nuclear physics simplified
DESCRIPTION
mass defect, radioactive decay, activity series, half-life, radiation effectsTRANSCRIPT
Nuclear Physics
Particle physics
Properties of Nuclei
• Every atom contains an extremely dense, positively charged nucleus
• The nucleus is much smaller than the overall size of the atom, but contains most of its total mass
Properties of Nuclei
• Nucleus as a sphere– R depends on the total number of
nucleons (neutrons and protons) in the nucleus
– A is the nucleon number
R= R0A1/3
R0= 1.2x10-15m
Properties of Nuclei
•A is also the mass number in u–Proton mass and neutron mass ~ 1u–1u = 1.6605402 x 10 -
27kg
Nuclides and Isotopes
The masses of the building blocks of the nucleus:
mp= 1.007276u= 1.67263x10-27kg
mn= 1.008665u= 1.674929x10-
27kgme= 0.000548580u= 9.10939x10-
31kg
Nuclides and Isotopes
Z is the number of
protons in the nucleus
N is the number of neutrons
A is the sum
Nuclides and Isotopes
NuclideA single nuclear species having specific values of both Z and N
Same Z but
different N
Isotopes
Nuclear Binding Energy
• The energy that must be added to separate the nucleons
EB
• The magnitude of the energy by which the nucleons are bound together
EB
• (ZMH + Nmn - Z
M)c2
EB
A
MH= mass of protons and electrons mn= mass of neutron
Z M= mass of neutral atom c2= 931.5MeV/u
A
Example
• Deuterium has a mass number 2, an isotope of H. Its nucleus consists of a proton and a neutron bound together to form a particle called the deuteron. Deuterium has an atomic mass of 2.014102u. What is the binding energy of deuteron?
• 2.224MeV
Nuclear Binding Energy
• From the example, what is the binding energy per nucleon?• 1.112MeVper nucleon
Nuclear Force
• The force that binds protons and neutron together in the nucleus, despite the electrical repulsion of the protons
• An example of strong interaction
Nuclear Force
Does not depend on charge
1 It has short range (10-15m)
2
Interaction is within immediate vicinity
3 Favors binding of pairs
4
Nuclear Structure• Liquid Drop
Model• Proposed by
George Gamow• Suggests that
all nuclei have nearly the same density
Nuclear Structure• Nuclear forces
show saturation
• Nucleons on the surface of the nucleus are less tightly bound
• N is close to Z for small A
Nuclear Structure• Nuclear force
favors pairing of protons and neutrons
• Positive (more binding) if both Z and N are even
Nuclear Structure
• Summarizing these estimates into five terms:
• C1= 15.75MeV• C2= 17.80MeV• C3= 0.7100MeV• C4= 23.69MeV• C5= 39MeV
EB= C1A-C2A2/3-C3[(Z(Z-1))/A1/3]- C4[(A-2Z)2/A]±C5A-4/3
Z M= ZMH + Nmn – EB/c2
(Semi-empirical mass formula)
A
Example
• Considering the nuclide 62Ni28. a.) Calculate the five terms in the binding energy and the total estimated binding energy. b.) Find its neutral atomic mass using the semi-empirical mass formula
a. C1A= 976.5MeV -C2A2/3= -278.8MeV -C3[(Z(Z-1))/A1/3]= -
135.6MeV-C4[(A-2Z)2/A] = -13.8MeV+C5A-4/3 = 0.2MeV b. 61.925u
Nuclear Structure• The Shell
Model• Uses the
concept of filled shells and subshells and their relation to stability
Nuclear Structure• In atomic
