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Chem 4502 & Spectroscopy (Jason Goodpaster) Fig. 1.7 p. 13 Atom Chapter 6: Hydrogen Atom (Lectures 19) Lecture 19: Review Chap. 1 Sect. 4, 5, 8: Rydberg formula, Chap. 6, Sect. 1,2 : Schrödinger equation for the H atom Separate into radial and angular parts: Ψ(r,θ,φ) = R(r) Θ(θ) Φ(φ) Angular solutions (spherical harmonics) are also the RR wave functions. ˆL 2 ; ℓ For the H atom (as for the RR), the magnitude of the angular momentum is well-defined.

The observed emitted (lines) were found to fit the Rydberg formula: -1 2 2 ν = 1 / λ (cm) = 109,680 cm ( 1 / n1 – 1 / n2 )

1 where n1 and n2 are integers (n2 > n1) 2

Bohrs Model of the H Atom Bohr Model of the H Atom Total energy (E) of the in the H atom: 2 2 E = ½ mev + (-e / (4πεo r)) • The electron can orbit around the nucleus only at kinetic potential

discrete values of r, the orbital radius. 2 Same as: E = -e / (8πεo r) 2 2 2 r • These stationary orbits must Substitute: r = εoh n / π m e (from quantized angular mom.) satisfy the condition: 4 2 2 2 Obtain: E = [ - m e / (8 εo h ) ] / n quantum 2 mev r = n ℏ where n = 1, 2, 3,.. number Quantized energy levels; energies go as 1/n where n (quantum number) is an integer (1, 2, 3, ...) ( me v r ) is the angular momentum (L) of the electron - (with me = e mass, v = velocity, r = orbital radius). Recall: Rydberg Formula for H atom ∼ -1 2 2 ν = 1 / λ (cm) = 109,680 cm ( 1 / n1 – 1 / n2 ) Thus, Bohr assumed that the angular momentum empirical formula based on measurements is quantized. 3 4 Bohr Model of the H Atom Fig. 1.10 p. 21 Bohr Model of the Hydrogen Atom

4 2 2 2 -1 Bohr prediction: E = [ - m e / (8 εo h ) ] / n n=∞, 0 cm Evaluate the constants in [brackets] (use m = 9.1044 x 10-31 kg, reduced mass of electron and ) n=2, -109,680/4 cm-1 -18 2 2 E = - 2.18 x 10 J / n = - 13.6 eV / n r2 = 0.529*4=2.12 Å I.e., 13.6/n2 eV is the energy (with respect to that of the Ionization - separated e and nucleus) by which an electron in a state with potential of H quantum number n is bound to the nucleus. atom: 13.6 eV Convert to (and use more sig figs): n=1, -1 Bohr: E / hc = ν∼ = - 109,680 cm-1 / n2 -109,680 cm SAME # !! (-13.6 eV) Recall: Rydberg Formula r = 0.529 -1 2 2 1 Å ν = 1 / λ (cm) = 109,680 cm ( 1 / n1 – 1 / n2 ) 5 6

Bohr Model of the H Atom H atom: Schrödinger equation

Despite its success in predicting the H atom emission Ĥ Ψ = E Ψ Ĥ = Kˆ + Vˆ Kinetic + Potential spectrum, the Bohr model had major problems: Must set up as 3-D problem (not 2-D as in Bohr model). ☹ did not predict the effects of external Planar (in an inherently 3-D system) would violate magnetic fields on the H atom (e.g., z=0, pz=0 so ∆z ∆pz=0).

☹ could not be successfully extended to atoms Kinetic energy operator: (same as for rigid rotator) with more than 1 electron µ is reduced mass Kˆ of e- and proton ☹ incompatible with uncertainty principle (text uses m, mass of electron) ∇2 LaPlacian Operator in Cartesian coordinates In Chap. 6, we will revisit the H atom spectrum

using the Schrödinger equation. 7 8 H atom: Schrödinger equation H atom: Schrödinger equation

Potential energy operator: Now the S. equation is: ˆ V(r) = -e2 / (4πε r) Coulomb attraction between electron o (charge -e) and proton (charge +e) r separated by distance r 1.11 x 10-10 C2 J-1 m-1 4πεo - permittivity (SI units)

Convenient to use r rather than (x2+y2+z2)½ Is this equation separable? equation # in text In Cartesian coordinates, S. equation isnt separable. Ĥ(r,θ,φ ) Ĥ(r) + Ĥ(θ) + Ĥ(φ) see pp. 192-3 So, must have ∇2 in spherical coordinates also: Therefore, we can write: Ψ(r,θ,φ) = R(r) Θ(θ) Φ(φ)

In contrast to the particle-in-a-cube, here the three 1-D wave functions have different functional forms. 9 10

Compare: separation constant H atom: Schrödinger equation H atom (θ, φ) β First, separate the S. equation into 2 parts: Radial equation: β = separation constant Rigid Rotor β - ĤRR ERR These 2 eigenvalue equations differ only by a constant. Solutions give R(r), the radial part of the So, they have the same eigenfunctions, Y(θ,φ) and quantum number n, which determines the total energy. (the spherical harmonics).

