2B24 Atomic and Molecular Physics Part 2 of 5 (UCL)

2 ONE-ELECTRON ATOMS

2

1

ONE-ELECTRON ATOMS

2.1

THE BOHR ATOM

The postulates: 1. The electron moves in circular classical orbits about the nucleus without radiating 2. The angular momentum L is quantized.

L = nÂŻ h

h ÂŻ =h/2Ď€=and n=1,2,3,.... ,

(1)

Coulomb force between nucleus and electron is balanced by the centripetal force. Assuming the nucleus is infinitely heavy Ze2 me v 2 = .

4Ď€ 0 R2 R

(2)

From the quantization of L, L = nÂŻ h we get 4Ď€ 0h ÂŻ 2 n2 rn = .

Ze2 me

nÂŻ h v= me R

(3)

4Ď€ 0h ÂŻ2 we can now work out the For n = 1, rn=1 = . For Z = 1 (hydrogen) Ze2 me this is called the BOHR RADIUS. We can eliminate r in the equation for the velocity: Ze2 vn = .

(4) 4Ď€ 0 nÂŻ h Knowing v and r we can work out the total kinetic energy, T , for each quantum state, since T = 21 me v 2 and the potential energy, V , since V = Ze2 − : Put these, with r and v given above, in TOTAL ENERGY: 4Ď€ 0 r E=T + V

(5)

you get 2

me  Ze2  1 (6) En = − 2 4Ď€ 0 n2 2ÂŻ h for n = 1, 2, 3, . . . . Note that the energy is negative, since we have a bound state. 

2 ONE-ELECTRON ATOMS 2.1.1

2

TRANSITIONS

If the electron jumps from an initial state i with ni to a final state j with nj , we have: hνif =

Eni

−

Enf

=

  

2 2

m e  Ze   1

2

2ÂŻ 4Ď€ 0 nf 2

h

−



1 

ni 2

(7)

ni < nf , the photon will be absorbed and the process is called excitation, ni > nf the photon will be emited and the process is de - excitation.

The wavelenght of the photon will be: 1 νif = = Νif c

  

2 2

m Ze 1 e

  

3

4Ď&#x20AC;ÂŻ nf 2 h c 4Ď&#x20AC; 0

â&#x2C6;&#x2019;



1 

ni 2

(8)

Similar to formula empirically determined by Rydberg for the H spectrum: 1 =R Îťif





1 1   â&#x2C6;&#x2019; nf 2 ni 2

(9)

R=109677.58 cmâ&#x2C6;&#x2019;1 is the observed Rydberg constant, note the dimensions. The Bohr model predicts: Râ&#x2C6;&#x17E;

me = 4Ď&#x20AC;ÂŻ h3 c

2

e2   = 109737cmâ&#x2C6;&#x2019;1 4Ď&#x20AC; 0 

(10)

(the symbol â&#x2C6;&#x17E; makes reference to the fact that we are assuming the mass of the nucleus to be infinite). The discrepancy is ' 5x10â&#x2C6;&#x2019;4 .

2.2

LYMAN, BALMER AND PASCHEN SPECTRAL SERIES

â&#x20AC;˘ Lyman series n = 2, 3, 4, 5... â&#x2020;&#x2019;n = 1

(u.v.)

â&#x20AC;˘ Balmer series n = 3, 4, 5... â&#x2020;&#x2019;n = 2

(optical)

â&#x20AC;˘ Paschen series n = 4, 5... â&#x2020;&#x2019;n = 3

(Infra Red)

See figure 2.1. The limit of the Lyman series is 912 Ë&#x161; A (in u.v.) corresponding to the ionisation energy of hydrogen Râ&#x2C6;&#x17E; = 109737cmâ&#x2C6;&#x2019;1 .

2 ONE-ELECTRON ATOMS 2.2.1

3

THE CORRESPONDENCE PRINCIPLE

In the limit of large quantum numbers n → ∞, QM results should tend to the classical results.

2.2.2

EFFECTS OF FINITE NUCLEAR MASS

The mass of the nucleus is not infinite. Corrections due to the finite mass are important, e.g. the spectral line frequencies differ for D and H. So, we consider the REDUCED MASS, µ, in the Bohr atom.

DEFINITION: the reduced mass, µ is the effective mass of a system once the Centre of Mass motion is separated off.

Consider the proton and electron system as a rigid rotor, separation r and note that Mp /me = 1836 and r = re − Rp . The Centre of Mass remains at rest. It is at the origin, so taking moments, Mp R p + m e r e = 0 , me r and m e + Mp Then, for the angular momentum, L:

Mp r m e + Mp

(12)

L = me ωre 2 + Mp ωRp 2 = µωr2 = µvr .

