7-1

Chapter 7

Quantum Theory and Atomic Structure

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Quantum Theory and Atomic Structure

7.1 The Nature of Light

7.2 Atomic Spectra

7.3 The Wave-Particle Duality of Matter and Energy

7.4 The Quantum-Mechanical Model of the Atom

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The Wave Nature of Light

Wave properties are described by two interdependent variables:

Frequency: (nu) = the number of cycles the wave undergoes

per second (units of s-1 or hertz (Hz)) (cycles/s) Wavelength: (lambda) = the distance between any point on awave and a corresponding point on the next wave (the distancethe wave travels during one cycle) (units of meters (m), nanometers (10-9 m), picometers (10-12 m) or angstroms (Å, 10-10 m) per cycle)(m/cycle)

Speed of a wave = cycles/s x m/cycle = m/s

c = speed of light in a vacuum = =3.00 x 108 m/s

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Figure 7.1

Frequency and

Wavelength

7-5Figure 7.2

Amplitude (intensity)of a Wave

(a measure of thestrength of the wave’selectric and magneticfields)

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Figure 7.3

Regions of the Electromagnetic Spectrum

Light waves all travel at the same speed through a vacuum butdiffer in frequency and, therefore, in wavelength.

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Sample Problem 7.1

SOLUTION:

PLAN:

Interconverting Wavelength and Frequency

PROBLEM: A dental hygienist uses x-rays (= 1.00 Å) to take a series of dental radiographs while the patient listens to a radio station ( = 325 cm) and looks out the window at the blue sky (= 473 nm). What is the frequency (in s-1) of the electromagnetic radiation from each source? Assume that the radiation travels at the speed of light, 3.00 x 108

m/s.

Use c = 1.00 Å

325 cm

473 nm

10-10 m1 Å

10-2 m1 cm

10-9 m1 nm

= 1.00 x 10-10 m

= 325 x 10-2 m

= 473 x 10-9 m

=3.00 x 108 m/s

1.00 x 10-10 m= 3 x 1018 s-1

=

=

3.00 x 108 m/s

325 x 10-2 m= 9.23 x 107 s-1

3.00 x 108 m/s

473 x 10-9 m= 6.34 x 1014 s-1

x

x

x

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Distinguishing Between a Wave and a Particle

Refraction: the change in the speed of a wave when itpasses from one transparent medium to another

Diffraction: the bending of a wave when it strikes theedge of an object, as when it passes through a slit;an interference pattern develops if the wave passes through two adjacent slits

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Figure 7.4

Different

behaviors of

waves and

particles

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Figure 7.5

The diffraction pattern caused by light passing through two adjacent slits

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Figure 7.6

Blackbody Radiation

E = nh

Changes in the intensity andwavelength of emitted light whenan object is heated at differenttemperatures

Planck’s equation

E = energy of radiationh = Planck’s constant = frequencyn = positive integer (a quantum number)

h = 6.626 x 10-34 J.s

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If an atom can emit only certain quantities of energy, then the atom can have only certain quantities of energy. Thus, the energy of an atom is quantized.

Each energy packet is called a quantum and hasenergy equal to h.

An atom changes its energy state by absorbing oremitting one or more quanta of energy.

Eatom = Eemitted (or absorbed) radiation = nh

E = h (n = 1)energy change between adjacent energy states

The Notion of Quantized Energy

7-13Figure 7.7

Demonstration of the photoelectric effect

Wave model could not explain the (a) presence of a threshold frequency,and (b) the absence of a time lag.

Led to Einstein’s photon theory of light:

Ephoton = h = Eatom

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Sample Problem 7.2

SOLUTION:

PLAN:

Calculating the Energy of Radiation from its Wavelength

PROBLEM: A cook uses a microwave oven to heat a meal. The wavelength of the radiation is 1.20 cm. What is the energy of one photon of this microwave radiation?

After converting cm to m, we use the energy equation, E = h combined with = c/ to find the energy.

E = hc/

E = (6.626 x 10-34J.s) (3.00 x 108 m/s)

1.20 cm 10-2m

cm

x= 1.66 x 10-23 J

x

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Figure 7.8

Line spectra of several elements

Atomic Spectra

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= RRydberg equation: -1

1

n22

1

n12

R is the Rydberg constant = 1.096776 x 107 m-1

Figure 7.9

Three series of spectral lines of atomic hydrogen

for the visible series, n1 = 2 and n2 = 3, 4, 5, ...

