Transcript CHAPTER 40 : INTRODUCTION TO QUANTUM PHYSICS 40.2) The Photoelectric Effect Light incident on certain metal surfaces caused electrons to be emitted from the surfaces.

CHAPTER 40 : INTRODUCTION TO
QUANTUM PHYSICS
40.2) The Photoelectric Effect
Light incident on certain metal surfaces
caused electrons to be emitted from the
surfaces = photoelectric effect
The emitted electrons = photoelectrons
Figure (40.6)
A diagram of an apparatus in which the photoelectric effect can occur
An evacuated glass or quartz tube contains a metallic plate E connected
to the negative terminal of a battery and another metallic plate C is
connected to the positive terminal of the battery
When the tube is kept in the dark – the ammeter reads zero – no current
in the circuit
When plate E is illuminated by light by light having a wavelength
shorter than some particular wavelength that depends on the metal used
to make plate E – a current is detected by the ammeter – a flow of
charges across the gap between plates E and C
This current arises from photoelectrons emitted from the negative plat
(the emitter) and collected at the positive plate (the collector)
Figure (40.7)
A plot of photoelectric current versus
potential difference V between plates E and
C for two light intensities
At large values of V – the current reaches a
maximum value – the current increases as
the intensity of the incident light increases
When V is negative (when battery in the
circuit is reversed to make plate E positive
and plate C negative) – the current drops to
a very low value because most of the emitted
photoelectrons are repelled by the now
negative plate C
Only those photoelectrons having a kinetic energy greater than the
magnitude of eV reach plate C – where e = the charge on the electron
When V is equal to or more negative than
- V s (the stopping potential) – no
photoelectrons reach C and the current is
zero
The stopping potential – independent of the radiation intensity
The maximum kinetic energy of the photoelectrons is
related to the stopping potential through the relationship:
K max  eVs
(40.7)
Features of the photoelectric effect – could not be
explained by classical physics or by the wave theory of
light
No photoelectrons are emitted if the frequency of the
incident light falls below some cutoff frequency fc (is
characteristic of the material being illuminated) (wave theory –
predicts that the photoelectric effect should occur at any frequency,
provided the light intensity is suffiently high)
The maximum kinetic energy of the photo-electrons is
independent of light intensity (wave theory – light of higher
intensity should carry more energy into the metal per unit time and eject
photoelectrons having higher kinetic energies)
The maximum kinetic energy of the photo-electrons
increses with increasing light frequency (wave theory – prdicts
no relationship between photoelectron energy and incident light frequency)
Photoelectrons are emitted from the surface almost
instantaneously (less that 10-9 s after the surface is
illuminated) – even at low light intensities (the photoelectrons
are expected to require some time to absorb the incident radiation before
they acquire enough kinetic energy to escape from the metal)
Successful explanation by Einstein
Assumed that light (or any other
electromagnetic wave) of frequency f can be
considered a stream of photons
Each photons has an energy E = hf
Figure (40.8)
In Einstein’s model – a photon is so localized
that it gives all its energy hf to a single
electron in the metal
According to Einstein – the maximum kinetic
energy for these liberated photoelectrons is :
K max  hf  
(40.8)
Photoelectric effect
equation
Where  = work function of metal ( the minimum energy with
which an electron is bound in the metal and is on the order of a
few electron volts) – Table (40.1)
Photon theory of light – explain the features of the
photoelectric effect that cannot be understood using
the concepts of classical physics :
• The effect is not observed below a cutoff frequency –
the energy of the photon must be greater than or equal
to .
• If the energy of the incoming photon does not satisfy
this condition – the electrons are never ejected from the
surface, regardless of the light intensity.
• Kmax is independent of light intensity – If the light
intensity is doubled, the number of photons is doubled –
doubles the number of photoelectrons emitted.
• Their maximum kinetic energy (= hf – ) – depends
only on the light frequency and the work function, not
on the light intensity
• Kmax increases with increasing frequency is easily
understood with Equation (40.8)
• Photoelectrons are emitted almost instantaneously – the
incident energy arrives at the surface in small packets
and there is a one-to-one interaction between photons
and photoelectrons.
• In this interaction the photon’s energy is imparted to an
electron that then has enough energy to leave the metal
– constrast to the wave theory, in which the incident
energy is distributed uniformly over a large area of the
metal surface
Final confirmation of Einstein’s theory
Experiment observation of a linear relationship
between Kmax and f –Figure (40.9)
Kmax
Figure (40.9)
fc
f
The intercept on the horizontal axis – the cutoff frequency below
which no photoelectrons are emitted, regardless of light intensity
The frequency is related to the work function
through the relationship fc =  / h
The cutoff frequency corresponds to a cutoff
wavelength of :
c 
c
c
hc