Angular part: same as above 2 Their eigenvalues are related: β = (2I / ћ ) ERR We already know the energies (E ) of the rigid rotator: β RR 2 2 ERR = ћ J (J+1) where I = µr moment of inertia Solutions give Y(θ,φ) = Θ(θ) Φ(φ), the angular part of the wave function, 2I J = 0, 1, 2, ... and quantum numbers ℓ and m, which determine the angular momentum. So, β = J (J+1) Ψ(r,θ,φ) = R(r) Y(θ,φ) For the H atom, we use ℓ instead of J : β = ℓ(ℓ +1) 11 12 We had: ℓ(ℓ +1) H atom H atom (θ, φ) β We had: Rigid Rotor ˆL 2 Y(θ,φ) = ћ2 ℓ (ℓ +1) Y(θ,φ) ĤRR ERR (r, , ) = R(r) Y( , ) ^ 2 operator for square of angular momentum For the H atom, Ψ θ φ θ φ Also recall: L = 2 I ĤRR Is Ψ(r,θ,φ) also an eigenfunction of ˆL 2 ?

To ^L 2 (θ,φ), R(r) is like a constant - no θ,φ dependence.

ˆ 2 2 2 So, Y(θ,φ) are also eigenfunctions of L with eigenvalues: ˆL R(r) Y(θ,φ) = ћ ℓ (ℓ +1) R(r) Y(θ,φ) 2 L = 2I ERR So, for the H atom wave functions (as for the RR), We had: E = ( ћ2 / 2I ) ℓ (ℓ +1) the magnitude of the angular momentum is well-defined: RR → → = length of L So: ˆL 2 Y(θ,φ) = ћ2 ℓ (ℓ +1) Y(θ,φ) │L│ = √ℓ (ℓ +1) ћ ang. mom. vector 13 14

H atom We had: ℓ(ℓ +1) Chapter 6: Hydrogen Atom Ψ(r,θ,φ) = R(r) Θ(θ) Φ(φ) H atom (θ, φ) β

Lecture 20: Chap. 6, Sect. 2, 3 To solve, separate θ and φ : Y(θ,φ) = Θ(θ) Φ(φ) Angular parts of the H atom wave function: Φ(φ) Lˆ z operator; quantum number m; spatial quantization Angular momentum vector diagrams separation Projection of the angular momentum on the z axis (i.e., any space-fixed axis) is constant well-defined m2

Lecture 21: Chap. 6, Sections 2, 3, 6 The other angular part of the H atom wave function: Θ(θ) Spherical harmonics + m2 Φ(φ) = 0 Our favorite diff eq ! Angular parts of the s, p, d orbitals px, py orbitals as linear combinations of the spherical harmonics

15 16 U(x,t) = Ψ(x) T(t) Classical Wave Equation Particle-in-a-Box (1 Dimension)

Solve the S. equation to obtain Ψ(x) inside the box. Therefore, k must be negative. Look at time-dependent part: positive

2 2 Let ω = - k v Then, 0 Ψ(x) = 0 Again, general solution is: k2 Same as: Solutions:

17 18

H atom H atom Ψ(r,θ,φ) = R(r) Θ(θ) Φ(φ) ℓ(ℓ +1) We had: m2 = H atom (θ, φ) β We had: separation constant To solve, separate and : Y( , ) = ( ) ( ) θ φ θ φ Θ θ Φ φ Apply boundary condition:

separation Φ(φ) must be single-valued, so constant Φ(φ) = Φ(φ+2π) m2 imφ im(φ+2π) imφ i2πm Φ(φ) = Ame = Ame = Ame e φ So this requires ei2πm = 1 2 + m Φ(φ) = 0 Our favorite diff eq ! -imφ -im(φ+2π) -imφ -i2πm Φ(φ) = A-me = A-me = A-me e requires e-i2πm = 1 Solutions: 1 0 Recall Eulers formula: e ± i 2πm = cos(2πm) ± i sin(2πm) Here, leave in complex exponential form. So, ± i 2πm 19 e = 1 requires m = 0, ± 1, ± 2, ... 20 H atom Ψ(r,θ,φ) = R(r) Θ(θ) Φ(φ) Differential volume element in spherical coordinates We had: dV = (r sinθ dφ) (r dθ) dr = r2 sinθ dr dθ dφ where m = 0, ± 1, ± 2, ... quantum number

