Bai giang ve tuong tac

1
Hànội , 20-2.2006

Interaction of Radiation with matter
Nguyễn Triệu Tú
8261730
0904505414
2
M.Planck
Einstein
N.Bohr
M.Curie
3
cell
Atom
Atomic nucleus
Radius of Earth
Radius of observable universe
15000.000.000.light -years
4
Wide Range of Radiation
Energy and Intensity
Accelerator, Cosmic rays
Radioisotopes
5
interaction of particles and radiation with matter
The measurement of nuclear radiation is based on its interaction with the detector.
1* The function of nuclear radiation detectors
2* The absorption phenomena in
the measurement of the radiation
3* With respect to radiation protection.
 We shall deal with the most important mechanisms of the interaction between nuclear radiation and matter in their basic features.
In order to understand:
6
To organize the discussions that
follow, it is convenient to arrange
the four major categories of
radiations into the following matrix:
Heavy charged particles
(characteristic distance 10-5m)
Fast electrons
(characteristic distance ?10-3m)
Neutrons
(characteristic length ?10-1m)
X-rays and  rays
(characteristic length 10-1m)
7
Mảnh giấy
Lá nhôm
Tấm chì
8
Interaction of particles and radiation with matter
Ionization Losses Due to Collisions of Charged Particles - Stopping power

Bohr`s Formula for Specific Ionization. Relativistic Effects and the Density Effect

Dependence of ionization Losses on the Medium
ionization Losses on the Medium
. Radiation Losses for Electrons.
. Cherenkov Radiation
. n and ? -radiation Interaction with matter


9
Interaction of Gamma Radiation with Matter
1. Photoelectriceffect
2 . Compton scattering
3. Electron-positron Pair production
10
h
Te = E? - Ii
Photoelectric effect cannot take place for a free electron (not associated
with an atom).
Photoelectric effect
A.E :Auger electron
X-ray
K
L
M
11
12
Photoelectric effect is therefore possible only for bound electrons.
13
the photoelectric effect cross section
1/ With decreasing E? (increasing ratio of the electron binding to the
photon energy IK/E? ), the cross section increases first as 1/E?, and later
(as E? approaches Ik) more rapidly as 1/ E?7/2 .
2/ The probability of photoeffect depends very strongly on the charge Z of the atom in which the effect is observed: ?phot ? Z5.
?phot ? Z5/E? for E? >> IK,
?phot ? Z5/E?7/2 for E? > IK.
* Photoeffect is especially signifficant for heavy materials where the probability is considerable even for high energies of ?-quanta.
* In light materials, this effect becomes significant only for relatively
low energies of ?-quanta.
14


h
h ’
In Compton scattering the incoming gamma-ray photon is deflect through an angle ? with respect to its original direction. The photon transfers a portion of its energy to the electron (assumed to be initially at rest), which is then known as a recoil electron.

(Free electron)

incoming ? -ray
scattered ? -ray

( recoil electron).
Compton scattering
15


h
h ’
Because all angles of scattering are possible, the energy transferred to the electron can vary from zero to a large fraction of the gamma-ray energy.

Free electron

incoming ? -ray
scattered ? -ray

recoil electron.

Compton scattering
16


17
Compton scattering
.A photon interacts with an electron, giving a partial energy, and scatters for different direction.
.Energies of the scattered photon and the secondary electron are calculated by:
Scattered photon:
h?`= h? / { 1 + ? ( 1 -cos?) }
? = h? / m0c2
Secondary electron :
Ee = h ? - h ?`
18
?= h/mec = 2.42 x 10-10cm: The Compton wavelength for electron
19
1.The wavelength ? of the displaced line increases with the scattering angle ? in such a way that:
2. However, for scattering at a given angle ?, the quantity ?? is independent of ?.
?? is determined only by ? and is independent of ?
20
3. The energy of a quantum scattered at an angle ?:
4. The kinetic energy of the recoil electron:
21
Electron-positron Pair production
- Discovered by Dirac in 1928
h


- The process of pair production cannot occur in vacuum and requires
a nucleus or an electron in the proximity.
22
It follows from formula (1) that:
However, this inequality cannot be true,
since in accordance with formula
(2) these vectors form a triangle.


