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Dragica Vasileska Arizona State University Importance of

• Hydrogen is the simplest atom • The quantum numbers used to characterize the allowed states of hydrogen can also be used to describe (approximately) the allowed states of more complex – This enables us to understand the periodic table • The hydrogen atom is an ideal system for performing precise comparisons of theory and experiment – Also for improving our understanding of atomic structure • Much of what we know about the hydrogen atom can be extended to other single- – For example, He+ and Li2+ Early Models of the Atom

• ’ J.J. Thomson s model of the atom – A volume of positive charge – embedded throughout the volume • ’ A change from Newton s model of the atom as a tiny, hard, indestructible sphere

“ ” watermelon model Experimental tests

Expect:

1. Mostly small angle scattering 2. No backward scattering events

Results:

1. Mostly small scattering events 2. Several backward scatterings!!! Early Models of the Atom

• ’ Rutherford s model

– Planetary model – Based on results of thin foil experiments – Positive charge is concentrated in the center of the atom, called the nucleus – Electrons orbit the nucleus like planets orbit the ’ Problem: Rutherford s model

“ ” ’ × – The size of the atom in Rutherford s model is about 1.0 10 10 m. (a) Determine the attractive electrical force between an electron and a separated by this distance. (b) Determine (in eV) the electrical potential energy of the atom. “ ” ’ × – The size of the atom in Rutherford s model is about 1.0 10 10 m. (a) Determine the attractive electrical force between an electron and a proton separated by this distance. (b) Determine (in eV) the electrical potential energy of the atom.

Electron and proton interact via the Coulomb force Given:  2 –   9  2 2   19  × 10 8.99 10 N m C 1.60 10 C r = 1.0 10 m  q1q2  F ke  2 2   10  r 1.0 10 m    8 2.3 10 N

Potential energy is Find:

   (a) F = ?  q1q2    18 1eV      (b) PE = ? PE ke 2.3 10 J  19 14 eV r 1.6 10 J  Difficulties with the • Atoms emit certain discrete characteristic frequencies of electromagnetic radiation – The Rutherford model is unable to explain this phenomena

• ’ Rutherford s electrons are undergoing a centripetal acceleration and so should radiate electromagnetic waves of the same frequency – The radius should steadily decrease as this radiation is given off – The electron should eventually spiral into the nucleus • ’ It doesn t Emission Spectra • A gas at low pressure has a voltage applied to it • A gas emits light characteristic of the gas • When the emitted light is analyzed with a spectrometer, a series of discrete bright lines is observed – Each line has a different wavelength and color – This series of lines is called an emission spectrum Emission Spectrum of Hydrogen

• ’ The wavelengths of hydrogen s spectral lines can be found from   1   1  1   RH 2 2  2 n  – R is the •H R = 1.0973732 x 107 m-1 – H … n is an integer, n = 1, 2, 3, – The spectral lines correspond to different values of n • A.k.a. • Examples of spectral lines – n = 3, ë = 656.3 nm – n = 4, ë = 486.1 nm Absorption Spectra

• An element can also absorb light at specific wavelengths • An absorption spectrum can be obtained by passing a continuous radiation spectrum through a vapor of the gas • The absorption spectrum consists of a series of dark lines superimposed on the otherwise continuous spectrum – The dark lines of the absorption spectrum coincide with the bright lines of the emission spectrum Applications of Absorption Spectrum

• The continuous spectrum emitted by the Sun passes ’ through the cooler gases of the Sun s atmosphere – The various absorption lines can be used to identify elements in the solar atmosphere – Led to the discovery of ’ Recall Bohr s Assumptions

• Only certain electron orbits are stable. Radiation is “ ” emitted by the atom when the electron jumps from a more energetic initial state to a lower state   Ei E f hf

• The size of the allowed electron orbits is determined by a ’ condition imposed on the electron s orbital   mevr n, n 1,2,3,...

