Lesson 9
• X-rays
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Roentgen’s Early Experiment
How are X-rays Produced
Hard and Soft X-RAYS
X-ray Spectra
X-ray spectra according to Bohr’s Theory
Properties of X Rays
Uses of X Rays
Dangers of X Rays
Tutorials
• Photoelectron Effect
• Energy Levels
Roentgen’s Early Experiment
• Wilhelm K. Roentgen, who in 1901 was the first man to receive the
Nobel Prize in physics, first observed x rays in 1895. He was studying
the light produced when electricity was passed through a gas in a
tube at low pressure.
• He noted that a paper screen coated with a fluorescent material
glowed when it was in the vicinity of the tube under operation. We now
know that x rays are produced if electrons are accelerated through a
potential of the order of 104 or more volts and then allowed to strike a
metal target.
• Classical electromagnetic theory indicates that the deceleration of an
electric charge causes it to radiate energy. In this case the form of
electromagnetic radiation is called x rays.
• Early observations showed that this newly detected radiation had greater
penetration power than any other electromagnetic radiation known at
that time. It was also observed that these x rays affected a
photographic film and would ionize atoms.
• These effects are utilized in detecting devices for x rays. Since x rays are
electromagnetic waves, they are not deflected by either an electric or a
magnetic field.
Roentgen’s Early Experiment
How are X-rays Produced?
Filament X-Ray Tube
Invented by Coolidge in 1913. The most widely-used laboratory X-Ray source. Major
components are a water-cooled target (anode) and a tungsten filament (cathode) that
emits electrons. A high potential (up to 60 kV) is maintained between the filament and the
anode, accelerating the electrons into the anode and generating X-Rays. Cooling water is
circulated through the anode to keep it from melting (>99% of input power generates
heat). Interior of the tube is evacuated for the electron beam; thin beryllium windows
transmit the X-Rays
Ceramic Diffraction X-Ray Tube Schematic View
How are X-rays Produced?
• Filament is heated,
releasing
electrons
via
thermionic emission (Vf ~
10V, If ~ 4A, resulting in
T>2000oC)
• X rays are produced by
high-speed
electrons
bombarding the target
• Typically < 1% of energy is
converted to x rays; the rest
is heat
Schematic diagram of x-ray tube and circuit
How are X-rays Produced?
• The electron beam produced and controlled by the current
that is passed through the filament.
• Stable high voltage and filament current power supplies are
needed (old-style transformers →high frequency supplies).
• Power rating: applied potential ×electron beam current
(example: 50 kV and 40 mA →2 kW).
• Maximum power determined by the rate of heat removal
(without water, a tube can be destroyed in seconds → flow
interlocks).
• The anode is electrically grounded, while the filament is
kept at negative kV’s (the water-cooled anode won’t short
out, and the filament is protected by glass insulation).
How are X-rays Produced?
Hard and Soft X-RAYS
• The wavelength of the x rays is controlled by the applied
voltage between the cathode and anode. For the higher
potential differences (short wavelengths) the term hard
x rays is used and for the lower potential differences
(long wavelengths) the term soft x rays is used to
describe the quality of the radiation.
• Since the tube is highly evacuated, the electron beam
current (usually in the 10 milliampere range) is
determined by the filament current. When these
electrons strike the metal anode target some of them
generate x rays as they are abruptly brought to rest.
• This deceleration radiation is called
Bremsstrahlung
(braking radiation) which appears as the continuous xray background shown in Figure
X-ray Spectra
The Bremsstrahlung Spectrum
X-ray Spectra
• Much of the electron energy goes into heating up the anode. Some
of the electrons in the beam interact with the innermost, most
tightly, bound electrons in the target and "knock" them into
excited states. When these excited atoms return to their ground
state, photons are emitted. Since the target material is a
metal with many electrons, the innermost electron energy
levels are of the order of thousands of electron volts (keV).
• The photons emitted due to this excitation are characteristic of
the target material and are called the characteristic spectra .
