Chapter 7
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Chapter 7. Quantum Theory and Atomic Structure. 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. The Wave Nature of Light.

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Chapter 7

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Chapter 7

Chapter 7

Quantum Theory and Atomic Structure


Chapter 7

Quantum Theory and Atomic Structure

7.1 The Nature of Light

7.2Atomic Spectra

7.3 The Wave-Particle Duality of Matter and Energy

7.4The Quantum-Mechanical Model of the Atom


Chapter 7

The Wave Nature of Light

Wave properties are described by two interdependent variables:

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

per second (units of s-1 or hertz (Hz)) (cycles/s)

Wavelength: l (lambda) = the distance between any point on a

wave and a corresponding point on the next wave (the distance

the 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 = nl =3.00 x 108 m/s


Chapter 7

Figure 7.1

Frequency and Wavelength


Chapter 7

Amplitude (intensity)

of a Wave

(a measure of the

strength of the wave’s

electric and magnetic

fields)

Figure 7.2


Chapter 7

Regions of the Electromagnetic Spectrum

Figure 7.3

Light waves all travel at the same speed through a vacuum but

differ in frequency and, therefore, in wavelength.


Chapter 7

PROBLEM:

A dental hygienist uses x-rays (l = 1.00 Å) to take a series of dental radiographs while the patient listens to a radio station (l = 325 cm) and looks out the window at the blue sky (l = 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 108m/s.

1.00 Å

10-10m

1 Å

325 cm

10-2m

1 cm

10-9m

473 nm

1 nm

3.00 x 108m/s

473 x 10-9m

Sample Problem 7.1

Interconverting Wavelength and Frequency

PLAN:

Use c = ln

x

= 1.00 x 10-10m

3.00 x 108m/s

SOLUTION:

n =

= 3 x 1018s-1

1.00 x 10-10m

x

= 325 x 10-2m

3.00 x 108m/s

n =

= 9.23 x 107s-1

325 x 10-2m

= 473 x 10-9m

x

n =

= 6.34 x 1014s-1


Chapter 7

Distinguishing Between a Wave and a Particle

Refraction: the change in the speed of a wave when it

passes from one transparent medium to another

Diffraction: the bending of a wave when it strikes the

edge of an object, as when it passes through a slit;

an interference pattern develops if the wave passes

through two adjacent slits


Chapter 7

Different behaviors of waves and particles

Figure 7.4


Chapter 7

The diffraction pattern caused by light passing through two adjacent slits

Figure 7.5


Chapter 7

Blackbody Radiation

Changes in the intensity and

wavelength of emitted light when

an object is heated at different

temperatures

Figure 7.6

Planck’s equation

E = nhn

E = energy of radiation

h = Planck’s constant

n = frequency

n = positive integer (a

quantum number)

h = 6.626 x 10-34 J.s


Chapter 7

The Notion of Quantized Energy

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 has

energy equal to hn.

An atom changes its energy state by absorbing or

emitting one or more quanta of energy.

DEatom = Eemitted(or absorbed) radiation= Dnhn

DE = hn (Dn = 1)energy change between adjacent energy states


Chapter 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 = hn = DEatom

Figure 7.7


Chapter 7

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?

10-2m

cm

Calculating the Energy of Radiation from its Wavelength

Sample Problem 7.2

PLAN:

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

SOLUTION:

E = hc/l

(6.626 x 10-34J.s)

x

(3.00 x 108m/s)

E =

= 1.66 x 10-23J

1.20 cm

x


Chapter 7

Atomic Spectra

Line spectra of several elements

Figure 7.8


Chapter 7

1

1

1

l

n12

n22

=

R

-

Rydberg equation:

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

Three series of spectral lines of atomic hydrogen

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

Figure 7.9


Chapter 7

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 by

absorbing 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 electron

orbit; the lower the n value, the smaller the radius of the orbit and

the lower the energy level.

ground state and excited state


Chapter 7

Quantum staircase

Figure 7.10


Chapter 7

Limitations of the Bohr Model

Can only predict spectral lines for the hydrogen atom

(a one electron model)

For systems having >1 electron, there are additional

nucleus-electron attractions and electron-electron

repulsions

Electrons do not travel in fixed orbits


Chapter 7

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

Figure 7.11


Chapter 7

Figure B7.1

Flame tests

strontium 38Sr

copper 29Cu

Emission and absorption spectra of sodium atoms

Figure B7.2


Chapter 7

Fireworks emissions

from compounds

containing specific

elements


Chapter 7

Samplein compartment absorbs characteristic amount of each incoming wavelength.

Computer converts signal into displayed data.

Lenses/slits/collimaters narrow and align beam.

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

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

Detector converts transmitted radiation into amplified electrical signal.

Figure B7.3

The main components of a typical spectrophotometer


Chapter 7

The Absorption

Spectrum of

Chlorophyll a

Absorbs red and

blue wavelengths

but no green and

yellow wavelengths;

leaf appears green.


Chapter 7

The Wave-Particle Duality of Matter and Energy

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

wavelike!

Electrons have wavelike motion and are restricted to orbits

of fixed radius; explains why they will have only certain possible

frequencies and energies.