Why is that? Modifications of the Bohr – Theory Elliptical Orbits • Sommerfeld extended the results to include elliptical orbits – Retained the principle , n – Added the orbital quantum number, ℓ • ℓ ranges from 0 to n-1 in integer steps – All states with the same principle quantum number are said to form a shell – The states with given values of n and ℓ are said to form a subshell – Modifications of the Bohr Theory and • Another modification was needed to account for the Zeeman effect – The Zeeman effect is the splitting of spectral lines in a strong – This indicates that the energy of an electron is slightly modified when the atom is immersed in a magnetic field – A new quantum number, m ℓ, called the orbital , had to be introduced • ℓ ℓ m ℓ can vary from - to + in integer steps

• High resolution spectrometers show that spectral lines are, in fact, two very closely spaced lines, even in the absence of a magnetic field – This splitting is called fine structure – Another quantum number, ms, called the magnetic quantum number, was introduced to explain the fine structure de Broglie Waves

• ’ One of Bohr s postulates was the angular momentum of the electron is quantized, but there was no explanation why the restriction occurred • de Broglie assumed that the electron orbit would be stable only if it contained an integral number of electron wavelengths de Broglie Waves in the Hydrogen Atom • In this example, three complete wavelengths are contained in the circumference of the orbit • In general, the circumference must equal some integer number of wavelengths      2 r n , 1,2,3,...   h

mev   mevr n, n 1,2,3,... This was the first convincing argument that the wave of matter was at the heart of the behavior of atomic systems and the Hydrogen Atom • One of the first great achievements of quantum mechanics was the solution of the wave equation for the hydrogen atom • The significance of quantum mechanics is that the quantum numbers and the restrictions placed on their values arise directly from the mathematics and not from any assumptions made to make the theory agree with experiments Problem: wavelength of the electron

Determine the wavelength of an electron in the third excited orbit of the hydrogen atom, with n = 4. Determine the wavelength of an electron in the third excited orbit of the hydrogen atom, with n = 4.

’ Recall that de Broglie s wavelength of electron  Given: depends on its momentum, = h/(mev). Let us find it,

n = 4   n mevrn n, so mev rn

 2  h Recall that rn n a0 , so mev    2 a0 n Find:   h           Thus, 2 a n 8 0.0529nm 1.33nm e = ? 0 mev Quantum Number Summary

• The values of n can increase from 1 in integer steps • The values of ℓ can range from 0 to n-1 in integer steps • ℓ ℓ The values of m ℓ can range from - to in integer steps Spin Magnetic Quantum Number • It is convenient to think of the electron as spinning on its axis – The electron is not physically spinning • There are two directions for the spin – ½ Spin up, m = – s ½ Spin down, m = - • s There is a slight energy difference between the two spins and this accounts for the Zeeman effect Electron Clouds

• The graph shows the solution to the wave equation for hydrogen in the – The curve peaks at the – The electron is not confined to a particular orbital distance from the nucleus • The probability of finding the electron at the Bohr radius is a maximum Electron Clouds

• The for hydrogen in the ground state is symmetric – The electron can be found in a spherical region surrounding the nucleus • The result is interpreted by viewing the electron as a cloud surrounding the nucleus – The densest regions of the cloud represent the highest probability for finding the electron

Mathematical Details

Roadmap for solution of Hydrogen-like atoms

• Start from 3D TISE for electron in Coulomb potential of nucleus • Separate variables to give 1D radial problem and angular problem Solution of angular part already known in terms of . • Simplify 1D radial problem using substitutions and atomic units • Solve radial problem Extract asymptotic solution at large r Use Frobenius method Find eigenvalues by requiring normalizable solutions Reminder: SE in three dimensions ö H-atom is our first example of the 3D Schr dinger equation          Wavefunction and potential energy are now x r x, y, z         functions of three spatial coordinates: V x V r V x, y, z

2 2 2  2  2 Kinetic energy involves three px  p  px py pz components of momentum 2m 2m 2m 2 2 2 2  2 2 2       2         2  2  2  2 2m x 2m 2m  x y z 

Interpretation of 3    2    2 wavefunction: d r r,t r,t probability of finding particle in a probability density at r volume element centred on r (probability per unit volume) ö Time-independent Schr dinger equation 2 ˆ           2            H r r E r r V r r E r 2m The Hamiltonian for a hydrogenic atom

In a hydrogenic atom or with nuclear charge +Ze there is the Coulomb -e attraction between electron and nucleus. – This has spherical symmetry potential only depends on r. This is known as a   Ze2 r CENTRAL POTENTIAL V (r)  4 0r +Ze

