Chapter 7

Quantum Theory and Atomic Structure

Electro-Magnetic radiation includes visible light, microwave, TV, radio, x-ray, etc.

Radiation is a combination of vibrating electric and magnetic fields in repeatable waveforms

Wavelength, λ (lambda): distance from crest to crest

Frequency, (nu): # crests to pass a point in 1 second, units are #/s or Hz

Wave velocity = λ All E-M radiation in a vacuum has constant velocity All E-M radiation in a vacuum has constant velocity

called the speed of light: called the speed of light: c = 2.998 x 10c = 2.998 x 1088 m/s m/s..Therefore Therefore c = λ c = λ (Memorize formula and c)(Memorize formula and c)

Long λ => short & vice-versa

Figure 7.1

Frequency and Wavelength

c =

The Wave Nature The Wave Nature ofof

LightLight

Figure 7.2 Amplitude (intensity) of a wave.

Figure 7.3 Regions of the electromagnetic spectrum.

Wavelength above in nanometers,Frequency below in Hertz or #/second

Frequency unit is #/s or s-1.

All waves travel at the same speed through vacuum but differ in frequency and wavelength

Sample Problem 7.1a

SOLUTION:PLAN:

Interconverting Wavelength and Frequency

wavelength in units given

wavelength in m

frequency (s-1 or Hz)

= c/

Use c = = 1.000 x 10-10 m

=2.998 x 108 m/s

1.000 x 10-10 m

= 2.998 x 1018 s-1

PROBLEM: A dental hygienist uses x-rays (= 1.000A) to take a series of dental radiographs. What is the frequency (in s-1) of the electromagnetic radiation? (Assume that the radiation travels at the speed of light, 2.998 x 108 m/s.)

1 A = 10-10 m1 cm = 10-2 m1 nm = 10-9 m

o

1.000Ao 10-10 m

1Ao

E-M radiation was considered to be a wave/energy phenomenon and not matter

Max Planck developed a new physics when classical physics could not be used to interpret data from blackbody radiation ( Blackbody is an object that absorbs all radiation incident on it)

Blackbody radiation is emitted by solid bodies that are heated to high T and become incandescent

Classical physics had to assume continuous radiation, and it could not resolve the data that there was discrete radiation

Planck developed theory of Packets of Energy called quanta

The energy associated with quanta was proportional to the frequency of the radiation: E = h h = Planck’s constant 6.626 x 10-34 J.s

If c = , and E = h, then with rearranging and substituting:

E = hc/

What is the energy of a photon with a wavelength of 399.0 nm?

Figure 7.6 Blackbody radiation (4th ed.)E = n h

E = n h

E = h when n = 1

smoldering coal

electric heating element light bulb filament

Solid heated To 1000 K It emits visible light)

At 1500 KAt 2000K

Figure 7.6

Demonstration of the photoelectric effect.

Wave model could not explain photoelectric effect.Flow of current when monochromatic light of suff

frequency falls on metal plate.Photoelectric Effect: electrons are ejected from a

metal's surface if it is exposed to uv radiationEach metal required a characteristic minimum uv

frequency to start ejecting e-s Called Threshold freq, o

- As increases more e-s ejected with higher vel (KE)These data also defied classical physical explanationEinstein reviewed data, recalled Planck's Einstein reviewed data, recalled Planck's quantaquantaThe "incident" radiation consists of quanta of energy, E

= h, called photons(small bundle of electromagnetic energy) - thus the PHOTOELECTRIC Effect

In order to eject an e-, a min KE is required, E = ho If E>ho then excess KE is supplied to the e-, increasing

its velocityFor Na metal, o = 5.51 x 1014 Hz

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 can use the energy equation, E = h combined with = c/ to find the energy.

E = hc/

E =(6.626 x 10-34 J*s) 2.998 x 108 m/s

1.20 cm x 10-2 m

cm

= 1.66 x 10-23 J

Experiments with "excited" atoms of H produced emission spectra - always a discrete set of lines at certain wavelengths

White light dispersed by a prism or diffraction grating: - we see ROYGBIV – a continuous spectrum from 750 nm to 400 nm

When a gas-filled tube is charged with current, only certain EM 's are detected - called a line spectrum or emission spectrum

The gas particles split into individual atomsThe e-s are excited by the current into a higher energy

level.When they drop down, they emit energy of a certain λ, with energy gaps at distinct intervals

Figure 7.7

The line spectra of

several elements.

Hydrogen atomic line spectra – also called emission spectra – - worked out mathematically (by several scientists) to define the energy of the light emitted & relationships between the lines

Balmer: red, green, blue, and violet lines (656.3, 486.1, 434.0, 410.1nm)

1/λ = R1/λ = Ryy(1/2(1/222 - 1/n - 1/n22)) If n > 2 Ry = Rydberg constant

= 1.096776 x 107/mIf n = 3 get red, n = 4 get green, n = 5 get

blue, n = 6 get violet

= RRydberg equation* -1

1

n22

1

n12

R is the Rydberg constant = 1.096776x107 m-1

Figure 7.8 Three series of spectral lines of atomic hydrogen. Balmer is in the visible region, and the other series, which have names also, are in uv or ir area.

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

*Memorize!

1. H atoms have only certain allowable energy levels called stationary states.

2. Atom does not radiate energy while in a stationary state.

3. Atoms changes to another stationary state by absorbing or emitting a photon.

Energy=EstateA-EstateB=h

Bohr found En = - Rhc/n2 R = 1.097 x 107/m h = 6.626 x 10-34 J.sc = speed of lightRhc = 2.178 x 10-18 J (since they are all constant)Then En= -2.178 x 10-18 J/n2

All E is therefore < 0, and has discrete values onlyNucleus (proton) & e- are so far apart there's no

attraction anymoreNegative E is more stable than zero energyn = 1 is the ground state, all above are excited

states

Figure 7.9

Quantum staircase.

