Chapter 18 - The Hydrogen Atom - Grandinetti · 2020-04-02 · Chapter 18 The Hydrogen Atom P. J....

Chapter 18The Hydrogen Atom

P. J. Grandinetti

Chem. 4300

P. J. Grandinetti Chapter 18: The Hydrogen Atom

The Hydrogen Atom

Hydrogen atom is simplest atomic system where Schrödinger equation can be solvedanalytically and compared to experimental measurements.

Analytical solution serve as basis for obtaining approximate solutions for multi-electronatoms and molecules, where no analytical solution exists.

Warning: Working through analytical solution of H-atom may cause drowsiness. Do notoperate heavy machinery during this derivation.


Two Particle Problem

x

y

z2 particles with mass mp and me at rp and re.Total mass, center of mass, and inter-particle distance vector is

M = mp + me, R =mprp + mere

M, and r = re − rp

Express rp and re in terms of M, R and r as

rp = R −meM

r and re = R +mp

Mr

Express individual momenta of 2 particles as

pp = mpdrp

dtand pe = me

dredt

then in terms of M, R and r as

pp = mp

(dRdt

−meM

drdt

)and pe = me

(dRdt

+mp

Mdrdt

)P. J. Grandinetti Chapter 18: The Hydrogen Atom

Two Particle ProblemNext consider total energy

E =pp

2

2mp+

pe2

2me+ V(r)

V(r) depends only on distance between 2 particles.Write energy in terms of M, R and r and obtain

E = 12

M(

dRdt

)2

+ 12𝜇(

drdt

)2+ V(r)

𝜇 is reduced mass, given by1𝜇= 1

mp+ 1

me

Define 2 new momenta associated with center of mass and reduced mass,

pR = M dRdt, and pr = 𝜇dr

dtP. J. Grandinetti Chapter 18: The Hydrogen Atom

Two Particle Problem

Total energy becomes

E =pR

2

2M+

pr2

2𝜇+ V(r)

1st term is translational energy of center of mass

2nd term is kinetic energy due to relative motion of 2 particles

Translating to quantum mechanics we writetime independent Schrödinger equation for 2 particle system as[

pR2

2M+

pr2

2𝜇+ V(r)

]f (R, r) =

[− ℏ2

2M∇2

R − ℏ2

2𝜇∇2

r + V(r)]

f (R, r) = Ef (R, r)


Two Particle Problem[− ℏ2

2M∇2

R − ℏ2

2𝜇∇2

r + V(r)]

f (R, r) = Ef (R, r)

Wave function can be separated into product of two wave functions

f (R, r) = 𝜒(R)𝜓(r)

𝜒(R) depending only on center of mass𝜓(r) depending only on relative motion of 2 particlesSubstitute product into Schrödinger Eq above we obtain 2 wave equations

− ℏ2

2M∇2

R𝜒(R) = ER𝜒(R) and[−ℏ

2

2𝜇∇2

r + V(r)]𝜓(r) = Er𝜓(r)

whereE = ER + Er

On left is wave equation for translational motion of free particle of mass MOn right is wave equation for particle with mass 𝜇 in potential V(r)For electron bound positively charged nucleus we focus on PDE for 𝜓(r)


Schrödinger Equation in Spherical Coordinates

Focus on PDE for 𝜓(r): Electron bound to positively charged nucleus[−ℏ

2

2𝜇∇2 + V(r)

]𝜓(r) = E𝜓(r)

Coulomb potential is

V(r) = −Zq2

e4𝜋𝜖0r

Central potential with 1∕rdependence.V → −∞ as r → 0V → 0 as r → ∞


Schrödinger Equation in Spherical CoordinatesSince V(r) only depends on r we adopt spherical coordinates

−ℏ2

2𝜇

[𝜕2

𝜕r2 + 2r𝜕𝜕r

+ 1r2

(1

sin 𝜃𝜕𝜕𝜃

(sin 𝜃 𝜕

𝜕𝜃

)+ 1

sin2 𝜃𝜕2

𝜕𝜙2

)⏟⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏟⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏟

−L2∕ℏ2

]

⏟⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏟⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏟∇2

𝜓 + V(r)𝜓 = E𝜓

Take term in parentheses as −L2∕ℏ2, rearrange, and simplify to

r2 𝜕2𝜓(r, 𝜃, 𝜙)𝜕r2 + 2r

𝜕𝜓(r, 𝜃, 𝜙)𝜕r

+2𝜇r2

ℏ2

(E − V(r)