23
? The threshold energy :
(3)
in the Coulomb field of a nucleus
E0 = 4mec2 = 2.04 MeV (4)
in the Coulomb field of an electron
24
Pair production
A Photon produces an electron (e-) and a positron (e+) near the nucleus, and total kinetic energy of both electrons is :

Ee- + Ee+ = h ν -2 m0c2
25
Pair production
Positron combines with an electron nearby, after losing kinetic energy, then the electron and positron pair annihilates and emits two photons ( annihilation photon; m0c2 =511keV). This process is called positron annihilation.
26
17
Độ quan trọng tương đối của ba hiệu ứng phụ thuộc vào năng lượng và điện tích Z của chất hấp thụ
Z
27
1/. Hiệu ứng quang điện trong chì. 2/. Hiệu ứng Compton trong chì.3/.Hiệu ứng tạo cặp trong chì.4/. Hệ số suy giảm toàn phần đối với chì. 5/. Hệ số suy giảm toàn phần đối với thiếc. 6/Hệ số suy giảm toàn phần đối với đồng. 7/.đối với nhôm
28
The total cross section
17
29
Energy distribution of secondary electrons
30
Energy distribution of secondary electrons
A) E = h ν
FEA Full energy absorption peak
B) E = hν - hν’
CS

Energy distribution of
secondary electrons
Energy of secondary electrons E (keV )
FPE
SPE
DPE
CS
D
Material
A
B
γ-rays
C
h ν
31
Energy distribution of secondary electrons
Energy distribution of
secondary electrons
Energy of secondary electrons E ( keV )
FPE
SPE
DPE
CS
D
Material
A
B
γ-rays
C
h ν
32
The crosS section for Ge and Si
/ ? )
?
T
33
Stopping power
1/ The effect for one electron
34
Độ hao iôn hoá do va chạm của các hạt tích điện
35
2/ The Effect of all electrons

Vne = 2ned dx
V = 2d dx
36
Calculation of ?min & ?max
It is well known that the maximum energy that a heavy particle moving with a velocity v<< c can impart to a fixed electron is :
Condition for min:

37


However, it can be shown that the integration limits are not 0 and  but have some finite values min and max. Consequently



The total specific loss is obtained by integrating
over all the possible values of the impact parameter 
(from 0 to ).
Difficulties are encountered for  = 0, since  is
in the denominator of the expression (5),
and for = since the integral is divergent.
38
The condition for max is obtained from classical considerations by taking into account the fact that electrons are bound in the atom. For large values of the impact parameter , the transmitted energy T becomes comparable with the binding energy of electrons in the atom. Electrons can no longer be treated as free, and for quite large  the transmitted energy may not be sufficient to excite the atom.Consequently, max must be associated with the average ionization potential of the atom. Finally, to calculate ln(max/min), we must take into account the relativistic effects.
Tmax = 2mev2/(1 - 2)
Exact calculations lead to the following formula for specific ionization losses :