(13)

Rp = −

where µ =

(11)

re =

m e Mp is the reduced mass. (me + Mp )

In the Bohr atom expressions we replace me by µ.

In general we can approximate µ ∼ me , but the difference accounts for the difference between R and R∞ , and can be important. We have for each atom a different value of R. 2.2.3

ASSESMENT OF THE BOHR MODEL

Predictions for energy levels of H and other one-electron (hydrogenic) atoms are good. Could not explain quantitatively transitions between states. It could not be extended to other atoms, even helium, but it does work well for hydrogenic atoms, with only one electron such as He+ or Li2+ .

2 ONE-ELECTRON ATOMS

2.3

4

ATOMIC UNITS AND WAVENUMBERS

2.3.1

ATOMIC UNITS â&#x20AC;&#x153;a.u.â&#x20AC;?

When working with atoms it is convenient to use a new set of units.

We define: â&#x20AC;˘ unit of mass = me â&#x20AC;˘ unit of charge = e â&#x20AC;˘ unit of length=Bohr radius = a0 So h ÂŻ = 4Ď&#x20AC; 0 = a0 = e = me = 1 .

Put these in eqs above and find that: n2 rn = .

Z

(14)

Z = 1 and n = 1 we have r1 = a0 = 1, the BOHR RADIUS, a0 = 0.529x10â&#x2C6;&#x2019;10 a.u. The energy is given by Z2 En = â&#x2C6;&#x2019; 2 .

2n

(15)

The ground state energy of hydrogen for infinite mass is: En=1 = â&#x2C6;&#x2019;1/2 a.u. . . . or â&#x2C6;&#x2019;13.605 eV. . . or â&#x2C6;&#x2019;2.18 Ă&#x2014; 10â&#x2C6;&#x2019;18 J. . . .

The atomic unit of energy is also called the Hartree: 1 Hartree= 1 a.u. =27.21 eV The speed of the eâ&#x2C6;&#x2019; in the ground (lowest) state of H is: v=

e2 = 1 a.u. = Îąc 4Ď&#x20AC; 0h ÂŻ

1 e2 ' is fine structure constant. where Îą = 4Ď&#x20AC; 0h ÂŻc 137 The speed of light is ' 137 a.u..

(16)

2 ONE-ELECTRON ATOMS

5

Can also use: R∞ = |En=1 |, the RYDBERG CONSTANT. The ∞ refers to infinite mass. Then Z2 En = −R∞ 2 n

 

µ Z2  strictly, En = −R∞ . me n2 

(17)

Note: the ionisation energy of hydrogen is |En=1 | since it is the energy which must be supplied to the atom in its ground state to remove the electron.

Another energy unit can be defined: Rydberg = 12 Hartree 2.3.2

SPECTROSCOPIC UNITS: WAVENUMBERS

Wavenumbers or cm−1 are the units favoured by spectroscopists.

Have dimension Length−1 .

Advantage: if the energies of quantum states are given in this unit one can rapidly infer the wavelength.

Using E = hν=hc/λ we get 

λ−1



˜∞  1 − 1  . = Z 2R n2f n2i

(18)

˜ ∞ defines the unit of the WAVENUMBER. We usually drop the Now R tilde.

To get the energy in different units we can use En = R∞ Z 2 /n2 with: • R∞ = 109737 cm−1 • R∞ =

1 2

a.u.

• R∞ = 13.605 eV So 1 a.u. of energy = 27.2 eV = 219475 cm−1 1 eV = 0.0368 a.u. = 8065.48 cm−1 1 cm−1 = 1.239 ×10−4 eV= 0.0456 ×10−4 a.u.

2 ONE-ELECTRON ATOMS

2.4

6

REVIEW OF QUANTUM ANGULAR MOMENTUM AND SPHERICAL HARMONICS

Classically, orbital angular momentum, L, is given by L = r × p so that Lx = ypz − zpy ,

etc ...

(19)

Quantum mechanically we use the momentum operator pˆ = −i¯ h∇ and ˆ whose components and length are given by rˆ which give L ∂ ∂ ˆ x = −i¯ L h y −z , ∂z ∂y !

