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Chemistry-140 Lecture 17

- Chapter 7 Atomic Structure

- Chapter Highlights
- electromagnetic radiation
- photons Plancks constant
- Bohr model of the atom
- Rydberg equation
- quantum mechanics
- orbitals
- Heisenberg uncertainty principle
- quantum numbers

Chemistry-140 Lecture 17

- Electronic Structure Electromagnetic Radiation

- Electronic structure of an atom detailed

description of the arrangement of electrons in

the atom - Electromagnetic radiation electrical and

magnetic waves traveling at 2.9979 x 108 m/s

(speed of light, c). Includes visible light,

radio waves, microwaves, infrared

(heat),ultraviolet, X-ray, and g-ray radiation.

Chemistry-140 Lecture 17

- Electromagnetic Radiation

- Wavelength, l distance between two successive

peaks or troughs of a wave. Units are length

(m). - Frequency, n number of complete waveforms that

pass through a point in one second. Units of

s-1, /s, or hertz (Hz). - Relationship
- speed of light (wavelength) x (frequency)

c ln

Chemistry-140 Lecture 17

Chemistry-140 Lecture 17

- Electromagnetic Radiation

Question Yellow light of a sodium vapour

lamp has a wavelength of 589 nm. What is the

frequency of this light?

Chemistry-140 Lecture 17

Answer Since we know

then n

c ln

5.09 x 1014 s-1

Chemistry-140 Lecture 17

- Quantum Effects Photons

- Max Planck proposed that radiation is not

continuous, but rather consists of small pieces

known as quanta (a quantum). - Frequencies, n, of these quanta were

whole-number multiples of a fundamental

frequency. - Energies E hn, 2hn, 3hn,...
- where h Planck's constant
- h 6.626 x 10-34 J-s.
- and

E hn

Chemistry-140 Lecture 17

- Quantum Effects Photons

Question A laser emits light energy in short

pulses with frequency 4.69 x 1014 Hz and deposits

1.3 x 10-2 J for each pulse. How many quanta of

energy does each pulse deposit ?

Chemistry-140 Lecture 17

Answer Step 1 Determine the energy of one

quantum (photon). E hn

(6.63 x 10-34 J-s) (4.69 x 1014 s-1)

3.11 x 10-19 J

Chemistry-140 Lecture 17

Step 2 Determine how many quanta are in a

laser pulse. Number

4.18 x 1016 quanta

Chemistry-140 Lecture 17

- Photoelectric Effect

- Photoelectric effect metallic surfaces produce

electricity (electrons are ejected) when exposed

to light. - There is a minimum frequency below which
- no electricity is produced.
- Above the minimum frequency
- i) number of electrons ejected depends only on

light intensity, - ii) energy of the ejected electrons depends only

on the frequency of the light.

Chemistry-140 Lecture 17

Chemistry-140 Lecture 17

- Photoelectric Effect

The packet of energy sufficient to eject an

electron is called a photon. The kinetic energy

EK of the electrons is given by EB

binding energy EP photon energy hn

EK EP - EB

Chemistry-140 Lecture 17

- Photoelectric Effect

Question Potassium metal must absorb

radiation with a minimum frequency of 5.57 x 1014

Hz before it can emit electrons from its surface

via the photoelectric effect. If K(s) is

irradiated with light of wavelength 510 nm, what

is the maximum possible velocity of an emitted

electron?

Chemistry-140 Lecture 17

Answer Step 1 Convert threshold frequency to

binding energy. Eb hn

(6.63 x 10-34 J-s) (5.57 x 1014 s-1)

3.69 x 10-19 J

Chemistry-140 Lecture 17

Step 2 Determine the photon energy of 510 nm

light. EP

3.90 x 10-19 J

Chemistry-140 Lecture 17

Step 3 Determine the kinetic energy of the

emitted electrons EK EP -

EB (3.90 x 10-19 J) - (3.69 x 10-19

J)

2.10 x 10-20 J

Chemistry-140 Lecture 17

Step 4 Calculate the velocity of the emitted

electrons EK mv2 2.10 x

10-20 J v

2.15 x 105 m/s

Chemistry-140 Lecture 17

- Bohrs Model of the Hydrogen Atom

- A spectrum is produced when radiation from a

source is separated into its component

wavelengths. - Bohr used Planck's quantum theory to interpret

the line spectrum of hydrogen. - Bohr's model of the hydrogen atom described a

nucleus surrounded orbits of fixed (quantized)

radius, - numbered n 1, 2, 3,...