The condition for ?max ?
39
where I is the average ionization potential of the absorber atoms,? and U are terms taking into account the density effect and the fact that K and L- electrons are bound
The main result following from formula ?
Such a regularity can be illustrated by considering the Bragg curve for specific ionization of  particles.
(9)
40
Formula (9) shows that with increasing particle energy, the specific ionization losses first decrease very rapidly (ininverse proportion to energy), but the decrease becomes slower as the particle velocity approaches the velocity of light. At a certain value of energy, the specific energy loss on ionization attains its lowest value. This means that the denominator in formula (9) contains a nearly constant quantity v2 ? c2. However, an examination of terms in the brackets shows that starting from a certain quite high particle energy, the magnitude of dT/dx again starts increasing slowly (logarithmically) and reaches a certain plateau.The suppression of the logarithmic growth of dT/dx is associated with the polarization of atoms near the particle trajectory, which leads to a decrease in the electromagnetic field acting on remote electrons. This effect is proportional to the density of the substance (more exactly, to the density of electrons) and is therefore called the density effect.
41
The density effect sets in earlier in condensed media than in rarefied ones. The last term in formula (9), denoted by U, is the correction factor for relatively low energies of the ionizing particle. The formula for calculating the ionization losses of electrons :
where Te is the relativistic kinetic energy of an electron, ne is the electron density in the medium,? is the correction for the density effect.
28
42
Hạt tích điện mất năng lượng trong vật chất
thông qua các quá trình iôn hoá và kích thích , tạo thành
nhiều êlectrôn và iôn (trong chất khí ) hoặc lỗ trống
( trong chất rắn). Năng lượng trung bình tạo cặp êlectrôn
iôn trong chất khí bằng 30?5 eV, trong bán dẫn ~3eV
Êlectron ? có động năng đủ để iôn hoá các nguyên tử khác. Iôn âm sẽ được tạo thành nếu êlectrôn bị .một nguyên tử trung hoà chiếm bắt
43
đường cong Bragg
44
Stopping power for α-rays (calculated)
Khả năng hãm :- dE/dX (Stopping power )
Energy loss per unit path length in material
xm(g/cm2) = x(cm) .?(g/cm3)
45
23
Absorption Characteristics of β-rays
46
Dependence of ionization Losses on the Medium
Let us suppose that two particles with the same charge (say, a proton and a deuteron) move in the same medium (ne = const).
*In this case, the value of dT/dx will be the same for both particles in the regions of equal velocities
Similarly, we can calculate the value of dT/dx for particles with other values of z (z  1).
In this case, we must remember that a particle with z  1 has z2 times higher value of dT/dx than a particle with z = 1 moving at the same velocity.
47
For example, values of dT/dx for an -particle and a proton are connected through the relation
Calculations for another medium can be easily carried out if we
remember that dT/dx is a linear function of the electron number
density ne in the medium.
It is well known that the electron number density ne in a medium is equal to nnucZ, where Z is the charge of the nuclei constituting the medium and nnuc is their number density.
However, nnuc const for all media, and therefore we must
introduce for recalculations the factor Z2/Z1, where Z1 and Z2 are the nuclear charges for the first and the second medium respectively.
48
The ionization losses for a particle moving in lead will be 14 times higher than the ionization losses for the same particle if it were to move in carbon
Thus, dT/dx varies strongly as we go over from one medium to another.
Hence we sometimes use the term
specific ionization losses dT/d,
which is calculated not per unit
length x (in cm), but per unit "density" ,
expressing the thickness in g/cm2.
Obviously,  = x, where  is the
density of the medium. This gives
  Z  dT/dx  Z,  (dT/dx) (1/)  const.
Thus, the quantity dT/d is practically constant for all media and is therefore moreconvenient that dT/dx for quick rough calculations.
49
For the sake of comparison, the values of dT/d for protons moving in air and in lead are given in
table 1 for different energies.
Table 1
50
Radiation Losses for Electrons
The rapid deceleration of a charged particle in the electric field of the atomic nucleus and atomic electrons results in radiation losses) (bremsstrahlung).
The loss of energy (dT/dx)rad by radiation is
proportional to the square of acceleration
Bremsstrahlung
51
Since the force F of Coulomb interaction of nuclei with particles having identical charges z is the same, we obtain:
52
The radiation losses for particles with equal charges are inversely proportional to the square of the particle mass.
These losses are especially significant for
light charged particles,for example, electrons.
53
A well-known example of radiation losses for low-energy electrons (Te << mec2) is the continuous X-ray spectrum which is created when electrons are stopped by the anticathode in the X-ray tube.
X-Ray tube
54
The energy dependence of this spectrum is given by the law
N(v)  1/v
The intensity of radiation has a
maximum in the direction
perpendicular to the direction
of motion of electrons.
55
A complete analysis of radiation losses for an electron was
carried out by Bethe and Heitler.
We shall now consider some formulas for calculating the energy losses due to radiation over a unit path length for different electron energies:
where n is the number density
of atoms, Z is the nuclear charge, re = e2/mec2
and Te<(3)
56
where mec2 << Te << 137mec2Z-1/3
? where Te >> 137mec2Z-1/3 (complete screening).
(4)
(5)
57
? It follows from Eqs. (3)- (5) that the energy losses due to radiation arproportional to the square of the charge Z of the nuclei in the medium,the number density of atoms n, and the kinetic energy Te of the electrons[neglecting the logarithmic dependence in Eq (4)]:
Comparing formula (6) with formula for ionization losses of electrons written in the simplified form
(6)
(7)
58
If Te is measured in MeV, we obtain:
we get 
(8)
(9)
59
This means that the radiation losses in water (Z=8) become comparable with ionization losses at Te ? 100MeV
Such a situation arises in lead at Te ?10MeV.
The energy at which radiation losses become comparable with ionization losses is called the critical energy.
(10)
60
The radiation length for water (and air), aluminium and lead is approximately equal to 36, 24 and 6g/cm2 respectively).