ˆ 2x + L ˆ 2y + L ˆ 2z ˆ2 = L etc ....... and L

(20)

ˆ i, L ˆ j ] = i¯ ˆ k , i, j, k stand for x, y and z, The commutation relations are [L hL ˆ x, L ˆ y ] = i¯ ˆ z so that the components of L ˆ cannot be assigned i.e. [L hL ˆ =0). definite values simultaneously (unless L ˆ 2, L ˆ j ] = 0, i.e [L ˆ 2, L ˆ x ] = 0 so that we can have simultaneous However, [L 2 ˆ and one component of L ˆ ( note in general we eigenfunctions of L ˆ z ). To understand this, picture L ˆ precessing about the z-axis choose L (can always rotate reference frame for this to be the case).

It is easier to use spherical polar coordinates, r, θ, φ (see text books for diagram, but note mathematics books often interchange θ and φ!). x = r sin θ cos φ y = r sin θ sin φ z = r cos θ

0≤r≤∞ 0≤θ≤π 0 ≤ φ ≤ 2π

which gives ∂ ˆ z = −i¯ L h ∂φ

and

∂ ∂ 1 ∂2  2 1 ˆ L = −¯ sin θ + h sin θ ∂θ ∂θ sin2 θ ∂ 2 φ 2



!



(21)

ˆ z are called spherical harˆ 2 and L The simultaneous eigenfunctions of L monics, Ylm (θ, φ) and the eigenvalue equations are ˆ 2 Ylm (θ, φ) = l(l + 1)¯ L h2 Ylm (θ, φ)

ˆ z Ylm (θ, φ) = m¯ L hYlm (θ, φ)

(22)

where l = 0, 1, 2, . . . are the orbital angular momentum quantum numbers and m(= ml ) = 0, ±1, ±2, ±3, . . . , ±l are the orbital angular momentum magnetic quantum numbers and there are 2l + 1 values.

2 ONE-ELECTRON ATOMS

7

So we find that angular momentum is quantized but not in the same way as in the Bohr picture were we had L = n¯ h2 .

The Spherical Harmonics:Ylm (θφ) = Θlm (θ)Φm (φ).

See figures 2.3, 2.4, 2.5 and 2.8. are separable in functions of θ and φ, where Φm (φ) = √12π eimφ and Θlm (θ) is proportional to Plm (cos θ), the Associated Legendre Polynomial.

The parity of Ylm (θ, φ). The parity operator, Pˆ , corresponds to an inversion of a vector r through the origin and is defined by the relation Pˆ f (r) = f (−r)

(23)

If Pˆ f (r) = +f (r) EVEN function.

If Pˆ f (r) = −f (r) ODD function.

The parity of Ylm (θ, φ) is (−1)l and Ylm (θ, φ) is EVEN if l is even and ODD is l is odd.

2.5

REVIEW OF QUANTUM TREATMENT OF ONE-ELECTRON ATOMS

Time-independent Schr¨odinger eq. for one electron-atoms (in a. u.) ˆ nlm (r) = − 1 ∇2 − Z Ψnlm (r) = En Ψnlm (r) HΨ 2 r "

#

(24)

where Ψnlm (r) = Rnl (r)Ylm (θ, φ) can be written as a product of the radial ˆ and angular eigenfunction of H. Radial Wavefunction 2Z r Rnl (r) = Nnl e−ρ/2 ρl Lnl (ρ) where we used scaled variables: ρ = n Nnl is a constant for given n, l and Z. Lnl (ρ) are the Associated Laguerre Polynomials.

2 ONE-ELECTRON ATOMS

8

For large Ď the exponential term, eâ&#x2C6;&#x2019;Ď /2 , dominates so that as Ď â&#x2020;&#x2019; â&#x2C6;&#x17E; we have Rnl â&#x2020;&#x2019; 0. So: â&#x20AC;˘ as r â&#x2020;&#x2019; â&#x2C6;&#x17E; we have Rnl â&#x2020;&#x2019; 0, as expected since the eâ&#x2C6;&#x2019; is bound to the atom. â&#x20AC;˘ The larger Z the faster the exponential decay, the eâ&#x2C6;&#x2019; is closer to the nucleus. â&#x20AC;˘ The larger n, the slower the decay. Excited states are more extended. â&#x20AC;˘ For normalized wavefunction Z

2

|Ψnlm | dĎ&#x201E; =

Z â&#x2C6;&#x17E; 0

2 2

|Rnl (r)| r dr

Z 2Ď&#x20AC; Z Ď&#x20AC; 0

0

|Ylm (θĎ&#x2020;)|2 sinθdθdĎ&#x2020; = 1

(25)

2 â&#x20AC;˘ r2 Rnl gives the probability density of finding the eâ&#x2C6;&#x2019; at a distance r from the nucleus. So to â&#x20AC;&#x2122;visiualizeâ&#x20AC;&#x2122; the radial function we can look at the integrand of the radial integral as shown in figures for H atom (figure 2.3).