fc  / h 
(40.9)
c = speed of light
Wavelengths greater than c incident on a material having a work
function  do not result in the emission of photoelectrons
40.3) The Compton Effect
Compton and his co-workers – the classical
wave theory of light failed to explain the
scattering of x-rays from electrons
Classical theory
Electromagnetic waves of frequency fo incident on
electrons should have two effects (Figure (40.10a)) :
Radiation pressure should
caused the electrons to
accelerate in the direction
of progagation of the
waves
The oscillating electric
field of the incident
radiation should set the
electrons into oscillation at
the apparent frequency f’
f’ = the frequency in the frame of
the moving electrons
Frequency f’ is different from the
frequency fo of the incident
radiation because of the Doppler
effect : Each electron first absorbs
as a moving particle and then
reradiates as a moving particle –
exhibiting two Doppler shifts in the
frequency of radiation
Because different electrons will move at
different speeds after the interaction –
depending on the amount of energy absorbed
from the electromagnetic waves – the scattered
wave frequency at a given angle should show a
distribution of Doppler-shifted values
Compton’s experiment
At a given angle – only one frequency of
radiation was observed
Could explain these experiment by treating photons not as waves but
as point-like particles having enegy hf and momentum hf/c and by
assuming that the energy and momentum of any colliding
photo-electron pair are conserved
Compton effect – adopting a particle model
for wave (a scattering phenomenon)
Figure (40.10b) – the quantum picture of the
exchange of momentum and energy between
an individual x-ray photon and an electron
Figure (40.10b) :
• In the classical model – the electron is pushed
along the direction of prpagation of the
incident x-ray by radiation pressure.
• In the quantum model – the electron is
scattered through an angle  with respect to
this direction – a billiard-ball type collision.
Figure (40.11a)
A schematic diagram of the apparatuse used by Compton
The x-rays, scattered from a graphite target –
were analyzed with a rotating crystal
spectrometer
The intensity was measured with an ionization
chamber that generated a current proportional
to the intensity
The incident beam consisted of monochromatic
x-rays of wavelength
o = 0.071 nm.
Figure (40.11b) – the experimental intensity-versuswavelength plots observed by Compton for four
scattering angles (corresponding to  in Fig. (40.10)
The graphs for the three nonzero angles show two peaks
At o
At ’ > o
The shifted peak at ’ is caused by the
scattering of x-rays from free electrons, and it
was predicted by Compton to depend on
scattering angle as :
h
1  cos
' o 
mec
(40.10)
Compton shift equation
Where me = the mass of the electron
h
mec
= Compton wavelength c of the electron
c 
h
 0.00243nm
mec
The unshifted peak at o (Figure (40.11b)) – is caused
by x-rays scattered from electrons tightly bound to the
target atoms
This unshifted peak also is predicted by Eq. (40.10)
if the electron mass is replaced with the mass of a
carbon atom, which is about 23 000 times the mass
of the electron
There is a wavelength shift for scattering from an
electron bound to an atom – but it is so small that it was
undetectable in Compton’s experiment
Compton’s measurements were in excellent agreement
with the predictions of Equation (40.10)
Derivation of the Compton Shift Equation
By assuming that the photon behaves like a
particle and collides elastically with a free
electron initially at rest – Figure (40.12a)
The photon is treated as a particle having
energy E = hf = hc/ and mass zero
In the scattering process – the total energy and
total linear momentum of the system must be
conserved
Applying the principle of conservation of energy
to this process gives :
hc hc
  Ke
 o '
hc/o = the energy of the incident
photon, hc/’ = the energy of the
scattered photon, and Ke = the kinetic
energy of the recoiling electron
Because the electron may recoil at speeds
comparable to the speed of light – use the
relativistic expression Ke = mec2 – mec2
hc hc
  me c 2  me c 2
 o '
where   1 / 1  v2 / c2
(40.11)
Apply the law of conservation of momentum to this
collision – noting that both the x and y components of
momentum are conserved
The momentum of a photon has a magnitude
p = E/c and E = hf
p = hf/c
Substituting f for c gives p = h/
Because the relativistic expression for the momentum
of the recoiling electron is pe = mev : we obtain the
following expression for the x and y components of
linear momentum, where the angles are as described in
Fig. (40.12b) :
x component:
h
h
 cos  me v cos
 o '
h
y component : 0  sin   m e v sin 
'
(40.12)
(40.13)
Eliminating v and  from Eq. (40.11) to (40.13) – a single
expression that relates the remaining three variables
(’, o, and ) – the Compton shift equation :
  ' o 
h
1  cos
mec
40.5) Bohr’s Quantum Model of the Atom
The basic ideas of the Bohr theory as it applies to the
hydrogen atom:
• The electron moves in circular orbits around the proton under the
influence of the Coulomb force of attraction
– Figure (40.15)
• Only certain electron orbits are stable – the electron does not emit
energy in the form of radiation – the total energy of the atom remains
constant – and classical mechanics can be used to describe the
electron’s motion
• Radiation is emitted by the atom when the electron “jumps” from a
more energetic initial orbit to a lower-energy orbit.
• The frequency f of the photon emitted in the jump is related to the
change in the atom’s energy and is independent of the frequency of
the electron’s orbital motion
• The frequency of the emitted radiation is found from the energyconservation expression :
Ei – Ef = hf
(40.18)
where Ei = the energy of the initial state, Ef = the energy of the final
state, and Ei > Ef
• The size of the allowed electron orbits is determined by a condition
imposed on the electron’s orbital angular momentum : The allowed
orbits are those for which the electron’s orbital angular momentum
about thenucleus is an integral multiple of ħ = h/2 :
mevr = nħ
n = 1, 2, 3, …
(40.19)
Using these four assumptions
Calculate the
allowed energy
levels
emission
wavelengths of the
hydrogen atom
Electric potential energy of the system
(Fig. (40.15)) : U = keq1q2/r = – kee2/r
where ke is the Coulomb constant and the negative sign arises
from the charge – e on the electron
The total energy of the atom which contains
both kinetic and potential energy terms :
2
1
e
E  K  U  me v 2  k e
2
r
(40.20)
Newton’s second law :
The Coulomb attractive force kee2/r2
exerted on the electron must equal the
mass times the centripetal acceleration
(a = v2/r) of the electron
k ee2 me v 2