structure, noble gases are stable
• A comparable effect occurs in nuclear structure
Nuclear Structure• An unusually
stable structure results when number of protons or number of neutrons is 2, 8, 20, 28, 50, 82, or 126
• MAGIC NUMBERS
Nuclear Stability & Radiation
• Radioactivity –The decay of unstable structures to
form other nuclides by emitting particles and electromagnetic radiation
–Alpha decay–Beta decay–Gamma decay
Nuclear Stability & Radiation
Alpha decay
Emission of alpha particle
4He nucleus
Emission occurs with nuclei that
are too large
N and Z, both decrease by 2, A decreases by
4
Nuclear Stability & Radiation
Beta minus decay occurs
When N/Z ratio is too
large
Neutral atomic mass is larger
than that of the final atom
Nuclear Stability & Radiation
Beta minus is
an electron
Emission involves
transformation of a neutron to
a:
Proton, electron, and
an antineutrino
N decreases and Z
increases by 1, A doesn’t change
Nuclear Stability & Radiation
Beta plus decay occurs
When N/Z ratio is too
small
Neutral atomic mass is at least 2 electron masses larger than that of the
final atom
Nuclear Stability & Radiation
Beta plus is a
positron
Identical to an electron but with
opp. charge
A proton is converted to a
neutron, a positron, and an
antineutrino
N increases and Z
decreases by 1, A remains
the same
Nuclear Stability & Radiation
Beta decay
Electron capture
Electron combines with a proton to form a neutron
and an antineutrino
The neutron stays in the
nucleus and the antineutrino is
emitted
N increases and Z decreases by
1, A remains the same
Nuclear Stability & Radiation
Bombardment of high-energy
particles and or by radioactive transformation
One or more photons are emitted from the nucleus
Gamma rays or gamma ray photons
(10keV – 5MeV)
Gamma ray decay
The element does
not change
Activities and Half-lives
• The decay rate• -dN(t)/dt
Activity
• Decay constant• Activity is prop. to the number
of radioactive nucleiλ
•=λN(t)-dN(t)/dt
Activities and Half-lives
• The number of remaining nucleiN(t)
• Decay constant• Activity is prop. to the number
of radioactive nucleiλ
•=N0e-λt N(t)
Activities and Half-lives
Units for activity– Ci or curie or Becquerel or Bq–1Ci= 3.70x1010Bq= 3.70x1010decays/s
Activities and Half-lives• Time required for
the no. of radioactive nuclei to decrease to half N0
Half life
•= 0.693/λT1/2
• =1/λ=T1/2/ln2=T1/2/0.693Tmean
Example
• The radioactive isotope 57Co decays by electron capture with a half-life of 272 days.
• A. find the decay constant and the lifetime
• B. if you have a radiation source containing 57Co, with activity 2.00μCi, how many radioactive nuclei does it contain?
• C. what will be the activity of your source after one year?
λ= 2.95x10-8/s Tmean= 3.39x107s
N(t)= 2.51x1012 nuclei
N(t)= 0.394N0 -dN(t)/dt= 0.788μCi
Radioactive Dating
A small proportion of the unstable 14C is present in CO2 in
the atmosphere
Plants that obtain their
carbon from this source contains the same prop.
When a plant dies, it stops taking carbon
14C β- decays to 14 N with a half-life of 5730 yrs. The
remaining 14C determines the age of the organism.