• The maximum energy x rays correspond to the conversion of the
maximum electron beam energy into a photon, electron beam
energy = photon energy maximum
• eV = hfmax = hc/ λmin
(30.1)
• where V is the accelerating voltage for the tube, fmax is
maximum x-ray frequency, h is Planck's constant, and e is the
change of the electron.
X-ray Spectra
• The x-ray photons interact with matter through the
photoelectric effect and Compton scattering. The
photoelectric effect increases rapidly with Z, the atomic
number of the target material. The photoelectric
interaction probability is proportional to Z and is the
greatest for low-energy photons.
• The Compton effect is relatively independent of energy
and the atomic number of the absorber. These two
processes determine the absorption coefficient of the
material for x rays.
• The intensity x rays passing through a material of
thickness x can be expressed as an exponential function,
I is the x-ray intensity at a distance x in the material, µ is absorption coefficient of
the material, Δx is the thickness of the material, and IO is the incident intensity of the x rays
in joules per unit area per second.
Properties of X Rays
i.
ii.
iii.
iv.
Like radio waves and other electromagnetic radiation, X rays have a wide range of
wavelengths (Obey the relationship . Strong, deeply penetrating, and highly
destructive rays with short wavelengths are called hard X rays. Those with longer
wavelengths and less penetrating power—the type used in medical and dental
diagnosis—are known as soft X rays.
X rays can penetrate some substances more easily than others. For example, they
penetrate flesh more easily than bone, and bone more easily than lead. Thus they
make it possible to see bones within flesh and a bullet embedded in bone. The ability
of X rays to penetrate depends not only on their wavelength, but also on the density
and thickness of the substance.
X rays affect photographic film in the same way as light rays do. An X-ray photograph
is made by passing a beam of X rays through the subject onto photographic film. In a
more recent technique, called xeroradiography, an electrostatic ally charged metal
plate is substituted for photographic film. When X rays pass through an object and
strike the plate, they discharge it in proportion to the density of the object. To bring
out the electrostatic image thus formed, the plate is sprayed with a powder that
adheres to the charged area. The powder gathers more thickly in heavily charged
areas than in lightly charged spots, producing a detailed picture.
X rays cause certain substances to glow, or fluoresce.
Properties of X Rays
• Bragg's Law
• According to W. L. Bragg, X-ray diffraction can be viewed as a process that is
similar to reflection from planes of atoms in the crystal. In Bragg's construct,
the planes in the crystal are exposed to a radiation source at a glancing angle
θ and X rays are scattered with an angle of reflection also equal to θ. The
incident and diffracted rays are in the same plane as the normal to the crystal
planes.
• Constructive interference occurs only when the path length difference
between rays scattered from parallel crystal planes would be an integer
number of wavelengths of the radiation. When the crystal planes are
separated by a distance d, the path length difference would be 2dsin θ.
Properties of X Rays
• Thus, for constructive interference to occur the following relation must
hold true.
• n λ = 2 d sin θ
• This relation is known as Bragg's Law. Thus for a given d spacing and
wavelength, the first order peak (n = 1) will occur at a particular θ value.
Similarly, the θ values for the second (n = 2) and higher order (n > 2) peaks
can be predicted.
1 2
mv eV
2
2eV
and
m
From
And
v
p
then
2d sin
n
mv
h
2eV
m
m
P
mv,
h
so
nh
2eVm
where p
h
h
mv
where
h
2eVm
is
the
de Broglie
wavelength
• The above derivation assumes that phase differences between wavelets
scattered at different points depend only on path length differences.
• It is assumed that there is no intrinsic phase change between the incident
and scattered beams or that this phase change is constant for all
scattering events. However this is not always the case (see anomalous
scattering elsewhere).
Properties of X Rays
Uses of X Rays
• In Medicine
• X rays help dentists detect diseases of the teeth. Doctors use X rays to
locate bullets and other foreign objects within the body; to guide them in
setting broken bones; and to detect cancer, ulcers, kidney stones, and
other abnormalities.
• Various types of X-ray scanners have been developed that allow highly
detailed views of a particular section of the body. One type, known as a CT
(computerized tomography) scanner, sends narrow beams of X rays at
various angles through a patient's body. The information obtained from
the X rays is processed by a computer to produce an image of a crosssection of the body. The image shows much more detail than an ordinary
X-ray picture. A section of the body can be studied in three dimensions by
producing a series of adjacent cross-sectional images.