)𝜓(r, 𝜃, 𝜙) = 1

ℏ2 L2𝜓(r, 𝜃, 𝜙)

To solve PDE use separation of variables

𝜓(r, 𝜃, 𝜙) = R(r)Y(𝜃, 𝜙)


Schrödinger Equation in Spherical CoordinatesSubstitute 𝜓(r, 𝜃, 𝜙) into PDE and dividing both sides by 𝜓(r, 𝜃, 𝜙)

r2

R(r)𝜕2R(r)𝜕r2 + 2r

R(r)𝜕R(r)𝜕r

+2𝜇r2

ℏ2

(E − V(r)

)= 1

Y(𝜃, 𝜙)1ℏ2 L2Y(𝜃, 𝜙)

Left side depends only on rRight side depends only on 𝜃 and 𝜙.We have turned one PDE into two ODEs.

1 We know eigenfunctions and eigenvalues of L2 on right side,

L2Y𝓁,m(𝜃, 𝜙) = 𝓁(𝓁 + 1)ℏ2Y𝓁,m(𝜃, 𝜙)

2 Define 𝓁(𝓁 + 1) as separation constant and obtain ODE for radial part

r2

R(r)d2R(r)

dr2 + 2rR(r)

dR(r)dr

+2𝜇r2

ℏ2

(E − V(r)

)= 𝓁(𝓁 + 1)


Schrödinger Equation in Spherical CoordinatesExpanding and rearranging ODE for radial part we obtain[

−ℏ2

2𝜇

(d2

𝜕r2 + 2r

ddr

)+ ℏ2

2𝜇𝓁(𝓁 + 1)

r2 + V(r)⏟⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏟⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏟

Veff(r)

]R(r) = ER(r)

Looks like 1D time independent Schrödinger Eq with effective potential

Veff(r) =ℏ2

2𝜇𝓁(𝓁 + 1)

r2⏟⏞⏞⏞⏞⏟⏞⏞⏞⏞⏟

Centrifugal Term

−Zq2

e4𝜋𝜖0r

Total angular momentum,√𝓁(𝓁 + 1)ℏ, creates centrifugal force that pushes electron away

from nucleus.Notice 2 terms in Veff(r) always have opposite signs.


Veff(r): Effective Potential of H-Like AtomSwitching to Atomic Units: Convenient units for atomic physicsAtomic unit of length, also called Bohr radius, is defined as

a0 ≡ 4𝜋𝜖0ℏ2

q2eme

= 52.9177210526763 pm

Atomic unit of energy is

1Eh ≡ ℏ2

mea20

=q2

e4𝜋𝜖0a0

= 27.21138602818051 eV

Dividing effective potential by Eh, set Z = 1, 𝜇 ≈ me, and obtain

Veff(r)∕Eh = 𝓁(𝓁 + 1)2

me𝜇

a20

r2 −a0r

≈ 𝓁(𝓁 + 1)2𝜚2 − 1

𝜚

Define dimensionless radius as 𝜚 = r∕a0P. J. Grandinetti Chapter 18: The Hydrogen Atom

Plot of Veff for 𝓁 = 0, 1, 2, and 3.

20 4 6 8 10 12 14-0.4

-0.3

-0.2

-0.1

0.0

0.1

0.2

Note: potential minimum shifts to higher radii with increasing angular momentum (i.e., centrifugalforce).


Solving the Wave Equation for the Radial Part

[−ℏ

2

2𝜇

(d2

𝜕r2 + 2r

ddr

)+ ℏ2

2𝜇𝓁(𝓁 + 1)

r2 −Zq2

e4𝜋𝜖0r

]R(r) = ER(r)

pour yourself a third cup of coffee...