The Hamiltonian operator is ˆ 2 2    2  Ze H  2me 4 0r

NB: for greater accuracy we should use the reduced mass not the electron mass here. This accounts for the relative motion of the electron and the nucleus (since the nucleus does not remain precisely fixed):

   memN   me  ; me electron mass, mN nuclear mass me mN Hamiltonian for hydrogenic atoms

The natural coordinate system is spherical polars. In this case the Laplacian operator becomes (see 2B72):

    2      2  2  1 2  L ˆ2   2 1   1  r  L  sin   2     2 2        2   2  r r r  r sin sin

So the Hamiltonian is ˆ 2 2 2     2 2 ˆ     2  Ze   2  L  Ze H r    r   2     2  2me 4 0r 2mer r r 2mer 4 0r

ˆ         TISE for H-like atom is H r r E r (m = me from now on) ˆ 2     2 2  2 (r)  L (r)  Ze      r  (r) E (r) 2     2  2mr r r 2mr 4 0r Angular momentum and the H atom

ˆ 2     2 2 ˆ     2  L  Ze ˆ H r   r  d L2 2   2  ˆ ˆ2    2mr r  r  2mr 4 r [H, L ] 0 0 0 dt ˆ  ˆ d L Reminder: d Q  i  ˆ ˆ   Q ˆ ˆ   z  ’ H,Q [H, Lz ] 0 0 Ehrenfest s theorem  dt dt  t

CONCLUSIONS • The angular momentum about any axis and the total angular momentum commute with the Hamiltonian • TheSy oa rwe eth ewreilflo rleo obokt hf ocor nas esroveludt qiouna notitfi etshe form • We can have simultaneous eigenfunctions of these operators and the Hamiltonian • We can have well-defined values of these quantities and the energy at the same time The angular wavefunction

This suggests we look for      separated solutions of the form (r) (r ) R(r)Ylm ( )

The angular part are the eigenfunctions of the total angular momentum operator L2. These are the spherical harmonics, so we already know the corresponding § eigenvalues and eigenfunctions (see 5): ˆ    1        Y00 ( , )  LzYlm , mYlm , 4 ˆ2        2    L Y , l l 1 Y ,     3   lm  lm Y11( , )  sin exp(i ) 8 ˆ2  2  Eigenvalues of L are l(l 1) , with l 0,1,2,    3  ˆ    Y10 ( , )  cos Eigenvalues of Lz are m, with l m l 4    3     l = principal angular momentum quantum number. Y1 1( , )  sin exp( i ) 8 m = magnetic quantum number (2l+1 possible values).

Note: this argument works for any spherically-symmetric potential V(r), not just the Coulomb potential. The radial equation ö Substitute separated solution into the time-independent Schr dinger equation ˆ       2     2 2 (r, , ) R(r)Y ( , )  2  L  Ze    lm   r  (r) E (r) 2     2  2mr r r 2mr 4 0r

Get radial equation 2    2 2  d 2 dR  l(l 1)  Ze    r   R R ER 2   2  2mr dr dr 2mr 4 0r

Note that this depends on l but not on m: R(r) and E therefore involve the magnitude of the angular momentum but not its orientation. The radial equation (2)

Define a new radial  2    2 2  (r)  d 2 dR  l(l 1)  Ze  ÷   r   R R ER function (r) by: R(r) 2   2  r 2mr dr dr 2mr 4 0r

÷ 2 2    2 2  Get radial equation for (r)  d  l(l 1)  Ze        E 2  2   2m dr 2mr 4 0r The effective potential

New radial equation looks like the 2 2    2 2  ö  d  l(l 1)  Ze    1D Schr dinger equation with an     E 2  2   effective potential 2m dr 2mr 4 0r

 2 2 2 2  l(l 1)  Ze  d     Veff (r) 2   2 Veff (r) E 2mr 4 0r 2m dr

2 2 1D TISE  d     V(r)  V (x) E 2m dx2

 2 l(l 1) r  2mr 2

is known as the centrifugal barrier potential The centrifugal barrier

Where does the centrifugal barrier come from?  2  l(l 1) Vcb (r) 2 CLASSICAL ARGUMENT 2mr  Fixed l corresponds to fixed angular momentum for the electron. L mvr so as r becomes small,v  must increase in order to maintain L. This causes an ‘ ’ increase in the apparent outward force (the centrifugal force).

