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Chem 1st LT

Jun 16, 2015

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Technology

Rio Cañal

Chem 1: 3rd Quarter. 1st LT Coverage
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Page 1: Chem 1st LT
Page 2: Chem 1st LT
Page 3: Chem 1st LT

Properties of Waves

Wavelength () is the distance between identical points on successive waves.

Amplitude is the vertical distance from the midline of a wave to the peak or trough.

7.1

Page 4: Chem 1st LT

Properties of Waves

Frequency () is the number of waves that pass through a particular point in 1 second (Hz = 1 cycle/s).

The speed (u) of the wave = x

7.1

Page 5: Chem 1st LT

Maxwell (1873), proposed that visible light consists of electromagnetic waves.

Electromagnetic radiation is the emission and transmission of energy in the form of electromagnetic waves.

Speed of light (c) in vacuum = 3.00 x 108 m/s

All electromagnetic radiation x c

7.1

Page 6: Chem 1st LT

Electric field componentz

y

x

Magnetic field component

Page 7: Chem 1st LT

x

Page 8: Chem 1st LT

Electric field componentz

x

Page 9: Chem 1st LT

Electric field componentz

x

Page 10: Chem 1st LT

x

Page 11: Chem 1st LT

y

x

Magnetic field component

Page 12: Chem 1st LT

y

x

Magnetic field component

Page 13: Chem 1st LT

Electric field componentz

y

x

Magnetic field component

Page 14: Chem 1st LT

7.1

Page 15: Chem 1st LT

x = c = c/ = 3.00 x 108 m/s / 6.0 x 104 Hz

= 5.0 x 103 m

Radio wave

A photon has a frequency of 6.0 x 104 Hz. Convertthis frequency into wavelength (nm). Does this frequencyfall in the visible region?

= 5.0 x 1012 nm

7.1

Page 16: Chem 1st LT

Mystery #1, “Black Body Problem”Solved by Planck in 1900

Energy (light) is emitted or absorbed in discrete units (quantum).

E = h x Planck’s constant (h)h = 6.63 x 10-34 J•s

7.1

Page 17: Chem 1st LT

Light has both:1. wave nature2. particle nature

Mystery #2, “Photoelectric Effect”Solved by Einstein in 1905

Photon is a “particle” of light

h

KE e-

7.2

Page 18: Chem 1st LT

E = h x

E = 6.63 x 10-34 (J•s) x 3.00 x 10 8 (m/s) / 0.154 x 10-9 (m)

E = 1.29 x 10 -15 J

E = h x c /

7.2

When copper is bombarded with high-energy electrons, X rays are emitted. Calculate the energy (in joules) associated with the photons if the wavelength of the X rays is 0.154 nm.

Page 19: Chem 1st LT

7.3

Line Emission Spectrum of Hydrogen Atoms

Page 20: Chem 1st LT

Fig. 7.p267middle

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7.3

Page 22: Chem 1st LT

1. e- can only have specific (quantized) energy values

2. light is emitted as e- moves from one energy level to a lower energy level

Bohr’s Model of the Atom (1913)

En = -RH ( )1n2

n (principal quantum number) = 1,2,3,…

RH (Rydberg constant) = 2.18 x 10-18J

7.3

Page 23: Chem 1st LT

E = h

E = h

7.3

Page 24: Chem 1st LT

De Broglie (1924) reasoned that e- is both particle and wave.

2r = n = h/mu

u = velocity of e-

m = mass of e-

Why is e- energy quantized?

7.4

Page 25: Chem 1st LT

= h/mu

= 6.63 x 10-34 / (2.5 x 10-3 x 15.6) = 1.7 x 10-32 m = 1.7 x 10-23 nm

What is the de Broglie wavelength (in nm) associated with a 2.5 g Ping-Pong ball traveling at 15.6 m/s?

m in kgh in J•s u in (m/s)

7.4

Page 26: Chem 1st LT

Chemistry in Action: Element from the Sun

In 1868, Pierre Janssen detected a new dark line in the solar emission spectrum that did not match known emission lines

In 1895, William Ramsey discovered helium in a mineral of uranium (from alpha decay).

