An Overview of the Atom

Home , Electron rest mass, Rydberg constant

Chapter 2

We begin our discussion by considering a hydrogen-like, one-electron atom, in the absence of an external field. This atom will be described by a family of stationary states of well defined energies, that can be ascertained from the time-independent Schrödinger equation

Hψ (r)=Eψ (r) . (2.1)

Equation 2.1 is obtained from Eq. 1.1 by letting Ψ (r, t) →e−iEt/ψ (r). The Hamiltonian is taken as the quantum mechanical analog of the classical 2 Hamiltonian given by T + V where T = p /2me and the potential, V (r), 2 is just the Coulomb potential, −Ze /4πεor,withZ being the charge on the nucleus (Z = 1 for neutral atoms). Since p →−i∇, the quantum mechanical Hamiltonian for this single-particle non-relativistic case takes the simple form 2 Ze2 H = − ∇2 − , (2.2) 2me 4πεor where me is the mass of the electron, e is the elementary charge, ,is Planck’s constant divided by 2π and εo is the vacuum permittivity.

2.1 The Bohr Atom

Before solving Eq. 2.1 with the Hamiltonian given in Eq. 2.2, it is interesting to see that it is possible to determine the energy spectrum classically by assuming the existence of stationary states – Bohr’s hypothesis that the electron does not radiate when circling the nucleus. For stationary states 2 2 to exist, the Coulomb force (−Ze /4πεor ) must equal the centripetal force

11 12 CHAPTER 2. AN OVERVIEW OF THE ATOM

2 (−mev /r, where v is the trangential velocity). This leads to a kinetic energy of 2 1 2 1 Ze T = mev = , (2.3) 2 2 4πεor and a total energy 1 Ze2 E = T + V = − . (2.4) 2 4πεor Here, we adopt the usual convention that negative energy means the system is bound. Prior to the development of quantum mechanics, it was known that atoms emitted light at specific wavelengths. In the late 19th century it was determined that the energy of the light, in wavenumbers (see Eq. A.4 and Appendix A), associated with a transition between energy levels n and Find the reference to n was found to obeyed1 the Rydberg and Balmer 1 1 ν R − footnote. ñ n = n2 n2 , (2.5) where R is a constant that is related to the Rydberg constant. If one further 2 postulates that the energy levels are quantized such that r → n r1, where r1 is the minimum stable radius, Eq. 2.4 becomes Ze2 E −1 n = 2 . (2.6) 2 4πεon r1 It follows that 1 Ze2 1 1 hcν E − E − ñ n = n n = 2 2 . (2.7) 2 4πεor1 n n If we identify the constant as 1 Ze2 R = , (2.8) 2hc 4πεor1 the energy spectrum for hydrogen and hydrogen-like ions 2 takes this simple form: E −hcR 1 . n = n2 (2.9) Since the potential is not harmonic, the energy levels are not equally spaced. Figure ?? shows that the second level, the first excited state, is three-

1 This general formula was proposed by Rydberg in 1889. Prior to this date, Balmer used the empirial formula, (4/B) 1/22 − 1/n2 to ﬁt the visible hydrogen spectral lines he observed. With n =3, 4, 5, 6, ..., he extracted a value of 3645.6 A˚ for B. 2Hydrogen-like ions are ions with a single electron plus the nucleus. These include H, He+, as well as all multiply-charged ions that have all but one electron removed; Ca19+ would be an example. 2.1. THE BOHR ATOM 13

Figure 2.1: Bound energy levels of a classical Bohr atom.

quarters of the way to the ionization limit while subsequent levels are more and more closely spaced. It is possible to determine R by ﬁtting the spectrum to Eq. 2.5 from which we can estimate r1. Furthermore, it is possible to predict R,andr1 theoretically from the Correspondence Principle.3 When the radius of the electron orbit gets large, n gets large and the circulating electron will radiate classically with a frequency given by v . 2 (2.10) 2πn r1 Quantum mechanically, radiation is a result of transitions, as implied by Eq. 2.5. To remain classical, radiation must be between levels with large n. Consider a transition between two adjacent levels where n = n +1. Equation 2.5 then becomes 1 1 2R ν˜nn = R − 2 . (2.11) n2 (n +1) n3

