ECE44 Nanoelectronics Lecture 7 Atomic Orbitals
Atoms and atomic orbitals It is instructive to compare the simple model of a spherically symmetrical potential for r R V ( r) for r R and the simplest hydrogen atom. This atom consists of the positive proton (nucleus) and the negatively charged electron, which interacts according to the Coulomb law: V () r 2 1 e 4 r (3.63) where ε is permittivity of free space, e is elementary electrical charge, and r is the distance between proton and electron. The negative sign in Eq.(3.63) indicates that electron and proton are attracted to each other.
2 15 1 5 5 1 15 2 2 1 e V () r. 4 r E(eV) r (Å) 1 5 n = 3 n = 2 4 2 2 me r E. 1 e me 2 4 1 2 2 4 2 r (4 ) 2 n = 1 E n E n 1 2 15 Fig. 3.15. Hydrogen atom potential well.
The potential of a hydrogen atom, reminds us of the quantum dot with impenetrable walls. But, there are two major differences: (i) the profile of the potential and (ii) the reference point for zero energy. Namely the zero energy was located in quantum dot and electron should have infinite energy to escape into the potential walls, while in hydrogen atom the energy of free electron is taken to be zero at r = ± and the energy at r =is equal to -. This analogy allows us to suppose that the energy of electron in the hydrogen atom would be quantized. For further analysis of hydrogen atom model, the free electron mass, m,is used in this section because the motion of nuclei is neglected due to the fact that the mass of nucleus is almost 2 times greater than the free electron mass. Hidden
The ground state of the hydrogen atom may be described by the so called radially symmetric function, ψ(r): () r 1 rr e (3.64) where r is the characteristic radius of the ground state. r can be estimated as follows. Let us evaluate the kinetic energy of the electron confined in the r sphere of radius r by using the uncertainty conditions of p x h p y h p z h p r x y z evaluated as: : ; then, the kinetic energy can be 2 2 p 2m 2m r 2 Thus, the total electron energy, H, at r = r is: 2 2 2 p 1 e Hr ( ) Vr ( ) 2 2m 4 r 2mr To find r that corresponds to the minimum of the total energy, H, we find first derivative of H(r ) with respect to r and equate the result to zero: dh ( r ) 1 e 2 dr 2 2 2 3 4 r 2m r
The solution of the derivative gives the formula for r : 4 2 2 me r (3.68) which is called the Bohr radius. Now we can find the energy of the ground state E 1 : E 1 e me 2 4 1 2 2 4 2 r (4 ) 2 Numerical estimates give the following values for the energy and the radius of the ground state of hydrogen atom: E 1 = -13.5 ev and r =.529 Å (1Å=1-1 m =.1 nm). An arbitrary atomic state is characterized by the following three quantum numbers, n, l,andm: principal number n 1 2 3 orbital number l 1 2 n 1 or l s p d magnetic number m 1 2 l (3.69)
The series of the energy levels for the hydrogen atom is the following: 1 s, 2s,2p, 3s, 3p, 3d,... The general expression for the n th energy level of hydrogen atom is given by: E n E1 2 (3.7) n Importantly, the energy spectrum of the hydrogen atom depends on the only one discrete number n and does not depend on the orbital and magnetic numbers l and m. This means, that some states of the atom can have the same energy. This situation is called degeneracy of the states. A number of detailed optical experiments has supported this classification of energy spectra of hydrogen atoms, as well as confirmed the energy values with high accuracy. However, it turns out that nature is more complicated. Additional non optical experiments revealed new property of quantum systems, which have no analogue in classical physics.
Indeed, in classical physics a particle with angular momentum l can have a magnetic dipole moment M =γl with the coefficient called the gyromagnetic ratio. The latter parameter is associated with the fundamental parameters of the electrons: e 2mc where e, m,andc are elementary charge, free electron mass, and speed of light, respectively. If an external magnetic field, B, is applied, additional potential energy of the particle in the magnetic field, B, arises: (3.71) U BM (3.72) M Let the magnetic field be non uniform, say it depends on z coordinate. Then a force acting on the particle will be: f z U z B z M Mz (3.73)
As a result, a beam of atoms with the magnetic moment will split into components corresponding to different values of M z. Special experiments, known as Stern Gerlach experiments, with hydrogen atoms in the ground s state, when l = m =and thus M z =,havedetectedasplittingofthebeam into two symmetrically deflected components. It provided evidence of the existence of a dipole magnetic moment, which we have not been taken into account yet. It has been assumed, that the electron has an intrinsic angular momentum which is just responsible for the results of the experiment. This intrinsic angular momentum S is called spin. Projection of the spin, S z,ofan electron in the magnetic field, B, takes on the half integer values ½ and -½. The spin effect explains also a number of other experiments with the electrons. Moreover, it has been found similarly that some spin characteristic can be attributed to any particle. For example, the photons light quanta have a spin equal to 1.
