of the Atom

Chapter 7 Lecture

Lecture Presentation

Chapter 7

The Quantum-

Mechanical Model of the Atom

Sherril Soman

Grand Valley State University

授課教師：林秀美

辦公室：綜合二館302B

TEL: 5562 email:hmlin@mail.ntou.edu.tw

• 評量方式：

• (一) 課堂互動討論（占總成績比例10%）：由授課教師決定。

• (二) 會考（占總成績比例90%）：

• 額外加分：維基百科3000字元一分，最多5分；可以問星期三的課輔助教，以潔與嘉珊；或

是參加維基台北寫作聚 每月第二個禮拜六。 。

•

• 1. 日期：第一次會考：10月23日(三)第二次會考：11月13日(三) 第三次會考：12月18日(三): Ch7,8,9

第四次會考：01月15日(三): Ch10,11,12

• 2. 時間：18:00~19:30 PM

• 3. 地點：延平技術大樓（考試當週於網站公布試場及座位）

• 4. 注意事項：如普通化學課程會考施行細則（CHEM003）。

• 5. 會考成績：會考結束七日內公告（http://academic.ntou.edu.tw/exam）

• (三) 積極性補強教學出席率、化學課後作業（10%加分制）：由授課教師決定。

• 詳參1021海洋大學整合型普通化學課程積極性補強教學施行細則。（CHEM004）。

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Pg. 329

Pg. 328

7.1 Schrodinger’s Cat

http://www.youtube.com/watch?v=BHEEdp79rIY

薛丁格的貓

The Beginnings of Quantum Mechanics量子 力學的開端

• Until the beginning of the twentieth century it was believed that all physical phenomena were

deterministic.直到二十世紀開始，所有的物理現象都

具有確定性。

• Work done at that time by many famous physicists discovered that for sub-atomic particles, the present condition does not determine the future condition.

– Albert Einstein, Neils Bohr, Louis de Broglie, Max Planck, Werner Heisenberg, P. A. M. Dirac, and Erwin Schrödinger

• 當時許多著名的物理學家發現了次原子粒子，目前 的狀況無法確定未來的狀況。

The Beginnings of Quantum Mechanics

• Quantum mechanics forms the foundation of chemistry量子力學奠定了基礎化學

– Explaining the periodic table

– The behavior of the elements in chemical bonding

– Provides the practical basis for lasers,

computers, and countless other applications 解釋週期表

的行為的元素的化學鍵

雷射，計算機，以及其他無數的應用提供了實踐基 礎

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The Behavior of the Very Small非常小的 行為

• Electrons are incredibly small.

電子是小得令人難以置信。

– A single speck of dust has more electrons than the number of people who have ever lived on Earth.

– 一個點塵埃的電子數比曾經生活在地球上的人的數量都還多

。

• Electron behavior determines much of the behavior of atoms. 電子行為決定大部分原子的行為。

• Directly observing electrons in the atom is

impossible; the electron is so small that observing it changes its behavior.

– Even shining a light on the electron would affect it.

直接觀察的原子中的電子是不可能的，電子是如此之小，觀察其行為改 變。

即使對電子發光光會影響它。

A Theory That Explains Electron Behavior 一個理論來解釋的電子行為

• The quantum-mechanical model explains the manner in which electrons exist and behave in atoms.

量子力學模型解釋原子中電子的存在和行為？

• It helps us understand and predict the properties of atoms that are directly related to the behavior of the electrons:它可以幫助我們了解和預測的原子的電子的行為有直接關係的屬性：

– Why some elements are metals and others are nonmetals – Why some elements gain one electron when forming an

anion, whereas others gain two

– Why some elements are very reactive, while others are practically inert

– Why in other periodic patterns we see in the properties of the elements

7.2 The Nature of Light: Its Wave Nature

• Light: a form of electromagnetic radiation

– Composed of perpendicular oscillating waves, one for the electric field and one for the

magnetic field

• An electric field is a region where an electrically charged particle experiences a force.

• A magnetic field is a region where a magnetized particle experiences a force.

• All electromagnetic waves move through space at the same, constant speed.

– 3.00 × 10⁸ m/s = the speed of light

Electromagnetic Radiation

7.2 The Nature of Light: Its Wave Nature

Composed of perpendicular oscillating waves, one for the electric field and one for the magnetic field An electric field is a region where an electrically charged particle experiences a force.

A magnetic field is a region where a magnetized particle experiences a force.

