chapter 7: quantum theory of the atom

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Vanessa N. Prasad-PermaulValencia Community College

CHM 1045

Frequency, : The number of wave peaks that pass a given point per unit time (1/s)

Wavelength, : The distance from one wave peak to the next (nm or m)

Amplitude: Height of wave

Wavelength x Frequency = Speed(m) x (s-1) = c (m/s)

The speed of light waves in a vacuum in a constant

c = 3.00 x 108 m/s

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THE WAVE NATURE OF LIGHT

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EXAMPLE 7.1 : WHAT IS THE WAVELENGTH OF THE YELLOW SODIUM EMISSION, WHICH HAS A FREQUENCY OF 5.09 X 1014 S-1?

c = =c

= 3.00 x 108 m/s 5.09 x 1014 s-1

= 5.89 x 10 -7 m = 589 x 10-9 m

= 589 nm

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EXERCISE 7.1 : The frequency of the strong red line in the spectrum of potassium is 3.91 x 1014 s-1. What is the wavelength of this light in nanometers?

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EXAMPLE 7.2 : WHAT IS THE FREQUENCY OF VIOLET LIGHT WITH A WAVELENGTH OF 408nm?

c = =c

= 3.00 x 108 m/s 408 X 10-9 m = 7.35 x 1014 s-1

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EXERCISE 7.2 : The element cesium was discovered in 1860 by Robert Bunsen and Gustav Kirchoff, who found to bright blue lines in the spectrum of a substance isolated from a mineral water. One of the spectral lines of cesium has a wavelength of 456nm. What is the frequency?

Several types of electromagnetic radiation make up the electromagnetic spectrum

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THE ELECTROMAGNETIC SPECTRUM

Atoms of a solid oscillate of vibrate with adefinite frequencyE = h E = hc / h = Planck’s constant, 6.626 x 10-34 J sE = energy1 J = 1 kg m2/s2

When a photon hits the metal, it’s energy (h) istaken up by the electron. The photon no longerexists as a particle and it is said to be absorbed

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QUANTUM EFFECTS & PHOTONS

Max Planck (1858–1947): proposed the energy is only emitted in discrete packets called quanta (now called photons).

The amount of energy depends on the frequency:

E = energy = frequency = wavelength c = speed of light h = planck’s constant

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E h

hc h 6.626 10 34J s


Albert Einstein (1879–1955): Used the idea of quanta to explain the photoelectric

effect. He proposed that light behaves as a stream of

particles called photons A photon’s energy must exceed a minimum

threshold for electrons to be ejected. Energy of a photon depends only on the frequency.

E = h

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THE PHOTOELECTRIC EFFECT:

The ejection of electrons from the surface of a metal or from a material when light shines on it


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EXAMPLE 7.3 : THE RED SPECTRAL LINE OF LITHIUUM OCCURS AT 671nm (6.71 x 10-7m). CALCULATE THE ENERGY OF ONE PHOTON OF THIS LIGHT.

= c = 3.00 x 108 m/s = 4.47 x 1014 s-1

6.71 x 10-7 m

E = h = 6.63 x 10-34 J.s * 4.47 x 1014 s-1 = 2.96 x 10-19 J


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EXERCISE 7.3 : The following are representative wavelengths in the infrared, ultraviolet and x-ray regions of the electromagnetic spectrum, respectively:1.0 x 10-6 m, 1.0 x 10-8 m and 1.0 x 10-10 m.

• What is the energy of a photon of each radiation?• Which has the greatest amount of energy per photon?• Which has the least?


Atomic spectra: Result from excited atoms emitting light.

Line spectra: Result from electron transitions between specific energy levels.

Blackbody radiation is the visible glow that solid objects emit when heated.

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THE BOHR THEORY OF THE HYDROGEN ATOM

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BOHR’S POSTULATE•The stability of the atom (H2)•The line spectrum of the atom

ENERGY-LEVEL POSTULATE: An electron can only have specific energy level values in an atom called ENERGY LEVELS E = RH where n = 1, 2, 3 n2

RH = 2.179 x 10-18 J

n = principle quantum number

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BOHR’S POSTULATE•The stability of the atom (H2)•The line spectrum of the atom

TRANSITIONS BETWEEN ENERGY LEVELS: An electron in an atom can change energy only by going from one energy level to another energy level. By doing so, the electron undergoes a transition.