• X rays can halt the growth of cells and even destroy them altogether. They
are therefore used to destroy benign and malignant tumors. X rays have
also been used in the treatment of leukemia and bursitis.
Uses of X Rays
• In Industry
• X rays are used to inspect canned goods and other
packaged products. A conveyor carries the goods past a
beam of X rays. If a container is improperly filled, or if it
contains a foreign substance, the X rays set off an alarm or
set into action a device that removes the container from
the conveyor. X rays are similarly used to separate beryl
from granite and to inspect airplane and automobile parts,
rubber goods, plastics, metal castings, and a variety of
other products.
• When used as a target in an X-ray tube, every element gives
off X rays of specific wavelengths. These characteristic rays
are used to analyze metal alloys, paint pigments, and other
substances.
Uses of X Rays
• In Science (crystallography)
• Scientists have learned much about the structure of matter
by means of X rays. Among other things, they have learned
how atoms are arranged in crystals. The average
wavelength of X- rays is about equal to the distance
between the atoms in crystals. Crystals therefore act as
diffraction gratings for X rays. That is, they scatter X- rays in
a pattern that shows the positions of their atoms.
• When the patterns of specific crystalline substances are
known, technicians can use X- rays to analyze substances of
which they are a part. Petroleum products, metal alloys,
and other substances are thus analyzed.
Uses of X Rays
• Other Uses
• Airport security and other personnel use X- rays in
examining luggage and packages to check for weapons or
smuggled articles.
• X-rays show whether pearls are natural or cultured, and
whether gemstones are natural or synthetic.
• X-rays have also been used to learn whether paintings
attributed to noted painters are authentic. Sometimes they
have revealed changes made in the original work, or an
earlier painting under the one that appears on the surface.
• In geology labs, the most common targets used are copper
and cobalt.
Dangers of X- Rays
• Because X rays can kill living cells, they must be used
with extreme care. When improperly used they can
cause severe burns, cancer, leukemia, and cataracts.
They can speed aging, reduce immunity to disease, and
bring about disastrous changes in the reproductive
cells.
• Lead screens, sheets of lead-impregnated rubber, and
leaded glass are used to shield patients and technicians
from undesired radiation.
• The effect of X radiation is cumulative. That is, a
number of minor doses over a number of years is
equivalent to a large dose at one time.
Tutorials
1. Electrons are accelerated from rest through a p.d of 10Kv in an x ray tube. Calculate:
(i)
The resultant energy of the electrons in eV.
(104ev)
(ii)
The wavelength of the associated electron waves.
(1.23x10-11m)
(iii)
The maximum energy and the minimum wavelength of the x ray radiation
generated (assume, me,e, h=6.62x10-34Js, c=3x108m/s): (1.6x10-15J, 1.24x10-10m)
h
where p
P
h
Hint:use the de Broglie equation
and
min
2eVme
2eVm
hc
1.24 x10
eV
10
m
2. The energy of an x ray photon is hf. X rays are emitted from a target bombarded by
electrons which have been accelerated from rest through 105V. Calculate the minimum
possible wavelength of the X rays assuming that the corresponding energy is equal to the
whole of the kinetic energy of one electron. ( assume h, e, c )
ans, 1.243x10-11m
(use: hf = eV),
Tutorials
3. a) State the de Broglie relationship in words.
b) Show that the speed of electrons which have been accelerated from rest through a p.d
of 2Kv is 2.6x107m/s and calculate the wavelength with a beam of these electrons.