Solving the Radial Wave EquationTo obtain radial wave function, R(r), we must solve[

−ℏ2

2𝜇

(d2

𝜕r2 + 2r

ddr

)+ ℏ2

2𝜇𝓁(𝓁 + 1)

r2 −Zq2

e4𝜋𝜖0r

]R(r) = ER(r)

Begin by dividing and both sides by E and obtain

− ℏ2

2𝜇E

[d2

𝜕r2 + 2r

ddr

]R(r) +

(ℏ2

2𝜇E𝓁(𝓁 + 1)

r2 −Zq2

e4𝜋𝜖0rE

− 1

)R(r) = 0

Define𝜅2 ≡ −

2𝜇Eℏ2 or E = −ℏ

2𝜅2

2𝜇𝜅 is in wave numbers. Rearrange potential energy expression to

−Zq2

e4𝜋𝜖0rE

=Zq2

e4𝜋𝜖0r

2𝜇ℏ2𝜅2 =

Zq2e𝜇

2𝜋𝜖0ℏ2𝜅1𝜅r

=2𝜌0𝜅r

𝜌0 is a dimensionless quantity: 𝜌0 ≡ Zq2e𝜇

2𝜋𝜖0ℏ2𝜅P. J. Grandinetti Chapter 18: The Hydrogen Atom

Solving the Radial Wave EquationSubstitute expression for potential energy into ODE

1𝜅2

[d2

𝜕r2 + 2r

ddr

]R(r) −

(1 + 𝓁(𝓁 + 1)

(𝜅r)2+

2𝜌0𝜅r

)R(r) = 0

Define dimensionless variable: 𝜌 ≡ 2𝜅rRewrite radial part as function of 𝜌 as[

d2

d𝜌2 + 2𝜌

dd𝜌

]R(𝜌) −

(1 + 𝓁(𝓁 + 1)

𝜌2 +2𝜌0𝜌

)R(𝜌) = 0

Further simplify ODE by defining

u(𝜌) = 𝜌R(𝜌),du(𝜌)

d𝜌= 𝜌

dR(𝜌)d𝜌

+ R(𝜌),d2u(𝜌)

d𝜌2 = 𝜌d2R(𝜌)

d𝜌2 + 2dR(𝜌)

d𝜌

Re-express ODE asd2u(𝜌)

d𝜌2 −(

14−𝜌02𝜌

+ 𝓁(𝓁 + 1)𝜌2

)u(𝜌) = 0

𝜌 varies from 0 to ∞. 1st look at asymptotic solutions: (1) 𝜌→ ∞ and (2) 𝜌→ 0P. J. Grandinetti Chapter 18: The Hydrogen Atom

Solving the Radial Wave Equation, 𝜌→ ∞

Starting withd2u(𝜌)

d𝜌2 −(

14−𝜌02𝜌

+ 𝓁(𝓁 + 1)𝜌2

)u(𝜌) = 0

(1) 𝜌 → ∞. At large values of 𝜌 approximate ODE as

d2u(𝜌)d𝜌2 −

u(𝜌)4

≈ 0

Solutions to this ODE have formu(𝜌) ∼ Ae−𝜌∕2 + Be𝜌∕2

Reject positive exponent since require that solution be finite everywhere,

u(𝜌) ∼ Ae−𝜌∕2 in limit that 𝜌 → ∞


Solving the Radial Wave Equation, 𝜌→ 0Starting with

d2u(𝜌)d𝜌2 −

(14−𝜌02𝜌

+ 𝓁(𝓁 + 1)𝜌2

)u(𝜌) = 0.

(2) 𝜌 → 0. At small values of 𝜌 approximate ODE as

d2u(𝜌)d𝜌2 − 𝓁(𝓁 + 1)

𝜌2 u(𝜌) ≈ 0.

Solutions to this ODE have formu(𝜌) ∼ A𝜌𝓁+1 + B𝜌−𝓁 .

Reject B term again because require that solution be finite at 𝜌 = 0.

u(𝜌) ∼ A𝜌𝓁+1 in limit that 𝜌→ 0


Solving the Radial Wave Equation

u(𝜌) ∼ Ae−𝜌∕2 in limit that 𝜌 → ∞

u(𝜌) ∼ A𝜌𝓁+1 in limit that 𝜌 → 0

Now, back tod2u(𝜌)

d𝜌2 −(

14−𝜌02𝜌

+ 𝓁(𝓁 + 1)𝜌2

)u(𝜌) = 0

we propose general solution of form

u(𝜌) = 𝜌𝓁+1 L(𝜌) e−𝜌∕2

This has correct behavior in two limits.

Only need to determine L(𝜌) to get behavior for all 𝜌.