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The Bohr Model of the Hydrogen Atom

Postulates:

1. The H atom has only certain allowable energy levels

2. The atom does not radiate energy while in one of its stationary states

3. The atom changes to another stationary state(i.e., the electron moves to another orbit) only byabsorbing or emitting a photon whose energy equals the difference in energy between the two states

The quantum number is associated with the radius of an electronorbit; the lower the n value, the smaller the radius of the orbit andthe lower the energy level.

ground state and excited state

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Figure 7.10

Quantum staircase

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Limitations of the Bohr Model

Can only predict spectral lines for the hydrogen atom (a one electron model)

Electrons do not travel in fixed orbits

For systems having >1 electron, there are additionalnucleus-electron attractions and electron-electronrepulsions

7-20Figure 7.11

The Bohr explanation of the three series of spectral lines for atomic hydrogen

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Figure B7.2

Figure B7.1

Emission and absorption spectra of sodium atoms

Flame tests

strontium 38Sr copper 29Cu

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Fireworks emissions from compounds containing specificelements

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Figure B7.3 The main components of a typical spectrophotometer

Monochromator (wavelength selector) disperses incoming radiation into continuum of component wavelengths that are scanned or individually selected.

Sample in compartment absorbs characteristic amount of each incoming wavelength.

Computer converts signal into displayed data.

Source produces radiation in region of interest. Must be stable and reproducible. In most cases, the source emits many wavelengths.

Lenses/slits/collimaters narrow and align beam.

Detector converts transmitted radiation into amplified electrical signal.

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QuickTime™ and aPhoto - JPEG decompressor

are needed to see this picture.

The AbsorptionSpectrum ofChlorophyll a

Absorbs red andblue wavelengthsbut no green andyellow wavelengths;leaf appears green.

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The Wave-Particle Duality of Matter and Energy

De Broglie: If energy is particle-like, perhaps matter iswavelike!

Electrons have wavelike motion and are restricted to orbitsof fixed radius; explains why they will have only certain possiblefrequencies and energies.

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Figure 7.13

Wave motion in restricted systems

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Table 7.1 The de Broglie Wavelengths of Several Objects

Substance Mass (g) Speed (m/s) (m)

slow electron

fast electron

alpha particle

one-gram mass

baseball

Earth

9 x 10-28

9 x 10-28

6.6 x 10-24

1.0

142

6.0 x 1027

1.0

5.9 x 106

1.5 x 107

0.01

25.0

3.0 x 104

7 x 10-4

1 x 10-10

7 x 10-15

7 x 10-29

2 x 10-34

4 x 10-63

h /mu

The de Broglie Wavelength

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Sample Problem 7.3

SOLUTION:

PLAN:

Calculating the de Broglie Wavelength of an Electron

PROBLEM: Find the deBroglie wavelength of an electron with a speed of 1.00 x 106m/s (electron mass = 9.11 x 10-31 kg; h = 6.626 x 10-34 kg.m2/s).

Knowing the mass and the speed of the electron allows use of the equation, = h/mu, to find the wavelength.

= 6.626 x 10-34 kg.m2/s

9.11 x 10-31 kg x 1.00 x 106 m/s= 7.27 x 10-10 m

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Figure 7.14

Comparing the diffraction patterns of x-rays and electrons

Electrons - particles with mass and charge - create diffractionpatterns in a manner similar to electromagnetic waves!

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CLASSICAL THEORYCLASSICAL THEORY

Matter particulate,

massive

Energy continuous,

wavelike

Since matter is discontinuous and particulate, perhaps energy is discontinuous and particulate.

Observation Theory

Planck: Energy is quantized; only certain values are allowed

blackbody radiation

Einstein: Light has particulate behavior (photons)photoelectric effect

Bohr: Energy of atoms is quantized; photon emitted when electron changes orbit.

atomic line spectra

Figure 7.15

Summary of the major observations and theories leading from classical theory to quantum theory

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Since energy is wavelike, perhaps matter is wavelike.

Observation Theory

deBroglie: All matter travels in waves; energy of atom is quantized due to wave motion of electrons

Davisson/Germer: electron diffraction by metal crystal

Since matter has mass, perhaps energy has mass

Observation Theory

Einstein/deBroglie: Mass and energy are equivalent; particles have wavelength and photons have momentum.