(2.7x10-11m) .
c) Suggest why electron diffraction is a useful tool for studying the arrangement of atoms
in a crystals where the separation of atoms is typically 0.3nm (λ is less than 0.03nm so
diffraction is observed)
4. Electrons are accelerated through a p.d of 50V. Calculate,
(a) The speed acquired.(4.123x106m/s)
(b) Their momentum (P = mv)(37.52x10-25kgm/s)
(c) The de Broglie wavelength associated with them. (h/p = 0.17nm)
Tutorials
5. What is the k.e of an electron with a de Broglie wavelength of 0.1nm.through what p.d
should it be accelerated to achieve this value? Assume e, me, h.
h
p
h
,v
me v
(use sin ce v
also
2
eV
h
m2
p
mv
but
k .e
1 2
mv
2
h2
m2e
k .e,
1
h2
me 2
2
m e
2
)
2
so
V
h2
2eme
2
, V
0.151Kv
6. An x-ray operates at 30Kv and the current through it is 2.0mA.Calulate:
(i) The electrical power output
(ii) The number of electrons striking the target per second.
(iii)The speed of the electrons when they hit the target
(iv) The lower wavelength limit of the x-rays emitted.
Tutorials
Solution.
(i )
p
VI
30 x10 3 x 2.0 x10
3
60w
(ii)
of
I ne where n
is
electrons
stricking
I
e
n
(i)
(iii)
number
t arg et per sec ond
2 x10 3
1.3 x1016
19
1.6 x10
1
mv 2 eV
2
v
min
the
the
hc
eV
2eV
m
0.41x10
2 x1.6 x10 16
9 x10 31
10
1x108
m
s
m
7. Monochromatic X-rays of wavelength 1.2x10-10m are incident on a crystal. The ist order
diffraction maximum is observed at when the angle between the incident beam and the
atomic plane is 120
(i)
What is the separation of the atomic planes responsible for the diffraction?
(d = 2.89x10-10m)
(ii)
What is the highest order Bragg diffraction observable? ( max n = 4)
Tutorials
8. An x-ray machine can accelerate electrons of energies 4.8x10-15J. The shortest
wavelength of the x- rays produced by the machine is found to be 4.1x10-11m.
Use this information to estimate the value of the plank constant. ( use,
E
hf
hc
,
h
6.56x10
34
Js
min
9. Molybdemumka x-rays have wavelength 7x10-11m. Find:
(i)
The minimum x ray potential difference that can produce these x rays. (17.7Kv)
(ii)
Their photon energy in e V (17.7 Kv x e = 17.7KeV)
10. The spacing between
Principal planes of Nacl crystal is 2.82Å . It is found that the first
order Bragg diffraction occurs at an angle of 100. What is the wavelength of the x rays?
2d sin
n
2d sin / n
9.8 x10
11
m
Tutorials
11. Bragg’s spectrometer is set for the first order reflection to be received by the dectector at
a glancing angle of 9.80. calculate the angle through which the detector is rotated to
receive 2nd order reflection from the same face of the crystal.
2d sin
1
and
2d sin
2
where
first
glamcing
angles for
the
2d sin 9.3
and
2
2d sin
and
1
and
2
sec ond
are
order
2
This gives θ2= 18.80 and therefore
2
1
18.8 9.3
9 .5 0
Photoelectron Effect
The photoelectric effect relates to the following phenomena: if a metal surface is
illuminated by visible or ultraviolet light radiation, electrons are released provided that
the frequency of the radiation exceeds a critical threshold.
In 1902 Lernard found that the velocity or kinetic energy of the electron emitted from
an illuminated metal was independent of the intensity of the particular incident
monochromatic light. It appeared to vary only with the wavelength or the frequency of
the incident light.
Quantum Theory of Radiation, Planck’s Constant
In 1902 Planck had shown that the experimental observations in black body heat
radiation could be explained on the basis that the energy from the body was emitted in
separate packets of energy. Each packet was called quantum of energy and the amount
of energy E carried is given by
E =hf
Where h = 6.63 × 10-34 Js called the Planck’s constant and f is the frequency of
radiation
Photoelectron Effect
Work Function
The least or minimum amount of work or energy necessary to take free electron out of a metal
against the attractive forces of surrounding positive ions is called the work function of the metal,
symbol w . The work functions of cesium, sodium and beryllium are respectively about 1.9 eV,
2.0 eV and 3.9 eV . Electron volt (eV) is a unit of energy equal to 1 electron charge e × 1 volt,
which is
1.6 × 10-19 C × 1V = 1.6 × 10-19 J
Hence for example the work of sodium, w =2eV = 2×1.6×10-19J = 3.2×10-19J.