2 2  For circular motion  mv  L L mvr F r mr3

  dV   L2 F V dr 2mr 2

Alternatively, we can say that the energy required to supply the extra angular speed must come from the radial motion so this decreases as if a corresponding outward force was being applied. Roadmap for solution of radial equation    2 2    2 2  (r) R(r)Y ( )   d  l(l 1)  Ze     nlm lm 2 2  E  2m dr  2mr 4 r   (r) 0 R(r) r • Simplify equation using atomic units • Solve equation in the asymptotic limit (large r)    (r)  exp( r) Gives a decaying exponential solution r • Define new radial function by factoring out asymptotic solution    (r) F(r)exp( r) •  Solve equation for F(r) using the series (Frobenius) method   p s F(r) apr p • Find that solution is not normalizable unless series terminates. This only happens if the eigenvalues have certain special values. Hence we find the eigenvalues and eigenstates of the H-atom. Atomic units

There are a lot of physical constants in these expressions. It makes atomic problems simpler to adopt a system of units in which as many as possible of these constants are one. In atomic units we set:

  2 1  1 (dimensions [ML T ])  Electron mass me 1 (dimensions [M ]) 2  e  3 2 Constant apearing in Coulomb's law  1 (dimensions [ML T ]) 4 0 It follows that:  2    4   11 Unit of length =  0   5.29177 10 m=Bohr radius, a  2  0 e me 2  2   e m   18  Unit of energy =   e 4.35974 10 J 27.21159eV=Hartree, E    2 h 4 0 

In these units the radial equation becomes

2 2    2 2  2      d  l(l 1)  Ze      1 d  l(l 1)  Z        E   E 2  2   2  2  2m dr 2mr 4 0r 2 dr 2r r Asymptotic solution of radial equation (large r)

Consider the radial equation at very large 2     distances from the nucleus, when both terms  1 d  l(l 1)  Z      E in the effective potential can be neglected. 2 dr 2  2r 2 r  We are looking for bound states of the atom where the electron does not have enough energy to escape to infinity (i.e. E < 0):

 2   Put E 2

This gives 2  d   2      (r) exp( r) dr 2

For normalizable solutions we must take the decaying solution

Inspired by this, rewrite the solution in terms    of yet another unknown function, F(r): (r) F(r)exp( r) Differential equation for F

2     Derive equation for F  1 d  l(l 1)  Z      E 2 dr 2  2r 2 r     (r) F(r)exp( r)

Differential equation for F:  2   d   d  l(l 1)  2Z     2  F r 0 dr 2 dr r 2 r  Series solution (1) Look for a power-series solution (Frobenius method). The point r = 0 is a regular singular point of the equation so at least one well-behaved series solution should exist (see 2B72).  2    d   d  l(l 1)  2Z      p s   Substitute F(r) a r 2 2 2 F r 0 p dr dr r r  p

       p s 2             p s 1         apr p s p s 1 l l 1 apr 2 p s 2Z 0   p 0 p 0 Series solution (2)

The indicial equation that fixes s comes from equating coefficients of the lowest power of r which is s - 2

       p s 2             p s 1         apr p s p s 1 l l 1 apr 2 p s 2Z 0   p 0 p 0     s(s 1) l(l 1) 0  2     s s l(l 1) 0         s l s (l 1) 0     s l, or s l 1

We need the regular solution that will be →   well-behaved as r 0, so take s l 1 Series solution (3)

General recursion relation comes from equating coefficients of r to the power p+l

      p l 1             p l          apr p l 1 p l l l 1 apr 2 p l 1 2Z 0   p 0 p 0

          ap 1  2 p l 1 Z            ap p l 2 p l 1 l l 1 Series solution (4) →∞ For p we find:           ap 1  2 p l 1 Z              ap p l 2 p l 1 l l 1 ap 1  2  p ap p

  p l 1 Compare with: (remember ap is coefficient of r in the expansion)

  n   n    (2 r)   n  (2 ) exp(2 r) bnr with bn .   n 0 n! n 0 n!   p l 1   Coefficient of r would be bp l 1.     p l 2    p l 2     Coefft. of r  bp l 2  (2 )  ( p l 1)!  2  2      p l 1  p l 1     p   Coefft. of r bp l 1 (2 ) ( p l 2)! ( p l 2) p