Multiplying Eq. 2.11 by c, the speed of light, and equating it with Eq. 2.10, using Eq. 2.3 to express v and Eq. 2.4 to express R, we can show that πε 2 r 4 o 1 = 2 (2.12) e Zme

3The Correspondence Principle states that in the limit of large sizes the quantum description must approach the classical description 14 CHAPTER 2. AN OVERVIEW OF THE ATOM from which we identify the minimum radius for Z=1 πε 2 a 4 o , o = 2 (2.13) e me called the Bohr radius, and the Rydberg constant for an inﬁnitely massive nucleus 1 e2 1 R∞ = . (2.14) 4π 4πεo cao Finally, we can express the energy spectrum as e2 Z2 Z2 e2 2 Z2m E −1 −hcR −1 e . n = 2 = ∞ 2 = 2 2 (2.15) 2 4πεo n ao n 2 4πεo n

2.2 The Schr¨odinger 1-Electron Atom

Although we were able to determine the spectrum for the simple hydrogen- like ions from our classical approach, to study any of the details of the sturcture (energy levels) requires the time-independent Schr¨odinger equation (Eq. 2.1) to be solved. In this section we will look at exact solutions in the nonrelativistic approximation for hydrogen-like ions.

2.2.1 Angular Wavefunctions

When we generalize the ﬁeld free Hamiltonian (Eq. 2.2) by letting me → µ, the reduced mass, Schr¨odinger’s equation takes the form µ Ze2 ∇2ψ r 2 E ψ r ( )+ 2 + ( ) = 0. (2.16) 4πεor

Again, Z is the residual charge on the nucleus. That is, Z =1forH, →− Z =2forHe+, etc. The Coulomb potential only depends on | r | and, thus, falls into the class of centrally symmetric potentials. It is natural to seek solutions to Eq. 2.16 in sphrical coordinates in this case. If we write the Laplacian in spherical coordinates, ∂2 2 ∂ 1 ∂ ∂ 1 ∂2 ∇2 → + + sin ϑ + , (2.17) ∂r2 r ∂r r2 sin ϑ ∂ϑ ∂ϑ r2 sin2 ϑ ∂ϕ2 we can look for solutions of the form

ψ (r)=Rnl(r)Ylm (ϑ, ϕ) , (2.18) A.4. ATOMIC ENERGY UNIT (HARTREE) 113

A.4 Atomic Energy Unit (Hartree)

The atomic energy (au), also know as a hartree, is e times the electric potential associated with two elementary charges (e) separated by the Bohr radius (ao), e αc 1au=e = =4.35974381(34) × 10−18 J. (A.6) 4πεoao ao 2 1 In Eq. A.6, α ≡ e /4πεoc ∼ 137 is the ﬁne structure constant, is Planck’s Constant and c is the speed of light. These units are equivalent to setting e = = me = 1, where me is the electron rest mass.

A.5 Rydberg Energy Unit

The Rydberg energy units is given by 1 1Ry= Hartree = 13.60569172(53) eV, (A.7) 2 and is the binding energy of the hydrogen atom with an inﬁnitely massive 2 nucleus. This unit is equivalent to setting = e /2=2me = 1. The atomic energy unit (a.u.) and Rydberg energy unit (cm−1) are alos related:

1au=4πcR∞, (A.8) where R∞ is the Rydberg constant 2 1 e α −1 R∞ ≡ = = 109737.31568549(83) cm .(A.9) 2 4πεoaohc 4πao

A.6 eV Energy Unit

Since the quantity in parentheses in Eq. A.6 has units of volts, we can also deﬁne a new energy unit called an electron volt, eV. An atomic unit of enery is related to the eV through: 1 au = 1 Hartree = 27.2113834(11) eV. (A.10) The conversion between Rydberg energy unit and electron-Volt units (eV) is: 1 eV = 8065.54477(32) cm−1. (A.11) It is helpful to know that the NIST web site also oﬀers a conversion factor applet that was used here to convert from au to eV.