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Quantum Physics
where the light hits the electron
And things get very weird.
Plank’s Constant
In his studies of black-body radiation, Maxwell
Planck discovered that electromagnetic energy is
emitted or absorbed in discrete quantities.
Planck’s
Equation:
E = hf
(h = 6.626 x 10-34 J s)
Apparently, light consists of
tiny bundles of energy called
photons, each having a welldefined quantum of energy.
Photon
E = hf
Energy in Electron-volts
Photon energies are so small that the energy is
better expressed in terms of the electron-volt.
One electron-volt (eV) is the energy of an
electron when accelerated through a potential
difference of one volt.
1 eV = 1.60 x 10-19 J
1 keV = 1.6 x 10-16 J
1 MeV = 1.6 x 10-13 J
Example 1: What is the energy of a photon
of yellow-green light (l = 555 nm)?
First we find f from wave equation: c = fl
f 
c
l
;
E  hf 
34
hc
l
(6.626 x 10 J  s)(3 x 10 m/s)
E
-9
555 x 10 m
E = 3.58 x 10-19 J
8
Or
E = 2.24 eV
Since 1 eV = 1.60 x 10-19 J
Useful Energy Conversion
Since light is often described by its wavelength in
nanometers (nm) and its energy E is given in eV, a
conversion formula is useful. (1 nm = 1 x 10-9 m)
E (in Joules) 
hc
l
; 1 eV  1.60 x 10-19 J
hc(1 x 109 nm/m)
E (in eV) 
(1.6 x 10-19 J/eV)l
If l is in nm, the energy in eV is found from:
E
1240
l
Verify the answer
in Example 1 . . .
The Photo-Electric Effect
Incident light
Cathode
Anode
A
C
-
+
Ammeter
A
When light shines on
the cathode C of a
photocell, electrons are
ejected from A and
attracted by the positive
potential due to battery.
There is a certain threshold energy, called the
work function W, that must be overcome
before any electrons can be emitted.
Photo-Electric Equation
Incident light
Cathode
Anode
A
C
-
+
Ammeter
A
E
hc
l
 W  12 mv 2
hc
Threshold
W
wavelength lo
l0
The conservation of energy demands that the
energy of the incoming light hc/l be equal to the
work function W of the surface plus the kinetic
energy ½mv2 of the emitted electrons.
Example 2: The threshold wavelength of
light for a given surface is 600 nm. What is
the kinetic energy of emitted electrons if
light of wavelength 450 nm shines on the
metal?
l = 600 nm
hc
W  K
l
hc hc

