Atomic and Nuclear Physics Dr David Roelant Atomic and Molecular Weight Problem 1 Using the data in the table below compute the atomic weigh of naturally occurring oxygen Isotope Abundance ID: 528847
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Slide1
Absorption of Nuclear Radiation & Radiation Effects on Matter:Atomic and Nuclear Physics
Dr. David
RoelantSlide2
Atomic and Molecular WeightSlide3
Problem 1Using the data in the table below, compute the atomic weigh of naturally occurring oxygen.
Isotope
Abundance
(%)
Atomic Weight
16
O
99.759
15.99492
17
O
0.037
16.99913
18
O
0.204
17.99916Slide4
Solution to Problem 1
Using the data in the table below, compute the atomic weigh of naturally occurring oxygen.
Isotope
Abundance
(%)
Atomic Weight
16
O
99.759
15.99492
17O0.03716.9991318O0.20417.99916Slide5
Atomic and Nuclear RadiiSlide6
Increase in Mass Relative to Observer of a Moving MassSlide7
Particle WavelengthsSlide8
Energy levels of hydrogen atom
Energy,
eV
E =0
eV
E =10.19
eV
E = 12.07
eV
E =13.58
eVSlide9
Decay Scheme of 60Co
Introduction to Nuclear Engineering by J.R.
LamarshSlide10
Radioactivity CalculationsSlide11
Decay Chain RadioactivitySlide12
Fundamental Laws Governing Nuclear Reactions
Conservation of nucleons
Conservation of charge
Conservation of momentumConservation of energySlide13
Balancing Nuclear EquationsSlide14
Balancing Nuclear EquationsSlide15
Problem 2One of the reactions that occurs when
3
H (tritium) is bombarded by deuterons (
2H nuclei) is 3H (
d,n
)
4
He
where,
d refers to the bombarding deuteron.
Compute the Q-value of this reaction. Slide16
Solution to Problem 2The Q-value is obtained from the following neutral atomic masses (in
amu
):
The Q-value in
amu
is Q = 5.030151 – 5.011269 = 0.018882
amu
. Since 1
amu
= 931.481 MeV, Q = 0.018882 x 931.481 = 17.588 MeV
. The Q-value is positive and so this reaction is exothermic. This means, for instance, that when stationary 3H atoms are bombarded by 1 – MeV deuterons, the sum of the kinetic energies of the emergent α–particle (4He) and neutron is 17.588 + 1 = 18.588 MeVSlide17
Binding Energy per Nucleon as a Function of Atomic Mass Number
Introduction to Nuclear Engineering by J.R.
LamarshSlide18
Atomic DensitySlide19
Nuclear radiation absorption p. 165Counting efficiency (self, abs s-d, det, geom.)Ionization, excitation, bremsstrahlung, positron annihilation, Cerenkov (.6MeV) (fig. 7.9)Slide20
Problem 3
The density of sodium is 0.97 g/cm
3
. Calculate its atomic density.Slide21
Solution to Problem 3
It is usual to express atomic densities as a factor x 10
24
The atomic weight of Na is 22.990.
The density of sodium is 0.97 g/cm
3
. Calculate its atomic density.Slide22
Interaction of Radiation with Matter
(in entire target area)Slide23
Problem 4
A beam of 1-MeV neutrons of intensity 5 x 10
8
neutrons/cm•s
strikes a thin
12
C target. The area of the target is 0.5 cm
2
and it is 0.05 cm thick. The beam has a cross-sectional area of 0.1 cm
2
. At 1 MeV, the total cross section of 12C is 2.6 b.At what rate do interactions take place with the target?What is the probability that a neutron in the beam will have a collision in the target?Slide24
Solution to Problem 4
In 1 sec, a total of
IA
= 5 x 108 • 0.1 = 5 x 10
7
neutrons strike the target. Of these, 5.2 x 10
5
interact. The probability that a neutron interacts in the target is therefore
: 5.2 x
10
5 / 5 x 107 = 1.04 x 10-2. Thus, only about 1 neutron in 100 has a collision while traversing the target.