Since e i (+m) φ = e –i (-m) φ, can write solutions as:

Evaluate normalization constant:

φ=2π ∫ Φm* Φm d-what? = 1 Math φ=0 Chapter D

What to use for φ-dependent part of 3-D volume element ? Fig. D.2 p. 148 21 22

H atom Ψ(r,θ,φ) = R(r) Θ(θ) Φ(φ) H atom Ψ(r,θ,φ) = R(r) Θ(θ) Φ(φ) We had: We have seen that : Y(θ,φ)

solving the angular part of the Schrödinger equation Evaluate normalization constant: yields two quantum numbers: ℓ and m

φ=2π ℓ determines the magnitude of the angular momentum ∫ Φm* Φm dφ = 1 φ=0 → → │L│ = √ℓ (ℓ +1) ћ L =2 φ π ˆ 2 2 (A * e- imφ) (A e imφ) dφ = 1 L R(r) Y(θ,φ) = ћ ℓ (ℓ +1) R(r) Y(θ,φ) ∫ m m 2 π φ=0 φ=2π φ=2π A * A e- imφ e imφ d = 1 A * A m m ∫ φ m m ∫ dφ = 1 What is the meaning of quantum number m ? φ=0 φ=0 e0=1

½ Am = 1/(2 π) 23 24 H atom Ψ(r,θ,φ) = R(r) Θ(θ) Φ(φ) H atom Ψ(r,θ,φ) = R(r) Θ(θ) Φ(φ) Y(θ,φ) Y(θ,φ) What is the meaning of quantum number m ? ^ Are Φm(φ) eigenfunctions of the Lz operator ? ^ Clue Φm(φ) are eigenfunctions of the Lz operator If so, what are the corresponding eigenvalues?

Recall: (MathChapter C pp. 111-112) We had: → → → L = r x p ^ ½ imφ In terms of Cartesian components, classically: Lz Φm(φ) = (- i ћ ∂/∂φ ) 1/(2π) e → → → → L = (y pz - z py) i + (z px - x pz) j + (x py - y px) k = - i ћ (im) 1/(2π)½ eimφ unit vector along z QM Operator corresponding to Lz : ^ Lz Φm(φ) = m ћ Φm(φ)

^ are eigenfunctions of Lz Convert to spherical coordinates: (Problem 6-11) with eigenvalues m ћ

25 26

H atom Ψ(r,θ,φ) = R(r) Θ(θ) Φ(φ) H atom Ψ(r,θ,φ) = R(r) Θ(θ) Φ(φ) Y(θ,φ) Y(θ,φ) ^ We ^ We had: Each Φm(φ) is an eigenfunction of Lz , had: Ψ(r,θ,φ) is an eigenfunction of Lz , with eigenvalue mћ. with eigenvalue mћ. This means that Are the 3-D H atom wavefunctions Ψ(r,θ,φ) also the projection of the angular momentum on the z-axis ^ eigenfunctions of Lz ? is well-defined, and has value m ћ.

Yes: Since , and R(r) Θ(θ) has no φ-dependence Example: for the Y(θ,φ) state with ℓ = 1, m = 1: The length of the angular momentum vector is ^ → Lz Ψ(r,θ,φ) = m ћ Ψ(r,θ,φ) │L│ = √ℓ (ℓ +1) ћ = √2 ћ

And its projection on the z axis is: Lz = m ћ = ћ The z axis is any space-fixed axis. (We call it z rather ^ than x or y because mathematical form of Lz is simpler.) 27 28 H atom Ψ(r,θ,φ) = R(r) Θ(θ) Φ(φ) H atom Ψ(r,θ,φ) = R(r) Θ(θ) Φ(φ) Y(θ,φ) Y(θ,φ) We We had: For the Y(θ,φ) state with ℓ = 1, m = 1: had: For the Y(θ,φ) state with ℓ = 1, m = 1: → → │L│ = √ℓ (ℓ +1) ћ = √2 ћ Lz = m ћ = ћ │L│ = √ℓ (ℓ +1) ћ = √2 ћ Lz = m ћ = ћ Draw the corresponding angular momentum vector diagram. Draw the corresponding angular momentum vector diagram.

