l, only for shorter than certain wavelength
Current
-
3I
Vacuum tube
0
Fixed wavelength
Varying intensity
2I
I
V
Vo
Maximum electron energy
V
Current
l1 < l2 < l3
Q. Why the maximum electron energy
depend on the wavelength (frequency)?
Why strong beams of long wavelength
cannot knock out electrons?
l1
l2
Classical EM theory (wave theory) tells us
that the max electron energy should depend
on only the intensity of the light.
l3
V
Huge water
wave, large
amplitude sound
wave
In 1900, Max Planck postulated that electromagnetic energy is emitted
in discrete packets, or quanta. The energy of these
quanta is proportional to the frequency of the radiation.
E = hf = hc/l
h = 6.6 x 10-34 Js (Planck’s constant)
In explaining the photoelectric effect, Einstein picked up the idea
of Planck, and proposed in 1905 that light was not only emitted
by bundles of energy E = hf, but it was also absorbed in such bundles.
Light Quanta = Photons
Kinetic Energy of an electron = Ephoton - Ethreshold
1 2
mv  0
2
Ephoton = hc/l > Ethreshold
Work function: depends on
material ~ eV
light quantum: photon
-
In this case, light can be considered as a massless particle, photon
with energy solely determined by its wavelength (frequency).
There is no conflict with Einstein’s relativistic mechanics:
massless particle can have speed of light!
E = hf = hc/l
h = 6.6 x 10-34 Js (Planck’s constant)
l = 580 nm photon (yellow light) carries
E = (6.6 x 10-34 Js)(3 x 108 m/s)/(580 x 10-9 m)
= 3.4 x 10-19 J
How do we understand the intensity of light?
Intensity of light (EM radiation) ≈
number of photons
with same wavelength or frequency
Wave behaves like a particle.
Does a particle behaves like wave?
It would seem that the basic idea of the quantum theory is
the impossibility of imagining an isolated quantity of energy
without associating with it a certain frequency
de Broglie in 1923 as a graduate student
light quantum (photon) E = hf = hc/l
Relativistic case
E  mo c 4  p 2 c 2
2
for mo = 0
E = pc = hc/l
l = h/p
wave quantity
particle quantity
l = h/p
De Broglie proposed that all particles (electrons) should have
wavelength associated with their momentum in exactly the
same manner.
particle
wave
Q. What is the de Broglie wavelength of an electron that has a
kinetic energy of 100 eV?
After an electron is accelerated in 100 V potential difference, its
kinetic energy is 100 eV.
eV unit has to be converted into SI unit, Joule.
1 eV = 1.6 x 10-19 J
Ek = (1/2)mov2 = 1.6 x 10-17 J
v2 = 2Ek/mo = 2(1.6 x 10-17 J)/(9.1 x 10-31 kg)
= 3.52 x 1013 m2/s2
v = 5.93 x 106 m/s
low speed: no need to use relativistic
l = h/p
= h/mov = (6.6 x 10-34 Js)/(9.1 x 10-31 kg x 5.93 x 106 m/s)
= 1.23 x 10-10 m = 0.123 nm
C.J. Davisson and L.H Germer scattered low energy electrons off
The Ni crystal and observed a peculiar pattern of scattered electrons.
Diffraction pattern from the regular
Crystalline structure of order of
0.1 nm size.
Hydrogen Spectrum
ABSORPTION SPECTRUM
Emission SPECTRUM
Bohr’s Hydrogen Atom and the Electron as a Wave
In 1913, Rutherford’s atom received a quantitative description
from Niels Bohr who explained experimentally observed discrete
nature of atomic spectrum of Hydrogen. In spite of its immediate
success in providing theoretical account of the spectrum and other
nature of Hydrogen atom, a complete understanding of Bohr’s
atom came only after de Broglie’s conjecture (1923) that electrons
should display wave properties.
+
Niels Bohr (1885 – 1962)
Louis de Broglie (1892 – 1987)
Bohr’s Hydrogen Atom
(1 proton and 1 electron)
centripetal force = Coulomb force
mv2
e2
k 2
r
r
+
-
-
-
electron as a wave
(de Broglie)
2pr = l, 2l, 3l, …
l = h/mv
This is the necessary condition for electron to maintain an orbit.
2
2pr = l, 2l, 3l, …
l = h/mv
2
mv
e
k 2
rn
rn
2
2prn = nh/mv
2
nh
rn 
2
2
4p ke m
2pke
vn 
nh
2
rn = (5.3 x 10-11)n2 (m); vn = 2 x106/n (m/s)
n
rn (nm)
vn (m/s)
2
0.212
1 x 106
3
0.477
5 x 105
1
0.053
2 x 106
Now, let’s think about the total energy of the electron in the nth orbit.
Etot
2
2
(1/2)mv
= Potential
+ Kinetic Energy
-ke /rEnergy
n
n
2
2pke
vn 
nh
2
2
nh
rn 
4p 2 ke2 m
2 2 4
2
p
k em
13.6
n
Etot  
  2 eV
2 2
nh
n
n
1
En (eV)
-13.6
2
-3.4
3
-1.5
∞
0
Etot
0
n=∞
-1.5 eV
n=3
-3.4 eV
n=2
+
-
-
-
Ionized state of Hydrogen: proton
n2h2
rn 
2
2
4p ke m
2pke
vn 
nh
2
-13.6 eV
n=1
2 2 2
2
p
k em
13.6
n
Etot  
  2 eV
2 2
nh
n
Electron Energy diagram of Hydrogen atom
Etot
0
n=∞
-1.5 eV
n=3
-3.4 eV
n=2
Electron has to absorb 12.1 eV energy for this
Electron has to absorb 10.2 eV energy for this
-13.6 eV
-
n=1
-
2p k e m
13.6
E 
  2 eV
2 2
nh
n
2
n
tot
2 2
lowest energy:
ground state
2 2 2
2
p
k em
13.6
n
Etot  
  2 eV
2 2
nE
h = hc/l n
1 
 1
E (n  m)  En  Em  13.6 2  2 
n 
m
in eV
1 
1
 1
 R 2  2 
 
n 
 l  nm
m
R  1.097 10 m
7
E(nm) or l < 0
E (nm) or l > 0
1
(Rydberg const.)
Absorb photons of a given l
Emit photons of a given l
Etot
E = hc/l
0
n=∞
-1.5 eV
n=3
-3.4 eV
n=2
E (12) = 10.2 eV
= 10.2 x (1.6 x 10-19 J/eV)
= 1.63 x 10-18 J
 l = 121 nm
Ultraviolet range
-13.6 eV
-
n=1
Q What is the longest wavelength em radiation that can ionize
unexcited hydrogen atom?
ground state (n = 1)
smallest energy
n=1m=∞
1 
1
 1
 R 2  2 
 
n 
 l  nm
m
R  1.097 107 m 1
1/l = (1.097 x 107) x (0 – 1)
= - 1.097 x 107 (m-1)
l = -9.12 x 10-8 (m)
= -91.2 (nm) (- means absorption)
1 
1
1
 R 2  2 
 
 l  m3
3 m 
1 
1
1

R

 
 2
2 
 l  m 2
2 m 
1 
1

 R1  2 
 
 l  m1
 m 
Q. What is the shortest wavelength for the Balmer series?
Largest energy difference in Balmer series
From n = ∞ to m = 2 (Balmer)
1 
1
 1
 R 2  2 
 
n 
 l  nm
m
R  1.097 107 m 1
l = (R x (1/4 – 0))-1
= 365 nm
The Balmer series are in the visible range.