Einstein’s Particle (Photon) Theory
In 1905 Einstein suggested that the experimental results in photoelectricity could be explained by
applying a quantum theory of light. He assumed that light of frequency f contains packets or
quanta of energy hf. On this basis, light consists of particles called photons. The number of
photons per unit area of cross section of the beam of light per second is proportional to its
intensity. But the energy of a photon is proportional to its frequency and is independent of the
light intensity.
Photoelectron Effect
Considering the case of sodium whose work function is 2eV or 3.2×10 -19J. According to
the particle theory, if the quantum energy in the incident light is 3.2×10 -19J, then the
electrons are just liberated from the metal. The particular frequency f is called the
c / f . Hence the
threshold frequency of the metal. The threshold wavelength
threshold frequency f and threshold for sodium is E=hf w ,
f
c
f
w
h
3.2 10
6.6 10
3 108
4.8 1014
19
34
4.8 1014 Hz
6.2 10 7 m
So electrons are not liberated from sodium if the incident light has a frequency less than 4.8 ×
1014 Hz or a wavelength longer than 6.2 × 10-7m.
Photoelectron Effect
Einstein’s Photoelectric Equation
If the quantum of energy hf in the light incident on a metal is say 4.2 × 10-19 J, and the work
function wo of the metal is 3.2 × 10-19J, the maximum energy of the liberated electrons is (4.2 ×
10-19 J - 3.2 × 10-19J) = 1.0 × 10-19J, because wo is the least energy to liberate electrons from the
metal. The maximum kinetic energy of the liberated electrons is given by E max or
E max
hf
hf
w
E max
w
The above equations are called the Einstein’s photoelectric equations.
1
me vm2 .
2
Photoelectron Effect
Example
Calculate the maximum energy and kinetic energy of emitted electron when sodium is
illuminated by radiation of wavelength 150 nm. (wo for sodium is 2.0eV)
Photon energy is given by E = hf
h
So maximum kinetic energy = hf
w
The threshold frequency is given by f
c
6.63 10 34 3 108
150 10 9
13.2 10
w
h
19
13.2 10
2 1.6 10
2 1.6 10 19 J
6.63 10-34 Js
19
19
J=10-18J
4.8 1014 Hz
J
Photoelectron Effect
Stopping Potential
Stopping potential is defined as the negative potential difference (p.d.) which can stop liberated
electrons with maximum energy. Stopping potential is denoted by Vs, hence
E max =eVs
hf
w
hc
w
Example
Caesium has a work function of 1.9eV. Find i) its threshold wavelength; ii) maximum energy of
liberated electrons when th metal is illuminated by light of wavelength 4.5 × 10 -7m; iii) stopping
potential. (take 1eV = 1.6×10-19J, c = 3×108m/s, h = 6.63×10-34Js).
i) Threshold wavelength
c
f
3 108 6.63 10
1.9 1.6 10 19
c
w /h
ch
w
34
6.5 10 7 m
recall that E=hf
w
Photoelectron Effect
ii) Maximum energy of liberated electrons =hf-wo where f
Max. Energy
hc
w
c/
6.63 10 34 3 108
1.9 10
7
4.5 10
hence
19
1.4 10
19
J
iii) the stopping potential Vs is given by eVs = 1.4×10-19J where e = 1.6×10-19C recall Emax=eVs
1.4 10-19
therefore V=
1.6 10-19
0.9V (approx)
Homework
1 Light of frequency 5.0×1014Hz liberates electrons with energy 2.31×10-19J from a certain
metallic surface. What is the wavelength of ultraviolet light which liberates electrons of
energy 8.93×10-19J from the same surface? Take c =3.0×108m/s, h=6.63×10-34Js.
2 When light of frequency 5.4×1014Hz is shone on to a metal surface the maximum energy
of the electrons emitted is 1.2×10-19J. If the same surface is illuminated with light of
frequency 6.6×1014Hz the maximum energy of the electrons emitted is 2.0×10 -19J. Use
this data to calculate a value for the Planck’s constant.