ê So, our series behaves for large p just like exp(2 r). Series solution (5)

So, if the series continues to arbitrarily    large p, the overall solution becomes (r) F(r)exp( r)       (r)  exp(2 r) exp( r) exp( r) (not normalizable)

To prevent this the series must terminate after a finite number of terms. This only happens if

Z      ( p l 1) for some integer p 0,1,2      n where n is a positive integer l : n l 1,l 2

 2 2 So finally the energy is     Z with n > l E 2 2n2

n is known as the . “ ” It defines the shell structure of the atom. Summary of solution so far

Each solution of the time-independent    ö Schr dinger equation is defined by nlm (r) Rnl (r)Ylm ( ) three quantum numbers n,l,m   (r)  nl Y ( ) r lm  The radial solution depends on  F (r)e Zr / n  n and l but not m nl Y ( ) r lm

            l 1 p ap 1  2Z p l 1 n Fnl (r) r apr               p 0 ap n p l 2 p l 1 l l 1

The energy only depends on the   Z 2     principal quantum number n En 2 , n 1,2,3 n l which is bigger than l 2n The hydrogen energy spectrum

  1 E n 2n2    In Hartrees Eh 27.2eV In eV ground state energy = -13.6eV = - ionisation energy

This simple formula agrees with observed frequencies to within 6 parts in ten thousand

Traditional spectroscopic nomenclature: n > l so “ ” l = 0: s states (from sharp spectral lines)   “ ” n 1 l 0 l = 1: p states ( principal ) “ ”   l = 2: d states ( diffuse ) n 2 l 0,1 “ ” l = 3: f states ( fine )   … … n 3 l 0,1,2 and so on alphabetically (g,h,i etc) The energy spectrum: degeneracy

… For each value of n = 1,2,3 we have a   Z 2 definite energy: E (in atomic units) n 2n2

… For each value of n, we can have n l = 0, 1, 2, , n-1 possible values of the total angular momentum quantum number l:      For each value of l and n we can have 2l+1 m l, (l 1), 0, (l 1),l values of the magnetic quantum number m:  

 The total number of states (statistical weight) n 1    2 associated with a given energy En is therefore: (2l 1) n .  (This neglects electron spin. See Section 7.) l 0

The fact that the energy is independent of m is a feature of all spherically symmetric systems and hence of all atoms. The independence on l is a special feature of the Coulomb potential, and hence just of hydrogenic atoms. This is known as accidental degeneracy.    nlm (r) Rnl (r)Ylm ( )  The radial wavefunctions  Zr / n  nl (r)  Fnl (r)e Rnl (r) Rnl(r) depends on n and l but not on m r r

 3/ 2  Z  R (r) 2  exp( Zr / a ) 10   0 a0  3/ 2       1 Z Zr Zr R (r)    exp  21       3 2a0 a0 2a0  3/ 2       Z  Zr Zr R (r) 2  1 exp  20       2a0 2a0 2a0  3/ 2  2     4 Z Zr Zr R (r)     exp  32       27 10 3a0 a0 3a0  3/ 2        4 2 Z  Zr Zr Zr R (r)   1  exp  31        9 3a0 6a0 a0 3a0 3/ 2    2 2      Z  2Zr  2Z r Zr R (r) 2  1 exp  30    2    3a0 3a0 27a0 3a0

For atomic units set a0 = 1 The radial wavefunctions (2)

 3/ 2  Z  Full wavefunctions are: R (r) 2  exp( Zr / a ) 10  a  0 0     3/ 2      (r) R (r)Y ( )  1 Z Zr Zr nlm nl lm R (r)    exp  21 3  2a   a   2a  0 0 0 Normalization chosen so that:  3/ 2       Z  Zr Zr R (r) 2  1 exp  20        2a0 2a0 2a0 2 2     drr R r 1  3/ 2  2    0 nl  4 Z Zr Zr R (r)     exp  32 27 10  3a   a   3a  0 0 0 Asymptotic solution  3/ 2        4 2  Z    Zr  Zr   Zr  R31(r) 1 exp 9  3a   6a  a   3a   0 0 0 0   n 1  3/ 2 R (r)  r exp( Zr / n)    2 2     nl r  Z  2Zr  2Z r Zr R (r) 2  1 exp  30  3a   3a 27a2   3a  0 0 0 0 Solution near r = 0

