Transcript Waves & Bohr’s Theory Chapter 7 §1-4 Waves Wavelength, λ, in meters (m) The length of a wave from crest to crest or trough to trough. Frequency,

Waves
&
Bohr’s Theory
Chapter 7 §1-4
Waves
Wavelength, λ, in meters (m)
The length of a wave from crest to crest or trough to trough.
Frequency, υ, in inverse seconds (s-1)
The number of waves that pass in a given amount of time.
Light’s Wave Characteristics
Light’s Wave Characteristics
 Speed of light equation
c  
c = speed of light, 3.00x108 m/s
λ = wavelength, in m
ν = frequency, in Hz or s–1
Light’s Wave Characteristics
 What is the frequency of red light
having a wavelength of 681 nm?
681 nm = 6.81x10–7 m
3.0x108 m/s = (6.81x10–7 m)(υ)
υ = 4.41x1014 s–1
Max Planck
 Quantized
Planck’s Equation
E  nh
n = integer other than zero, no unit, represents the
energy level of the atom
h = Planck’s constant, 6.63x10–34 J•s
ν = frequency, in Hz or s–1
E = energy of a quantum – amount of energy to move
an e– from its present energy to its next higher one
The Photoelectric Effect
 Although the photoelectric effect was
first discovered by Heinrich Hertz in
1887, Albert Einstein incorporated
Planck’s ideas into the explanation of
the photoelectric effect.
The Photoelectric Effect
 Einstein stated that electrons could
move within their atoms if a minimum
amount of energy were reached.
Let’s Practice
 The blue-green line of the hydrogen
atom spectrum has a wavelength of
486 nm. What is the energy of a
photon of this light?
486 nm = 4.86x10–7 m
3.0x108 m/s = (4.86x10–7 m) (υ)
υ = 6.17x1014 s–1
E = (1)(6.63x10–34 J•s)(6.17x1014 s–1)
E = 4.09x10–19 J
Bohr’s Line Spectra
 Bohr noticed that elements emitted a
line spectrum.
Bohr’s Line Spectra
 He hypothesized that each line in the
spectra were created when an electron
fell from a higher energy level to a
lower one within the atom.
Bohr’s Line Spectra
 Here, the
electron of the
hydrogen atom
is shown moving
between the
various energy
levels.
Bohr’s Postulates
 Bohr felt the need to explain two main
issues:
 1st – If electrons are negative and protons
are positive…..
Bohr’s Postulates
 Bohr felt the need to explain two main
issues:
 2nd – How are the line spectra being
created?
Bohr’s Postulates
 Postulate #1:
 Electrons only have specific energy values
and energy levels.
RH
E 2
n
 = the energy of a particular e– energy level
 = the Rydberg constant = 2.18x10–18 J
 = integral value representing the energy level
(principal quantum number
Bohr’s Postulates
 Postulate #2:
 Electrons become excited by collisions of
atoms or absorption of energy – think of
anything colored… it absorbs light to
later emit (reflect) other colors
 Electrons can change energy only by
going from one energy level to another –
making a transition.
Bohr’s Postulates
 Postulate #2:
 The electron
absorbing energy
Bohr’s Postulates
 Postulate #2:
 When an electron falls from a higher
energy level to a lower energy level, it
emits a photon of light.
Bohr’s Postulates
 Postulate #2:
 This energy of this photon can be found
by:
RH RH
E  En f  Eni   2  2
nf
ni
Bohr’s Postulates
 Postulate #2:
 Remember that the energy of a photon
can be determined by:
RH RH
E  En f  Eni   2  2  h 
nf
ni
Bohr’s Postulates
 Postulate #2:
 Which can be related to wavelength:
RH RH
c

E  En f  Eni   2  2  h   h  
nf
ni

Bohr’s Postulates
 Postulate #2:
 Which leads to the Balmer equation:
 1
1 
 1.097 x10 m   2  2 

 2 ni 
1
7
1
 This equation can calculate the
wavelength of any electron falling to the
2nd energy level – emitting visible light.
Bohr’s Postulates
 Postulate #2:
 By the end of Experiment #13, you
should be able to identify the Balmer
Series, along with the Paschen Series
and the Lyman Series.
Let’s Practice
 What is the wavelength of the light
emitted when the electron in a
hydrogen atom undergoes a transition
from energy level n = 6 to level n = 3?
Let’s Practice
 What is the wavelength of the light
emitted when the electron in a
hydrogen atom undergoes a transition
from energy level n = 6 to level n = 3?
2.18 1018 J 2.18 1018 J
19
∆E = E3 – E6 =



1.82
x
10
J
2
2
3
6
1.82 x1019 J   6.636 x10 34 J s   
υ = 2.74x1014 s–1
c = λυ → 3.0x108 m/s = λ (2.74x1014 s–1)
λ = 1.09x10–6 m = 1090 nm (Infrared)