The distance xo at which the mean electron energy is reduced to 1/e of its value on account of radiation losses is called the radiation length.
61
In 1934 P.A. Cherenkov,
a post graduate student of
Acad. S. I. Vavilov investigated the luminescence of
uranyl salts under the action of ?-rays from radium and
discovered a new type of luminescence which could
not be explained by the ordinary fluorescence mechanism.
Cherenkov Radiation
Bức xạ Cherenkov
62
The polarization of the luminescence changes sharply when a magneticfield is applied. This means that the luminescence is caused by charged particles rather than by ?-quanta. In Cherenkov`s experiment, these particles could be electrons produced by the interation of ?-quanta with the medium due to the photoelectric effect or the Compton effect
63
The intensity of the radiation is independent of the charge
Z of the medium; hence it cannot be of radiative origin.
3/ The radiation is at a certain angle to the
direction of motion of charged particles.
64
*The Cherenkov radiation was explained in 1937 by Frank and Tamm on the basic of classical electrodynamics.They observed that the statement of classical electrodynamics concerning the impossibility of energy loss by radiation for a charged particle moving uniformly and rectilinearly in vacuum is no longer valid if we go over from vacuum to a medium with a refractive index n>1.
65
* The conclusion drawn by Frank and Tamm can be
illustrated with the help of the following arguments
based on the laws of conservation of energy and
momentum.
* Suppose that a charged particle uniformly moving in a straight line can lose energy and momentum through radiation. In this case, the following equality must be satisfied:
66
*It can be easily seen that this equality cannot be satisfied for vacuum, but may be valid for a medium with n>1.
* In deed, on the one hand, the
total energy E of a particle having
a mass m? 0 and moving freely
in vacuum with a momentum p
(velocity v) is)
67
* On the other hand, we have for electromagnetic radiation in vacuum:
Erad = pc
Since v68
* Thus, the laws of conservation of energy and momentum prevent a charged particle moving uniformly and rectilinearly in vacuum from giving away its energy and momentum in the form of electromagnetic radiation).
* However, this restriction is removed when the particle moves in a medium with a refractiveindex n>1. In this case, the volocity of light in the medium is c` = c/n < c (7) and the velocity v of the particle may not only become equal to the velocity of light c` in the medium, but even exceed it): v? c` = c/n (8)
69
* Obviously for v= c`, the condition (1) will
be satisfied for electromagnetic radiation emitted
strictly in the direction of motion of the particle (?=0o).
For v>c`, the condition (1) will be
satisfied for a direction ? along which v`= c`, where
v`= v cos ? is the projection of the particle velocity
v on this direction.
70
Thus, in a medium with n>1, the conservation laws allow a charged particle moving uniformly and rectilinearly with a volocity v?c`= c/n to lose fractions dE and dp of its energy and momentum, that can be carried away by electromagnetic radiation propagating in this medium at an angle ? to the direction of motion of the particle
(9)
71
* Arguments based on the conservation laws do not provide any idea about the mechanism of the energy and momentum losses. However, it is clear that this mechanism must be associated with certain processes occurring in the medium (n>1), since the losses under consideration can not occur in vacuum.
*The Cherenkov radiation is of the same nature as certain
other processes observed in various media when bodies
move in these media at a velocity exceeding that of wave
propagation.
72
1/ The electromagnetic interation between
neutrons and electrons is determined by the
interaction between their magnetic moments.
2/. The cross section of ionization losses for a neutron is
equal to 10-22 cm2, which is about a millionth part
of the corresponding value for a charged particle.
Tưương tác của nơtrôn với vật chất
Interaction of neutron with matter
73
3/. Besides the magnetic interaction between neutrons and electrons, their electric interaction must also be observed.
4/. However, this interaction is even weaker than the magnetic interaction.
74
(9)
Năng lượng
Nơtrôn chậm
Nơtrôn
trung gian
Nơtrôn nhanh
Phân loại nơtrôn theo năng lượng
75
* The main type of interaction of neutrons with matter
is their interaction with atomic nuclei. Depending on
whether a neutron falls into a nucleus or not, its interaction
with nuclei can be divided into two classes:
(1) Elastic potential scattering
by nuclear forces (n,n)pot.; in this case
the neutron does not enter the nucleus;
76
(2) ? Different types of nuclear reactions,
viz. (n, ?), (n, p), (n, ?) and fission reactions;
inelastic scattering (n, n`); elastic scattering
involving the entry of the neutron into
the nucleus, i.e. the elastic resonance
scattering (n,n)res.
77
*?The relative role of each process is defined by the cross section corresponding to it. In some materials, the role of elastic scattering is comparatively significant, and in this case a fast neutron loses its energy in a series of successive elastic collisions with the
nuclei of these materials (moderation of neutrons).
78
* The process of moderation continues until
the kinetic energy of the neutron becomes
comparable with the thermal energy of the atoms
of the moderating material (moderator). Such
neutrons are called thermal neutrons.
79
* Subsequent collisions of thermal neutrons with the moderator atoms practically do not change the energy of the neutron and result only in their displacement in the material (diffusion of thermal neutrons) which continues until the neutron is absorbed by the nucleus.
80
* ? The interaction of fast and slow neutrons with matter is of utmost importance invarious problems of neutron physics, and in particular in the design of nuclear reactors. Some of these problems, for example, the moderation of fast neutrons, should have been considered on the basis of the momentum diagram.
81
* However, a close relation between all the above-mentioned processes
involving the interaction of neutrons
with matter make it expedient to
consider these questions together.
82
* Khi đi qua vật chất, nơtrôn không trực
tiếp iôn hoá các nguyên tử . Do đó, để ghi
nơtrôn, người ta sử dụng các hạt tích điện
thứ cấp, xuất hiện khi nơtrôn tương tác với
hạt nhân của vật chất.
* Việc ghi nhận các nơtrôn chậm được tiến
hành nhờ các hạt tích điện, xuất hiện trong
kết quả phản ứng hạt nhân, gây ra bởi những
nơtrôn này. Để ghi nơtrôn chậm, người ta
thường hay sử dụng phản ứng :