Solving the Radial Wave EquationPlug u(𝜌) = 𝜌𝓁+1 L(𝜌) e−𝜌∕2, where 𝜌 ≡ 2𝜅r, into

d2u(𝜌)d𝜌2 −

(14−𝜌02𝜌

+ 𝓁(𝓁 + 1)𝜌2

)u(𝜌) = 0

gives

𝜌d2L(𝜌)

d𝜌2 +dL(𝜌)

d𝜌(2(𝓁 + 1) − 𝜌) +

(𝜌02

− (𝓁 + 1))

L(𝜌) = 0

Set j = 𝜌0∕2 − 𝓁 − 1, and k = 2𝓁 + 1 and this ODE is recognized as

𝜌d2Lk

j (𝜌)

d𝜌2 + (k + 1 − 𝜌)dLk

j (𝜌)

d𝜌+ jLk

j (𝜌) = 0, associated Laguerre differential equation

Has nonsingular solutions only if j is non-negative integers, j = 0, 1, 2,…These solutions, Lk

j (𝜌), are called the associated Laguerre polynomials.P. J. Grandinetti Chapter 18: The Hydrogen Atom

Selected associated Laguerre polynomialsn 𝓁 L2𝓁+1

n−𝓁−1(𝜌) Polynomial Roots1 0 L1

0(𝜌) 1 -2 0 L1

1(𝜌) 2 − 𝜌 2 - -2 1 L3

0(𝜌) 1 - -3 0 L1

2(𝜌)12

(6 − 6𝜌 + 𝜌2) 1.26795 4.73205 -

3 1 L31(𝜌) 4 − 𝜌 4 - -

3 2 L50(𝜌) 1 - - -

4 0 L13(𝜌)

16

(24 − 36𝜌 + 12𝜌2 − 𝜌3) 7.75877 0.935822 3.30541

4 1 L32(𝜌)

12

(20 − 10𝜌 + 𝜌2) 2.76393 7.23607 -

4 2 L51(𝜌) 6 − 𝜌 6 - -

4 3 L70(𝜌) 1 - - -

5 0 L14(𝜌)

124

(120 − 240𝜌 + 120𝜌2 − 20𝜌3 + 𝜌4) 0.743292 2.57164 5.73118 10.9539

5 1 L33(𝜌)

16

(120 − 90𝜌 + 18𝜌2 − 𝜌3) 2.14122 5.31552 10.5433 -

5 2 L52(𝜌)

12

(42 − 14𝜌 + 𝜌2) 4.35425 9.64575 - -

5 3 L71(𝜌) 8 − 𝜌 8 - - -

5 4 L90(𝜌) 1 - - - -

Associated Laguerre polynomials following definition where Lkj (x) = (−1)k dk

dxk Lj+k(x).

Laguerre polynomial Lj(x) is defined by Rodrigues formula: Lj(x) =1n!

ex dj

dxj

(xje−x).


Solving the Radial Wave EquationSince 𝓁 = 0, 1, 2,… , and j = 0, 1, 2,… , then

𝜌02

= j + 𝓁 + 1

can only take on integer values of 𝜌0∕2 = 1, 2,….We define this as the principal quantum number

n ≡ 𝜌0∕2 = j + 𝓁 + 1,

which can only take on values of

n = 1, 2, 3,… .

Wave functions with same n value form set called a shell.Special letters are sometimes assigned to each n value

n = 1 2 3 4 5 ← numerical valueK L M N O ← symbol


Solving the Radial Wave Equation𝓁 is called the azimuthal quantum numberRearranging n = j + 𝓁 + 1 for 𝓁 gives

𝓁 = n − j − 1

and we find that 𝓁 cannot exceed n − 1 (since lowest value of j is zero).Range of 𝓁 is

𝓁 = 0,… , n − 1.

Recall that azimuthal quantum number, 𝓁, defines total angular momentum of√𝓁(𝓁 + 1)ℏ.

m𝓁 is called the magnetic quantum number.m𝓁 has positive and negative integer values between −𝓁 and 𝓁.When x, y, or z component of the electron’s angular momentum is measured only values ofm𝓁ℏ are observed.