The tip of the angular First draft: z What is θ ? momentum vector z ћ 45o = cos-1 (1/√2) can be anywhere on ℓ = 1 ℓ = 1 ћ the circle, consistent What is φ, the with a complete θ m = 1 m = 1 y angle in the x-y uncertainty in φ. plane? y According to the uncertainty x What about Lx and Ly? principle, since Lz is exactly known, φ is completely uncertain. x 29 30

H atom Ψ(r,θ,φ) = R(r) Θ(θ) Φ(φ) H atom Ψ(r,θ,φ) = R(r) Θ(θ) Φ(φ) Y(θ,φ) Y(θ,φ) We We had: had: For the Y(θ,φ) state with ℓ = 1, m = 1: → │L│ = √ℓ (ℓ +1) ћ = √2 ћ Lz = m ћ = ћ Draw the corresponding angular momentum vector diagram.

What about Lx and Ly? z If Lz is known exactly, ћ ℓ = 1 Lx and Ly are uncertain. m = 1 Average values: Do these commute and what is the residual: y = = 0 ^ ^ ^ No two projections (Lx, Ly, Lz) commute. ? x ^L 2 commutes with ^L , ^L and ^L . x y z 31 32 H atom Ψ(r,θ,φ) = R(r) Θ(θ) Φ(φ) H atom Ψ(r,θ,φ) = R(r) Θ(θ) Φ(φ) Y(θ,φ) Y(θ,φ) We had: We had:

33 34

H atom Ψ(r,θ,φ) = R(r) Θ(θ) Φ(φ) H atom Ψ(r,θ,φ) = R(r) Θ(θ) Φ(φ) Y(θ,φ) Y(θ,φ) We had: We had:

Do these commute and what is the residual:

?

35 36 H atom Ψ(r,θ,φ) = R(r) Θ(θ) Φ(φ) H atom Ψ(r,θ,φ) = R(r) Θ(θ) Φ(φ) Y(θ,φ) Y(θ,φ) We had: We had:

37 38

H atom Ψ(r,θ,φ) = R(r) Θ(θ) Φ(φ) H atom Ψ(r,θ,φ) = R(r) Θ(θ) Φ(φ) Y(θ,φ) Y(θ,φ) We had: We had:

39 40 H atom Ψ(r,θ,φ) = R(r) Θ(θ) Φ(φ) H atom Ψ(r,θ,φ) = R(r) Θ(θ) Φ(φ) Y(θ,φ) Y(θ,φ) We had: We had:

41 42

H atom Ψ(r,θ,φ) = R(r) Θ(θ) Φ(φ) H atom Ψ(r,θ,φ) = R(r) Θ(θ) Φ(φ)

Y(θ,φ) We had: Y(θ,φ) We had: ± ℓ

ℓ(ℓ +1) H atom Schrödinger β equation (θ, φ)

When the angular part of the H atom Schrödinger equation is solved completely, for solutions to remain finite, require: m = 0, ±1, ±2, ... ± ℓ So, each ℓ state has 2 ℓ+1 different m components.

43 44 H atom Ψ(r,θ,φ) = R(r) Θ(θ) Φ(φ) H atom Ψ(r,θ,φ) = R(r) Θ(θ) Φ(φ)

We had: m = 0, ±1, ±2, ... ± ℓ Y(θ,φ) Y(θ,φ) So, each ℓ state has 2 ℓ+1 different m components. For the Y(θ,φ) state with ℓ = 1, m can be 0, 1 or -1. → │L│ = √ℓ (ℓ +1) ћ = √2 ћ Lz = m ћ = 0, ћ, -ћ Angular momentum labels for H wave functions (orbitals): Draw a diagram including all 3 angular momentum vectors. ℓ # of m states (2 ℓ+1) 0 s 1 ℓ = 1 z 1 p 3 ћ m=1 2 d 5 f 7 3 m=0 y

-ћ m = -1

45 46

H atom Ψ(r,θ,φ) = R(r) Θ(θ) Φ(φ) REVIEW: Postulates of Y(θ,φ) Postulate #3: In any (ideal) experimental measurement of For the Y(θ,φ) state with ℓ = 1, m can be 0, 1 or -1. → the property associated with operator Â, │L│ = √ℓ (ℓ +1) ћ = √2 ћ Lz = m ћ = 0, ћ, -ћ the only values that can be observed are the eigenvalues a, which satisfy: Â Ψ = a Ψ. Can the angular ℓ = 1 z momentum vector ever be Example: for the 1-D PIB of length a, Ĥ Ψn = En Ψn ћ m=1 aligned along the z axis? where the eigenfunctions Ψn = No: maximum m = ℓ of Ĥ are: m=0 y √ℓ (ℓ +1) > ℓ So for the state with n=3: │L│ = Lz would violate -ћ m = -1 Uncertainty Principle; would imply planar motion. will be the measured energy. 47 48 H atom Ψ(r,θ,φ) = R(r) Θ(θ) Φ(φ) Chapter 6: Hydrogen Atom (r, , ) = R(r) ( ) ( ) Y(θ,φ) Ψ θ φ Θ θ Φ φ

