General Chemistry I

INTRODUCTION TO QUANTUMMECHANICS

4.1 Preliminaries: Wave Motion and Light

4.2 Evidence for Energy Quantization in Atoms4.3 The Bohr Model: Predicting Discrete Energy

Levels in Atoms4.4 Evidence for Wave-Particle Duality

4.5 The Schrödinger Equation

4.6 Quantum mechanics of Particle-in-a-Box Models

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Nanometer-Sized Crystals of CdSe

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4.1 PRELIMINARIES: WAVE MOTION AND LIGHT

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- amplitude of the wave: the height or the displacement- wavelength, : the distance between two successive crests- frequency, : units of waves (or cycles) per second (s-1)

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- speed =

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Electromagnetic Radiation

- A beam of light consists of oscillating electric and magnetic fieldsoriented perpendicular to one another and to the direction in whichthe light is propagating.

- Amplitude of the electric field

E(x,t) = Emax cos[2 (x/ - t)]

- The speed, c, of light passing through a vacuum,

c = = 2.99792458 x 108 ms-1

is a universal constant; the same for all types of radiation.

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Interference of waves

- When two light waves pass through the same region of space, theyinterfere to create a new wave called the superposition of the two.

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4.2 EVIDENCE FOR ENERGY QUANTIZATION IN ATOMS

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- Every objects emits energy from its surface in the form of thermalradiation. This energy is carried by electromagnetic waves.

- The distribution of the wavelength depends on the temperature.

- The maximum in the radiation intensitydistribution moves to higher frequency(shorter wavelength) as T increases.

- The radiation intensity falls to zeroat extremely high frequencies forobjects heated to any temperature.

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expected from theclassical theory

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- From classical theory, = 8 23: intensity at , kB: Boltzmann constant, T: temperature (K)

- Predicting an infinite intensity at very short wavelengths

expected from theclassical theory

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Plank’s quantum hypothesis

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- The oscillator must gain and lose energy in quanta of magnitudeh , and that the total energy can take only discrete values:

Plank’s constant h = 6.62606896(3) x 10-34 J s

- Radiation intensity = /

OSC = nh

When / = [ ] = the classical result

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Physical meaning of Plank’s explanation

1. The energy of a system can take only discrete values.

2. A quantized oscillator can gain or lose energy only in discreteamounts E = h .

3. To emit energy from higher energy states, T must be sufficiently high.

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Light from an electrical discharge150

Ne Ar Hg

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Spectrograph150

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Balmer series for hydrogen atoms

= × 3.29 × 1015 s-1

n = 3n = 4n = 5n = 6

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Bohr’s explanationThe frequency of the light absorbed is connected to the energy ofthe initial and final states by the expression= or E = h

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The Frank-Hertz Experiments

Electrons of known energy collided with gaseous atoms, and theenergy lost from the electrons was measured.

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- The abrupt fall in the plot of I vs V at Vthrsuggested that the kinetic energy of theelectrons must reach a threshold eVthr totransfer energy to the gas atoms.=- Suggesting that atomsmust be quantized indiscrete state.

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4.3 THE BOHR MODEL: PREDICTING DISCRETE ENERGY LEVELS IN ATOMS

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- Starting from Rutherford’s planetary model of the atom- the assumption that an electron of mass me moves in a circularorbit of radius r about a fixed nucleus

Classical theory states are not stable.

Bohr model

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mev2

rZe2

4 0r2

- The total energy of the hydrogen atom: kinetic + potential

E = mev2 -12

Ze2

4 0r

- Coulomb force = centrifugal force

Ze2

4 0r2 = mev2

r

- Bohr’s postulation: angular momentum of the electron is quantized.

L = mevr = n h2 n = 1, 2, 3, …

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- Radius rn = 0n2h2

Ze2me= a0

n2

Z

a0 (Bohr radius) = 0h2

e2me= 0.529 Å

- Velocity vn =nh

2 mern=

Ze2

2 02nh

- Energy En =-Z2e4me

8 02n2h2 = -R Z2

n2

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n = 1, 2, 3, …

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Ionization energy: the minimum energy required to remove an electron from an atom

E = Efinal – Einitial = 0 – (-2.18 10-18 J) = 2.18 10-18 J

IE = NA x 2.18 10-18 J = 1310 kJ mol-1

EXAMPLE 4.3

Consider the n = 2 state of Li2+. Using the Bohr model, calculate r, v,and E of the ion relative to that of the nucleus and electron separatedby an infinite distance.

a0Z2

0.705 Å3

v = nh

2 mern

2h2 mern

3.28 106 m s-1a0

E2 = -R Z2

n2 = -R 94 = - 4.90 10-18 J

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Atomic spectra: interpretation by the Bohr model

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- Light is emitted to carry off the energy h by transition from Ei to Ef.

h = Z2e4me

8 02h2

1 1- Lines in the emission spectrum with frequencies,

= Z2e4me

8 02h3

1 13.29 1015 s-1) Z2

1 1ni > nf = 1, 2, 3, … (emission)

- Lines in the absorption spectrum with frequencies,

= Z2e4me

8 02h3

1 13.29 1015 s-1) Z2

1 1nf > ni = 1, 2, 3, … (absorption)

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4.4 EVIDENCE FOR WAVE-PARTICLE DUALITY157

- The particles sometimes behave as waves, and vice versa.