• All electromagnetic waves move through space at the same, constant speed.

– 3.00 × 10⁸ m/s = the speed of light

Speed of Energy Transmission

Wave Characteristics

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• The amplitude is the height of the wave.

– The distance from node to crest or node to trough

– The amplitude is a measure of light intensity—the larger the amplitude, the brighter the light.

• The wavelength (l) is a measure of the distance covered by the wave.

– The distance from one crest to the next

• The distance from one trough to the next, or the distance between alternate

nodes

Characterizing Waves

• The frequency (n) is the number of waves that pass a point in a given period of time.

– The number of waves = the number of cycles.

– Units are hertz (Hz) or cycles/s = s⁻¹ (1 Hz = 1 s⁻¹).

• The total energy is proportional to the amplitude of the waves and the frequency.

– The larger the amplitude, the more force it has.

– The more frequently the waves strike, the more total force there is.

**Frequency, Wavelength and Speed of Electromagnetic Radiation

• Frequency (

n

• Wavelength (λ) in meters—m.

•

^mm m nm Å pm

(10 ^-3 m) (10 ^-6 m) (10 ^-9 m) (10 ^-10 m) (10 ^-12 m)

• Velocity (c)—2.997925



c = λ n λ = c/ n n ^{= c/λ}

Color xx

• The color of light is determined by its wavelength or frequency.

• White light is a mixture of all the colors of visible light.

– A spectrum

– Red Orange Yellow Green Blue Indigo Violet

• When an object absorbs some of the

wavelengths of white light and reflects

others, it appears colored; t

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Chemistry: A Molecular Approach, 3rd Edition Nivaldo J. Tro

Solution

You are given the frequency of the light and asked to find its wavelength. Use Equation 7.1, which relates frequency to wavelength. You can convert the wavelength from meters to nanometers by using the conversion factor between the two (1 nm = 10^–9 m).

For Practice 7.1

A laser dazzles the audience in a rock concert by emitting green light with a wavelength of 515 nm. Calculate the frequency of the light.

Calculate the wavelength (in nm) of the red light emitted by a barcode scanner that has a frequency of 4.62 × 10¹⁴ s^–1.

Example 7.1 Wavelength and Frequency

**Electromagnetic Spectrum

• Visible light comprises only a small fraction of all the wavelengths of light, called the electromagnetic spectrum.

• Shorter wavelength (high-frequency) light has higher energy.

– Radio wave light has the lowest energy.

– Gamma ray light has the highest energy.

• High-energy electromagnetic radiation can potentially damage biological molecules.

– Ionizing radiation

Thermal Imaging using Infrared Light （IR）

Using High-Energy Radiation to Kill Cancer Cells

During radiation therapy, a tumor is

targeted from multiple directions in order to minimize the exposure of healthy cells, while maximizing the exposure of cancerous cells.

To produce a medical X-ray, the patient is exposed to short-

wavelength electromagnetic radiation that can pass

through the skin to create an image of bones and internal organs.

(X)

Interference(X)

建設性干涉

破壞性干涉

Diffraction(X)

• When traveling waves encounter an obstacle or opening in a barrier that is about the same size as the wavelength, they bend around it; this is called

– Traveling particles do not diffract.

• The diffraction of light

through two slits separated by a distance comparable to the wavelength results in an

diffracted waves.

• An interference pattern is a characteristic of all light waves.

http://www.dcfever.com/learn/viewglossary.php?glossary_id=19

Two-Slit Interference(X)

http://www.youtube.com/watch?v=ZImlf6S9WnE

**pg.302 The Particle Nature of Light:

The Photoelectric Effect

http://www.youtube.com/watch?v=ubkNGwu_66s

• It was observed that many metals emit electrons when a light shines on their surface.

– This is called the photoelectric effect.

• Classic wave theory attributed this effect to the light energy being transferred to the electron.

• According to this theory, if the wavelength of light is made shorter, or the light wave’s intensity made brighter, more electrons should be ejected.

– Remember that the energy of a wave is directly proportional to its amplitude and its frequency.

– This idea predicts if a dim light were used there would be a lag time before electrons were emitted.

• To give the electrons time to absorb enough energy

The Particle Nature of Light:

The Photoelectric Effect

The Photoelectric Effect: The Problem

• Experimental observations indicate the following:

– A minimum frequency was needed before electrons would be emitted regardless of the

intensity called the threshold frequency.

