Transcript Quantization of Charge, Light, and Energy

Quantization of Charge, Light, and
Energy
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1. Quantization of Electric Charge;
2. Blackbody Radiation;
3. The Photoelectric Effect
4. X-rays and the Compton Effect.
Three great quantization discoveries:
1. Quantization of Electrical Charge
2. Quantization of Light Energy
3. Quantization of energy of Oscillating
Mechanical System
Quantization of Electric Charge
Early measurements of e and e/m.
The first estimates of the magnitude of electric charges
found in atoms were obtained from Faraday law.
Faraday passed a direct current through weakly conducting
solutions and observed the subsequent liberation of the
components of the solution on electrodes.
Faraday discovered that the same quantity of electricity, F,
called latter one faraday, and equal to about 96,500 C,
always decomposed 1 gram-ionic weight of monovalent
ions.
Quantization of Electric Charge
1F (one faraday) = 96,500C
1F decomposed 1gram-ionic weight of
monovalent ions.
Example: If 96,500 C pass through a solution of
NaCl, 23g of Na appears at the cathode and 35.5g
of Cl appears at the anode.
1F =NAe - Faradays Law of Electrolysis
where NA – Avogadro’s number
e – minimum amount of charge, that was
called an electron
Discovery of electron: Thomson’s Experiment.
Many studies of electrical discharges in gases were done
in the late 19th century. It was found that the ions responsible
for gaseous conduction carried the same charge as did those
in electrolysis.
J.J. Thomson in 1897 used crossed electric and magnetic
fields in his famous experiment to deflect the cathode-rays.
In this way he verified that cathode-rays must consist of
charged particles. By measuring the deflection of these
particles Thomson showed that all the particles have the same
charge-to-mass ratio q/m. He also showed that particles with
this charge-to-mass ratio can be obtained using any material
for a source, which means that this particles, now called
electrons, are a fundamental consistent of all matter.
Thomson’s tube for measuring e/m.
Electrons from the cathode C pass through the slits at
A and B and strike a phosphorescent screen. The beam can
be deflected by an electric field between plates D and E or by
magnetic field. From measurements of the deflections
measured on a scale on the tube’s screen, e/m can be
determined.
FE=qE
_q
-
FB=qvB
FE=FB
qE = qvB
Crossed electric and magnetic fields. When a negative
particle moves to the right the particle experience a
downward magnetic force FB=qvB and an upward electric
force FE=qE. If these forces are balanced, the speed of the
particle is related to the field strengths by
v=E/B
Thomson’s Experiment
In his experiment Thomson adjusted E
that the particles were undeflected.
┴B
so
This allowed him determine the speed of the
particle u =E/B. He then turned off the B field
and measured the deflection of the particles on
the screen.
Deflection of the Electron Beam.
Deflection of the beam is shown with the top plate positive.
Thomson used up to 200 V between the plates. A magnetic
field was applied perpendicular to the plane of the diagram
directed into the page to bend the beam back down to its
undeflected position.
With the magnetic field turned off, the beam are deflected by an
amount y=y1+y2. y1 occurs while the electrons are between the
plates, y2 after electrons leave the region between the plates. Lets
x1 be the horizontal distance across the deflection plates. If the
electron moves horizontally with speed v0 when it enters the region
between the plates, the time spent between the plates is t1=x1/v0,
and the vertical component of velocity when it leaves the plates is
qE y
qE y x1
v y  a yt1 
t1 
m
m v0
where Ey is the upward component of electric field. The deflection
2
y1 is:
qE y 2
qE y  x1 
2
1
1
1
 
y1 
a t 
t 
2 y1
2 m 1
2 m v 
 0
The electron then travels an additional horizontal distance x2
in the field-free region from the deflection plates to the screen.
Since the velocity of the electron is constant in this region, the
time to reach the screen is t2=x2/v0 , and the additional vertical
deflection is:
y2  v y t2
qE y x1 x2