Solving the Radial Wave EquationWave functions with same value of n and 𝓁 form set called a sub-shell.Special letters are assigned to sub-shell with given 𝓁 value,

𝓁 = 0 1 2 3 4 5 6 7 8 9 10 11 12 13 ← numerical values p d f g h i k l m o q r t ← symbol

and continue afterwards in alphabetical order.1st four letters arose in pre-quantum atomic spectroscopy for classifying emission lines and standfor sharp, principal, diffuse, and fine.Shorthand n𝓁 notation uses principal quantum number with 𝓁 symbol,Wave function with

▶ n = 1, 𝓁 = 0 is referred to as 1s state▶ n = 2, 𝓁 = 1 is referred to as 2p state

Number of roots of L2𝓁+1n−𝓁−1 is n − 𝓁 − 1. Thus, number of radial nodes is equal to n − 𝓁 − 1.

Recall for Spherical Harmonics that number of angular nodes is 𝓁

Thus, total number of nodes is (n − 𝓁 − 1) + 𝓁 = n − 1.P. J. Grandinetti Chapter 18: The Hydrogen Atom

Energy of the Hydrogen-Like Atom

Given the constraint that 𝜌0∕2 = n we go back to

𝜌0 ≡ Zq2e𝜇

2𝜋𝜖0ℏ2𝜅= 2n and rearrange to 𝜅n =

Zq2e𝜇

4𝜋𝜖0ℏ2n

From

𝜅2 ≡ −2𝜇Eℏ2 rearranges to En = −

ℏ2𝜅2n

2𝜇= −

Z2q4e𝜇

32𝜋2𝜖20ℏ

2n2= −

Z2q4e𝜇

8𝜖20h2n2

En = −Z2q4

e𝜇

8𝜖20h2n2

Energy of H-like atom


Energy of the Hydrogen-Like Atom

En = −Z2q4

e𝜇

8𝜖20h2n2

Energy of H-like atom

e− bound to nucleus with charge Z energyonly depends on n and ZIf Z increases (holding n constant) thenEn decreases—becomes more negative.Higher Z means nucleus holds e− moretightlyIn limit that n → ∞ then E → 0 and e− isunbound


Energy of the Hydrogen-Like AtomWith 𝜇 ≈ me, the n = 1 energy is called Rydberg unit of energy (Ry)

1Ry = −E1 =q4

eme

8𝜖20h2

= 13.60569301218355 eV

With 𝜇 ≈ me approximation

En = −1 Ry

n2 Energy of H atom

Can also divide by atomic unit of energy, Eh, to obtain

En = −1 Eh

2n2 Energy of H atom

Given n, energy is identical for each 𝓁. Given 𝓁 energy is identical for 2𝓁 + 1 values of m𝓁.

Degeneracy of nth energy level is gn =n−1∑𝓁=0

(2𝓁 + 1) = n + 2n−1∑𝓁=0

𝓁 = n + n(n − 1) = n2


Energy levels of single electron bound to protonorbitals

UV light emission“Lyman series

Visible light emissionBalmer series”

degeneracy


Energy of the Hydrogen-Like AtomRemember Balmer series for hydrogen emission spectra obeys relation:

1𝜆= RH

( 122 − 1

n2

)where RH = 1.097 × 107 /m

Followed by Bohr’s theory of atom which gave

1𝜆=(

14𝜋𝜖0

)2(

meq4e

4𝜋ℏ3c0

)Z2

⏟⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏟⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏟perfect agreement with RH

(1n2

f

− 1n2

i

)

Taking h𝜈 = h𝜈𝜆= ΔE = Ef − Ei and solving for 1∕𝜆 from H-like atom energy gives exact

same result as Bohr’s theory for RH.At this point Schrödinger knew wave equation approach was working.After determining full wave function for H-like atom next step is solutions formulti-electron atoms where Bohr’s theory had failed.


Normalizing the Radial Wave FunctionSolutions, Lk

j (𝜌), are called associated Laguerre polynomials.

From these we construct radial part of wave function.

Need to retrace steps from u(𝜌) back to R(r) starting with

un,𝓁(𝜌n) = 𝜌𝓁+1n L2𝓁+1

n−𝓁−1(𝜌n) e−𝜌n∕2

going back through

Rn,𝓁(𝜌n) =un,𝓁(𝜌n)𝜌n

= 𝜌𝓁n L2𝓁+1n−𝓁−1(𝜌n) e−𝜌n∕2

and finally with 𝜌 = 2𝜅r returning to

Rn,𝓁(r) = (2𝜅nr)𝓁 L2𝓁+1n−𝓁−1(2𝜅nr) e−𝜅nr


Normalizing the Radial Wave FunctionTo normalization radial part we define x = 2𝜅nr and express radial part as