Biological Effects of Radiation
Biological Effects of Radiation
When radiation
pass through matter
They lose
energy
Break molecular
bonds
Create ions
Ionizing radiatio
n
Biological Effects of Radiation
Charged particles interact directly with the electrons in the
material
X rays and γ rays interact by the
photoelectric effect
Neutrons cause
ionization by:
Collisions with nuclei
Absorption by nuclei with subsequent
radioactive decay
Biological Effects of Radiation
Radiation dosimetry
The quantitative description of the
effect of radiation on living tissue
Absorbed dose
Energy delivered to the tissue per unit
mass
SI: J/kg, Gy(gray),
present: rad
1rad=0.01J/kg=0.01Gy
Biological Effects of RadiationRelative
Biological Effectiveness
(RBE)
The numerical factor describing
the different extents of
biological effects of different kinds
of radiation
Also called the Quality
factor
Xrays have the RBE of unit 1
Biological Effects of Radiation
Biological Effect
The product of the absorbed dose and
the RBE of the radiation
Biologically equivalent dose or equivalent
dose- SI: Sievert (Sv)
Equivalent dose (Sv)= RBE x
absorbed dose(Gy)
Biological Effects of Radiation
Biological Effect
More common unit, corresponding to the rad: Roentgen equivalent for man
(rem)
Equivalent dose (rem)= RBE x
absorbed dose(rad)
RBE units1Sv/Gy or 1rem/rad;
conversion 1rem= 0.01Sv
Biological Effects of Radiation
Radiation RBE ( Sv/Gy or rem/rad)
X rays and γ rays 1Electrons 1.0-1.5Slow neutrons 3-5Protons 10α particles 20Heavy ions 20
Example
• During a diagnostic x-ray examination, a 1.2kg portion of a broken leg received an equivalent dose of 0.40mSv. a) what is the equivalent dose in mrem? b)what is the absorbed dose in mrad and mGy? c)if the x-ray energy is 50keV, how many x-ray photons were absorbed? (1eV= 1.602x10-19J)
a. 40mremb. 40mrad; 0.40mGyc. 6.0x1010 photons
Example
• Calculate the total amount of energy (in joules) absorbed by a 750 g guinea pig, from an x-ray, at the lethal dose level of 4Sv.• 3 Joules
Example
• A typical cup of tea (250 mL) when consumed at 85oC will yield nearly 200 J of thermal energy to a human's stomach. What absorbed dose of radiation would be required to deliver the same amount of energy to a 60 kg human (by radiation)?
• 3.33 Gy
Nuclear Reactions
A process that alters the energy or structure or
composition of atomic nuclei.
Nuclear Reactions
Nuclear fissionA decay process
in which an unstable nucleus
splits into two fragments of
comparable mass
Nuclear ReactionsNuclear fission• Discovered by
the experiments of Otto Hahn and Fritz Strassman
• Bombardment of U with neutrons
Nuclear ReactionsNuclear fission• The resulting
radiation did not coincide with any of the know radioactive nuclide
• Lise Meitner found out its barium
Nuclear ReactionsNuclear fission• With Otto
Frisch, they discovered that U nuclei was splitting into 2 massive fragments called fission fragments
Nuclear ReactionsInduced fission• Fission resulting
from neutron absorption
• Rare Spontaneous fission occurs w/out initial neutron absorption
Nuclear Reactions: Fission
n
U-235
Ba-144
Kr-89
3n
Nuclear ReactionsNuclear Fusion• Occurs when 2
or more small light nuclei come together to form a larger nucleus
Nuclear ReactionsNuclear Fusion
1H1 + 1H1 2H1 + β+ + ve
2H1 + 1H1 3He2 + γ
3He2 + 3He2 4He2+1H1+1H1Proton-proton
chain
Nuclear ReactionsNuclear Fusion• 2 or more
nuclei must be w/in the range of nuclear force (approx. 2x10-15m)
Nuclear ReactionsNuclear Fusion• Thermonuclear
reactions• Proton-proton
reaction occurs at “only” 1.5x107K in the sun
• Cold fusion doesn’t require high temperatures
Reaction Energy
The difference between the
masses before and after the
reaction
When Q is pos., total
mass dec & the
total KE inc
EXOERGIC
Reaction
When Q is neg., total mass inc & the total KE
dec
ENDOERGIC Reaction
Q= (MA + MB – MC – MD) c2
Example
• When lithium (7Li) is bombarded by a proton, two alpha particles (4He) are produced. Find the reaction energy.
• Q= 17.35MeV
11H + 7
3Li 42He +
42He
H=1.007825u; Li= 7.016004u; He= 4.002603u
Example
• Calculate the reaction energy for the nuclear reaction represented below:
• Q= -1.192MeV
42He + 14
7N 178O
+ 11H
H=1.007825u; N= 14.003074u; He= 4.002603u; O=16.999132u
Nuclear Reactions
Nuclear Generator