Lectures 3-4: One-electron atoms

1 / 26

# Lectures 3-4: One-electron atoms - PowerPoint PPT Presentation

Lectures 3-4: One-electron atoms. Schrödinger equation for one-electron atom. Solving the Schrödinger equation. Wavefunctions and eigenvalues. Atomic orbitals. See Chapter 7 of Eisberg & Resnick. The Schrödinger equation. One-electron atom is simplest bound system in nature.

I am the owner, or an agent authorized to act on behalf of the owner, of the copyrighted work described.

## Lectures 3-4: One-electron atoms

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.

- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -
Presentation Transcript
Lectures 3-4: One-electron atoms
• Schrödinger equation for one-electron atom.
• Solving the Schrödinger equation.
• Wavefunctions and eigenvalues.
• Atomic orbitals.
• See Chapter 7 of Eisberg & Resnick.

PY3P05

The Schrödinger equation
• One-electron atom is simplest bound system in nature.
• Consists of positive and negative particles moving in 3D Coulomb potential:
• Z =1 for atomic hydrogen, Z =2 for ionized helium, etc.
• Electron in orbit about proton treated using reduced mass:
• Total energy of system is therefore,

PY3P05

The Schrödinger equation
• Using the Equivalence Principle, the classical dynamical quantities can be replaced with their associated differential operators:
• Substituting, we obtain the operator equation:
• Assuming electron can be described by a wavefunction of form,

can write

or

where, is the Laplacian operator.

PY3P05

The Schrödinger equation
• Since V(x,y,z) does not depend on time, is a solution to the Schrödinger equation and the eigenfunction is a solution of the time-independent Schrödinger equation:
• As V = V(r), convenient to use spherical polar coordinates.

where

• Can now use separation of variables to split the partial

differential equation into a set of ordinary differential equations.

(1)

PY3P05

Separation of the Schrödinger equation
• Assuming the eigenfunction is separable:
• Using the Laplacian, and substituting (2) and (1):
• Carrying out the differentiations,
• Note total derivatives now used, as R is a function of r alone, etc.
• Now multiply through by and taking transpose,

(2)

(3)

PY3P05

Separation of the Schrödinger equation
• As the LHS of Eqn 3does nor depend on r or  and RHS does not depend on  their
• common value cannot depend on any of these variables.
• Setting the LHS of Eqn 3 to a constant:
• and RHS becomes
• Both sides must equal a constant, which we choose as l(l+1):
• We have now separated the time-independent Schrödinger equation into three
• ordinary differential equations, which each only depend on one of  (4),  (5) and R(6). .

(4)

(5)

(6)

PY3P05

Summary of separation of Schrödinger equation
• Express electron wavefunction as product of three functions:
• As V ≠ V(t), attempt to solve time-independent Schrodinger equation.
• Separate into three ordinary differential equations for and .
• Eqn. 4 for () only has acceptable solutions for certain value of ml.
• Using these values for mlin Eqn. 5, () only has acceptable values for certain values of l.
• With these values for l in Eqn. 6, R(r) only has acceptable solutions for certain values of En.
• Schrödinger equation produces three quantum numbers!

PY3P05

Azimuthal solutions (())
• A particular solution of (4) is
• As the einegfunctions must be single valued, i.e.,  =>

and using Euler’s formula,

• This is only satisfied if ml = 0, ±1, ±2, ...
• Therefore, acceptable solutions to (4) only exist when ml can only have certain integer

values, i.e. it is a quantum number.

• ml is called the magnetic quantum number in spectroscopy.
• Called magnetic quantum number because plays role when atom interacts with magnetic fields.

PY3P05

Polar solutions (())
• Making change of variables (z = rcos, Eqn.5 transformed into an associated Legendre equation:
• Solutions to Eqn. 7 are of form

where are associated Legendre polynomial functions.

•  remains finite when = 0, 1, 2, 3, ...

ml = -l, -l+1, .., 0, .., l-1, l

• Can write the associated Legendre functions using quantum number subscripts:

00 = 1

10 = cos1±1 = (1-cos2)1/2

• 20 = 1-3cos22±1 = (1-cos2)1/2cos
• 2±2 = 1-cos2

(7)

PY3P05

Spherical harmonic solutions
• Customary to multiply () and () to form so called spherical harmonic functions
• which can be written as:

i.e., product of trigonometric and polynomial

functions.

• First few spherical harmonics are:

Y00= 1

Y10= cos Y1±1= (1-cos2)1/2 e±i

Y20= 1-3cos2 Y2±1= (1-cos2)1/2cos e±i

PY3P05

• What is the ground state of hydrogen (Z=1)? Assuming that the ground state has n = 1, l = 0 Eqn. 6can be written
• Taking the derivative

(7)

• Try solution , where A and a0are constants. Sub into Eqn. 7:
• To satisfy this Eqn. for any r, both expressions in brackets must equal zero. Setting the second expression to zero =>
• Setting first term to zero =>

Same as Bohr’s results

eV

PY3P05

has many solutions, one for each positive integer of n.