Rn,𝓁(x) = An,𝓁x𝓁 L2𝓁+1n−𝓁−1(x) e−x∕2

An,𝓁 is to-be-determined normalization constant. Normalization integral is

∫∞

0R∗

n,𝓁(r)Rn,𝓁(r)r2dr = 1(2𝜅n)3 ∫

∞

0R∗

n,𝓁(x)Rn,𝓁(x)x2dx = 1

Substituting Rn,𝓁(x) into integral gives

A2n,𝓁

(2𝜅n)3 ∫∞

0x2𝓁+2 [L2𝓁+1

n−𝓁−1(x)]2 e−xdx = 1

Look up the general integral for associated Laguerre polynomials

∫∞

0xk+1 [Lk

j (x)]2 e−xdx = (2j + k + 1)

(j + k)!j!

we findA2

n,𝓁

(2𝜅n)32n(n + 𝓁)!(n − 𝓁 − 1)!

= 1 or An,𝓁 = (2𝜅n)3∕2

√(n − 𝓁 − 1)!2n(n + 𝓁)!


Radial Wave FunctionTo further simplify Rn,𝓁(r) we introduce quantity analogous to Bohr radius,

a𝜇 ≡ 4𝜋𝜖0ℏ2

q2e𝜇

then we obtain𝜅n = Z

na𝜇and express 𝜌n as

𝜌n = 2𝜅nr = 2Zrna𝜇

Finally, we obtain radial part of wave function of hydrogen-like atom

Rn,𝓁(r) =(

2Zna𝜇

)𝓁+3∕2√

(n − 𝓁 − 1)!2n(n + 𝓁)!

L2𝓁+1n−𝓁−1

(2Zna𝜇

r)

r𝓁 e−Zr∕(na𝜇)


Radial part of H-like wave functions for n = 1 to n = 4.

R1,0 = 2(

Za𝜇

)3∕2

e−Zr∕a𝜇

R2,0 = 12√

2

(Za𝜇

)3∕2 (2 − Zr

a𝜇

)e−Zr∕(2a𝜇 )

R2,1 = 12√

6

(Za𝜇

)5∕2

re−Zr∕(2a𝜇 )

R3,0 =√

281

(Za𝜇

)3∕2(

6 − 6(

2Zr3a𝜇

)+(

2Zr3a𝜇

)2)

e−Zr∕(3a𝜇 )

R3,1 = 127

√23

(Za𝜇

)5∕2 (4 −

(2Zr3a𝜇

))re−Zr∕(3a𝜇 )

R3,2 = 281

√215

(Za𝜇

)7∕2

r2 e−Zr∕(3a𝜇 )

R4,0 = 116

(Za𝜇

)3∕2 16

(24 − 36

(Zr

2a𝜇

)+ 12

(Zr

2a𝜇

)2

−(

Zr2a𝜇

)3)

e−Zr∕(4a𝜇 )

R4,1 = 132

√1

15

(Za𝜇

)5∕2 12

(20 − 10

(Zr

2a𝜇

)+(

Zr2a𝜇

)2)

r e−Zr∕(4a𝜇 )

R4,2 = 1384

√1

35

(Za𝜇

)7∕2 (6 −

(Zr

2a𝜇

))r2 e−Zr∕(4a𝜇 )

R4,3 = 1768

√15

(Za𝜇

)9∕2

r3 e−Zr∕(4a𝜇 )


Rn,𝓁(r) for s orbitals (𝓁 = 0) of n = 1, 2, 3, 4, 5

20 4 6 8 10 12 14

0.5

0.0

1.0

1.5

2.01s

50 10 15 20 25 30

0.2

0.0

0.4

0.6 2s

10 20 30 40

0.1

0.0

0.2

0.3

0.43s

100 20 30 40 50 60

0.050.00

0.100.150.200.25

4s

100 20 30 40 50 60 70

0.05

0.00

0.10

0.15 5s

20 4 6 8 10 12 14

0.10.0

0.20.30.40.5 1s

0 5 10 15 20 25 30

0.05

0.00

0.10

0.15

0.202s

100 20 30 40 50 60

0.010.00

0.020.030.040.050.06 4s

100 20 30 40

0.020.040.060.080.10 3s

100 20 30 40 50 60 70

0.01

0.00

0.02

0.03

0.04 5s

Wave function extends further out in r, away fromthe nucleus, as n increasesAs with harmonic oscillator classically excludedpositions are displacements where E > V(r)For hydrogen atom classically excluded radii are

r∕a𝜇 > 2n2

Classically excluded regions are indicated by grayregions.