^臨界頻率

electron emission without any lag time.

**Einstein’s Explanation

• Einstein proposed that the light energy was

delivered to the atoms in packets, called quanta or photons.量子或光子。

• The energy of a photon of light is directly proportional to its frequen

cy.

– Inversely proportional to its wavelength

– The proportionality constant is called Planck’s

Constant, (h), and has the value 6.626 × 10

⁻³⁴ J ∙ s.

Ejected Electrons

• One photon at the threshold frequency gives the electron just enough energy for it to

escape the atom.克服原子核對電子的吸引力

– Binding energy, f

• When irradiated with a shorter wavelength photon, the electron absorbs more energy than is necessary to escape.

• This excess energy becomes kinetic energy of the ejected electron.

Kinetic Energy = E

– E

KE = hn − f

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Chemistry: A Molecular Approach, 3rd Edition Nivaldo J. Tro

Sort

You are given the wavelength and total energy of a light pulse and asked to find the number of photons it contains.

Given:

Find: number of photons

Strategize

In the first part of the conceptual plan, calculate the energy of an individual photon from its wavelength.

In the second part, divide the total energy of the pulse by the energy of a photon to get the number of photons in the pulse.

Conceptual Plan

A nitrogen gas laser pulse with a wavelength of 337 nm contains 3.83 mJ of energy. How many photons does it contain?

Example 7.2 Photon Energy

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Chemistry: A Molecular Approach, 3rd Edition

Relationships Used

E = hc/λ (Equation 7.3)

Solve

To execute the first part of the conceptual plan, convert the wavelength to meters and substitute it into the equation to calculate the energy of a 337 nm photon.

To execute the second part of the conceptual plan, convert the energy of the pulse from mJ to J. Then divide the energy of the pulse by the energy of a photon to obtain the number of photons.

Solution Continued

Example 7.2 Photon Energy

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Chemistry: A Molecular Approach, 3rd Edition Nivaldo J. Tro

Check

The units of the answer, photons, are correct. The magnitude of the answer (10¹⁵) is reasonable. Photons are small particles and any macroscopic collection should contain a large number of them.

For Practice 7.2

A 100-watt lightbulb radiates energy at a rate of 100 J/s (The watt, a unit of power, or energy over time, is defined as 1 J/s.) If all of the light emitted has a wavelength of 525 nm, how many photons are emitted per second? (Assume three significant figures in this calculation.)

For More Practice 7.2

The energy required to dislodge electrons from sodium metal via the photoelectric effect is 275 kJ/mol. What wavelength in nm of light has sufficient energy per photon to dislodge an electron from the surface of sodium?

Continued

Example 7.2 Photon Energy

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Chemistry: A Molecular Approach, 3rd Edition

Solution

Examine Figure 7.5 and note that X-rays have the shortest wavelength, followed by visible light and then microwaves.

a. wavelength

X rays < visible < microwaves

Since frequency and wavelength are inversely proportional—the longer the wavelength the shorter the frequency—the ordering with respect to frequency is the reverse of the ordering with respect to wavelength.

b. frequency

microwaves < visible < X rays

Energy per photon decreases with increasing wavelength but increases with increasing frequency; therefore, the ordering with respect to energy per photon is the same as for frequency.

c. energy per photon

microwaves < visible < X rays

Arrange the three types of electromagnetic radiation—visible light, X-rays, and microwaves—in order of increasing a. wavelength. b. frequency. c. energy per photon.

Example 7.3 Wavelength, Energy, and Frequency

FIGURE 7.5 The Electromagnetic Spectrum The right side of the spectrum consists of high-energy, high-frequency, short-wavelength radiation. The left side consists of low-energy, low-frequency, long-wavelength radiation. Visible light constitutes a small segment in the middle.

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For Practice 7.3

Arrange these three colors of visible light—green, red, and blue—in order of increasing a. wavelength. b. frequency. c. energy per photon.

Continued

Example 7.3 Wavelength, Energy, and Frequency

7.3 Atomic Spectroscopy and the Bohr Model

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Spectra

Exciting Gas Atoms to Emit Light with Electrical Energy

When atoms or molecules absorb energy, that energy is often released as light energy,

Fireworks, neon lights, etc.