• Solutions are of the form (see Appendix N of Eisberg & Resnick):

where a0is the Bohr radius. Bound-state solutions are only acceptable if

where n is the principal quantum number, defined by n = l +1, l +2, l +3, …

• Enonly depends on n: all l states for a given n are degenerate (i.e. have the same energy).

eV

PY3P05

• Gnl(Zr/a0) are called associated Laguerre polynomials, which depend on n and l.
• Several resultant radial wavefunctions (Rnl( r )) for the hydrogen atom are given below

PY3P05

• The radial probability function Pnl(r ), is the probability that the electron is found between r and r + dr:
• Some representative radial probability functions are given at right:
• Some points to note:
• The r2factor makes the radial probability density

vanish at the origin, even for l = 0 states.

• For each state (given n and l), there are n - l - 1

nodes in the distribution.

• The distribution for states with l = 0, have n maxima,

which increase in amplitude with distance from origin.

PY3P05

• Radial probability distributions for an electron in several of the low energy orbitals of hydrogen.
• The abscissa is the radius

in units of a0.

s orbitals

p orbitals

d orbitals

PY3P05

Hydrogen eigenfunctions
• Eigenfunctions for the state described by the quantum numbers (n, l, ml) are therefore of form:

and depend on quantum numbers:

n = 1, 2, 3, …

l = 0, 1, 2, …, n-1

ml = -l, -l+1, …, 0, …, l-1, l

• Energy of state on dependent on n:
• Usually more than one state has same

energy, i.e., are degenerate.

PY3P05

Born interpretation of the wavefunction
• Principle of QM: the wavefunction contains all the dynamical information about the system it describes.
• Born interpretationof the wavefunction: The probability (P(x,t)) of finding a particle at a position between x and x+dx is proportional to |(x,t)|2dx:

P(x,t) = *(x,t) (x,t) = |(x,t)|2

• P(x,t) is the probability density.
• Immediately implies that sign of wavefunction has no

direct physical significance.

(x,t)

P(x,t)

PY3P05

Born interpretation of the wavefunction
• In H-atom, ground state orbital has the same sign everywhere => sign of orbital must be all positive or all negative.
• Other orbitals vary in sign. Where orbital changes sign,  = 0 (called a node) => probability of finding electron is zero.
• Consider first excited state of hydrogen: sign of

wavefunction is insignificant (P = 2 = (-)2).

PY3P05

Born interpretation of the wavefunction
• Next excited state of H-atom is asymmetric about origin. Wavefunction has opposite sign on opposite sides of nucleus.
• The square of the wavefunction is identical on

opposite sides, representing equal distribution

of electron density on both side of nucleus.

PY3P05

Atomic orbitals
• Quantum mechanical equivalent of orbits in Bohr model.

PY3P05

s orbitals
• Named from “sharp” spectroscopic lines.
• l = 0, ml = 0
• n,0,m = Rn,0 (r ) Y0,m (, )
• Angular solution:
• Value of Y0,0is constant over sphere.
• For n = 0, l = 0, ml = 0 => 1s orbital
• The probability density is

PY3P05

p orbitals
• Named from “principal” spectroscopic lines.
• l = 1, ml = -1, 0, +1 (n must therefore be >1)
• n,1,m = Rn1 (r ) Y1,m (, )
• Angular solution:
• A node passes through the nucleus and separates the two lobes of each orbital.
• Dark/light areas denote opposite sign of the wavefunction.
• Three p-orbitals denoted px, py , pz

PY3P05

d orbitals
• Named from “diffuse” spectroscopic lines.
• l = 2, ml = -2, -1, 0, +1, +2 (n must therefore be >2)
• n,2,m = Rn1 (r ) Y2,m (, )
• Angular solution:
• There are five d-orbitals, denoted
• m = 0 is z2. Two orbitals of m = -1 and +1 are xz and yz. Two orbitals with m = -2 and +2 are designated xy and x2-y2.

PY3P05

Quantum numbers and spectroscopic notation
• Angular momentum quantum number:
• l = 0 (s subshell)
• l = 1 (p subshell)
• l = 2 (d subshell)
• l = 3 (f subshell)
• Principal quantum number:
• n = 1 (K shell)
• n = 2 (L shell)
• n = 3 (M shell)
• If n = 1 and l = 0 = > the state is designated 1s. n = 3, l = 2 => 3d state.
• Three quantum numbers arise because time-independent Schrödinger equation contains three independent variables, one for each space coordinate.
• The eigenvalues of the one-electron atom depend only on n, by the eigenfunctions depend on n, l and ml, since they are the product of Rnl(r ), lml () and ml().
• For given n, there are generally several values of l and ml => degenerate eigenfunctions.

PY3P05

Orbital transitions for hydrogen
• Transition between different energy levels of the hydrogenic atom must follow the following selection rules:

l = ±1

m = 0, ±1

• A Grotrian diagram or a term diagram shows the allowed transitions.
• The thicker the line at right, the more probable and hence more intense the transitions.
• The intensity of emission/absorption lines could not be explained via Bohr model.

PY3P05

Schrödinger vs. Bohr models
• Schrodinger’s QM treatment had a number of advantages over semi-classical Bohr model:
• Probability density orbitals do not violate the Heisenberg Uncertainty Principle.
• Orbital angular momentum correctly accounted for.
• Electron spin can be properly treaded.
• Electron transition rates can be explained.

PY3P05