Chemistry-140 Lecture 17

Chemistry-140 Lecture 17

Chemistry-140 Lecture 17

- Bohrs Model of the Hydrogen Atom

- Bohr concluded
- the energy of the electron in an orbit of

hydrogen is quantized - the energy difference between two orbits must

also be quantized - The frequency of a line in the spectrum

corresponds to the energy difference between two

orbits

DE hn

Chemistry-140 Lecture 17

- Bohrs Model of the Hydrogen Atom

- The energy of a Bohr orbit (and an electron in

it) is given by - where RH is the Rydberg constant 2.179 x

10-18 J

Chemistry-140 Lecture 17

Transitions in the Bohr Hydrogen Atom

Chemistry-140 Lecture 17

- Transitions and the Rydberg Equation

- An electron in the lowest energy orbit, n 1, is

in the ground state - An electron in any orbit other than n 1, is in

an - excited state
- The energy of a line is the difference in the

energies of the two orbits involved in the

transition - DE Efinal - Einitial

Chemistry-140 Lecture 17

- Transitions in the Bohr Hydrogen Atom

- The radius of a Bohr orbit is given by
- r n2(5.30 x 10-11 m)
- The ionization energy of hydrogen is the energy

required to remove the electron from the atom,

that is - the energy of the n 1 to n transition

Chemistry-140 Lecture 17

- Transitions in the Bohr Hydrogen Atom

Question (similar to example 7.4) Calculate

the wavelength of light that corresponds to the

transition of the electron from the n 4 to the

n 2 state of the hydrogen atom. Is the light

absorbed or emitted by the atom?

Chemistry-140 Lecture 17

Answer Step 1 Use the Rydberg equation with

ni 4 and nf 2. n

-6.17 x 1014 s-1

Chemistry-140 Lecture 17

Step 2 Convert to wavelength of

light l

4.86 x 10-7 m

c ln

486 nm

Chemistry-140 Lecture 18

- Chapter 7 Atomic Structure

- Chapter Highlights
- electromagnetic radiation
- photons Plancks constant
- Bohr model of the atom
- Rydberg equation
- quantum mechanics
- Heisenberg uncertainty principle
- orbitals
- quantum numbers

Chemistry-140 Lecture 18

- Bohrs Model of the Hydrogen Atom

- The energy of a Bohr orbit (and an electron in

it) is given by - where RH is the Rydberg constant 2.179 x

10-18 J

Chemistry-140 Lecture 18

- Transitions and the Rydberg Equation

- An electron in the lowest energy orbit, n 1, is

in the ground state - An electron in any orbit other than n 1, is in

an - excited state
- The energy of a line is the difference in the

energies of the two orbits involved in the

transition - DE Efinal - Einitial

Chemistry-140 Lecture 18

Transitions in the Bohr Hydrogen Atom

Chemistry-140 Lecture 18

- Transitions in the Bohr Hydrogen Atom

- The radius of a Bohr orbit is given by
- r n2(5.30 x 10-11 m)
- The ionization energy of hydrogen is the energy

required to remove the electron from the atom,

that is - the energy of the n 1 to n transition

Chemistry-140 Lecture 18

- Transitions in the Bohr Hydrogen Atom

Question Calculate the wavelength of light

that corresponds to the transition of the

electron from the n 4 to the n 2 state of the

hydrogen atom. Is the light absorbed or emitted

by the atom?

Chemistry-140 Lecture 18

Answer Step 1 Use the Rydberg equation with

ni 4 and nf 2. n

-6.17 x 1014 s-1

Chemistry-140 Lecture 18

Step 2 Convert to wavelength of

light l

4.86 x 10-7 m

c ln

486 nm

Chemistry-140 Lecture 18

- Dual Nature of the Electron

- De Broglie proposed that particles may behave as

if they were waves. Similar to the idea that

light may behave as if it was a particle. Matter

wave is the term used by de Broglie where - where momentum mv, (m is mass v is

velocity)

Chemistry-140 Lecture 18

- Dual Nature of the Electron

Question What is the characteristic

wavelength of an electron with velocity 5.97 x

106 m/s ? (mass of an electron is 9.11 x 10-28 g)

Chemistry-140 Lecture 18

Answer Use de Broglie's equation for matter

waves.