When that emitted light is passed through a prism, a pattern of particular wavelengths of light is seen that is unique to that type of atom or molecule – the pattern is called an emission

Noncontinuous

Can be used to identify the material Flame tests

Na K Li Ba

Continuous Spectrum

量子 Emission Spectra

古典 Continuous Spectrum

Examples of Spectra

Bohr Model of H Atoms

Rydberg’s Spectrum Analysis

Johannes Rydberg (1854–1919)

The Bohr Model of the Atom

• Bohr’s major idea was that the energy of the atom was quantized, and that the amount of energy in the atom was related to the electron’s position in the atom.

– Quantized means that the atom could only

have very specific amounts of energy.

Neils Bohr (1885–1962)

• The electrons travel in orbits that are at a fixed distance from the nucleus.

– Stationary states

– Therefore, the energy of the electron was proportional to the distance the orbit was from the nucleus.

• Electrons emit radiation when they “jump” from an orbit with higher energy down to an orbit with lower energy.

– The emitted radiation was a photon of light.

– The distance between the orbits

determined the energy of the photon of light

produced.

101

Fireworks typically contain the salts of such metals as

sodium, calcium, strontium, barium, and copper.

Emissions from these elements produce the

brilliant colors of pyrotechnic

displays.

Emission versus Absorption Spectra

Spectra of Mercury

Emission

Absorption

7.4 The Wave Nature of Matter: The de Broglie Wavelength, the Uncertainty Principle, and

Indeterminacy

• de Broglie predicted that the wavelength of a particle was inversely proportional to its

momentum.

• Because it is so small, the wave character of electrons is significant.

Louis de Broglie (1892–1987)

Wave Behavior of Electrons, m = 9.1 x 10

https://www.youtube.com/watch?v=4dHu9aJuDCs

xxElectron Diffraction

• Proof that the electron had wave nature came a few years later with the demonstration that a beam of

electrons would produce an interference pattern the same as waves do.

However, electrons

actually present an

interference pattern,

demonstrating they

behave like waves.

xxElectron Diffraction

• Proof that the electron had wave nature came a few years later with the demonstration that a beam of

electrons would produce an interference pattern the same as waves do.

If electrons behave only like particles, there should only be two bright spots on the target.

Complementary Properties互補屬性 粒子與波不會同時存在

• When you try to observe the wave nature of the electron, you cannot observe its particle nature, and vice versa.

– Wave nature = interference pattern

– Particle nature = position, which slit it is passing through

• The wave and particle nature of the electron are complementary properties.

– As you know more about one you know less about the other.

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Chemistry: A Molecular Approach, 3rd Edition

Sort

You are given the speed of an electron and asked to calculate its wavelength.

Given: v = 2.65 × 10⁶ m/s Find: λ

Strategize

The conceptual plan shows how the de Broglie relation relates the wavelength of an electron to its mass and velocity.

Conceptual Plan

Relationships Used

λ = h/mv (de Broglie relation, Equation 7.4)

Solve

Substitute the velocity, Planck’s constant, and the mass of an electron to calculate the electron’s wavelength. To correctly cancel the units, break down the J in Planck’s constant into its SI base units (1 J = 1 kg · m²/s²).

Calculate the wavelength of an electron traveling with a speed of 2.65 × 10⁶ m/s.

Example 7.4 De Broglie Wavelength

© 2014 Pearson Education, Inc.

Chemistry: A Molecular Approach, 3rd Edition Nivaldo J. Tro

Solution

Check

The units of the answer (m) are correct. The magnitude of the answer is very small, as expected for the wavelength of an electron.

For Practice 7.4

What is the velocity of an electron that has a de Broglie wavelength approximately the length of a chemical bond?

Assume this length to be 1.2 × 10^–10 m.

Continued

Example 7.4 De Broglie Wavelength

Uncertainty Principle

• Heisenberg stated that the product of the

uncertainties in both the position and speed of a particle was inversely proportional to its mass.

– x = position, Dx = uncertainty in position – v = velocity, Dv = uncertainty in velocity – m = mass

• This means that the more accurately you know the position of a small particle, such as an

electron, the less you know about its speed, and vice versa.

xxUncertainty Principle Demonstration

Any experiment designed to observe the electron results in detection of a single

electron particle and no interference pattern.

Xx Determinacy versus Indeterminacy

• According to classical physics, particles move in a path determined by the particle’s velocity,

position, and forces acting on it.

– Determinacy = definite, predictable future

• Because we cannot know both the position and velocity of an electron, we cannot predict the path it will follow.