Mystery element was named Helium

Page 27: Chem 1st LT

Schrodinger Wave EquationIn 1926 Schrodinger wrote an equation that described both the particle and wave nature of the e-

Wave function () describes:

1. energy of e- with a given

2. probability of finding e- in a volume of space

Schrodinger’s equation can only be solved exactly for the hydrogen atom. Must approximate its solution for multi-electron systems.

7.5

Page 28: Chem 1st LT

Schrodinger Wave Equation

fn(n, l, ml, ms)

principal quantum number n

n = 1, 2, 3, 4, ….

n=1 n=2 n=3

7.6

distance of e- from the nucleus

Page 29: Chem 1st LT

e- density (1s orbital) falls off rapidly as distance from nucleus increases

Where 90% of thee- density is foundfor the 1s orbital

7.6

Page 30: Chem 1st LT

= fn(n, l, ml, ms)

angular momentum quantum number l

for a given value of n, l = 0, 1, 2, 3, … n-1

n = 1, l = 0n = 2, l = 0 or 1

n = 3, l = 0, 1, or 2

Shape of the “volume” of space that the e- occupies

l = 0 s orbitall = 1 p orbitall = 2 d orbitall = 3 f orbital

Schrodinger Wave Equation

7.6

Page 31: Chem 1st LT

l = 0 (s orbitals)

l = 1 (p orbitals)

7.6

Page 32: Chem 1st LT

l = 2 (d orbitals)

7.6

Page 33: Chem 1st LT

= fn(n, l, ml, ms)

magnetic quantum number ml

for a given value of lml = -l, …., 0, …. +l

orientation of the orbital in space

if l = 1 (p orbital), ml = -1, 0, or 1if l = 2 (d orbital), ml = -2, -1, 0, 1, or 2

Schrodinger Wave Equation

7.6

Page 34: Chem 1st LT

ml = -1 ml = 0 ml = 1

ml = -2 ml = -1 ml = 0 ml = 1 ml = 27.6

Page 35: Chem 1st LT

= fn(n, l, ml, ms)

spin quantum number ms

ms = +½ or -½

Schrodinger Wave Equation

ms = -½ms = +½

7.6

Page 36: Chem 1st LT

Detecting screen

Slit screen

Magnet

Oven

Page 37: Chem 1st LT

Detecting screen

Slit screen

Magnet

Oven

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Detecting screen

Slit screen

Magnet

Atom beamOven

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Detecting screen

Slit screen

Magnet

Atom beam

12

ms = + –

Oven

Page 40: Chem 1st LT

Detecting screen

Slit screen

Magnet

ms = – –12

Atom beamOven

Page 41: Chem 1st LT

Detecting screen

Slit screen

Magnet

ms = – –12

12

ms = + –

Atom beamOven

Page 42: Chem 1st LT

Existence (and energy) of electron in atom is described by its unique wave function .

Pauli exclusion principle - no two electrons in an atomcan have the same four quantum numbers.

Schrodinger Wave Equation= fn(n, l, ml, ms)

Each seat is uniquely identified (E, R12, S8)Each seat can hold only one individual at a time

7.6

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7.6

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Schrodinger Wave Equation= fn(n, l, ml, ms)

Shell – electrons with the same value of n

Subshell – electrons with the same values of n and l

Orbital – electrons with the same values of n, l, and ml

How many electrons can an orbital hold?

If n, l, and ml are fixed, then ms = ½ or - ½

= (n, l, ml, ½) or= (n, l, ml, -½)An orbital can hold 2 electrons 7.6

Page 45: Chem 1st LT

How many 2p orbitals are there in an atom?

2p

n=2

l = 1

If l = 1, then ml = -1, 0, or +1

3 orbitals

How many electrons can be placed in the 3d subshell?

3d

n=3

l = 2

If l = 2, then ml = -2, -1, 0, +1, or +2

5 orbitals which can hold a total of 10 e-

7.6