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The Quantum The Quantum Mechanical Model Mechanical Model

of the Atomof the Atom

Chapter 7

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The nature of lightThe nature of light

Light is electromagnetic radiation: a wave of oscillating electric & magnetic fields.

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Wave propertiesWave properties

A wave has wavelength, frequency, and amplitude

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Wave propertiesWave properties

Wavelength = (lambda), in mFrequency = (nu), in cycles/sec or s-1 Wavelength & frequency related by wave speed:

Speed of light c = 3.00 x 108 m s-1

c

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Example1Example1

Calculate the frequency of light with = 589 nm

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Example 1Example 1

Calculate the frequency of light with = 589 nm = c c = 3.00 x 108 m s-1

Must convert nm to m so units agree (n = 10-9)

c

c

3.00 108 ms 1

589 10 9 m5.09 1014 s 1

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Example 2Example 2

Calculate the wavelength in nm of light with = 9.83 x 1014 s-1

c

c

3.00 108 ms 1

9.831014 s 13.05 10 7 m

305 10 9 m  305nm

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Electromagnetic spectrum = all Electromagnetic spectrum = all wavelengths of electromagnetic radiationwavelengths of electromagnetic radiation

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Photoelectric effectPhotoelectric effectLight shining on metal surface can cause metal to

emit e- (measured as electric current)

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Photoelectric effectPhotoelectric effect

Classical theorymore e- emitted if light either brighter (amplitude) or

more energetic (shorter )e– still emitted in dim light if given enough time for e-

to gather enough energy to escape

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Photoelectric effectPhotoelectric effect

Observations do not match classical predictions!A threshold frequency exists for e- emission: no e-

emitted below that regardless of brightnessAbove threshold , e- emitted even with dim lightNo lag time for e- emission in high , dim light

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Albert Einstein explainsAlbert Einstein explainsphotoelectric effectphotoelectric effect

Light comes in packets or particles called photonsAmount of energy in a photon related to its

frequency

h = 6.626 x 10-34 J s

E h hc

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Example 3Example 3

What is the energy of a photon with wavelength 242.4 nm?

E h hc

E 6.626 10 34 Js 3.00 108 ms 1

242.4 10 9 m8.20 10 19 J

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Example 4Example 4

A Cl2 molecule has bond energy = 243 kJ/mol. Calculate the minimum photon frequency required to dissociate a Cl2 molecule.

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Example 4Example 4243103 J

mol

1 mol

6.02 1023 photons

4.04 10 19 J

1  photon

E h

E

h

4.04 10 19 J

6.626 10 34 Js6.10 1014 s 1

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Example 4Example 4

What color is this photon?

A photon with wavelength 492 nm is blue

c

3.00 108 ms 1

6.10 1014 s 14.92 10 7 m  492 10 9 m

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Example 4Example 4

A Cl2 molecule has bond energy = 243 kJ/mol. Calculate minimum photon frequency to dissociate a Cl2 molecule. What color is this photon?

A photon with wavelength 492 nm is blue

243103 J

mol

1 mol

6.02 1023 photons

4.04 10 19 J

1  photon

E h

E

h

4.04 10 19 J

6.626 10 34 Js6.10 1014 s 1

c

3.00 108 ms 1

6.10 1014 s 14.92 10 7 m  492 10 9 m

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Emission spectraEmission spectra

When atom absorbs energy it may re-emit the energy as light

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Emission spectraEmission spectra

White light spectrum is continuous

Atomic emission spectrum is discontinuous

Each substance has a unique line pattern

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Hydrogen emission spectrumHydrogen emission spectrum

Visible lines at410 nm (far violet)434 nm (violet)486 nm (blue-green)656 nm (red)

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Emission spectraEmission spectra

Classical theories could not explain Why atomic emission spectra were not continuousWhy electron doesn’t continuously emit energy as it

spirals into the nucleus

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Bohr modelBohr model

Niels Bohr’s model to explain atomic spectraelectron = particle in circular orbit around nucleusOnly certain orbits (called stationary states) can exist

rn = orbit radius, n = positive integer, a0 = 53 pm

Electron in stationary state has constant energyRH = 2.179 x 10–18 J

Bohr model is quantized

rn n2a0

En RH

n2

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Bohr modelBohr model

e– can pass only from one allowed orbit to anotherWhen making a transition, a fixed quantum of

energy is involved

E electron E final E initial

E electron RH

n f2

RH

ni2 2.179 10 18 J

1

ni2

1

n f2

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Bohr modelBohr model

Calculate the wavelength of light emitted when the hydrogen electron jumps from n=4 to n=2

Photon energy is an absolute amount of energyElectron absorbs photon, ∆Eelectron is +