m v0 v0
The total deflection at screen is therefore

 qE y  2  qE y 
 E y  x12 
q


y  y1  y2  1  2  x1   2  x1x2    2    x1x2 
2  mv 
 mv 
 v  2 
m



 0
 0
 0  
This equation cab be used to determine the charge tomass ratio q/m from measured deflection y.
Example
Electrons pass undeflected through the plates of Thomson’s
apparatus when the electric field is 3000 V/m and there is a
crossed magnetic field of 1.40 G. If the plates are 4-cm long
and the ends of the plates are 30 cm from the screen, find
the deflection on the screen when the magnetic field is
turned off.
me=9.11 x 10-31kg;
q=e=-1.6 x 10-19C
The Mass Spectrometer
The mass spectrometer, first designed by Francis
William Aston in 1919, was developed as a means
of measuring the masses of isotopes.
Such measurements are important in determining
both the presence of isotopes and their abundance
in nature.
For example, natural magnesium has been found
to consist of 78.7 % 24Mg; 10.1% 25Mg; and 11.2%
26Mg. These isotopes have masses in the
approximate ratio 24:25:26.
Schematic drawing of a mass spectrometer.
Positive ions from an ion source
are accelerated through a
potential difference ΔV and enter
a uniform magnetic field.
The magnetic field is out of the
plane of the page as indicated by
the dots.
The ions are bent into a circular
arc and emerge at P2 (a plane of
photographic plate or another ion
detector). The radius of the circle
is proportional to the mass of the
ion
The Mass Spectrometer
In the ion source positive ions are
formed by bombarding neutral atoms
with X-rays or a beam of electrons.
(Electrons are knock out of the atoms by
the X-rays or bombarding electrons).
These ions are accelerating by an
electric field and enter a uniform
magnetic field. If the positive ions start
from rest and move through a potential
difference ΔV, the ions kinetic energy
when they enter the magnetic field
equals their loss in potential energy,
q|ΔV|:
1 2
mv  q V
2
F  ma
2
v
qvB  m
R
The ions move in a semicircle
of radius R. The velocity of the
particle is perpendicular to the
magnetic field. The magnetic
force provides the centripetal
force v2/R in circular motion.
We will use Newton’s second
low to relate the radius of
semicircle to the magnetic
field and the speed of the
particle. If the velocity of the
particle is v, the magnitude of
the net force is qvB, since v
and B are perpendicular.
mv
R
qB
The Mass Spectrometer
The speed v can be eliminate from equations
1 2
mv  q V
2
mv
R
qB
and
2
2
R q B
v 
2
m
2
2
Substituting this for v2:
1 R 2q 2 B 2
m
 q V
2
2
m
Simplifying and solving for (m/q):
2
2
m B R