For the Y(θ,φ) state with ℓ = 1, m can be 0, 1 or -1. → Lecture 21: Chap. 6, Sections 2, 3, 6 │L│ = √ℓ (ℓ +1) ћ = √2 ћ Lz = m ћ = 0, ћ, -ћ The other angular part of the H atom wave function: Θ(θ) Spherical harmonics Angular parts of the s, p, d orbitals ℓ = 1 z Spatial Quantization px, py orbitals as linear combinations of the spherical harmonics

For a given ideal ћ m=1 Lecture 22: Chap. 6, Sections 4, 5 measurement H atom wave functions: combine angular and radial parts regardless of the Calculate for the wave function (1s orbital): m=0 probabilities y direction of the z axis radial distribution function there are only 2ℓ +1 average radius (r) at which the electron will be found most probable value of r possible values of Lz -ћ m = -1 value of r within which there is a 90% probability to find the electron Tunneling

49 50

H atom Ψ(r,θ,φ) = R(r) Θ(θ) Φ(φ) H atom Ψ(r,θ,φ) = R(r) Θ(θ) Φ(φ) Y(θ,φ) Y(θ,φ) We had: ℓ(ℓ +1) Spherical harmonics: H atom (θ, φ) β m |m | imφ page 198 Yℓ (θ,φ) = N Pℓ e where: s To solve, separate θ and φ : Y(θ,φ) = Θ(θ) Φ(φ) |m | Pℓ = separation associated Legendre function; constant function of cos θ; m2 depends on ℓ, |m | N = normalization constant depends on ℓ, |m |

0 = ½ Y0 1/(4π) Spherically symmetric; no angular mom.; s orbital (when multiply by R(r)) What about Θ(θ) ?

51 52 H atom Ψ(r,θ,φ) = R(r) Θ(θ) Φ(φ) H atom Ψ(r,θ,φ) = R(r) Θ(θ) Φ(φ) Y(θ,φ) Y(θ,φ) Spherical harmonics: Spherical harmonics:

m |m | imφ page 198 m |m | imφ page 198 Yℓ (θ,φ) = N Pℓ e Yℓ (θ,φ) = N Pℓ e

pz pz Homework problem: 0 Phase of the Y1 wave Draw the angular part function (= sign of cos θ): of a pz orbital + for 0 < θ < π/2 Recall: in spherical coords, - for π/2 < θ < π z = r cos θ z z (θ is angle from z axis) pz Doesnt affect probabilities or average pz values for a given wave function. Tangent spheres. + + Crucial when combining wave functions Angular node at θ = π/2 (i.e., in x-y plane) - (e.g., add atomic orbitals to form - So, how does the e- get from one side to the other? bonding and anti-bonding MOs) Wave function replaces idea of particles trajectory Superposition principle; interference 53 54

H atom Ψ(r,θ,φ) = R(r) Θ(θ) Φ(φ) Recall from Chap 4: Postulates of Quantum Mechanics Y(θ,φ) For a linear operator, if 2 eigenfunctions have the same Spherical harmonics: page 198 eigenvalue (are degenerate ), m |m | imφ Yℓ (θ,φ) = N Pℓ e then any linear combination of those eigenfunctions is also an eigenfunction, with the same eigenvalue. Where are the familiar Proof: Assume  f (x) = a f (x) and  f (x) = a f (x) p and p functions? 1 1 2 2 x y If  is linear, we also have: Recall:  (c1 f1(x) + c2 f2(x)) = c1  f1(x) + c2  f2(x) x = r sin θ cos φ = y = r sin θ sin φ c1 a f1(x) + c2 a f2(x)

z = r cos θ Â (c1 f1(x) + c2 f2(x)) = a ( c1 f1(x) + c2 f2(x) )

Eulers Formula: 1 -1 A linear combination of Y1 and Y1 will be an eigenfunction e ± iφ = cos φ ± i sin φ of L2, with the same magnitude of the angular momentum.