B10 + n ? He4 + Li7

Phản ứng trên toả ra năng lượng 2,78 MeV ,
đủ lớn để iôn hoấ chất khí BF3 của đêtectơ .
83
Tiết diện ? của phản ứng tỷ lệ nghịch với tốc độ ? của nơtrôn : ? ~ 1/ ?
Đối với nơtrôn chậm ? = 600 barn.
(1 barn = 10-24cm2)
84
Nơtrôn chậm cũng tương tác với Li6 theo phản ứng:
Li 6 + n ? He4 + H3,
trong đó toả ra năng lượng bằng 4,63 MeV.Như vậy, trong phản ứng này sẽ xuất hiện hai hạt tích điện với năng lượng lớn .
Một phản ứng khác thường được sử dụng để ghi nhận nơtrôn là phản ứng phân chia U235 và Pu239.Các mảnh phân chia trong những trường hợp này có tổng động năng bằng 160 MeV.
Trong trường hợp các nơtrôn nhanh,để ghi nhận và xác định năng lượng của chúng,người ta sử dụng hiệu ứng tán xạ đàn hồi của nơtrôn trên hạt nhân.Giả sử nơtrôn với khối lượng Mn = 1, năng lượng E0 bị tán xạ một góc? bởi hạt nhân khối lượng M
85
Năng lượng của hạt nhân giật lùi :
E = E0.4M cos2?/(M +1)2.
Khi ? = 0, cos? = 1, hạt nhân giật lùi thu được động năng cực đại
Emax = 4ME0/(M+1)2
Đôi khi, để ghi nơtrôn nhanh người ta còn sử dụng các chất phân chi có ngưỡng tương đối lớn. Ví dụ:
U238 (ngưỡng phân chia1,1MeV), Th232(ngưỡng phân chia 1,17 MeV)
Np237( ngưỡng phân chia 0,4 MeV).
Biểu thức trên chỉ ra rằng hạt nhân giật lùi sẽ thu được năng lượng lớn nhất khi các hạt va chạm có khối lượng như nhau, nghĩa là khi nơtrôn va chạm với nguyên tử hyđrô
Trong thực tế,để ghi nhận và xác định năng lượng của các nơtrôn nhanh người ta thường hay sử dụng nhiều các chất chứa hyđrô
86
Problems
Các câu hỏi và bài tập
1/ Compare the range of electrons, alpha
particles and gamma rays in matter ,
if they have the same energy 1 MeV.
2/ Calculate the energy of a 1 MeVgamma-ray photon after
Compton scattering through 900.
3/ Give a rough estimate of the ratio of the probability per atom
for photoelectric absorption in Silicon to that in Germanium.
87
4/Indicate which of three major interaction processes
(photoelectric absorption,Compton scattering, pair
production) is dominant in the following situations:
a/ 1 MeV gamma rays in Al,
b/ 100keV gamma rays in hydrogen,
c/ 10MeV gamma rays in lead.
5/A gamma ray photon after Compton scattering through an angle of 900 has an energy of 0,5 MeV. Find its energy befor the scattering.
6/ Explain the reason why:
a/ The process of pair production cannot occur in vacuum
b/ The photoelectric effect cannot take place for a free electron
88
Answer by writing one of the heading
alphabets into the answer sheet.
Which statement is wrong?
A)? and ? -rays impart their energy to matter mainly through
ionizations and excitations .
B) The penetration range of? - rays and that of fast electrons
in a certain substance are of the same order when their energy are
the same.
C) The atomic cross section for photoelectric absorption is less in
Ge than in Si at the same photon energy.
d) The cross section or the probability for Compton scattering
increases with incident-photon energy.
89
Tín hiệu
Phương pháp ghi nhận bức xạ hạt nhân
?
?
?
?
90
Thank you very
much for attention

Cảm ơn các đ/c đã
chú ý theo dõi bài giảng