q 2 V
Separating Isotopes of Nickel
• A 58Ni ion charge +e and mass 9.62 x 10-26kg is
accelerated through a potential drop of 3 kV and
deflected in a magnetic field of 0.12T.
• (a) Find the radius of curvature of the orbit of the
ion.
• (b) Find the difference in the radii of curvature of
58Ni ions and 60Ni ions. (assume that the mass
ratio is 58:60).
Blackbody Radiation
• One unsolved puzzle in physics in late nineteen
century was the spectral distribution of so called
cavity radiation, also referred to as blackbody
radiation.
• It was shown by Kirchhoff that the most efficient
radiator of electromagnetic waves was also a most
efficient absorber. A “perfect” absorber would be
one that absorbed all incident radiation.
• Since no light would be reflected it is called a
blackbody.
A small hole in the wall of the cavity approximating
an ideal blackbody. Electromagnetic radiation (for
example, light) entering the hole has little chance of
leaving before it is completely adsorbed within the
cavity.
Blackbody Radiation
• As the walls of the cavity absorb this incoming radiation ,
their temperature rises and begin to irradiate.
• In 1879, the Austrian physicist J.Stefan first measured the
total amount of radiation emitted by blackbody at all
wavelengths and found it varied with absolute temperature.
• It was latter explained through a theoretical derivation by
Boltzman, so the result became known as the StefanBoltzman radiation Law:
R = σT4
where R is the power radiated per unit time and per unit
area of blackbody; T is the Kelvin temperature; and σ is the
Stefan-Boltzman constant, σ=5.672 x 10-8 W/m2K4.
Blackbody Radiation
R = σT4
• Note that the power per unit area radiated by blackbody
depends only on the temperature, and not of other
characteristic of the object, such as its color or the material, of
which it is composed.
• R tells as the rate at which energy is emitted by the object. For
example, doubling the absolute temperature of an object
increases the energy flows out of the object by factor of 24=16.
• An object at room temperature (300 K) will double the rate at
which it radiates energy as a result of temperature increase of
only 570.
Radiation emitted by the object at temperature T that passed
through the slit is dispersed according to its wavelength. The
prism shown would be an appropriate device for that part of
the emitted radiation in the visible region. In other spectral
regions other types of devices or wavelength-sensitive
detectors would be used.
Spectral distribution function R(λ) measured at different
temperatures. The R(λ) axis is in arbitrary units for comparison
only. Notice the range in λ of the visible spectrum. The Sun
emits radiation very close to that of a blackbody at 5800 K. λm
is indicated for the 5000-K and 6000-K curves.
Blackbody Radiation
The German physicist W.Wien derived a relationship for
maximum wavelength and absolute temperature, known as
Wien’s displacement law:
λmT = constant=(2.898 x 10-3m)K
Example: The wavelength at the peak of the spectral distribution
for a blackbody at 4300 K is 674 nm (red). At what temperature
would the peak be 420 nm (violet)?
Solution: From the Wien’s law, we have
λ1T1 = λ2T2
(674 x 10-9m)(4300 K) = (420 x 10-9m)(T2)
T2=6900 K
•
This law is used to determine the surface
temperatures of stars by analyzing their
radiation. It can also be used to map out the
variation in temperature over different regions of
the surface of an object. Such a map is called
thermograph.
• For example thermograph can be used to detect
cancer because cancerous tissue results in
increased circulation which produce a slight
increase in skin temperature.
Radiation From the Sun. The radiation emitted by
the surface of the sun emits maximum power at
wavelength of about 500 nm. Assuming the sun to
be a blackbody emitter, (a) what is it surface
temperature? (b) Calculate λmax for a blackbody
at room temperature, T=300 K.
The calculation of the
distribution function R(λ)
involves the calculation of
the energy density of
electromagnetic waves in
the cavity. The power
radiated out of the hole is
proportional to the total
energy density U (energy
per unit volume) of the
radiation in the cavity. The
proportionality constant can be shown to be c/4, where c is
the speed of the light:
R = (1/4)cU
Similarly, the spectral distribution of the power proportional to
the spectral distribution of the energy density in the cavity.
• If u(λ)dλ is the fraction of
the energy per unit
volume in the cavity in the
range dλ, then u(λ) and
R(λ) are related by
R(λ)=(1/4)cu(λ)
The energy density
distribution function u(λ)
can be calculated from
classical physics. We can
find the number of modes
of oscillation of the
electromagnetic field in
the cavity with wavelength λ in the interval dλ and multiply it
by average energy per mode. The result is that the number
of modes of oscillation per unit volume, n(λ), is independent
of the shape of cavity and is given by:
n(λ) = 8πcλ-4
Rayleigh-Jeans Equation
The number of modes of oscillation per unit volume:
n(λ) = 8πcλ-4
According to the classical kinetic theory, the average
energy per mode of oscillation is kT, the same as for a
one-dimensional harmonic oscillator, where k is the
Boltzman constant.
Classical theory thus predicts for the energy density
spectral distribution function
u(λ) = kTn(λ) = 8πckTλ-4
Rayleigh-Jeans Equation
u(λ) = kTn(λ) = 8πckTλ-4
This prediction, initially derived by Rayleigh, is called the
Rayleigh-Jeans Law.
At very long wavelength the Rayleigh-Jeans law agrees with
experimentally determined spectral distribution, but at short
wavelength this law predicts that u(λ) becomes large,
approaching infinity as λ→0, whereas experiment shows that
the distribution actually approaches zero as λ→0.
This enormous disagreement between experimental
measurements and classical theory for short wavelength was
called the ultraviolet catastrophe.
Comparison of the Rayleigh-Jeans Law with experimental
data at T=1600 K. The u(λ) axis is linear.
In 1900 the German
physicist Max Plank by
making some unusual
assumptions derived a
function u(λ) that agreed
with experimental data.
Classically, the
electromagnetic waves in
the cavity are produced
by accelerated electric
charges in the walls
vibrating like simple harmonic oscillators. The average energy for
simple harmonic oscillator is calculated classically from MaxwellBoltzman distribution function:
f(E) = Ae-E/kT
where A is a constant and f(E) is the fraction of oscillators with
energy E.
Maxwell-Boltzman distribution function
f(E) = Ae-E/kT
The average energy is then found, as is any weighted
average:

E   Ef E dE   EAe
E
kT
dE
Plank found that he could derive his empirical function
assuming the energy of oscillators, and hence the
radiation that they emitted, was a discrete variable that
could take only the values o, ε, 2ε, 3ε,…..,nε where n is
an integer, and that ε is proportional to the frequency of
the oscillators, and, thus, the radiation.
Plank therefore wrote the energy as
En=nε=nhf
n=0,1,2,…..
where h is a constant now called Plank constant.
The Maxwell-Boltzman distribution than becomes:
fn= Ae-E/kT = Ae-nε/kT
where A is determined by normalization condition
that the sum of all fractions fn must be equal 1.
The normalization condition
f
n
 A e

n
kT
1
The average energy of oscillation is then given by discretesum equivalent
E   En f n   En Ae

E
kT

hf
x

kT kT
To solve this equation let put
f
n
 A e
 nx
A(1  y  y  y  ....)  1
2
where y=e-x.
3
x
x 2
x 3
 A[e  e  (e )  (e )  ....] 
0
This sum is a series expansion of (1-y)-1, so
(1-y)-1=1+y+y2+y3+… ,
then Σfn=A(1-y)-1=1 gives A=1-y and

E   En Ae
0

E
kT

 A nhfe

nhf
kT

 Ahf  ne nx
0
0
d
   e nx . But
dx
 nx
1
e

(
1

y
)

Note that
 ne
 nx
so we have
d
d
dy 
 nx
 nx
1
2 
2
ne


e


(
1

y
)

(
1

y
)


y
(
1

y
)




dx
dx
 dx 
since
dy d  x

(e )   e  x   y
dx dx
Plank’s Law
Multiplying this sum by hf and using A=(1-y), the
average energy is:
E  hfA
 ne
 nx
x
hfy hfe
 hf (1  y) y(1  y) 

x
1 y 1 e
2
Multiplying the numerator and the denominator by
ex and substituting for x, we obtain:
E
hf
hf
 kT

 e  1





hc

hf
 kT

 e  1




Plank’s Law
Multiplying this result by the number of oscillators
per unit volume in the interval dλ given by
n(λ)=8πcλ-4 (the number of modes of oscillation
per unit volume) we obtain the energy distribution
function for the radiation in cavity:
8hc 
u ( ) 
hc
 kT

e


1




2
5
This function is called Plank’s Law.
Plank’s Law
The value of Plank’s constant h can be
determined by fitting the function
8hc 
u ( ) 
 hc