An electron goes from a higher energy level (Ei) to a lower energy level (Ef) emitting light:

-E = -(Ef - Ei)E = Ei - Ef

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ENERGY LEVEL DIAGRAM OF THE HYDROGEN ATOM

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EXAMPLE 7.4 : WHAT IS THE WAVELENGTH OF THE LIGHT EMITTED WHEN THE ELECTRON IN A HYDROGEN ATOM UNDERGOES A TRANSITION FROM ENERGY LEVEL n = 4 TO LEVEL n = 2.

Ei = -RH Ef = - RH 42 22

E = -RH - -RH 16 4

E = -4RH + 16RH = -RH + 4RH = 3RH = h 64 16 16

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= E = 3RH = 3 * 2.179 x 10-18 J = 6.17 x 1014 s-1

h 16* h 16 * 6.626 x 10-34 J.s

= c = 3.00 x 108 m/s = 4.86 x 10-7 m 6.17 x 10 14 s-1

= 486 nm

(the color is blue-green)

EXAMPLE 7.4 : Cont…

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EXERCISE 7.4 : Calculate the wavelength of light emitted from the hydrogen atom when the electron undergoes a transition from level 3 (n = 3) to level 1 (n = 1).

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EXERCISE 7.5 : What is the difference in energy levels of the sodium atom if emitted light has a wavelength of 589nm?

Louis de Broglie (1892–1987): Suggested waves can behave as particles and particles can behave as waves. This is called wave–particle duality.

m = mass in kg p = momentum (mc) or (mv)

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For Light : hmc

hp

For a Particle : hmv

hp

QUANTUM MECHANICS

The de Broglie relation

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QUANTUM MECHANICS

EXAMPLE 7.5 : A)CALCULATE THE (in m) OF THE WAVE ASSOCIATED WITH A 1.00 kg MASS MOVING AT 1.00km/hr.

v = 1.00 km x 1000m x 1hr x 1min = 0.278m/s hr 1km 60min 60 sec

= h = 6.626 x 10-34 kg.m2/s2.s = 2.38 x 10-33m mv 1.00kg * 0.278m/s

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QUANTUM MECHANICS

EXAMPLE 7.5 : cont… B) WHAT IS THE (in pm) ASSOCIATED WITH AN ELECTRON WHOSE MASS IS 9.11 x 10-31kg TRAVELING AT A SPEED OF 4.19 X 106 m/s ?

= h = 6.626 x 10-34 kg.m2/s2.s mv9.11 x 10-31kg * 4.19 x 106 m/s

= 1.74 x 10-10 m = 174pm

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QUANTUM MECHANICS

EXERCISE 7.6 : Calculate the (in pm) associated with an electron traveling at a speed of 2.19 x 106 m/s.

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QUANTUM MECHANICS

QUANTUM MECHANICS ( WAVE MECHANICS):The branch of physics that mathematically describes the wave properties of submicroscopic particles

UNCERTAINTY PRINCIPLE:A relation that states that the product of the uncertainty in position and the uncertainty in momentum (mass times speed) of a particle can be no smaller than Planck’s constant divided by 4.

SCHRODINGER’S EQUATION:gives the probability of finding the particle within a region of space

Quantum Mechanics

Niels Bohr (1885–1962): Described atom as electrons circling around a nucleus and concluded that electrons have specific energy levels.

Erwin Schrödinger (1887–1961): Proposed quantum mechanical model of atom, which focuses on wavelike properties of electrons.

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Quantum Mechanics

Werner Heisenberg (1901–1976): Showed that it is impossible to know (or measure) precisely both the position and velocity (or the momentum) at the same time.

The simple act of “seeing” an electron would change its energy and therefore its position.

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Quantum Mechanics

Erwin Schrödinger (1887–1961): Developed a compromise which calculates both the energy of an electron and the probability of finding an electron at any point in the molecule.

This is accomplished by solving the Schrödinger equation, resulting in the wave function

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QUANTUM NUMBERS

According to QUANTUM MECHANICS:Each electron in an atom is described by 4 different quantum numbers: (n, l, m1 and ms).The first 3 specify the wave function that gives the probability of finding the electron at various points in space. The 4th (ms) refers to a magnetic property of electrons called spin

ATOMIC ORBITAL: A wave function for an electron in an atom

Quantum Numbers

Wave functions describe the behavior of electrons.