K
l l0
hc
A
hc
1240
1240
K



;
l l0 450 nm 600 nm
K = 0.690 eV
Or
K = 2.76 eV – 2.07 eV
K = 1.10 x 10-19 J
Stopping Potential
A potentiometer is used
to vary to the voltage V
between the electrodes.
The stopping potential
is that voltage Vo that
just stops the emission
of electrons, and thus
equals their original K.E.
Photoelectric equation:
E  hf  W  eV0
Incident light
Cathode
Anode
V
A
+
-
Potentiometer
Kmax = eVo
W
h
V0    f 
e
e
Slope of a Straight Line (Review)
The general equation for
a straight line is:
y = mx + b
The x-intercept xo occurs
when line crosses x axis
or when y = 0.
The slope of the line is
the rise over the run:
The slope of a line:
y
Slope
y
x
xo
x
y
Slope 
m
x
Finding Planck’s Constant, h
Using the apparatus on the previous slide, we
determine the stopping potential for a number
of incident light frequencies, then plot a graph.
W
h
V0    f 
e
e
Finding h constant
V
h
Slope 
e
Note that the x-intercept fo
is the threshold frequency.
Stopping
potential
Slope
fo
y
x
Frequency
Example 3: In an experiment to determine
Planck’s constant, a plot of stopping
potential versus frequency is made. The
slope of the curve is 4.13 x 10-15 V/Hz.
What is Planck’s constant?
Stopping
W
h
V
potential
V0    f 
e
e
Slope
y
fo
x
Frequency
h
-15
Slope   4.13 x 10 V/Hz
e
h = e(slope) = (1.6 x 10-19C)(4.13 x 10-15 V/Hz)
Experimental Planck’s h = 6.61 x 10-34 J/Hz
Example 4: The threshold frequency for a given
surface is 1.09 x 1015 Hz. What is the stopping
potential for incident light whose photon energy is
8.48 x 10-19 J? Incident light
Photoelectric Equation:
Cathode
E  hf  W  eV0
V
eV0  E  W ; W  hf 0
Anode
A
+
-
W = (6.63 x 10-34 Js)(1.09 x 1015 Hz) =7.20 x 10-19 J
eV0  8.48 x 10 J  7.20 x 10 J  1.28 x 10 J
-19
1.28 x 10-19 J
V0 
1.6 x 10-19 J
-19
-19
Stopping
Vo = 0.800 V
potential:
Waves and Particles
We know that light behaves as both a wave and
a particle. The rest mass of a photon is zero, and
its wavelength can be found from momentum.
E  pc 
hc
l
Wavelength
of a photon:
h
l
p
All objects, not just EM waves, have wavelengths
which can be found from their momentum
de Broglie
Wavelength:
h
l
mv
Finding Momentum from K.E.
In working with particles of momentum p = mv,
it is often necessary to find the momentum from
the given kinetic energy K. Recall the formulas:
K = ½mv2 ;
Multiply first
Equation by m:
p = mv
mK = ½m2v2 = ½p2
Momentum from K:
p  2mK
Example 5: What is the de Broglie
wavelength of a 90-eV electron? (me = 9.1
-31 kg.)
x
10
-19


1.6 x 10 J
-17
K  90 eV 

1.44
x
10
J

 1 eV 
Next, we find momentum
from the kinetic energy:
p  2mK
p  2(9.1 x 10-31kg)(1.44 x 10-17 J)
p = 5.12 x 10-24 kg m/s
h
6.23 x 10-34 J
l 
-24
p 5.12 x 10 kg m/s
e- 90 eV
h
h
l 
p mv
l = 0.122 nm
Electron Orbits
Consider the planetary model for electrons which
move in a circle around the positive nucleus. The
figure below is for the hydrogen atom.
r
FC
-
e-
+
Nucleus
mv 2
e2