Thus, the complete description of an energy state of hydrogen atom should include additionally the spin number S z. Now the series of the states including the spin degeneracy can be written down as 1 s(2) 2 s(2) 2 p(6) 3 s(2) 3 p(6) 3 d(1) Numbers in the brackets indicate the degeneracy of the states. For example, the nomenclature 2p(6) describes 6 states of the 2p type ( l=1 )withthe same energy, but with different spins ( ±½ ) and different magnetic numbers ( m = -1,, 1 ). Spin is a very important characteristic. It is responsible for a number of quantum effects. Moreover, it determines the character of the population of different energy levels by particles: In a system consisting of identical electrons only two ( with different spins) or fewer electrons can be found in the same state at the same time. This is the so-called Pauli exclusion principle.thisisonemore quantum principle. The Pauli exclusion principle is especially important for many-electron systems.
The classification of energy levels used for the hydrogen atoms may be applied to many electron atoms. In the case of many electron atoms, we should take the nuclear charge as Ze,whereZ is the atomic number in the Periodic Table of Elements (Fig. 3.16), and take into account the Pauli exclusion principle. The electronic structure of complex atoms can be understood in the terms of filling up energy levels. One can introduce internal energy shells and external energy shells (or valence energy states). The latter are filled up by the outermost valence electrons and play the major role in chemical behavior of complex atoms. Hidden
The Periodic Table of Elements (copied from http://education.jlab.org/).
While the energy levels of the hydrogen atom exhibit some degree of degeneracy, i.e., they are independent on the angular momentum described by orbital number l and its projection described by magnetic number m, in many electron atoms, the energies depend not only on the principal quantum number n, but also on the orbital number l. Therefore, the electrons can be subdivided into subshells" corresponding to different orbital numbers l. The spectroscopic notations for these subshells are the same as above: s,p,d,f,... For a given l there are 2l 1values of m and two values of the spin S z = ±½. Thus, the total number of the electrons, which can be placed in the subshell is 2(2l 1). Periods of the Periodic Table of Elements are constructed according to filling of the shells. The hydrogen, H, is the first element of the Table. It has one electron occupying the 1s state, i.e., l =. The next atom is helium, He. Ithastwoelectrons which occupy the lowest s states in the same shell, but have opposite spins; this is referred to as the 1s 2 electron configuration. So, the 1s shell is completely filled by two electrons. The next period of the Table begins with lithium, Li. Li has three electrons with the third electron sitting on the 2s type level. Thus, we may immediately represent the electron configuration of Li as 1s 2 2s 1. After lithium follows beryllium, Be, with the configuration 1s 2 2s 2. In the next element boron, B, the fifth electron begins to fill up next subshell: 1s 2 2s 2 2p 1.Forthep shell, we have l =1and m = 1,, 1. Thus, the valence electron in Be can take three values of the magnetic quantum number, m, and two values of the spin, S z. In general, we may put 6 electrons in p shell,andso on. Atoms with completed subshells are stable and chemically inert. Examples are: helium, He, with the configuration 1s 2 2s 2 ; neon, Ne, with the configuration 1s 2 2s 2 2p 6 ;argon,ar, with the configuration 1s 2 2s 2 2p 6 3d 1 4s 2 4p 6,etc. Hidden
For nanoelectronics the elements of group IV and groups III and V composing semiconductor crystals are important: silicon (Si, atomic number and number of electrons is 14), germanium (Ge, atomicnumber32), gallium (Ga, atomic number 31), arsenic (As, atomic number 33), etc. Their electron configurations are as follows: Ne Ar group IV Si 1s 2s 2 p 3s 3p Ge 1s 2s 2 p 3s 3p 3d 4s 4p 2 2 6 2 2 2 2 6 2 6 1 2 2 group III Ga 1s 2s 2p 3s 3p 3d 4s 4p 2 2 6 2 6 1 2 1 2 2 6 2 6 1 2 3 group V As 1s 2s 2 p 3s 3p 3d 4s 4p (3.74) One can see that for both, Si and Ge, the outer valence electron configurations are very similar (the same number of electrons in the same p state). Thus, we can expect that their chemical and physical properties are also similar.
1A group # = # valence (outside) e 8A 1 2A 3A 4A 5A 6A 7A 2 Row = # shells 3 4 5 s 3B 4B 5B 6B 7B 8B 8B 8B 1B 2B d p 6 7 6 7 f
Electron Configuration 1s 1 row # shell # possibilities are 1 7 7 rows subshell possibilities are s, p, d, or f 4 subshells group # # valence e possibilities are: s: 1 or 2 p: 1 6 d: 1 1 f: 1 14 Total e should equal Atomic # What element has an electron configuration of 1s 1?