Rn,𝓁(r) for n = 5 and all possible values of 𝓁

0.00

0.01

0.02

0.03

0.04

0.05

0.00

0.01

0.02

0.03

0.04

0.05

0.00

0.01

0.02

0.03

0.04

0.05

0.00

0.01

0.02

0.03

0.04

0.05

0 20 40 60 800.00

0.01

0.02

0.03

0.04

0.05

0 20 40 60 80

-0.05

0.00

0.05

0.10

0.15

-0.04

-0.02

0.00

0.02

0.04

-0.04

-0.02

0.00

0.02

0.04

-0.04

-0.02

0.00

0.02

0.04

-0.04

-0.02

0.00

0.02

0.04

5s

5p

5d

5f

5g

5s

5p

5d

5f

5g

Radial function at origin, r = 0, is non-zero only fors states, where 𝓁 = 0.Maximum in wave function at constant n is pushedfurther away from nucleus as 𝓁 increases.This is consequence of centrifugal term in effectivepotential.

Veff(r) =ℏ2

2𝜇𝓁(𝓁 + 1)

r2⏟⏞⏞⏞⏞⏟⏞⏞⏞⏞⏟

Centrifugal Term

−Zq2

e4𝜋𝜖0r


Hydrogen-Like Atom Wave Functions

In summary, solutions to Schrödinger equation for single electron bound to positive chargenucleus are

𝜓n,𝓁,m𝓁(r, 𝜃, 𝜙) = Rn,𝓁(r)Y𝓁,m𝓁

(𝜃, 𝜙).

Rn,𝓁(r) is radial part of wave function and depends on quantum numbers n and 𝓁

Y𝓁,m𝓁(𝜃, 𝜙) is angular part of wave function and depends on quantum numbers 𝓁 and m𝓁.

Radial part combined with spherical harmonics give full wave function, also called orbitalsince wave function describes electron’s “orbit” around nucleus.


Hydrogen wave functions for n = 1 to n = 3

Orbital Wave Function

1s 𝜓1,0,0 = 2√4𝜋

(Za𝜇

)3∕2

e−Zr∕a𝜇

2s 𝜓2,0,0 = 1√32𝜋

(Za𝜇

)3∕2 (2 − Zr

a𝜇

)e−Zr∕(2a𝜇 )

2p0 𝜓2,1,0 = 1√32𝜋

(Za𝜇

)5∕2

re−Zr∕(2a𝜇 ) cos 𝜃

2p±1 𝜓2,1,±1 = ∓ 1√64𝜋

(Za𝜇

)5∕2

re−Zr∕(2a𝜇 ) sin 𝜃 e±i𝜙

Orbital Wave Function

3s 𝜓3,0,0 = 1√162𝜋

(Za𝜇

)3∕2(

6 − 6(

2Zr3a𝜇

)+(

2Zr3a𝜇

)2)

e−Zr∕(3a𝜇 )

3p0 𝜓3,1,0 = 127

1√2𝜋

(Za𝜇

)5∕2 (4 −

(2Zr3a𝜇

))re−Zr∕(3a𝜇 ) cos 𝜃

3p±1 𝜓3,1,±1 = ∓ 127

1√4𝜋

(Za𝜇

)5∕2 (4 −

(2Zr3a𝜇

))re−Zr∕(3a𝜇 ) sin 𝜃 e±i𝜙

3d0 𝜓3,2,0 = 281

1√6𝜋

(Za𝜇

)7∕2

r2 e−Zr∕(3a𝜇 ) 12(3 cos2 𝜃 − 1)

3d±1 𝜓3,2,±1 = ∓ 1243

1√𝜋

(Za𝜇

)7∕2

r2 e−Zr∕(3a𝜇 ) 3 cos 𝜃 sin 𝜃 e±i𝜙

3d±2 𝜓3,2,±2 = 1486

1√𝜋

(Za𝜇

)7∕2

r2 e−Zr∕(3a𝜇 ) 3 sin2 𝜃 e±i2𝜙


Hydrogen-Like Atom Wave Functions2D cross sections through x-z plane of 3D probability density of selected hydrogen atom wave functions.