Compton: photon wavelength increases (momentum decreases) after colliding with electron

Figure 7.15 (continued)

QUANTUM THEORY

Energy same as Matter:particulate, massive, wavelike

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The Heisenberg Uncertainty Principle

x x m u ≥ h

4

It is impossible to know simultaneously the exact position and momentum of a particle

x = the uncertainty in position

u = the uncertainty in speed

A smaller x dictates a larger u, and vice versa.

Implication: cannot assign fixed paths for electrons; can knowthe probability of finding an electron in a given region of space

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Sample Problem 7.4

SOLUTION:

PLAN:

Applying the Uncertainty Principle

PROBLEM: An electron moving near an atomic nucleus has a speed 6 x 106 ± 1% m/s. What is the uncertainty in its position (x)?

The uncertainty in the speed (u) is given as ±1% (0.01) of 6 x 106

m/s. Once we calculate this value, the uncertainty equation is used to calculate x.

u = (0.01)(6 x 106 m/s) = 6 x 104 m/s (the uncertainty in speed)

x x m u ≥ h

4

x ≥4 (9.11 x 10-31 kg)(6 x 104 m/s)

6.626 x 10-34 kg.m2/s ≥ 1 x 10-9 m

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The Schrödinger EquationH = E

d2dy2

d2dx2

d2dz2

+ +82m

h2(E-V(x,y,z)(x,y,z) = 0+

how changes in space

mass of electron

total quantized energy of the atomic system

potential energy at x,y,zwave function

Each solution to this equation is associated with a givenwave function, also called an atomic orbital

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Figure 7.16

Electron probability in the ground-state hydrogen atom

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Quantum Numbers and Atomic Orbitals

An atomic orbital is specified by three quantum numbers.

n the principal quantum number - a positive integer (energy level)

l the angular momentum quantum number - an integer from 0 to (n-1)

ml the magnetic moment quantum number - an integer from -l to +l

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Table 7.2 The Hierarchy of Quantum Numbers for Atomic Orbitals

Name, Symbol(Property) Allowed Values Quantum Numbers

Principal, n (size, energy)

Angular momentum, l (shape)

Magnetic, ml (orientation)

Positive integer (1, 2, 3, ...)

0 to n-1

-l,…,0,…,+l

1

0

0

2

0 1

0

3

0 1 2

0

0-1 +1 -1 0 +1

0 +1 +2-1-2

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Sample Problem 7.5

SOLUTION:

PLAN:

Determining Quantum Numbers for an Energy Level

PROBLEM: What values of the angular momentum (l) and magnetic (ml) quantum numbers are allowed for a principal quantum number (n) of 3? How many orbitals are allowed for n = 3?

Follow the rules for allowable quantum numbers.l values can be integers from 0 to (n-1); ml can be integers from -l through 0 to + l.

For n = 3, l = 0, 1, 2

For l = 0 ml = 0 (s sublevel)

For l = 1 ml = -1, 0, or +1 (p sublevel)

For l = 2 ml = -2, -1, 0, +1, or +2 (d sublevel)

There are 9 ml values and therefore 9 orbitals with n = 3

Anthony S. Serianni:

Anthony S. Serianni:

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Sample Problem 7.6

SOLUTION:

PLAN:

Determining Sublevel Names and Orbital Quantum Numbers

PROBLEM: Give the name, magnetic quantum numbers, and number of orbitals for each sublevel with the following quantum numbers:

(a) n = 3, l = 2 (b) n = 2, l = 0 (c) n = 5, l = 1 (d) n = 4, l = 3

Combine the n value and l designation to name the sublevel. Knowing l, find ml and the number of orbitals.

n l sublevel name possible ml values no. orbitals

(a)

(b)

(c)

(d)

3

2

5

4

2

0

1

3

3d

2s

5p

4f

-2, -1, 0, 1, 2

0

-1, 0, 1

-3, -2, -1, 0, 1, 2, 3

5

1

3

7

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Figure 7.17

1s2s 3s

S orbitals

sphericalnodes

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Figure 7.18

2p orbitals

nodal planes

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Figure 7.19

3d orbitals

perpendicular nodal planes

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Figure 7.19 (continued)

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Figure 7.20

One of the seven possible 4f orbitals

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Figure 7.21

The energy levelsin the hydrogen atom

The energy leveldepends only on then value of the orbital