1.22 x 10-10 m

0.122 nm

Chemistry-140 Lecture 18

- Quantum Mechanics

- Heisenberg Uncertainty Principle Werner

Heisenberg proposed the uncertainty principle,

which states that it is impossible for us to

know, simultaneously, both the exact momentum of

an electron and its exact location in space

Chemistry-140 Lecture 18

- Quantum Mechanics

- Schrödinger Wave Equation Erwin Schrödinger

proposed a mathematical model of the atom using

measured energies and known forces rather than a

preconceived "picture" of the atom's structure.

This is called quantum mechanics or wave

mechanics.

Chemistry-140 Lecture 18

- Wave Functions Probability

- Solutions to the wave equation are called wave

functions, symbolized y. Wave functions cannot

describe the exact position of an electron only

the probability of finding it in a given location.

- The probability of finding the electron in a

given location is the electron density and is

given by the square of the wave function for that

location, y2.

Chemistry-140 Lecture 18

- The Wave Equation Orbitals

- Solutions to the wave equation are called

orbitals..

Chemistry-140 Lecture 19

- Chapter 7 Atomic Structure

- Chapter Highlights
- electromagnetic radiation
- photons Plancks constant
- Bohr model of the atom
- Rydberg equation
- quantum mechanics
- Heisenberg uncertainty principle
- quantum numbers
- orbitals

Chemistry-140 Lecture 19

- The Wave Equation Orbitals

- Solutions to the wave equation are called

orbitals and - each has a characteristic energy.
- An orbital is a region for which there is a high

probability of finding the electron it is not a

path or trajectory.

Chemistry-140 Lecture 19

- The Wave Equation Quantum Numbers

- Variables in the wave equation are called quantum

numbers. The Bohr model used only one variable

or quantum number, n. The quantum mechanical

model uses three quantum numbers, n, l ml to

describe each orbital

y n, l, m (r, q, f) Rnl(r)Clm(q, f)

Chemistry-140 Lecture 19

- Quantum Numbers

- The principal quantum number (n) has possible

values of - It describes the relative size of the orbital

n 1, 2, 3,...

Chemistry-140 Lecture 19

- Quantum Numbers

- The angular momentum quantum number (l)
- has possible values of
- It describes the shape of the orbital.
- The value of l is often referred to by a letter

equivalent - 0 s, 1 p, 2 d, 3 f, .... (the rest are

alphabetical)

l 0, 1, 2, ...n-1

Chemistry-140 Lecture 19

- Quantum Numbers

- The magnetic quantum number ( ml ) has values
- It describes the orientation of the orbital in

space.

ml -l,... -1, 0, 1, ...l

Chemistry-140 Lecture 19

- Electronic Shells Sub-shells

- A collection of orbitals with the same value of n

- is called an electron shell
- A collection of orbitals with the same values of

n and l is called an electronic subshell. A

subshell can be referred to using n and the

letter equivalent of l,

1s, 2s, 2p, 3s, 3p 3d, 3f etc

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Chemistry-140 Lecture 19

- s Orbitals

- The s orbitals are those for which l 0. All s

orbitals are spherical. There is one s orbital

in each s subshell.

Chemistry-140 Lecture 19

- Probability Density in Orbitals

Chemistry-140 Lecture 19

- Radial Distribution in Orbitals

Chemistry-140 Lecture 19

- p Orbitals

- The p orbitals are those for which l 1. All p

orbitals are "dumbbell" or "figure-eight" shaped.

There are - three p orbitals in each p subshell..

Chemistry-140 Lecture 19

- d Orbitals

- The d orbitals are those for which l 2.
- There are five d orbitals in each d subshell.
- Four are "four-leaf clovers" the fifth looks

like a p orbital with the addition of a ring

around the centre

Chemistry-140 Lecture 19

- d Orbitals

- The d orbitals are those for which l 2.
- There are five d orbitals in each d subshell.
- Four are "four-leaf clovers" the fifth looks

like a p orbital with the addition of a ring

around the centre

Chemistry-140 Lecture 19

Nodal Planes in Orbitals

Chemistry-140 Lecture 19

- f Orbitals

- The f orbitals are those for which l 3.
- There are seven f orbitals in each f subshell.
- Each has 8 lobes

Chemistry-140 Lecture 19

- Textbook Questions From Chapter 7
- EM Radiation 22, 28, 30
- Photoelectric Effect 32
- Bohr Atom 35, 38, 40
- Matter Waves 42
- Quantum Mechanics 46, 48, 50, 57, 58, 60
- General Conceptual 68, 72, 77, 79, 91