 e kT  1




2
5
to the experimental data, although the direct
measurement is better, but more difficult. The
presently accepted value of Plank’ constant is:
h = 6.626 x 10-34 J·s = 4.136 x 10-15 eV·s
u ( ) 
8ckT
4
Comparison of Plank’s Law and the Rayleigh-Jeans Law
with experimental data at T=1600 K. The u(λ) axis is linear.
Plank’s Law
A dramatic example of an application of Planck’s law is the
test of the predictions of the so-called Big Bang theory of the
formation and expansion of the universe.
Current cosmological theory suggests that the universe
originated in an extremely high-temperature explosion, one
consequence of which is to fill the infant universe with
radiation that can be approximate with black body spectral
distribution.
In 1965, Arno Penzias and Robert Wilson discovered
radiation of wavelength 7.35 cm reaching the Earth with
same intensity from all directions in space. It was recognized
soon as a remnant of the Big Bang (relict radiation).
The energy density spectral distribution of the cosmic
microwave background radiation. The solid line is Plank’s Law
with T=2.735 K. The measurements were made by the
Cosmic Back Ground Exploder (COBE) satellite.
Problem 1. Thermal Radiation from the Human Body.
The temperature of the skin is approximately 35°C. What
is the wavelength at which the peak occurs in the
radiation emitted from the skin?
Problem 2. The Quantized Oscillator. A 2-kg mass is
attached to a massless spring of force constant
k=25N/m. The spring is stretched 0.4m from its
equilibrium position and released. (a) Find the total
energy and frequency of oscillation according to classical
calculations. (b) Assume that the energy is quantized
and find the quantum number, n, for the system. (c) How
much energy would be carried away in one-quantum
change?
Problem 3. The Energy of a “Yellow” Photon. What is
the energy carried by a quantum of light whose
frequency equals 6 x 1014 Hz yellow light? What is the
wavelength of this light?
The Photoelectric Effect
It is one of the ironies in the history of the science that
in the same famous experiment of Heinrich Hertz in1887 in
which he produced and detected electromagnetic waves,
thus confirmed Maxwell’s wave theory of light, he also
discovered the photoelectric effect led directly to particle
description of light.
It was found that negative charged particles were
emitted from a clean surface when exposed to light.
P.Lenard in 1900 detected them in a magnetic field
and found that they had a charge-to-mass ratio of the same
magnitude as that measured by Thompson for cathode rays:
the particles being emitted were electrons.
Schematic diagram of
the apparatus used by
P.Lenard to
demonstrate the
photoelectric effect and
to show that the
particles emitted in the
process were electrons.
Light from the source L
strikes the cathode C.
Photoelectrons going through the hole in anode A are
recorded by the electrometer connected to α. A magnetic
field, indicated by the circular pole piece, could deflect the
particles to an electrometer connected to β, enabling the
establishment of the sign of their charge and their q/m ratio.
If some of emitted
electrons that reaches
an anode A pass
through the small hole,
a current results in the
external electrometer
circuit connected to α.
The number of emitted
electrons, reaching the
anode, can be
increased by making the anode positive with respect to
cathode.
Letting V be the potential difference between A and C the
next picture shows the current versus V for two values of
the intensity of light incident on the cathode:
 Photocurrent i versus anode voltage V for light of frequency
f with two intensities I1 and I2 , where I2>I1 . At sufficiently large
V all emitted electrons reach the anode and the current
reaches its maximum value.
 From experiment it was observed that the maximum current
is proportional to the light intensity.
 An expected result – if the intensity of incident light doubled
the number of emitted electrons should also double. If
intensity of incident light is too low to provide electrons with
energy necessary to escape from the metal, no emission of
electron should be observed.
 However, there was no minimum intensity below which the
current was absent.
When V is negative, the electrons are repelled from the
anode and only electrons with initial kinetic energy mv2/2
grater than eV can reach the anode. From the graph we can
see if the voltage is less, than –V0 no electrons reach anode.
The potential V0 is called the stopping potential. It related to
the maximum kinetic energy as
(mv2/2)max = eV0
 The experimental result, that V0 is independent of the
incident light intensity was surprising – increasing the
rate of energy falling on the cathode does not increase
the maximum kinetic energy of the emitted electrons.
 In 1905, Einstein offered an explanation of this result:
he assumed, that energy quantization used by Plank
in the blackbody problem was a universal
characteristic of light. Light energy consist of discrete
quanta of energy hf.
 When one of this quanta, called photon, penetrates the
surface of cathode, all of its energy may be given
completely to electron. If Φ is the energy necessary to
remove an electron from the surface (Φ is called the work
function and is a characteristic of the metal), the
maximum kinetic energy of the electrons leaving the
surface will be (hf – Φ) and the stopping potential V0
should be given by
 mv2 

eV0  
 hf  
 2 

 max
 This equation is referred as the photoelectric effect
equation.
 As can be seen from
 mv2 

eV0  
 hf  
 2 

 max
the slope of the line on the graph V0 versus f should equal h/e.
 The minimum, or threshold, frequency for photoelectric effect,
ft and the corresponding threshold wavelength λt are related
to work function Φ by setting V0 = 0 :
  hft 
hc
t
 Photons of frequency lower than ft ( and therefore having
wavelength grater than λt ) do not have enough energy to
eject an electron from the metal.
For constant I Einstein’s explanation of the photoelectric effect
indicates that the magnitude of the stopping voltage should be
grater for f2 than f1, as observed, and that there should be a
threshold frequency ft below which no photoelectrons were
seen, also in agreement with experiment.
Millikan’s data for stopping potential versus frequency for the
photoelectric effect. The data falls on a straight line of slope
h/e, as predicted by Einstein. The intercept of the stopping
potential axis is –Ф/e.
Example: Photoelectric Effect in Potassium
The threshold wavelength of potassium is 558 nm. What is
the work function for potassium? What is the stopping
potential when light of wavelength 400 nm is used?
Example: Photoelectric Effect in Potassium
The threshold wavelength of potassium is 558 nm. What is the
work function for potassium? What is the stopping potential
when light of wavelength 400 nm is used?
• Solution:
1 2
eV0   mv   hf  
2
 max
hf 
V0 

e e
 hf t hc


e
e et
hc 1240eV  nm


 2.22eV
t
558nm
Example: Photoelectric Effect in Potassium
• The threshold wavelength of potassium is 558 nm.
What is the work function for potassium? What is
the stopping potential when light of wavelength
400 nm is used?
1240eV  nm
V0 
 