r
4 0 r 2
Coulomb’s law:
FC 
e
2
4 0 r
2
Radius of
Hydrogen atom
Centripetal FC:
mv
FC  2
r
r
2
e
2
4 0 mv
2
Failure of Classical Model
v
-
e-
+
The loss of energy should
cause the velocity v to decrease, sending the electron
crashing into the nucleus.
Nucleus
r
e
When an electron is accelerated by the central force, it
must radiate energy.
2
4 0 mv
2
This does NOT happen and
the Rutherford atom fails.
Atomic Spectra
Earlier, we learned that objects continually
emit and absorb electromagnetic radiation.
In an emission spectrum, light is separated
into characteristic wavelengths.
Gas
Emission Spectrum
l1
l2
Absorption Spectrum
In an absorption spectrum, a gas absorbs certain
wave lengths, which identify the element.
Emission Spectrum for H Atom
Characteristic wavelengths
n=3
653 nm
n=4
434 nm
n
6
n=5
486 nm
410 nm
Balmer worked out a mathematical formula,
called the Balmer series for predicting the
absorbed wavelengths from hydrogen gas.
 1 1 
 R  2  2  ; n  3, 4, 5, . . .
l
2 n 
1
R
1.097 x 107 m-1
Example 1: Use the Balmer equation to find the
wavelength of the first line (n = 3) in the Balmer
series. How can you find the energy?
 1 1 
 R  2  2  ; n  3 R = 1.097 x 107 m-1
l
2 n 
1
1
 1 1
 R  2  2   R(0.361); l 
l
0.361R
2 3 
1
l
l = 656 nm
7 -1
0.361(1.097 x 10 m )
1
The frequency and the energy are found from:
c = fl and E = hf
The Bohr Atom
Atomic spectra indicate that atoms emit or
absorb energy in discrete amounts. In 1913,
Neils Bohr explained that classical theory did not
apply to the Rutherford atom.
An electron can only
have certain orbits and
the atom must have
definite energy levels
which are analogous to
standing waves.
e+
Electron orbits
Wave Analysis of Orbits
n=4
e+
Electron orbits
Stable orbits exist for
integral multiples of de
Broglie wavelengths.
2r = nl n = 1,2,3, …
h
2 r  n
mv
Recalling that angular momentum is mvr, we write:
h
L  mvr  n
; n  1, 2,3, . . .
2
The Bohr Atom
An electron can have only
those orbits in which its
angular momentum is:
h
Ln
; n  1, 2,3, . . .
2
Energy levels, n
+
The Bohr atom
Bohr’s postulate: When an electron changes
from one orbit to another, it gains or loses
energy equal to the difference in energy
between initial and final levels.
Bohr’s Atom and Radiation
Emission
Absorption
When an electron drops
to a lower level, radiation
is emitted; when radiation
is absorbed, the electron
moves to a higher level.
Energy: hf = Ef - Ei
By combining the idea of energy levels with
classical theory, Bohr was able to predict the
radius of the hydrogen atom.
Radius of the Hydrogen
Atom
Radius as
function of
energy level:
Bohr’s
radius
h
L  mvr  n
; n  1, 2,3, . . .
2
nh
r
mv
2
e
Classical r 
2
radius
4 0 mv
By eliminating r from these equations, we find the
velocity v; elimination of v gives possible radii rn:
vn 
e
2
2 0 nh
n  0h
rn 
 me2
2
2
Example 2: Find the radius of the Hydrogen
atom in its most stable state (n = 1).
m = 9.1 x 10-31 kg
n  0h
rn 
 me2
2
2
r
2
(1) (8.85 x 10
e = 1.6 x 10-19 C
-12 Nm 2
C2
-31
 (9.1 x 10
r = 5.31 x 10-11 m
)(6.63 x 1034 J  s) 2
-19
kg)(1.6 x 10 C)
r = 53.1 pm
2
Total Energy of an Atom
The total energy at level n is the sum of the
kinetic and potential energies at that level.
E  K U;
K  12 mv 2 ;
But we recall that:
vn 
e
2
2 0 nh
n  0h
rn 
 me2
2
Total energy of
Hydrogen atom
for level n.
2
U
e2
4 0 r
Substitution for
v and r gives
expression for
total energy.
me4
En   2 2 2
8 0 n h
Energy for a Particular State
It will be useful to simplify the energy formula for
a particular state by substitution of constants.
m = 9.1 x
10-31
e = 1.6 x
10-19
kg
o = 8.85 x 10--12 C2/Nm2
C
h = 6.63 x 10-34 J s
me4
(9.1 x 10-31kg)(1.6 x 10-19C)4
En   2 2 2  
-12 C2 2 2
8 0 n h
8(8.85 x 10 Nm2 ) n (6.63 x 10-34 Js)2
2.17 x 10-18 J
En  
2
n
Or
13.6 eV
En 
n2
Balmer Revisited
Total energy of
me4
Hydrogen atom En   2 2 2
8 0 n h
for level n.
Negative because
outside energy to
raise n level.
When an electron moves from an initial state
ni to a final state nf, energy involved is:
4 4
1hc me 4  1 1 1 1  me 4 meme
E    E2 0 3 E2 f ; 2  2  ; If 2R2 2  2 3 2 2 2
l l 8 0 h cn f  n f l nihc  8 0 h n08 08h 0cn
h fn f
Balmer’s
Equation:
 1
1 
 R  2  2  ; R  1.097 x 107 m -1
n

l
n
f
0


1



Energy Levels
We can now visualize the hydrogen atom with an
electron at many possible energy levels.
Emission
Absorption
The energy of the atom increases
on absorption (nf > ni) and decreases on emission (nf < ni).
Energy of
nth level:
13.6 eV
E
n2
The change in energy of the atom can be given in
terms of initial ni and final nf levels:
 1
1 
E  13.6 eV  2  2 
n