Order of Electron Subshell Filling: It does not go in order 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 4p 6 5s 2 5p 6 6s 2 6p 6 7s 2 7p 6 3d 1 4d 1 5d 1 6d 1 4f 14 5f 14 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 1 4p 6 5s 2 4d 1 5p 6 6s 2 4f 14 5d 1 6p 6 7s 2 5f 14 7p 6 6d 1
Electron Configuration 1s 1 row # shell # possibilities are 1 7 7 rows subshell possibilities are s, p, d, or f 4 subshells group # # valence e possibilities are: s: 1 or 2 p: 1 6 d: 1 1 f: 1 14 Total e should equal Atomic # What element has an electron configuration of 1s 1?
For a symmetric function (the constructive interference of the orbitals), there is a finite probability of finding the electron between nuclei. The presence of electrons between two nuclei, A and B, attracts both nuclei and keeps them together. The corresponding state is called a bonding state of two atoms and corresponding wavefunction, ψ AB, is called the bonding orbital. The antisymmetric hybrid (the destructive interference of the orbitals) does not bind atoms and is called antibonding state. The simplest example of the binding effect is given by the two atom hydrogen molecule: the bonding state constructed from s states of the hydrogen atoms corresponds to an energy lower than the energy of two uncoupled hydrogen atoms. This results in stable H 2 molecule. The antibonding state corresponds to the energy greater than the energy of two free hydrogen atoms. In the antibonding state the atoms repel each other. ψ A (a) ψ B Antibonding orbital ψ AB Higher level ψ AB (b) A B Atomic orbitals ψ A and ψ B with equal energies of atoms A and B Lower level Fig. 3.17. Formation of bonding and antibonding atomic orbitals (two atoms). ψ AB (c) nucleus nucleus Bonding orbital ψ AB
In general, atoms with not completely filled up subshells, that is, those which have valence electrons, are expected to be chemically active, i.e., they can form chemical bonds. Coupling between atoms in molecules and solids is determined by the type of the electron wavefunctions. Often, these wavefunctions are called atomic orbitals. For example, the wavefunction of two atom molecule should be a combination (a hybrid) of two orbitals. Such a hybrid can be either a symmetric function, or an antisymmetric one.
Actually, formation of bonding and antibonding states of two hydrogen atoms is a particular example of more general and very important result: when two identical atoms come closer, their energies change and instead of a single atomic (degenerate) energy level, two levels arise, as illustrated below. This is the so called energy splitting. Thus, bonding and antibonding orbitals correspond to these two levels. Antibonding orbital ψ AB ψ A (a) ψ B Higher level ψ AB (b) A Atomic orbitals ψ A and ψ B with equal energies of atoms A and B B Lower level Fig. 3.17. Formation of bonding and antibonding atomic orbitals (two atoms). ψ AB (c) nucleus nucleus Bonding orbital ψ AB Three atoms drawn together are characterized by three close levels, etc.
Orbital Hybridization Now, consider in detail atoms of group IV. As presented above, for these atoms the outer shell has four electrons: two in s states and two in p states, i.e., their configuration can be thought of as a core the s 2 p 2 shell. Thus, atoms of the group IV can form four bonds with other atoms in molecules or solids (one bond for one electron). The wavefunctions of these outer shell states have very different spatial configurations.
x (a) z s - orbital ψ y x (b) z p y - orbital ψ y (c) x (d) z p z - orbital z y y y x p x - orbital y Fig. 3.18. Atomic orbitals: (a) s, (b) p y, (c) p z, and (d) p x.
z x x s - orbital z p y - orbital y y x (a) z sp 3 - orbital y (b) Fig. 3.19. Formation of hybridized sp 3 orbitals (a) and four sp 3 hybridized orbitals in Si (b).
Figure 3.18 illustratesthe wavefunctions (orbitals) for s state and p states with different angular momentum projection m. Thus, the s orbital is spherically symmetric without angular dependence. The 3p orbitals are anisotropic and can be considered as perpendicular" to each other. These hybrid orbitals forming bonds with other atoms have two electrons with opposite spins. In fact, because of interaction between electrons the s and p orbitals overlap significantly and realistic bonds are formed as a linear combination of s and p orbitals, which are called sp 3 hybridized orbitals. Importantly,whenthethree directed p orbitals are hybridized with the s orbital, the negative parts of the p orbitals are almost cancelled, so that mainly positive parts remain as shown in Fig. 3.19(a). Thus, the hybridized orbitals present the directed" bonds in space, as shown in Fig. 3.19(b). These directed bonds of atoms are responsible for the particular group IV crystal structures. Hidden