more tightly the electron is bound to a nucleus, and so these energies are more strongly negative for heavy atoms than for a hydrogen atom. 20 chapter 1 atoms: the quantum world The ionization energy is discussed further in Section 1.16. FIGURE 1.24 The spherical polar coordinates: r is the radius, 6 is the colatitude, and 4>, the "longitude," is the azimuth. Geographical latitudes are measured from the equator, not the poles. In an early model of the hydrogen atom proposed by Niels Bohr, the electron traveled in a circular orbit of radius aQ around the nucleus. The uncertainty principle rules out this model. Figure 1.22 shows the energy levels calculated from Fq. 16. From the minus signs in Fig. 1.22 and in Eq. 16, we can see that the energy of an electron in a hydrogen atom is always lower than that of a free electron. The principal quantum number, n, is an integer that labels these energy levels, from n = 1 for the first (lowest, most negative) level, n = 2 for the second, continuing to infinity. The lowest energy possible for an electron in a hydrogen atom, — A>3J., is obtained when w = 1. This lowest energy state is called the ground state of the atom. A hydrogen atom is normally found in its ground state, with its electron in the level with n = 1. The energy of the bound electron climbs up the ladder of levels as n increases. It reaches the top of the ladder, corresponding to K = 0 and freedom, when n reaches infinity. At that point, the electron has left the atom. This process is called ionization. The difference in energy between the ground state and the ionized state is the energy required to remove an electron from the neutral atom in its ground state. The energy levels of a hydrogen atom, Eq. 16, are defined by the principal quantum number, n = 1, 2, ..., and form a converging series, as shown in Fig. 1.22. 1.8 Atomic Orbitals Wavefunctions of electrons in atoms are called atomic orbitals. The name was chosen to suggest something less definite than an "orbit" of an electron around a nucleus and to take into account the wave nature of the electron. The mathematical expressions for atomic orbitals—which arc obtained as solutions of the Schrodinger equation—are more complicated than the sine functions for the particle in a box, but their essential features are quite simple. Moreover, we must never lose sight of their interpretation, that the square of a wavefunction tells us the probability density of an electron at each point, To visualize this probability density, imagine a cloud centered on the nucleus. The density of the cloud at each point represents the probability of finding an electron there. Denser regions of the cloud therefore represent locations where the electron is more likely to be found. To interpret the information in each atomic orbital, we need a way to identify the location of each point around a nucleus. It is most convenient to describe these locations in terms of spherical polar coordinates, in which each point is labeled with three coordinates: • r, the distance from the center of the atom; • 8 (theta), the angle from the positive z-axis (the "north pole"), which can be thought of as playing the role of the "latitude"; and • cj> (phi), the angle about the s-axis, the "longitude." These coordinates are shown in Fig. 1.24, Each wavefunction, which in general varies from point to point, can be written as a function of the coordinates, i|f(r,9,c|>). It turns out, however, that all the wavefunctions can be written as the product of a function that depends only on r and another function that depends only on the angles 6 and That is, Wr,M») = ■ (17)* The function R[r) is called the radial wavefunction, and the function Y(9,4>) is called the angular wavefunction. For example, the wavefunction corresponding to the ground state of the hydrogen atom is 2e"r/ö° lit1* The quantity a0 is called the Bohr radius; when the values of the fundamental constants are inserted, we find aa = 52.9 pm. For this wavefunction, the angular wavefunction Y is a constant. 1/2-it1'2, independent of the angles, and the radial wavefunction R decays exponentially toward zero as r increases. One of the wave-functions for the next higher energy level (at E1 = -jhR) is v(0,<|>) Wc shall see shortly that, with the exception of the ground state, there is more than one wavefunction for each energy level. l (iy2 iy_i VP sin 6 cos (b 4 Notice how this wavefunction depends on the angles 6 and . The expressions for a number of other atomic orbitals are shown in Table 1 2a (for K) and Table 1.2b (for Y). To understand these tables, we need to know that each wavefunction is labeled by three quantum numbers, n, I, and tnt. As we shall see, n is related to the size and energy of the orbital, I is related to its shape, and mi is related to its orientation in space. We have already encountered n, the principal quantum number, which specifies the energy of the orbital (through Eq. 16). In a hydrogen atom, all atomic orbitals with the same value of n have the same energy and are said to belong to the same shell of the atom. The second quantum number needed to specify an orbital is /, the orbital angular momentum quantum number. This quantum number can take the values / = 0, 1, 2,...,»- 1 There are n different values of / for a given value of n. For instance, when n = 3, / can have any of the three values 0, 1, and 2. The orbitals of a shell with principal quantum number n therefore fall into « groups (each identified by a different value of /), which are called subshclls of the shell. There is only one subshell in the n = 1 level (/ = 0), two in the n - 2 level (/ = 0 and 1), three in the H = 3 level TABLE 1.2 Hydrogen Wavefunctions (Atomic Orbitals), i)i = RY (a) Radial wavefunctions, R„j(r) Z % 2V1 27V6 4 -TU- «0/ v "J '4/2(3 - 2Zr "o Zr\ — 2 - c W V 3V 2ZV /ZWZr' 8 1 V'30 \a (b) Angular wavefunctions, Yh (B,) I m, Y/,„,«M>) 0 (iT X I 3 Y'2 — sin 0 cos (b \4tiJ y I 3 Y/2 z (ifcose xy 1 11 y'1 (——) sin26cos2 \16nJ yz — cos 6 sin B sin \4ti/ zx /15Y/2 — cos 8 sin 8 cos V4it/ f —) sin2 0 sin 2; so we say that they are spherically symmetrical (Fig. 1.26). The probability density of an electron at the point (r,9,2(a0,8,