 2.22eV

400nm
 3.10eV  2.22eV  0.88V
hc
Photoelectric Effect for Sodium
A sodium surface is illuminated with light of
wavelength 3x10-7m. The work function for sodium
is 2.28 eV. Find (a) the kinetic energy of the
ejected photoelectrons and (b) the cutoff
wavelength for sodium.
X Rays and the Compton Effect
Future evidence of the correctness of the photon concept
was given by Arthur Compton, who measured the scattering
of x-rays by free electrons.
The German physicist Wilhelm Röentgen discovered
x-rays in 1895 when he was working with a cathode-ray tube.
He found, that “rays”, originating from the point where the
cathode rays (electrons) hit the glass tube, or a target within the
tube, could pass through the materials opaque to light and
activate a fluorescent screen or photographic film. He found
that all materials were transparent to this rays to some degree,
depending of the density of this materials.
The slight diffraction of an x-ray beam after passing slit of a
few thousands of a mm wide indicated their wavelength in other
of 10-10m = 0.1nm.
(a)
(b)
(a) Early x-ray tube and (b) typical of the mid-twenties
century x-ray tube design.
Diagram of the components of a modern x-ray tube. Design
technology has advanced enormously, enabling very high
operating voltages, beam currents, and x-ray intensities, but
the essential elements of the tubes remain unchanged.
X-Ray’s Diffraction
Experiment soon confirmed that x-rays are a form
of electromagnetic radiation with wavelength of
about 0.01nm to 0.10 nm.
It was also known that atoms in crystals are
arranged in regular arrays that are spaced by about
same distances.
In 1912, Laue suggested that since the wavelength
of x-rays were on the same order of magnitude as
the spacing of atoms in a crystal, the regular array
of atoms in crystal might act as a three-dimensional
grating for diffraction of x-rays.
X-Ray’s Diffraction
• W.L.Bragg, in 1912, proposed a simple and and
convenient way of analyzing the diffraction of xrays due to scattering from various sets of
parallel planes of atoms, now called Bragg
planes.
• Two sets of Bragg planes are illustrated for
NaCl, which has a simple crystal structure called
face-centered cubic.
A face-centered cubic crystal of NaCl showing
two sets of Bragg planes.
• Waves scattered at equal angels from atoms in
two different planes will be in phase (constructive
interference) if the difference in path length is an
integer number of wavelength. This condition is
satisfied if
2d sin θ = mλ where
m = an integer
This equation called the Bragg condition.
Bragg scattering from two successive planes. The waves from
the two atoms shown have a path difference of 2dSinθ. They
will be in phase if the Bragg condition 2dSinθ = mλ is met.
Schematic sketch of the Laue experiment. The crystal acts as a
three-dimensional grating, which diffracts the x-ray beam and
produce a regular array of spots, called a Laue pattern, on a
photographic plate.
Modern Laue-type x-ray
diffraction pattern using a
niobium diboride crystal
and 20-kV molybdenum
x-rays. [General Electric
Company.]
Incident
X-ray
Beam
Scattered
X-rays
Schematic diagram of Bragg crystal spectrometer. A collimated
x-ray beam is incident on a crystal and scattered into an
ionization chamber. The crystal and ionization chamber can be
rotated to keep the angles of incidence and scattering equal as
both are varied. By measuring the ionization in the chamber as a
function of angle, the spectrum of the x-rays can be determined
using the Bragg condition 2dSinθ = mλ, where d is the
separation of the Bragg planes in the crystal. If the wavelength λ
is known, the spacing d can be determined.
Two typical x-ray spectra are produced by accelerating
electrons through two voltages V and bombarding a tungsten
target. I(λ) is the intensity emitted with the wavelength interval
dλ at each value of λ.
The spectrum consist of a series of sharp lines, called the
characteristic spectrum. The line spectrum is a
characteristic of target material and varies from element to
element.
The continuous spectrum has a sharp cutoff wavelength λm
which is independent of the target material but depends on
the energy of the bombarding electrons.
If the voltage of the x-ray tube is V in volts, the cutoff
wavelength was found empirically by
1.24 103
m 
nm
V
It was pointed out by Einstein that x-ray production by
electron bombardment was an inverse photoelectric effect and
equation
 mv2 