55 56 m |m | imφ H atom Ψ(r,θ,φ) = R(r) Θ(θ) Φ(φ) m |m | imφ H atom Ψ(r,θ,φ) = R(r) Θ(θ) Φ(φ) Yℓ (θ,φ) = N Pℓ e Yℓ (θ,φ) = N Pℓ e Y(θ,φ) Y(θ,φ) x = r sin cos x = r sin cos θ φ page 198 θ φ page 198 y = r sin θ sin φ y = r sin θ sin φ z = r cos θ z = r cos θ Construct the angular part of the p orbital Construct the x 1 -1 ½ iφ -iφ 1 -1 ½ iφ -iφ from a linear Y1 + Y1 = (3/8π) sin θ (e + e ) angular part of the Y1 - Y1 = (3/8π) sin θ (e - e ) combination of Y 1 1 py orbital from a -1 . cos + i sin + (cos - i sin ) cos + i sin - (cos - i sin ) and Y1 φ φ φ φ linear combination φ φ φ φ Eulers Formula: = 2 cos φ 1 -1 . = 2i sin φ of Y1 and Y1 Normalize: (1/2)½ (Y 1 + Y -1 ) Normalize: 1/(2½ i ) ( Y 1 - Y -1 ) e ± iφ = cos φ ± i sin φ 1 1 1 1 = (1/2)½ (3/8π)½ 2 sin θ cos φ = 1/(2½ i ) (3/8π)½ 2 i sin θ sin φ = (3/4π)½ sin θ cos φ = (3/4π)½ sin θ sin φ

px orbital (angular part) py orbital (angular part) 57 58

m |m | imφ H atom Ψ(r,θ,φ) = R(r) Y( , ) m |m | imφ H atom Ψ(r,θ,φ) = R(r) Y( , ) Yℓ (θ,φ) = N Pℓ e θ φ Yℓ (θ,φ) = N Pℓ e θ φ page 198 page 198 x = r sin θ cos φ Angular parts of d orbitals y = r sin θ sin φ z = r cos θ

d 2 Angular parts of p orbitals: z

½ 1 -1 ½ px (1/2) (Y1 + Y1 ) = (3/4π) sin θ cos φ ½ 1 -1 ½ py 1/(2 i ) ( Y1 - Y1 ) = (3/4π) sin θ sin φ p Y 0 = (3/4 )½ cos z 1 π θ # angular = ℓ nodes Which are eigenfunctions of the following operators: ^ˆ 2 p , p , p ^ L ? x y z Lz ? pz only s 0 0 ½ 1 -1 p 1 1 dxz (1/2) (Y2 + Y2 ) For experiments in which Lz (and quantum number m) 1 -1 d 2 2 affects observed properties, use Y1 or Y1 . etc. (p. 217) 59 60

REVIEW Chapter 6: Hydrogen Atom H atom: Schrödinger equation

Now the S. equation is: Lecture 22: Chap. 6, Sections 4, 5 H atom wave functions: combine angular and radial parts Calculate for the ground state wave function (1s orbital): probabilities radial distribution function average radius (r) at which the electron will be found - most probable value of r value of r within which there is a 90% probability to find the electron Is this equation separable? Tunneling

Ĥ(r,θ,φ ) Ĥ(r) + Ĥ(θ) + Ĥ(φ) Lecture 23: Chap. 6, Sections 4 - 6 see pp. 192-3 H atom: energy levels, nodes, degeneracies Radial wave functions for orbitals with angular momentum (ℓ > 0), e.g., p orbitals Therefore, we can write: Ψ(r,θ,φ) = R(r) Θ(θ) Φ(φ) Plot square of radial wave function: R(r)2 Plot radial distribution function: r2R(r)2 Combine radial and angular wave functions: In contrast to the particle-in-a-cube, here the three 1-D Probability density plots for H atom orbitals wave functions have different functional forms. 61 62

H atom REVIEW H atom: Schrödinger equation

First, separate the S. equation into 2 parts: Ground state (1s orbital):

Radial equation: β = separation constant = ℓ(ℓ +1) 0 = ½ Y0 1/(4π) Spherically symmetric; no angular mom.; no angular nodes β -

3/2 -r/ao R(r) = (2/ao ) e where ao = Bohr radius Solutions give R(r), the radial part of the wave function and quantum number n, which determines the total energy.

½ 3/2 -r/ao Ψ(r,θ,φ) = R(r) Y(θ,φ) = 1/(4π) (2/ao ) e Angular part: ℓ(ℓ +1) ½ 3/2 -r/a (r, , ) = (1/ a ) e o 1s orbital Ψ θ φ π o β

Solutions give Y(θ,φ) = Θ(θ) Φ(φ), the angular part of the wave function, Normalization and quantum numbers ℓ and m, which determine the angular momentum. constant

Ψ(r,θ,φ) = R(r) Y(θ,φ) 63 64 H atom: Ground state (1s) Ψ(r,θ,φ) = R(r) Y(θ,φ) H atom: Ground state (1s) Ψ(r,θ,φ) = R(r) Y(θ,φ)

½ 3/2 -r/ao ½ 3/2 -r/ao Ψ1s = (1/ π ao ) e 1s orbital Ψ1s = (1/ π ao ) e 1s orbital

2 Plot Ψ1s and Ψ 1s vs. r :

Decaying exponential. Some probability of finding Ψ1s Ψ1s Spherically symmetric. electron far from nucleus

(ao = 0.529 Å) (at r >> a0).