Cartesian (real) wave functionsm𝓁 ≠ 0 complex wave function are hard to visualize.Sometimes easier to work with real wave functions by taking sum and difference of |m𝓁| wavefunctions,

𝜓 (±) = c(𝜓|m𝓁| ± 𝜓−|m𝓁|

)Since 𝜓∗|m𝓁| = 𝜓−|m𝓁| coefficient c is adjusted to make 𝜓 (±) real and normalized.

𝜓px= 1√

2

(𝜓p+1

+ 𝜓p−1

)𝜓py

= 1√2i

(𝜓p+1

− 𝜓p−1

)Superposition principle tells us that these are also solutions to same wave equation.


Cartesian (real) wave functions: n = 2, 𝓁 = 1

2px, 2py, 2pzOrbitron Web Site


https://winter.group.shef.ac.uk/orbitron/AOs/2p/index.html

Cartesian (real) wave functions

Can also form real wave functions with d orbitals

ndz2 = 𝜓n,2,0

ndxz =1√2

(𝜓n,2,1 + 𝜓n,2,−1

)ndyz =

1√2i

(𝜓n,2,1 − 𝜓n,2,−1

)ndx2−y2 = 1√

2

(𝜓n,2,2 + 𝜓n,2,−2

)ndxy =

1√2i

(𝜓n,2,2 − 𝜓n,2,−2

)As long as 2 wave functions are degenerate they will also be stationary states.

Numerous illustrations of shape of the Cartesian H-orbitals can be found in elementarychemistry texts.


Cartesian (real) wave functions: n = 3, 𝓁 = 2

Top: 2dx2−y2 and 2dz2

Bottom 2dxy, 2dxz, and 2dyzOrbitron Web Site


https://winter.group.shef.ac.uk/orbitron/AOs/2p/index.html

Web Apps by Paul Falstad

Hydrogen atom orbitals


http://www.falstad.com/qmatom

H-like Atom Transition Selection RulesFor transitions between H-like atom energy states through absorption and emission of light we requirenon-zero electric dipole transition moment,

⟨𝜇⟩n,𝓁,m𝓁 ,n′,𝓁′,m′𝓁= ∫V

𝜓∗n,𝓁,m𝓁

(r, 𝜙, 𝜃) 𝜇 𝜓n′,𝓁′,m′𝓁(r, 𝜙, 𝜃)d𝜏

For H-like atom electric dipole moment is𝜇 =

∑i

qir = Zqe

rp − qere

Since re = r + rp we can write

𝜇 = Zqerp − qe(r + rp) = (Z − 1)qe

rp − qer

Substituting this into transition moment integral gives

⟨𝜇⟩n,𝓁,m𝓁 ,n′,𝓁′,m′𝓁=∫V

𝜓∗n,𝓁,m𝓁

(r, 𝜙, 𝜃) (Z − 1)qerp 𝜓n′,𝓁′,m′

𝓁(r, 𝜙, 𝜃)d𝜏

− ∫V𝜓∗

n,𝓁,m𝓁(r, 𝜙, 𝜃) qe

r𝜓n′,𝓁′,m′𝓁(r, 𝜙, 𝜃)d𝜏


H-like Atom Transition Selection Rules

⟨𝜇⟩n,𝓁,m𝓁 ,n′,𝓁′,m′𝓁=

��:0

∫V𝜓∗

n,𝓁,m𝓁(r, 𝜙, 𝜃) (Z − 1)qe

rp 𝜓n′,𝓁′,m′𝓁(r, 𝜙, 𝜃)d𝜏

− ∫V𝜓∗

n,𝓁,m𝓁(r, 𝜙, 𝜃) qe

r𝜓n′,𝓁′,m′𝓁(r, 𝜙, 𝜃)d𝜏

1st integral involving rp goes to zero for all transitions2nd integral gives transition dipole momentConvenient to express r in spherical coordinates

−qer = −qer

(sin 𝜃 cos𝜙ex + sin 𝜃 sin𝜙ey + cos 𝜃ez

)Using H-like wave functions find that Δn can have any value and

Δ𝓁 = ±1, Δm𝓁 = 0 for ��z, Δm𝓁 = ±1 for ��x and ��y


Web Apps by Paul Falstad

Atomic dipole transitions


http://www.falstad.com/qmatomrad

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