Transcript introduction to atomic structure

Electromagnetic Radiation
1
2
3
c=n.l
P
How many wavelengths
pass through point P in
one second? Frequency!
4
Electromagnetic Radiation
5
A radio operator broadcasts at a frequency of 14.2 MHz (megahertz).
What is the wavelength of the radio waves put out by the transmitter?
c n  l
3.0 108 m  s 1 3.0 108 m  s 1
l 

 21.1 m
6 1
n
14.2 MHz
14.2 10 s
c
- Solve example 7.2.
- Calculate the wavelengths of the electromagnetic radiation presented in
previous slide.
6
Atomic Spectra
Now, replace white light source with a hydrogen lamp.
7
cathode
H2 → 2H
H → H*(excited)
H*(excited) → H (Ground state) + light
anode
‫أ‬
‫طاق ة ال و ضع‬
Discharge tube
‫ب‬
8
Continuous spectrum
Line spectrum
Problem: No explanation provided by classical physics.
Scientists (such as Lyman, Balmer and Paschen) analyzed the observed lines
with respect to their wavelengths. Rydberg summarized their efforts in the
so-called Rydberg’s equation:
R=1.09678×10-2 nm-1
1
1
1
Rydberg’s constant
R  
l
 n2
 1
n22 
n: positive integer. 9
Calculate the wavelength, in nanometers, in the line of spectrum of hydrogen
Corresponding to n1=2 and to n2=4!
 1 1 
 R   2  2 
l
 n1 n2 
1
 1 1
1 1 
 109678 cm 1   2  2   109678 cm 1    
l
2 4 
 4 16 
1
3
 109678 cm 1   20564.6 cm 1
l
16
1
l
 4.863 10 5 cm
1
20564.6 cm
1 cm  ? nm
1
110  2 m  x 10 9 m
x  107
1 cm  107 nm
l  4.863 10 5 cm  4.863 10 5 107 nm  486.3 nm
Green light
10
Energy of Light
- Energy of electromagnetic waves
-However, Planck (Black body radiation)
Einstein (photoelectric effect)
hn
hn
hn
hn
hn
hn
hn
hn
hn
hn
Light composed of tiny particles, called quanta (photons)
Energy of each photon (quantum) = h × n
Number of photons determines light intensity.
Ephoton = h × n
h=6.6×10-34 J.s
Planck’s constant
11
Bohr’s Theory
Bohr’s Postulates:
•Electron moves in circular orbits around the nucleus.
•Electron can possess only certain energy values corresponding to
the orbit.
•Electron can “jump” from one orbit to another, the energy difference
will be emitted or absorbed in the form of light quanta.
12
1
En   A Z 2
n
2
A=2.18×10-18 J=13.6 eV
Z : atomic number
n : positive integer = 1, 2, 3, …
13
The larger n
- the larger is the orbit size, the farther is the
electron from nucleus
- the larger is the electron energy
Comparison to throwing stone upwards.
Negative sign means that the electron is under the influence of the nucleus.
Electron free from nucleus attracting force when n=∞.
14
Explanation of line spectrum
 A 
E  Ehigh  Elow   2    2 
nhigh  nlow 
A
 1

1
E  A 2  2   h n
n

n
low
high


A  1
1 
n  2  2 
h  nlow nhigh 
c n  l
nhigh
nlow
A  1
1 
 
 2
2

l c h  c  nlow nhigh 
1
n
A
R
hc
15
Lyman series
ends at n=1
UV
Balmer Series
ends at n=2
visible
Paschen Series
ends at n=3
Infra-red
16
Example 7.5
Calculate the energy, frequency and wavelength of the photon
emitted when an electron in the hydrogen atom drops from the
fifth to the second energy level.
 1
1 

E photon  E  A 2  2
n

n
low
high


 1 1
E photon  2.18 10 18  2  2  J  4.578 10 19 J
2 5 
E photon 4.578 10 19 J
14 1
n


6
.
94

10
s
34
h
6.6  10 J s
3.0 108 m s 1
7
l 

4
.
325

10
m  432.5 nm
14 1
n 6.94 10 s
c
17
De Broglie Hypothesis
Light behaves:
- as waves (electromagnetic waves)
Hertz experiment
- as particles (photons)
Photoelectric effect, Compton effect
Why should light be special??!!!!!
De Broglie suggested:
Generalization of dual nature (wave nature & particle nature) to all matter.
Any moving object can be considered to be a wave!!!
Energy of that object is E=mc2
If object considered to be a wave,
E=hn
18
h n  m  c 2
h
c
l
h
 m c2
h
l
mc
l
 mc
h
l
mv
h
l
p
De Broglie relation
De Broglie wavelength
Experimental evidence: electron diffraction
Diffraction is a phenomenon that only waves can undergo
Includes waves interference
19
Example: Calculate the wavelength of a football player weighing 60 kg
moving in the yard with 10 km/h velocity.
v  10 km / h 
10000 m
 2.778 m / s
3600 s
h
6.6  10 34 Js s
l

 3.96  10 35 m
m v 60  2.778 kg m
too small to be observed experimentally
Example: Calculate the wavelength of an electron moving with a
velocity of 1000 km/h.
h
6.6 1034 J s
s
10
l


7
.
25

10
m  7.25 Angstrom
31
m v (9.1110 ) 1000000 kg m
X-ray: diffraction on crystals
20
Wave mechanics
A version of quantum mechanics (modern concepts in physics)
Electron in atoms are considered to be standing waves.
Each electron in atom is described by a set of numbers called
Quantum numbers.
21