Ψ2 2 Ψ2 1s Ψ 1s : probability per unit volume 1s No well-defined orbital radius of finding the electron a distance as in the Bohr model. r from the nucleus. Maximum probability density at the nucleus. 65 66

½ 3/2 -r/ao ½ 3/2 -r/ao H atom: Ground state (1s) Ψ1s = (1/ π ao ) e H atom: Ground state (1s) Ψ1s = (1/ π ao ) e

What is the probability of finding the electron between r Graph the radial distribution function for the 1s orbital. and (r+dr) from the nucleus... P(r) = Ψ2 4πr2

...that is, within a thin spherical Probability is: shell of radius r, thickness dr, 2 2 area 4πr2 and volume 4πr2dr. Ψ 4πr dr P1s(r) ao r Radial distribution function

P (r) = 0 at nucleus (volume element → 0). P(r) = Ψ2 4πr2 1s As r increases, dV increases as r2 and compensates for decaying exponential,

so P1s(r) rises.

-2r/a At larger r, e o dominates and P(r) → 0. Fig. D3 p. 149 Fig. D3 p. 149 67 68 ½ 3/2 -r/ao ½ 3/2 -r/ao H atom: Ground state (1s) Ψ1s = (1/ π ao ) e H atom: Ground state (1s) Ψ1s = (1/π ao ) e a What is the most probable value of r ? ao What is the average value of r ? o

P1s(r) P1s(r) 3/2 ao Method: Set derivative of radial r r distribution function = 0 P(r) = Ψ2 4πr2 Recall: dV = 4 π r2 dr for a spherically symmetric orbital

Obtain (as in homework)

= 3/2 ao

=0 The most probable distance between electron and nucleus in the H atom ground state is the Bohr radius, ao (0.529 Ǻ). 69 70

½ 3/2 -r/a H atom: Ground state (1s) Ψ = (1/ π a ) e o 1s o Tunneling (e.g., 1s orbital) ao ao What is the radius within which P (r) 2.7 ao 2.7 a In terms of a (the Bohr radius), 1s there is a 90% probability to find the P1s(r) o o 1.5 a a = 2 µ e2 o electron? 1.5 ao o εo h / π r r

2 2 2ao En = - (1/n ) (e / (4πεo ) 2ao) classically forbidden 2 region (for V(r) = - e / (4πεo r ) 1s orbital) Evaluating integral and So for n=1 (ground state), the classical turning solving for r iteratively (as in point occurs at r = 2ao

homework) gives 2.7 ao = 1.4 Ǻ The 90% definition is Yet, electron has a substantial probability of being often used when plotting outside this region (∼20% in the 1s orbital). orbital contours. 71 72 H atom

When solve the H atom Schrödinger equation in general: Lecture 23: Hydrogen Atom Chap. 6, Sections 4 - 6 Allowed energies for all states (not just s orbitals) are: H atom: energy levels, nodes, degeneracies 2 4 2 2 Radial wave functions for orbitals with angular momentum (ℓ > 0); En = - (1/n ) (µ e / (8 εo h )) n = 1, 2, 3... e.g., p orbitals Plot square of radial wave function: R(r)2 same as in the Bohr model! principal quantum # Plot radial distribution function: r2R(r)2 Combine radial and angular wave functions: 2 2 Probability density plots for H atom orbitals In terms of ao (the Bohr radius), ao = εo h / π µ e

E = - (1/n2) (e2 / (4πε ) 2a ) Lecture 24: Chapter 7, Approximation Methods n o o Chap. 7.1 Variational Method Energy depends only on quantum number n

(not on ℓ or m) in absence of external field.

In H atom, e.g., 2s and 2p orbitals (n=2): same energy.

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H atom H atom

Degeneracy total # of nodes in wave function n-1 # of different states with the same energy

(sum of radial and angular nodes) possible values of ℓ ℓ = 0, 1, 2, ... , n-1

possible values of m m = 0, ±1, ±2, ..., ± ℓ # of angular nodes ℓ

Degeneracy (g) of states with a given n maximum value of quantum # ℓ n-1 no radial nodes in the absence of an external electric or magnetic field

minimum value of quantum # ℓ 0 ℓ=n-1 no angular nodes g = ∑ (2 ℓ + 1) = 1 + 3 + 5 + ... + 2n-1 = n2 n-1 radial nodes ℓ=0 What is the degeneracy of the n=3 ? 9 # of radial nodes n - ℓ - 1 What are all of the ℓ=0, m=0 ℓ=1, m=0 or -1 or +1 R(r) depends on quantum numbers n and ℓ ℓ, m states? ℓ=2, m=0 or -1 or +1 or -2 or +2 75 76 H atom H atom