Electron emits photon, ∆Eelectron is –

Eelectron 2.179 10 18 J1

ni2

1

n f2

Eelectron 2.179 10 18 J1

42

1

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   4.09 10 19 J

Ephoton 4.09 10 19 J

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Bohr modelBohr model

Calculate the wavelength of light emitted when the hydrogen electron jumps from n=4 to n=2

486 nm corresponds to the blue-green line

Eelectron 2.179 10 18 J1

ni2

1

n f2

Ephoton 4.09 10 19 J

Ephoton 4.09 10 19 J hc

6.626 10 34 Js 3.00 108 ms 1

4.86 10 7 m 486 10 9 m 486nm

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ExampleExample

What wavelength of light will cause the H electron to jump from n=1 to n=3? To what region of the electromagnetic spectrum does this photon belong?

Eelectron 2.179 10 18 J1

ni2

1

n f2

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ExampleExample

What wavelength of light will cause the H electron to pass from n=1 to n=3?

Eelectron 2.179 10 18 J1

12

1

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1.94 10 18 J

Ephoton 1.94 10 18 J

Eelectron 2.179 10 18 J1

ni2

1

n f2

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ExampleExample

What wavelength of light will cause the H electron to pass from n=1 to n=3?

The atom must absorb an ultraviolet photon with = 103 nm

Ephoton 1.94 10 18 J hc

6.626 10 34 Js 3.00 108 ms 1

1.0310 7 m 10310 9 m 103nm

Eelectron 2.179 10 18 J1

ni2

1

n f2

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Light has both particle & wave behaviorsLight has both particle & wave behaviors

Wave nature shown by diffraction

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Light has both particle & wave behaviorsLight has both particle & wave behaviors

Particle nature shown by photoelectric effect

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Electrons also have wave propertiesElectrons also have wave properties

Individual electrons exhibit diffraction, like waves

How can e– be both particle & wave?

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ComplementarityComplementarity

Without laser, single e– produces diffraction pattern (wave-like)

With laser, single e– makes a flash behind one slit or the other, indicating which slit it went through –– and diffraction pattern is gone (particle-like)

We can never simultaneously see the interference pattern and know which slit the e– goes through

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ComplementarityComplementarity

Complementary properties exclude each otherIf you know which slit the e– passes through (particle),

you lose the diffraction pattern (wave)If you see interference (wave), you lose information

about which slit the e– passes through (particle)Heisenberg uncertainty principle sets limit

on what we can know² x m² v h

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IndeterminancyIndeterminancy

Classical outcome is predictable from starting conditions Quantum-mechanical outcome not predictable but we can

describe probability region

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Electrons & probabilityElectrons & probability

Schrodinger applied wave mechanics to electronsEquation (wave function, ) describe e– energyEquation requires 3 integers (quantum numbers)Plot of 2 gives a probability distribution map of e–

location = orbitalSchrodinger wave functions successfully predict

energies and spectra for all atoms

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Quantum numbersQuantum numbers

Principal quantum number, nDetermines size and overall energy of orbitalPositive integer 1, 2, 3 . . .Corresponds to Bohr energy levels

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Quantum numbersQuantum numbers

Angular momentum quantum number, lDetermines shape of orbitalPositive integer 0, 1, 2 . . . (n–1)Corresponds to sublevels

l

letter

0 s

1 p

2 d

3 f

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Quantum numbersQuantum numbers

Magnetic quantum number, ml

Determines number of orbitals in a sublevel and orientation of each orbital in xyz space

integers –l . . . 0 . . . +l

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Shapes of orbitalsShapes of orbitals

s orbital (l = 0, ml = 0)

p orbitals (l = 1, ml = –1, 0, +1)

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Shapes of orbitalsShapes of orbitals

d orbitals (l = 2, ml = –2, –1, 0, +1, +2)

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What type of orbital is designated by each set of quantum numbers?n = 5, l = 1, ml = 0

n = 4, l = 2, ml = –2

n = 2, l = 0, ml = 0

Write a set of quantum numbers for each orbital4s3d5p

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What type of orbital is designated by each set of quantum numbers?n = 5, l = 1, ml = 0 5p

n = 4, l = 2, ml = –2 4d

n = 2, l = 0, ml = 0 2s

Write a set of quantum numbers for each orbital4s n = 4, l = 0, ml = 0

3d n = 3, l = 2, ml = –2, –1, 0, +1, or +2

5p n = 5, l = 1, ml = –1, 0, or +1

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Electron configurationsElectron configurations

Aufbau principle: e– takeslowest available energy

Hund’s rule: if there are 2 or more orbitals of equal energy (degenerate orbitals), e– will occupy all orbitals singly before pairing

Pauli principle: Adds a 4th quantum number, ms (spin) No two e– in an atom can have the same set of 4 quantum

numbers ⇒ 2 e– per orbital

ms 12  or   1

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