eV0  
 hf  
 2 

 max
should be applied. The λm simply correspond to a photon with
the maximum energy of the electrons, i.e. the photon emitted
when the electron losses all of its kinetic energy in a simple
collision.
• Since the kinetic energy of the electron in the x-ray tube is
20,000 eV or larger, the work function Φ is negligible by
comparison and equation
 mv2 

eV0  
 hf  
 2 

 max
becomes
eV  hf 
or
hc

hc 1240eV  nm


 1.24 103 nm
eV
eV
Thus, the x-ray spectrum can be explained by Plank’s
quantum hypothesis and λm can be used to determine h/e.
Compton Effect
Schematic sketch of Compton apparatus. X-rays from the tube
strike the carbon bloke R and are scattered into a Bragg-type
crystal spectrometer. In this diagram, the scattering angle is 300.
The beam was defined by slits S1 and S2. Although the entire
spectrum is being scattered by R, the spectrometer scanned the
region around Kα line of molybdenum.
Derivation of Compton’s Equation
Let λ1 and λ2 be the wavelengths of the incident and scattered
x rays, respectively. The corresponding momentum are
E1 hf1
h
P1 


c
c
1
and
E2
h
P2 

c
2
using fλ = c. Since Compton used the Kα line of molybdenum
( λ = 0.0711 nm), the energy of the incident x ray (17.4 keV) is
much greater than the binding energy of the valence electrons in
the carbon scattering block (about 11 eV); therefore, the carbon
electron can be considered to be free.
Conservation of momentum gives
  
p1  p2  pe
2
 
2
2
pe  p1  p2  2 p1  p2  p12  p 22  2 p1 p2 cos 
where pe is the momentum of the electron after the
collision and θ is the scattering angle for the photon,
measured as shown in Figure. The energy of the electron
before the collision is simply its rest energy E0 = mc2.
Diagram for Derivation of Compton’s Equation
After the collision, the energy of the electron is (E02 + pe2c2)1/2.
Conservation of energy gives
p1c + E0 = p2c + (E02 + pe2c2)1/2
Transposing the term p2c and squaring we obtain
E02+c2(p1-p2)2+2cE0(p1-p2)=E02+pe2c2
or
pe2=p12+p22-2p1p2+2Eo(p1-p2)/c
pe2=p12+p22-2p1p2+2Eo(p1-p2)/c
If we eliminate pe2 from the previous equations, we obtain
E0(p1-p2)/c = p1p2(1-cosθ)
Multiplying each term by hc/p1p2E0 and using λ=h/p, we
obtain Compton’s equation:
λ2 – λ1 = hc(1-cosθ)/E0 = hc(1-cos θ)/mc2
or
λ2 – λ1 = h(1-cos θ)/mc
Compton Wavelength
hc
  2  1 
(1  cos  )
me c
The increase in wavelengths is independent of the
wavelength λ1 of the incident photon. The quantity
h
has dimensions of length and is called the
me c
Compton Wavelength. Its value is
h
hc
1240eV  nm
12
C 



2
.
43

10
m  2.43 pm
2
5
me c me c
5.1110 eV
Compton Wavelength
Because λ2–λ1 is small it is difficult to observe
unless λ1 is so small that the fractional change
(λ2–λ1)/λ1 is appreciable. Compton used X-rays of
wavelength 71.1pm. The energy of a photon of this
wavelength is
1240eV  nm
E

 17.1keV

0.0711nm
hc
Since this is mush greater than the binding energy of
the valence electrons in most atoms, these electrons
can be considered to be essentially free. Compton’s
measurements confirmed the correctness of the
photon concept.
Reflection from Calcite
If the spacing between certain planes in crystal
of calcite is 0.314nm, find the grazing angles at
which first and third order interference will occur
for x-rays of wavelength 0.070nm.
Compton Scattering at 450
X-rays wavelength λ0=0.200 000 nm are
scattered from a block of material. The scattered
x-rays are observed at an angle of 450 to the
incident beam. Calculate the wavelength of the xrays scattered at this angle.
The Minimum X-ray Wavelength
Calculate the minimum x-rays wavelength
produced when electrons are accelerated through a
potential difference of 100 000V, a not-uncommon
voltage for an x-ray tube.
Photoelectric Effect in Lithium
Light of wavelength of 400nm is incident upon lithium
(Φ = 2.9eV). Calculate (a) the photon energy and (b)
the stopping potential V0.
What frequency of light is needed to produce electrons
of kinetic energy 3eV from illumination from Li?