Radial wave functions: (function of) Radial wave functions: (function of)

N = N = Normalization Normalization constant Laguerre function constant Laguerre function ( polynomial in r ) ( polynomial in r )

1 3/2 -r/ao For n=1, ℓ=0: L1 = -1 R1,0(r) = (2/ao ) e 1 For n=2, ℓ=0: L2 = - 4 + 2r/ao 2s orbital

½ 3/2 -r/ao Recall, we had: Ψ1s = (1/ π ao ) e

Ψ1s(r,θ,φ) = R(r) Y(θ,φ) decays more slowly than 1s; (1/4π)½ radial node 2s orbital more expanded at r = 2 ao 77 78

H atom REVIEW H atom: Schrödinger equation Radial wave functions: (function of) First, separate the S. equation into 2 parts: Radial equation: β = separation constant

ℓ(ℓ +1) N = - Normalization constant Laguerre function Solutions give R(r), the radial part of the wave function ( polynomial in r ) and quantum number n, which determines the total energy.

Angular part: ℓ(ℓ +1) order r n-1

β In general, the ns orbital has (n-1) radial nodes. What about states with ℓ > 0 ? Solutions give Y(θ,φ) = Θ(θ) Φ(φ), the angular part of the wave function, and quantum numbers ℓ and m, which determine the angular momentum. r ℓ → wave function will be zero at the nucleus Ψ(r,θ,φ) = R(r) Y(θ,φ) 79 80 H atom H atom Radial Equation Effective potential energy, V At small r, eff V repulsive centrifugal term ℓ(ℓ +1) dominates (V → ∞ as r → 0) - E R(r) eff electron is excluded from nucleus

repulsive (+) attractive (-) centrifugal term Coulomb term At large r, attractive Coulomb term dominates (as r → ∞, 1/r2 → 0 faster than 1/r) Classical picture (as for skaters arms): centrifugal effect tends to fling the electron away from Expect R(r) for ℓ =0 and ℓ ≠ 0 to be the nucleus in a state with angular momentum (ℓ>0). quite different near nucleus but similar far away.

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H atom H atom (function of) Radial wave functions: 3 For n=2, ℓ=1: L3 = - 6 2p orbital

N = Normalization constant Laguerre function R(r) = 0 at nucleus ( polynomial in r ) no radial nodes

3 1 For n=2, ℓ=1: L3 = - 6 2p orbital For n=2, ℓ=0: L2 = - 4 + 2r/ao 2s orbital

decays more slowly than 1s; R(r) = 0 at nucleus radial node 2s orbital more expanded at r = 2 ao 83 84 Plot with θ, φ fixed. r2R2 Radial Distribution Function s orbitals: ℓ =0; no angular nodes Text p. 210 ∝

r/ao Large probability to find As n (and energy) increase, electron near nucleus. probability distributions become more expanded # radial nodes = n - 1 2

R(r) changes sign at node. R 2 For wave functions with no ) r ) 3

o radial nodes (1s, 2p, 3d, etc.) 2 p, d orbitals (etc.): maximum occurs at n ao (1/a in agreement with Bohr model. Small probability to find electron near nucleus.

ℓ > 0 = # angular nodes

# radial nodes = n - ℓ - 1

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Text p. 210 r2R2 ∝ Radial Distribution Function Electron in 2s orbital has higher probability of being close to nucleus than electron in 2p.

2 Thus, a 2s electron shields a R

2 2p electron from the nuclear

) r ) charge more than it shields 3 o another 2s electron. (1/a

As a result, the 2s orbital has a lower energy than the 2p in multielectron atoms.

( Ditto for 3s < 3p < 3d ), etc.

87 88 Ψ(r,θ,φ) = R(r) Θ(θ) Φ(φ) Table 6.6 p. 218 Hydrogenic (1-electron) atomic wave functions expressed as real functions. z Y(θ,φ) pz Z = , σ = Zr / ao 2 2 + where ao is the Bohr radius (4πεo! / (mee ) = 0.5292 Å) - Angular part of a pz orbital, 0 ½ Y1 = (3/4π) cos θ

Probability density plot of the overall wave function. 2pz 3pz Density of dots represents the probability of finding the electron. p. 214 89 90

-1 2 2 Figure 6.5 p. 214 En = -109,680 cm / n = - 13.6 eV / n Probability density plots for some hydrogen atomic orbitals n=∞, 0 cm-1

n=2, -109,680/4 cm-1

Ionization potential of H atom: 13.6 eV

n=1, -109,680 cm-1 (-13.6 eV)

rmp = 0.529 Å 91 92