Figure 7.10 The Bohr explanation of the three series of spectral lines.

Instrumental techniques used to obtain information about atomic or molecular energy levels

Emission: electrons in an atom are excited to a higher energy state and then emit photons as they return to lower energy states

Absorption: electrons in an atom absorb photons of certain wavelengths and jump to higher energy states; photons NOT absorbed are observed!

See figure 7.11 in text: why chlorophyll looks green

Figure B7.2

Figure B7.1 (4th ed.)

Emission and absorption spectra of sodium atoms.

Flame tests.

strontium 38Sr copper 29Cu

A hydrogen atom has an e- excited up to level 4, and it drops back to level 2. (a) determine the wavelength of the photon emitted and (b) the energy difference.

(a) Use 1/ = Ry(1/22 – 1/42)

= 4.8617 x 10-7 m

(b) Use E = hc/E = 4.086 x 10-19 J

Follow-up: Answer the same questions for the e- excited up to level 6.

Stationary wave: - fixed at both ends- has "nodes" - never moves on those spots with distance = length/2

Only certain λ's are possible for a standing wave

Figure 7.12

Wave motion in restricted systems.

Einstein remembered for E = mc2 m = E/c2 = (hc/λ)/c2 = h/λc

This appears to say that a photon of a certain wavelength has mass!

Proved by Arthur Compton in 1922E-M radiation is both waves & little packets of

energy and matter called photonsDe Broglie 1923: if light has wave-particle duality,

then matter, which is particle-like, must also be wavelike under certain conditions

Rearranged m = h/c to get λ = h/mvThis is called the deBroglie wavelengthIt means that all matter exhibits both particle and

wave properties

Sample Problem 7.4

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 106 m/s (electron mass = 9.11 x 10-31 kg; h = 6.626 x 10-34 J.s).

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

= 6.626 x 10-34 (kg*m2/s2)s

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

Bohr’s Theory: 1 e- in H atom occupying certain energy states - a certain quanta

Spherical orbitals around the nucleusWith de Broglie's hypothesis: e- must have

a certain λ to make a complete revolution - like a standing wave

An integral # of complete λ's to fit the sphere's circumference

Circumference = 2 r, therefore nλ = 2 r, n = 1, 2, 3....

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 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 7.14

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

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 7.14 continued

QUANTUM THEORY

Energy same as Matterparticulate, massive, wavelike

It is impossible to know the exact position andmomentum of a particle simultaneously.

Uncertainty: (Δx)(mΔv) > h/4Δx is the location of the electron mΔv is its momentumMore accurately we know the position of the

particle less accurately we know the speed.Need Δ because we can’t know both at the

same time

Sample Problem 7.4 (4th ed)

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 (x) is given as ±1%(0.01) of 6 x 106 m/s. Once we calculate this, plug it into the uncertainty equation.

u = (0.01)(6 x 106 m/s) = 6 x 104 m/s

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/s2).s= 9.52 x 10-9 m

1 J = 1 kg*m2/s2

Schrodinger developed Wave Functions, Ψ(psi), where Ψ2 is the probability of finding e- in a given space

Led to 4 quantum numbers that describe the e-'s Led to 4 quantum numbers that describe the e-'s position in a complex equation:position in a complex equation:

1. Only certain wave functions are allowed2. Each Ψn corresponds to an allowed energy for e- in atom3. Thus energy of e- is quantized4. Ψ has no physical meaning, but Ψ2 give the probability density5. Allowed energy states are called orbitals6. 3 integer #'s req'd to solve Ψ2 for 3-D space: n, l, ml

An atomic orbital is specified by three quantum numbers.

n = principal quantum number, a positive integer = 1, 2, 3,...

- determines total E of e- in its electron shell- gives measure of prob distance from nucleus

(orbital size)- 2 or more e-s can be in same electron shell

l = angular momentum or shape = < n - 1 , = 0,1,2,...- subshells w/in main shell, characterized by certain

wave shapes0 = s, 1 = p, 2= d, 3 = f, etc.

ml = magnetic q.n. = +l, +l -1, +l - 2, … 0, ... -l- specifies which orbital w/in a subshell e- is in

(later we’ll do ms = spin q.n., +½ or -½ for each e-)

Watch: YouTube - The Quantum Number Rag

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

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 found in the text.

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

For l = 1 ml = -1, 0, or +1

For l = 2 ml = -2, -1, 0, +1, or +2

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

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, we can find ml and the number of orbitals.

n l sublevel name possible ml values # of 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

Ψ21s – the 1s orbital is spherical.

Ψ22s – the 2s orbital has some density close to nucleus and then another sphere farther away – a sphere within a sphere

Ψ22p – the 2p orbitals have no probabilty of e- at the nucleus - called nodal planeCan be oriented in 3 directions of 3-D graph - x, y, z. 2px, 2py, 2pz have the 3 ml “names” +1, 0 and -1

Ψ23d – the 3d orbitals have 5 ml values, and each has 2 nodal surfaces, so they are in four sections. 3dxy, 3dxz, 3dyz, 3dx2-y2, 3dz2

Ψ24f – the 4f orbitals have 7 ml values, 3 nodal surfaces

Figure 7.15

Electron probability in the ground-state H atom.

Figure 7.16 The 1s, 2s, and 3s orbitals

1s 2s 3s

Figure 7.17

The 2p orbitals.

Figure 7.18 The 3d orbitals.

Figure 7.18 continued

Figure 7.19 One of the seven possible 4f orbitals.

Be able to draw 1s, 2s, 2p, and 3d orbitals. Practice now!