Only s states (l = 0) are finite at the origin.   l R (r)  r Radial functions have (n-l-1) zeros nl r 0 (excluding r = 0). Radial probability density

Total probability density  2  2  2 | nlm (r) | Rnl (r) Ylm ( ) (Rnl is real)

 2 3 = probability of finding particle in a è ö | nlm (r) | d r volume element centred on (r, , )

Integrate over all angles using normalization of spherical harmonics

3  2       d r r drd d sin d d (solid angle element)

Radial probability density 2 2   2  2 2  r Rnl (r)dr Rnl (r) Ylm ( ) r drd   ,

2 2 = probability of finding the particle in a r Rnl (r)dr spherical shell centred on r, i.e. at any angle

2 2   2 r Rnl (r)dr nl (r)dr so this is analogous to the 1D case Angular probability density

Solid angle probability density

 2 | Y ( ) | = probability density of finding      lm particle in a solid angle element d sin d d

 2   | Ylm ( ) | d = probability of finding particle between è è è ö ö ö  2    and + d and and + d | Ylm ( ) | sin d d

Total probability density

 2 3   2 2    2     | nlm (r) | d r r Rnl (r)dr . |Ylm ( ) | sin d d

= (Radial probability) x (Angular probability) Radial probability density

Radial probability density 2 2 r Rnl (r)

Orbital n l (au)

1s 1 0 1.5

2s 2 0 6.0

2p 2 1 5.0

3s 3 0 13.5

  3  *  r d r nlm (r)r nlm (r)    3   2 dr r Rnl r 0 Comparison with

Bohr model Quantum mechanics

Angular momentum (about any axis) Angular momentum (about any axis) assumed to be quantized in units of shown to be quantized in units of ’ ’ Planck s constant: Planck s constant:      Lz n, n 1,2,3, Lz m, m l,,l Electron otherwise moves Electron wavefunction spread over according to classical all radii. Can show that the mechanics and has a single quantum mechanical expectation well-defined orbit with radius value of 1/r satisfies n2a  Z    0  1 1 , a Bohr radius rn , a0 Bohr radius r 2 r 0 Z n a0 n Energy quantized and Energy quantized, but determined solely by angular determined solely by principal momentum: quantum number n, not by 2 angular momentum:   Z  2 En 2 Eh , Eh Hartree   Z  2n En Eh , Eh Hartree 2n2 The remaining approximations These results are not exact because we have made several approximations. • We have neglected the motion of the nucleus. To fix this we should replace m ì e by the reduced mass . This improves agreement with experiment by an order of magnitude (simple formula gives spectral lines to within 4 parts in 100 thousand!)

   memp  me  and E(n) me me mp     1   1 Z 2 E(n) E(n)   E     2 h 1 me / mp 1 me / mp 2n • We have used a non-relativistic treatment of the electron and in particular have § neglected its spin (see 7). Including these effects give rise to “ ” fine structure ’ (from the interaction of the electron s with its spin) “ ” ’ (from the interaction of the electron s spin with the nuclear spin) • We have neglected the fact that the EM field between the nucleus and the “ ” electron is itself a quantum object. This to quantum electrodynamic “ ” (QED) corrections, and in particular to a small of the energy levels. Summary Energy levels in au   Z 2     En 2 , n 1,2,3 n l    2n In Hartrees Eh 27.2eV

Ground state energy = -1/2 au = -13.6eV = - ionisation energy

 n 1      2 Statistical weight g  2l 1 n  l 0

Wavefunction    l 1 p    Fnl (r) r apr  nlm (r) Rnl (r)Ylm ( ) p 0  Zr / n        Fnl (r)e     ap 1  2Z p l 1 n Ylm ( )            r ap n p l 2 p l 1 l l 1

Radial probability density 2 2 r Rnl (r) Permissible Quantum States Orbital energies of the hydrogen atom. Orbital Energies of Multielectron Atoms • All elements have the same number of orbitals (s,p, d, and etc.). • In hydrogen these orbitals all have the same energy. • In other elements there are slight orbital energy differences as a result of the presence of other electrons in the atom. • The presence of more than one electron changes the energy of the electron orbitals (click here) Shape of 1s Orbital Shape of 2p Orbital Shape of 3d Orbitals