Each wave function contains four variables called quantum numbers:

• Principal Quantum Number (n)

• Angular-Momentum Quantum Number (l)

• Magnetic Quantum Number (ml)

• Spin Quantum Number (ms)31

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QUANTUM NUMBERS

PRINCIPLE QUANTUM NUMBERS (n):This quantum number is the one on which the energy of the electron in an atom principally depends; it can have any positive value (1, 2, 3 etc..)

•The smaller n, the lower the energy.•The size of an orbital depends on n; the larger the value of n, the larger the orbital.•Orbitals of the same quantum number (n) belong to the same shell which have the following letters:

Letter: K L M N n: 1 2 3 4

Quantum Numbers

ANGULAR MOMENTUM QUANTUM NUMBER (l): Defines the three-dimensional shape of the orbital.

For an orbital of principal quantum number n, the value of l can have an integer value from

0 to n – 1.

This gives the subshell notation:

Letter: s p d f g l: 0 1 2 3 4

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Quantum Numbers

Magnetic Quantum Number (ml): Defines the spatial orientation of the orbital.

For orbital of angular-momentum quantum number, l, the value of ml has integer values from –l to +l.

This gives a spatial orientation of:l = 0 giving ml = 0l = 1 giving ml = –1, 0, +1l = 2 giving ml = –2, –1, 0, 1, 2, and so on…...

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Quantum Numbers

Magnetic Quantum Number (ml): –l to +l

S orbital 0

P orbital -1 0 1D orbital -2 -1 0 1 2

F orbital -3 -2 -1 0 1 2 3

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Quantum Numbers

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Table of Permissible Values of Quantum Numbers for Atomic Orbitals

Quantum Numbers

Spin Quantum Number: ms

The Pauli Exclusion Principle states that no two electrons can have the same four quantum numbers.

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QUANTUM MECHANICS

EXAMPLE 7.6 : State whether each of the following sets of quantum numbers is permissible for an electron in an atom. If a set is not permissible, explain. a)n = 1, l = 1, ml = 0, ms = +1/2NOT permissible: The l quantum number is equal to n.IT must be less than n.

b) n = 3, l = 1, ml = -2, ms = -1/2NOT permissible: The magnitude of the ml quantumnumber (that is the ml value, ignoring it’s sign) must begreater than l.

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QUANTUM MECHANICS

EXAMPLE 7.6 : cont… c)n = 2, l = 1, ml = 0, ms = +1/2Permissible

d)n = 2, l = 0, ml = 0, ms = +1NOT permissible: The ms quantum number can only be+1/2 or -1/2.

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QUANTUM MECHANICS

EXERCISE 7.7 : Explain why each of the following sets of quantum numbers is not permissible for an orbital:

a) n = 0, l = 1, ml = 0, ms = +1/2b) n = 2, l = 3, ml = 0, ms = -1/2c) n = 3, l = 2, ml = +3, ms = +1/2d) n = 3, l = 2, ml = +2, ms = 0

Electron Radial Distribution

s Orbital Shapes: Holds 2 electrons

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p Orbital Shapes: Holds 6 electrons, degenerate

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d and f Orbital Shapes: d holds 10 electrons and f holds 14 electrons, degenerate

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Effective Nuclear Charge

Electron shielding leads to energy differences among orbitals within a shell.

Net nuclear charge felt by an electron is called the effective nuclear charge (Zeff).

Zeff is lower than actual nuclear charge. Zeff increases toward nucleus

ns > np > nd > nf

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Effective Nuclear Charge

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Example1: Light and Electromagnetic Spectrum

The red light in a laser pointer comes from a diode laser that has a wavelength of about 630 nm. What is the frequency of the light? c = 3 x 108 m/s

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Example 2: Atomic Spectra

For red light with a wavelength of about 630 nm, what is the energy of a single photon and one mole of photons?

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Example 3: Wave–Particle Duality

How fast must an electron be moving if it has a de Broglie wavelength of 550 nm?

me = 9.109 x 10–31 kg

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Example 4: Quantum Numbers

Why can’t an electron have the following quantum numbers?

(a) n = 2, l = 2, ml = 1

(b) n = 3, l = 0, ml = 3

(c) n = 5, l = –2, ml = 1

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Example 5: Quantum Numbers

Give orbital notations for electrons with the following quantum numbers:

(a)n = 2, l = 1

(b) n = 4, l = 3

(c) n = 3, l = 2

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chapter 7: quantum theory of the atom

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