Chapter 7

Wave motion in restricted systems

Figure 7.13


Chapter 7

h

/mu

l

=

The de Broglie Wavelength

Table 7.1 The de Broglie Wavelengths of Several Objects

Substance

Mass (g)

Speed (m/s)

l (m)

slow electron

9 x 10-28

1.0

7 x 10-4

fast electron

9 x 10-28

5.9 x 106

1 x 10-10

alpha particle

6.6 x 10-24

1.5 x 107

7 x 10-15

one-gram mass

1.0

0.01

7 x 10-29

baseball

142

25.0

2 x 10-34

Earth

6.0 x 1027

3.0 x 104

4 x 10-63


Chapter 7

PROBLEM:

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

Calculating the de Broglie Wavelength of an Electron

Sample Problem 7.3

PLAN:

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

SOLUTION:

6.626 x 10-34kg.m2/s

= 7.27 x 10-10m

l =

9.11 x 10-31kg

x

1.00 x 106m/s


Chapter 7

Comparing the diffraction patterns of x-rays and electrons

Figure 7.14

Electrons - particles with mass and charge - create diffraction

patterns in a manner similar to electromagnetic waves!


Chapter 7

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

blackbody radiation

Planck: Energy is quantized; only certain values

are allowed

photoelectric effect

Einstein: Light has particulate behavior (photons)

atomic line spectra

Bohr: Energy of atoms is quantized; photon

emitted when electron changes orbit.

Figure 7.15

CLASSICAL THEORY

Matter particulate, massive

Energy continuous, wavelike

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

Observation

Theory


Chapter 7

Figure 7.15 (continued)

Since energy is wavelike, perhaps matter is wavelike.

Davisson/Germer:

electron diffraction

by metal crystal

deBroglie: All matter travels in waves; energy of

atom is quantized due to wave motion of

electrons

Since matter has mass, perhaps energy has mass

Observation

Theory

Compton: photon

wavelength increases (momentum decreases) after colliding with electron

Einstein/deBroglie: Mass and energy are

equivalent; particles have wavelength and

photons have momentum.

QUANTUM THEORY

Energy same as Matter:

particulate, massive, wavelike

Observation

Theory


Chapter 7

h

4p

The Heisenberg Uncertainty Principle

It is impossible to know simultaneously the exact

position and momentum of a particle

DxxmDu≥

Dx = the uncertainty in position

Du = the uncertainty in speed

A smaller Dx dictates a larger Du, and vice versa.

Implication: cannot assign fixed paths for electrons; can know

the probability of finding an electron in a given region of space


Chapter 7

PROBLEM:

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

h

D xx mD u ≥

4p

Sample Problem 7.4

Applying the Uncertainty Principle

PLAN:

The uncertainty in the speed (Du) is given as ±1% (0.01) of 6 x 106m/s. Once we calculate this value, the uncertainty equation is used to calculate Dx.

SOLUTION:

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

6.626 x 10-34kg.m2/s

≥ 1 x 10-9m

Dx ≥

4p (9.11 x 10-31kg)(6 x 104m/s)


Chapter 7

d2Y

d2Y

d2Y

wave function

mass of electron

potential energy at x,y,z

dx2

dy2

dz2

8p2mQ

+

+

+

(E-V(x,y,z)Y(x,y,z) = 0

h2

how y changes in space

total quantized energy of the atomic system

The Schrödinger Equation

HY = EY

Each solution to this equation is associated with a given

wave function, also called an atomic orbital


Chapter 7

Electron probability in the ground-state hydrogen atom

Figure 7.16


Chapter 7

Quantum Numbers and Atomic Orbitals

An atomic orbital is specified by three quantum numbers.

nthe principal quantum number - a positive integer (energy level)

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

mlthe magnetic moment quantum number - an integer from -l to +l


Chapter 7

0

1

2

0

-1

0

+1

-1

0

+1

-2

-1

0

+1

+2

Table 7.2 The Hierarchy of Quantum Numbers for Atomic Orbitals

Name, Symbol

(Property)

Allowed Values

Quantum Numbers

Principal, n (size, energy)

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

1

2

3

Angular momentum, l (shape)

0 to n-1

0

0

1

0

0

Magnetic, ml (orientation)

-l,…,0,…,+l


Chapter 7

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?

Anthony S. Serianni:

Sample Problem 7.5

Determining Quantum Numbers for an Energy Level

PLAN:

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.

SOLUTION:

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 mlvalues and therefore 9 orbitals with n = 3


Chapter 7

PROBLEM:

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

Sample Problem 7.6

Determining Sublevel Names and Orbital Quantum Numbers

(a) n = 3, l = 2

(b) n = 2, l = 0

(c) n = 5, l = 1

(d) n = 4, l = 3

PLAN:

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

SOLUTION:

n

l

sublevel name

possible ml values

no. orbitals

(a)

2

3d

-2, -1, 0, 1, 2

3

5

(b)

2

0

2s

0

1

(c)

5

1

5p

-1, 0, 1

3

(d)

4

3

4f

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

7


Chapter 7

S orbitals

1s

3s

2s

spherical

nodes

Figure 7.17


Chapter 7

2p orbitals

Figure 7.18

nodal planes


Chapter 7

3d orbitals

Figure 7.19

perpendicular nodal planes


Chapter 7

Figure 7.19 (continued)


Chapter 7

One of the seven possible 4f orbitals

Figure 7.20


Chapter 7

The energy levels

in the hydrogen atom

The energy level

depends only on the

n value of the orbital

Figure 7.21


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