– Indeterminacy = indefinite future, can only predict probability

• The best we can do is to describe the probability an electron will be found in a particular region

using statistical functions.

xxTrajectory versus Probability

xxElectron Energy

• Electron energy and position are complementary.

– KE = ½ mv²

• For an electron with a given energy, the best we can do is describe a region in the atom of high probability of finding it.

• Many of the properties of atoms are related to

the energies of the electrons.

xx7.5 Quantum Mechanics and the Atom

• Schrödinger’s equation allows us to calculate the probability of finding an

electron with a particular amount of energy at a particular location in the atom.

• Solutions to Schrödinger’s equation produce many wave functions, Y .

• A plot of distance versus Y

represents an orbital, a probability distribution map of a

region where the electron is likely to be found.

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Schrö dinger’s Equation

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Schrodinger Wave Equation Y = fn(n, l, m_l, m_s)

principal quantum number n n = 1, 2, 3, 4, ….

n=1 n=2 n=3

7.6

distance of e^- from the nucleus

Solutions to the Wave Function, Y

• These integers are called quantum numbers.

– Principal quantum number, n energy

– Angular momentum quantum number, l size, shape

– Magnetic quantum number, m_l

orientation

Principal Quantum Number, n :Energy

Bohr’s energy level

• n  1 integer.

E _n = (-13.6 eV) / n ²

Angular Momentum Quantum Number, l shape of the orbital

• l ＝ 0 to (n – 1).

• Each value of l is called by a particular letter that designates the shape of the orbital.

– l=0, s orbitals are spherical.

– l=1, p orbitals are like two balloons tied at the knots.

– l=2, d orbitals are mainly like four balloons tied at the knots.

– l=3, f orbitals are mainly like eight balloons tied at the knots.

Magnetic Quantum Number, m

orientation of the orbital

• m

= −l ~ +l

– Including 0

– Gives the number of orbitals of a particular shape

• When l = 2, m

= −2, −1, 0, +1, +2, 5 orbitals

Describing an Orbital

• Each set of n, l, and m

describes one orbital.

• n : principal energy level.

– Also called the principal shell

• l , m

: sublevel.

– Also called a subshell

Energy Levels and Sublevels

The n = 2 principal energy level contains two sublevels:

a.The l = 0: 2s sublevel with one orbital with m

= 0

b.The l = 1: 2p sublevel with three p orbitals with

m_l

= -1, 0, +1

Energy Levels and Sublevels

• In general,

– the number of sublevels within a level = n.

– the number of orbitals within a sublevel = 2l + 1.

– the number of orbitals in a level n².

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Chemistry: A Molecular Approach, 3rd Edition

Solution

First determine the possible values of l (from the given value of n). For a given value of n, the possible values of l are 0, 1, 2, . . . , (n – 1).

n = 4; therefore, l = 0, 1, 2, and 3

Next, determine the the possible values of m_l for each value of l. For a given value of l, the possible values of m_l are the integer values including zero ranging from –l to +l . The name of an orbital is its principal quantum number (n) followed by the letter corresponding to the value l. The total number of orbitals is given by n².

For Practice 7.5

List the quantum numbers associated with all of the 5d orbitals. How many 5d orbitals exist?

What are the quantum numbers and names (for example, 2s, 2p) of the orbitals in the n = 4 principal level? How many n = 4 orbitals exist?

Example 7.5 Quantum Numbers I

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Chemistry: A Molecular Approach, 3rd Edition Nivaldo J. Tro

Solution

Choice (d) is erroneous because for l = 1, the possible values of m_l are only –1, 0, and +1.

For Practice 7.6

Each set of quantum numbers is supposed to specify an orbital. However, each set contains one quantum number that is not allowed. Replace the quantum number that is not allowed with one that is allowed.

a. n = 3; l = 3; m_l = +2 b. n = 2; l = 1; m_l = –2 c. n = 1; l = 1; m_l = 0

These sets of quantum numbers are each supposed to specify an orbital. One set, however, is erroneous. Which one and why?

a. n = 3; l = 0; m_l = 0 b. n = 2; l = 1; m_l = –1 c. n = 1; l = 0; m_l = 0 d. n = 4; l = 1; m_l = –2

Example 7.6 Quantum Numbers II

Quantum Leaps

Hydrogen Energy Transitions and Radiation

DE_electron = Efinal state − Einitial state

Eemitted photon = −DE_electron

n = 32 656nm, red light

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Chemistry: A Molecular Approach, 3rd Edition

Sort

You are given the energy levels of an atomic transition and asked to find the wavelength of emitted light.