The Photoelectron Effect

- A beam of light shining onto a metal surface(photocathode) can ejectelectrons (photoelectrons)and cause an electriccurrent (photocurrent)to flow.

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Only light above 0 can ejectphotoelectrons from the surface.

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159- Einstein’s theory predicts that the maximum kinetic energy ofphotoelectrons emitted by light of frequency

Emax = m = h -

- Workfunction of the metal, , represents the binding energy that electronsmust overcome to escape from the metal surface after photon absorption.

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Standing Wave161

- Standing wave under a physical boundary condition

n = L n = 1, 2, 3, …

- n = 1, fundamental or firstharmonic oscillation

- node, at certain points wherethe amplitude is zero.

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De Broglie Waves162

- The electron with a circular standing wave oscillating about the nucleusof the atom.

n = 2 r n = 1, 2, 3, …

From Bohr’s assumption, mevr = n 2 = n =EXAMPLE 4.3

Calculate the de Broglie wavelengths of an electron moving withvelocity 1.0 x 106 m s-1.

7.3 Å

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Electron Diffraction163

- An electron with kinetic energy of 50 eV has a de Broglie wave lengthof 1.73 Å, comparable to the spacing between atomic planes.

T = eV = 2 = p = 2 = h/ 2 = 1.73 Å

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- The diffraction condition is

- For two dimensional surface with a along the x-axis and b alongthe y-axis

a a a b b b

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The Heisenberg Indeterminacy (uncertainty)165

- When we measure two properties A and B, the product of which hasdimensions of action, we will obtain a spread in results for eachidentified by A and B that will satisfy the following condition:/4

/- For the position x and the momentum p,

x and p cannot be determined simultaneously.

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For instance, an object of 1.0 g mass with v = 2.0 mm s-1,x = 2.6 10-29 m, very small

an electron confined to the diameter of a typical atom (200 pm),

v = 2.89 105 m s-1, very large

EXAMPLE 4.6

Suppose photons of green light (wavelength 5.3 x 10-7 m) are usedTo locate the position of the baseball with precision of one wavelength.Calculate the minimum speed in the speed of the baseball (0.145 kg).

= 9.9x10-29 kg m s-1= 6.8x10-28 m s-1

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4.5 THE SCHRÖDINGER EQUATION168

wave function ( , psi): mapping out the amplitude of a wave in threedimensions; it may be a function of time.

- The origins of the Schrödinger equation:

If the wave function is described as = sin= sin =

= = = 2

+ ( ) = E = T + V(x)

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Born interpretation: probability of finding theparticle in a region is proportional to the valueof 2

probability density (P(x)): the probability that the particle will be found in a small region divided bythe volume of the region

= probability

1) Probability density must be normalized.= ( ) 2 = 12) P(x) must be continuous at each point x.

3) ( ) must be bounded at large values of x. ( ) ±boundary conditions

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+ ( ) =How can we solve the Schrödinger equation?

The allowed energy values and wave functions

From the boundary conditions, energy quantization arises.Each energy value corresponds to one or more wave functions.

The wave functions describe the distribution of particleswhen the system has a specific energy value.

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4.6 QUANTUM MECHANICS OFPARTICLE-IN-A-BOX MODELS

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Particle in a box

- Mass m confined between two rigid walls a distance L apart- = 0 outside the box at the walls (boundary condition)

= 0 for 0 and

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=- Inside the box, where V = 0,

=

- From the boundary conditions, = 0 at = 0 and = L.= A sin kx = A sin kL = 0= n = 1, 2, 3, …= A sin = 1, 2, 3, …

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- For the normalization,

A2 sin2 = A2 = 1 A =

= = 1, 2, 3, …

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- The second derivative of the wave function:2 2 = = 8 2 2= = 1, 2, 3, …

Energy of the particle is quantized !

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has n - 1 nodes, and # of nodes increases with the energy.

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Zero-Point Energy175

The lowest value of n is 1, and the lowest energy is E1 = h2/8mL2,not zero.

According to quantum mechanics, aparticle can never be perfectly stillwhen it is confined between two walls:it must always possess an energy.

consistent with the indeterminacy principle/4when 0,

not confined within L

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Correspondence Principle175

- At n = 20, the probability becomesessentially uniform across the box, andthere is little evidence of quantization.

- The results of quantum mechanics reduceto those of classical mechanics forlarge values of the quantum numbers, n.

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Particle in a 3-D Box

Particle in a cubic box of length Lon each side,

- Energy: = + +nx = 1,2,3,…ny = 1,2,3,…

nz = 1,2,3,…

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- Wave functions ( ) /181

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EXAMPLE 4.7

Consider the following two systems:(a) An electron in a one-dimensional box of length 1.0 Å(b) A helium atom in a cube 30 cm on an edge.Calculate the energy difference between ground state and first excitedstate, expressing your answer in kJ mol-1.

(a) = 22 12 = 1.8 × 10 = 11,000(b) = 22 + 12 + 12) (12 + 12 + 12= 1.7 × 10

very close, no role of quantum effects

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10 ex-Problem Sets

For Chapter 4,

6, 10, 20, 28, 32, 38, 40, 48, 50, 58