n
f
0


Spectral Series for an Atom
The Lyman series is for transitions to n = 1 level.
The Balmer series is for transitions to n = 2 level.
The Pashen series
is for transitions
to n = 3 level.
The Brackett series
is for transitions to
n = 4 level.
n =1
n =2
n =3
n =4
n =5
n =6
 1
1 
E  13.6 eV  2  2 
n

n
f
0


Example 3: What is the energy of an
emitted photon if an electron drops from the
n = 3 level to the n = 1 level for the
hydrogen atom?
 1
1 
E  13.6 eV  2  2 
n

n
f
0


Change in
energy of the
atom.
1 1
E  13.6 eV  2  2   12.1 eV
1 3 
E = -12.1 eV
The energy of the atom decreases by 12.1 eV
as a photon of that energy is emitted.
You should show that 13.6 eV is required to
move an electron from n = 1 to n = .
Modern Theory of the Atom
The model of an electron as a point particle moving
in a circular orbit has undergone significant change.
• The quantum model now presents the
location of an electron as a probability
distribution - a cloud around the nucleus.
• Additional quantum numbers have been
added to describe such things as shape,
orientation, and magnetic spin.
• Pauli’s exclusion principle showed that no
two electrons in an atom can exist in the
exact same state.
Modern Atomic Theory (Cont.)
The n = 2 level of the
The Bohr atom for
Hydrogen atom is
Beryllium suggests a
shown here as a
planetary model which
is not strictly correct. probability distribution.
Summary
Bohr’s model of the atom assumed the electron to
follow a circular orbit around a positive nucleus.
r
FC
+
Nucleus
-
e-
Radius of
Hydrogen Atom
r
e
2
4 0 mv
2
Summary (Cont.)
In an emission spectrum, characteristic
wavelengths appear on a screen. For an
absorption spectrum, certain wavelengths
are omitted due to absorption.
Gas
Emission Spectrum
l1
l2
Absorption Spectrum
Summary (Cont.)
Spectrum for nf = 2 (Balmer)
n=3
n=4
434 nm
n=5
Emission spectrum
653 nm
486 nm
n
6
410 nm
The general equation for a change
from one level to another:
Balmer’s
Equation:
 1
1 
 R  2  2  ; R  1.097 x 107 m -1
n

l
n
f
0


1
Summary (Cont.)
Bohr’s model sees the hydrogen atom with an
electron at many possible energy levels.
Emission
Absorption
The energy of the atom increases
on absorption (nf > ni) and decreases on emission (nf < ni).
Energy of
nth level:
13.6 eV
E
n2
The change in energy of the atom can be given in
terms of initial ni and final nf levels:
 1
1 
E  13.6 eV  2  2 
n

n
f
0


Summary
Apparently, light consists of
tiny bundles of energy called
photons, each having a welldefined quantum of energy.
Planck’s
Equation:
E = hf
Photon
E = hf
(h = 6.626 x 10-34 J s)
The Electron-volt:
1 eV = 1.60 x 10-19 J
1 keV = 1.6 x 10-16 J
1 MeV = 1.6 x 10-13 J
Summary (Cont.)
Incident light
Cathode
C
-
Anode
A
+
Ammeter
A
E
hc
l
 W  12 mv 2
hc
Threshold
W
wavelength lo
l0
If l is in nm, the energy in eV is found from:
Wavelength in nm;
Energy in eV
E
1240
l
Summary (Cont.)
Planck’s Experiment:
Incident light
Cathode
Anode
Stopping
potential
V
Slope
fo
y
x
Frequency
V
A
+
-
Potentiometer
Kmax = eVo
W
h
V0    f 
e
e
h
Slope 
e
Summary (Cont.)
Quantum physics works for waves or particles:
For a particle with zero
momentum p = 0:
A light photon has
mo = 0, but it does
have momentum p:
Wavelength
l
of a photon:
h

p
E = m oc 2
E = pc
de Broglie
Wavelength:
h
l
mv