Given:

Find: λ

Strategize

In the first part of the conceptual plan, calculate the energy of the electron in the n = 6 and n = 5 orbitals using Equation 7.7 and subtract to find ΔE_atom.

In the second part, find E_photon by taking the negative of ΔE_atom and then calculate the wavelength corresponding to a photon of this energy using Equation 7.3. (The difference in sign between E_photon and ΔE_atom applies only to emission.

The energy of a photon must always be positive.) Conceptual Plan

Determine the wavelength of light emitted when an electron in a hydrogen atom makes a transition from an orbital in n = 6 to an orbital in n = 5.

Example 7.7 Wavelength of Light for a Transition in the Hydrogen Atom

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Chemistry: A Molecular Approach, 3rd Edition Nivaldo J. Tro

Relationships Used

E_n = –2.18 × 10^–18 J(1/n²) E = hc/λ

Solve

Follow the conceptual plan. Begin by calculating ΔE_atom. Calculate E_photon by changing the sign of ΔE_atom.

Solve the equation relating the energy of a photon to its wavelength for λ. Substitute the energy of the photon and calculate λ.

Solution

E_photon = –ΔE_atom = +2.6644 × 10^–20 J Continued

Example 7.7 Wavelength of Light for a Transition in the Hydrogen Atom

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Chemistry: A Molecular Approach, 3rd Edition

Check

The units of the answer (m) are correct for wavelength. The magnitude is reasonable because 10^–6 m is in the infrared region of the electromagnetic spectrum. We know that transitions from n = 3 or n = 4 to n = 2 lie in the visible region, so it makes sense that a transition between levels of higher n value (which are energetically closer to one another) would result in light of longer wavelength.

For Practice 7.7

Determine the wavelength of the light absorbed when an electron in a hydrogen atom makes a transition from an orbital in which n = 2 to an orbital in which n = 7.

For More Practice 7.7

An electron in the n = 6 level of the hydrogen atom relaxes to a lower energy level, emitting light of λ = 93.8 nm.

Find the principal level to which the electron relaxed.

Continued

Example 7.7 Wavelength of Light for a Transition in the Hydrogen Atom

7.6 The Shapes of Atomic Orbitals, l

•

y

– The probability of finding an electron at a particular point in space

– For s orbital maximum at the nucleus?

– Decreases as you move away from the nucleus

• The radial distribution function represents the total probability at a certain distance from the nucleus.

– Maximum at most probable radius

• Nodes in the functions are where the probability drops to 0.

Probability and Radial Distribution Functions

The Shapes of Atomic Orbitals

• The l quantum number primarily determines the shape of the orbital.

• l can have integer values from 0 to (n – 1).

• Each value of l is called by a particular letter that designates the shape of the orbital.

– s orbitals are spherical.

– p orbitals are like two balloons tied at the knots.

– d orbitals are mainly like four balloons tied at the knots.

– f orbitals are mainly like eight balloons tied at the knots.

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The 1s Orbital

Spherical symmetry;

probability of finding the electron is the same

in each direction.

• The 1s orbital (n = 1, l = 0, m

= 0) has spherical symmetry.

• An electron in this orbital spends most of its time near the nucleus.

The electron cloud doesn’t

“end” here …

… the electron just spends very little

time farther out.

Chapter Seven

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• The 2s orbital has two concentric, spherical regions of high electron probability.

• The region near the nucleus is separated from the outer region by a node—a region (a spherical shell in this case) in which the electron probability is zero.

The 2s Orbital, l=0

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The Three p Orbitals, l=1

Three values of m_l (-1, 0, 1)gives three p

orbitals in the p subshell.

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The Five d Orbitals, l=2

Five values of m_l (–

2, –1, 0, 1, 2) gives five d orbitals in the d

subshell.

The seven f Orbitals, l=3

seven values of m_l (–3, -2, –1, 0, 1, 2, 3) gives seven f orbitals in the

d subshell.

Phases

Orbitals are determined from mathematical wave functions.

A wave function can have positive or negative values.

As well as nodes where the wave function

= 0

The sign of the wave function is called its

When orbitals interact, their wave functions may be in phase (same sign) or out of phase

(opposite signs).

This is important in bonding as will be

examined in a later chapter.

Why Are Atoms Spherical?