It should be noted that the 10
-24
in the cross section cancels the 10
24
in atom density. This is the reason for writing atom densities in the form of a number x 10
24
Slide25
Collision DensitySlide26
Neutron AttenuationSlide27
Compound Nucleus Formation
56
Fe + n (elastic scattering)
56
Fe + n’ (inelastic scattering)
57
Fe +
γ (
radiative capture)
55
Fe + 2n (n, 2n reaction)Slide28
Elastic ScatteringSlide29
Energy Loss in Scattering Collisions
Introduction to Nuclear Engineering by J.R.
Lamarsh
Fig. 3.6 Elastic Scattering of a Neutron by a NucleusSlide30
The Energy Released in Fission
Introduction to Nuclear Engineering by J.R.
LamarshSlide31
ɣ-Ray Interactions with Matter
In nuclear engineering problems only three processes must be taken into account to understand how
Ɣ
-rays interact with matter.The Photoelectric Effect
Pair Production
Compton EffectSlide32
Photoelectric EffectIncident ɣ
-ray interacts with an entire atom, the
ɣ
-ray disappears, and one of the atomic electrons is ejected from the atom.The hole in the electronic structure is latter filled by a transition of one of the outer electrons into the vacant position.The electronic transition is accompanied by the emission of x-rays characteristic of the atom or by the ejection of an Auger electron.Slide33
Dependence of Z on Photoelectric Cross Section
where, n is a function of E shown in Fig. 3.14.
Because of the strong dependence of
σ
pe
on Z, the photoelectric effect is of greatest importance for the heavier atoms such as lead, especially at lower energies.
Introduction to Nuclear Engineering by J.R.
LamarshSlide34
Pair ProductionPhoton disappears and an
electron pair
– a positron and a negatron – is created.
This effect doesn’t occur unless the photon has at least 1.02 MeV of energy.Cross section for pair production (σ
pp
) increases steadily with increasing energy.
Pair production can take place only if vicinity of a Coulomb field.Slide35
Compton EffectElastic scattering of a photon by an electron, in which both energy and momentum are conserved.A Compton cross section per electron (
e
σ
C) decreases monotonically with increasing energy from a maximum value 0.665 b (essentially 2/3 of a barn) at E = 0, which is known as the Thompson cross section,
σ
T
. Slide36
Attenuation CoefficientsMacroscopic
ɣ
-ray cross sections are called attenuation coefficients.
Mass attenuation coefficient (μ/
ρ
)
Introduction to Nuclear Engineering by J.R.
LamarshSlide37
Introduction to Nuclear Engineering by J.R.
LamarshSlide38
Introduction to Nuclear Engineering by J.R.
LamarshSlide39
Introduction to Nuclear Engineering by J.R.
LamarshSlide40
Problem 5
It is proposed to store liquid radioactive waste in a steel container. If the intensity of
ɣ
-rays incident on the interior surface of the tank is estimated to be 3 x 1011
ɣ
-rays/cm
2
•sec and the average
ɣ
-ray energy is 0.8 MeV, at what rate is energy deposited at the surface of the container?Slide41
Solution to Problem 5
Steel is a mixture of mostly iron and elements such as nickel and chromium that have about the same atomic number as iron. So as far as
ɣ
-ray absorption is concerned, there, steel is essentially all iron. From table 3.8,
μ
o
/
ρ
for iron at 0.8 MeV is 0.0274 cm
2
/g. The rate of energy deposition is then:
It is proposed to store liquid radioactive waste in a steel container. If the intensity of
ɣ
-rays incident on the interior surface of the tank is estimated to be 3 x 10
11
ɣ
-rays/cm
2
•sec and the average
ɣ
-ray energy is 0.8 MeV, at what rate is energy deposited at the surface of the container?