Transcript The Quantum Model

The Quantum Model
Unit D: Lesson 5
Summary of the Bohr Model
Absorption (nf > ni). When an electron
"jumps" to a higher energy level, it must
absorb energy. Each different transition
requires a specific amount of energy. The
dark lines in an absorption spectrum
correspond to specific photon
wavelengths that are needed for an
electron to jump from lower to higher
energy levels

Emission (nf < ni). When an electron
"falls" to a lower energy level, energy is
emitted. Each different transition emits a
specific amount of energy. The lines that
are seen in an emission spectrum
correspond to specific photon
wavelengths that are emitted when an
electron jumps from higher to lower
energy levels.

Review Eg) In an atom the electron goes from
the first orbital E = -24.0 eV to the 5th orbital.
A photon with wavelength of 5.40 x 10-8m is
absorbed. Based on this information, what is
the energy of the 5th orbital for this atom?
E
hc

4.14  1015 eV s  3.00  108 m / s
E
8
5.40  10 m
E  23.0eV
E  E f  Ei
E f  E  Ei
E f  23.0eV  24.0eV  1.0eV
Quantum Model of the Atom

Recall de Broglie - particles have wave
properties because they have momentum
h
Particle: p = mv
Photon: p 

 de Broglie’s wavelength is
h

•
mv
de Broglie’s wave equation indicates
that particles (electrons) travel in a
wave-like pattern

Rutherford’s model implied that electrons
orbit a nucleus

de Broglie suggested that the electrons
travel in a standing wave pattern and
orbitals exist only where the standing
wave produces constructive interference

Consider the following: If the wave doesn’t fit,
it will add destructively to itself and collapse.
If the wave fits the radius, the electron is very
stable. It exists as a circular standing wave

the standing wave is how electrons can occupy
an orbit without giving off energy

the atom only has specific energy levels
indicating that the energies of the electrons
are quantized

Since electrons diffract this also indicates that
they travel in a wave-like pattern
circumference of the orbit =
whole number of wavelengths
NOTE: the number of standing wavelengths
corresponds to the orbit number
 n =1 means 1 wavelength
 n = 2 means 2 wavelengths and so on

Contemporary Model
the electron stops being a particle
orbiting the nucleus at a certain point
 Instead, an electron’s mass and charge can
be thought of as “spread out” as a
standing wave around the nucleus.

The electron is not really at any one position as
a particle, it's everywhere as a wave.
(Remember the video.)
Electron Cloud Model
The electrons have
become a “cloud” of
electrons.
 This is essentially the
model used today.

Example:
Determine the de Broglie wavelength for the
electron in the 2nd Bohr orbit for hydrogen
where the energy level is 3.40 eV.
Solution:
1. Convert energy into joules

Ek  3.40eV  1.60  10
 5.44  10
19
J
19
C
Solution:
2. Recognize that de Broglie’s wavelength
requires velocity so calculate velocity
1 2
from energy using Ek  mv
2
1 2
Ek  mv
2
1
31
2
5.44  10 J  (9.11  10 kg )v
2
6
v  1.09  10 m / s
19
Solution:
3. Solve for wavelength
p  mv and p 
h

mv
h

so mv 
h

Remember this formula is not on the formula sheet.
6.63  1034 Js
10

 6.66  10 m
31
6
9.11  10 kg (1.09  10 m)
Eg. 2 Find the  of an electron that was
accelerated by a potential difference of 500V.
1. Determine energy.
E
V 
q
E  Vq
E  500V  1.60  10
 8.00  10
17
J
19
C
2. Determine velocity.
1 2
Ek  mv
2
v
v
2E
m
17
2(8.00  10 J )
31
9.11  10 kg
v  13252591  1.33  10 m / s
7
3. Determine wavelength
p  mv and p 
h

so mv 
h

h

mv
6.63  1034 Js
11


5.49

10
m
31
7
9.11  10 kg (1.33  10 m)
Warning:

Do not find energy using Ek=1/2mv2 and then
use E = hc/ to find wavelength – it implies
the mass ‘stops’ and changes its energy into
EMR.