2
r
r
The kinetic energy of the electron is :
me v 2 k ee2
K

2
2r
(40.21)
Substituting this value of K into Eq. (40.20)
– the total energy of the atom is :
k ee2
E
2r
The total energy is negative –
indicationg a bound electronproton system
(40.22)
Energy in the amount of kee2/2r must
be added to the atom to remove the
electron and make the total energy
of the system zero
Obtain an expression for r, the radius of the
allowed orbits – by solving Equations (40.19)
and (40.21) for v and equationg the results :
n 2h 2 k ee2
v  2 2
me r
me r
2
n 2h 2
rn 
me k ee2
n = 1, 2, 3, …
(40.23)
The radii have discrete values – they are quantized
The result is based on the assumption that the
electron can exist only in certain allowed orbits
determined by the integer n
The orbit with the smallest radius = Bohr
radius ao (corresponds to n = 1) :
h2
ao 
 0.0529nm
2
me k e e
(40.24)
A general expression for the radius of any
orbit in the hydrogen atom by substituting
Equation (40.24) into Equation (40.23) :
rn  n 2a o  n 2 (0.0529nm)
(40.25)
Radii of Bohr orbits in hydrogen
Figure (40.16) – the first three circular Bohr
orbits of the hydrogen atom
The quantization of orbit radii immediately
leads to energy quantization
Substituting rn = n2ao into Equation (40.22)
– obtained the allowed energy levels of
hydrogen atom :
k ee2  1 
En  
 2
2a o  n 
n = 1, 2, 3, …
(40.26)
Inserting numerical values :
13 .606
En  
eV
2
n
Energy levels
n = 1, 2, 3, …
(40.27)
Only energies satisfying this
equation are permitted
Ground state = the lowest allowed energy
level (n = 1, energy E1 = – 13.606 eV)
First excited state = the next energy level
(n = 2, energy E2 = E1/22 = – 3.401 eV)
Figure (40.17) – an energy level diagram
showing the energies of these discrete energy
states and the corresponding quantum
numbers n
The uppermost level – corresponding to n = 
(or r = ) and E = 0 : represent the state for
which the electron is removed from the atom
Ionization energy = the minimum energy
required to ionize the atom (to completely
remove an electron in the ground state from
the proton’s influence)
Figure (40.17) – the ionization energy for
hydrogen in the ground state (based on Bohr’s
calculation = 13.6 eV
Eqs. (40.18) and (40.26) – to calculate the
frequency of the photon emitted when the
electron jumps form an outer orbit to an inner
orbit :
Ei  E f k ee2  1
1
 2  2 
f

h
2a o h  n f n i 
(40.28)
Frequency of a photon emitted from hydrogen
Because the quantity measured
experimentaly is wavelength – use c = f
to convert frequency to wavelength :
k ee2  1
1 f
1 
 2  2 
 
 c 2a o hc  n f n i 
 1 1
1
 R H  2  2 

 nf ni 
(40.29)
(40.30)
Identical to relationships discovered by Balmer and Rydberg
(Eqs. (40.14) to (40.17)
The constant kee2/2aohc = Rydberg constant, RH = 1.097 373 2 x 107m-1
Bohr extended his model for hydrogen to
other elements
In general – to describe a singel electron
orbiting a fixed nucleus of charge +Ze (where
Z = the atomic number of the element),
Bohr’s theory gives :
ao
rn  (n )
Z
(40.31)
 Z2 
 2 
n 
n = 1, 2, 3, … (40.32)
2
k ee2
En  
2a o