We do not have a well defined orbit as in the Bohr model.

2 â&#x20AC;˘ Number of peaks in r2 Rnl is equal to n â&#x2C6;&#x2019; l ( From Laguerre Polynomial).

â&#x20AC;˘ For a given n the probability of finding the eâ&#x2C6;&#x2019; near the nucleus decreases as l increases, because the centrifugal barrier pushes the eâ&#x2C6;&#x2019; out. So the low-l orbitals are called penetrating.

The energy eigenvalues obtained by solving the Schr¨odinger eq. are 1  Ze2  Âľ En = â&#x2C6;&#x2019; 2 2n 4Ď&#x20AC; 0 h ÂŻ2 



(26)

The result is the same as in the Bohr model, En depends on n, but for each value of n we have n values of l ( =0,1,2,. . . ,n-1) and for each each value of l we have (2l + 1) values of m (0, Âą1, Âą2, Âą3, . . . , Âąl). In quantum mechanics solution for H, we have for each energy several quantum statesâ&#x2020;&#x2019; Degeneracy.

Hence the total number of states with the same En , i.e. total degeneracy with respect to l and m is nâ&#x2C6;&#x2019;1 X

(2l + 1) = n2

l=0

(27)

2 ONE-ELECTRON ATOMS

9

Spectroscopic notation. Each level is labelled by nl where l is represented by a letter l = 0 1 2 3 4 ... s p d f g s stands for ’sharp’, p for ’principal’, d for ’diffuse’ and f for ’fundamental’.

So the ground state (g.s.) of H is labelled 1s, the first excited states 2s and 2p. Transitions between states with any ∆n are allowed but as we will see there is a restriction on l and m; ∆l = ±1 and ∆m = 0, ±1 (called electric-dipole selection rule). Also we label the n quantum numbers by n= 1 2 3 4 . . .

K L M N

2.6

ELECTRON SPIN AND ANTIPARTICLES

As experimental techniques improved, many transitions were found to exhibit what became known as fine structure e.g., the 2p→1s transition in atomic hydrogen was found to be a doublet. This lead to the assignment of an additional quantum number to the electron, spin, s, associated with its intrinsic angular momentum, Sˆ (remember these are operators), such ˆ we obtain the eigenvalue expressions: that (in analogy with L) Sˆ2 χsms = s(s + 1)¯ h2 χsms

with s =

1 2

1 with ms = ± .

2

Sˆz χsms = msh ¯ χsms

(28) (29)

and the spin eigenfunctions χsms are called spinors.

So, the total wavefunction of the H atom is: Ψnlml ms (r) = Rnl (r) Ylml (θ, φ) χms . |

{z

}|

{z

} | {z }

radial angular spin

(30)

2 ONE-ELECTRON ATOMS

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The theoretical justification for spin eventually emerged through Dirac’s relativistic theory of quantum mechanics. The same theory lead to the prediction of antiparticles.

In order to make the Schr¨odinger equation consistent with the relativistic relation E 2 = c2 p2 + m2 c4 and to allow for the conservation of angular momentum in a central potential, V (r), Dirac found that he had to allow the electron an extra ˆ such that the total angular momentum, angular momentum, the spin S, ˆ + S, ˆ is a constant in a central field of force.

Jˆ = L Dirac’s theory (1930) also predicted that the allowed values of the total relativistic energy for a free electron are: q

E = ± c2 p2 + m2 c4 ,

(31)

where c is the speed of light, p and m the momentum and rest mass of the electron, respectively.

If these negative energy states do exist, what prevents (E > 0) electrons from jumping to such lower energy states ? Dirac removed this difficulty by proposing that the negative energy states are usually full, so that the Pauli Exclusion Principle (see next section) prevents further transitions.

The normal state of the vacuum then consists of an infinite density of negative energy electrons. If an electron with E < 0 is excited to a state with E > 0, the hole left behind (or absence of a negatively charged electron with negative mass and negative kinetic energy) would manifest itself as a particle with E > 0, charge +e, and mass equal to the mass of an electron. This antiparticle of the electron—the positron—was observed a few years later (Anderson 1932, Blackett and Occhialini 1933).

The excitation of an electron from a state with E < 0 to one with E > 0 state is known as e− –e+ pair production; it is an endothermic process requiring E ≥ 2mc2 . The reverse is known as annihilation This is an exothermic process releasing energy, most commonly through the emission of two γ-rays with E ≥ 511 keV each.