3. Nuclear Masses Flashcards
Z
- proton number / atomic number
- determines the place of an element in the periodic table
N
-neutron number
A
-mass number or nucleon number
A = Z + N
Nuclide
Definition
-a specific set of numbers Z, N, A determines a nuclide
Isobar
Definition
-nuclides with the same mass number, A, are isobars
Isotope
Definition
-nuclides with the same proton/atomic number, Z, are isotopes
Isotone
Definition
-nuclides with the same neutron number, N, are isotones
Why is this formula for nuclear mass incorrect?
M = Zmp + Nmn
-due to the strong force, we cannot simply calculate the nuclear mass in this way as the binding energy makes a large contribution
Nuclide Map
- a plot of neutron number N on the x axis and proton number Z on the y axis
- there is a line that is most stable and a region of lower stability but where nuclides can still exist, this is known as the valley of stability
- outside of this nuclides are so unstable that they cannot exist
- to decay a nuclide needs to transition to one of slightly lower mass close to it on the nuclide map
- i.e. if a nuclide is surrounded by higher mass nuclides on the map then it won’t decay
Binding Energy
Definition
-the energy required to release a nucleon from the nucleus
Eb = ∫ F ds
-where the integral is taken from the nuclear radius r to infinity
Nuclear Mass
Equation
M = Zmp + Nmn - Eb/c²
Binding Energy vs Ionisation Energy
- Eb accounts for around 1% of the nuclear mass
- comparing with the analogous electron ionisation energy:
- electron ionisation energy accounts for only ~10^(-9)% of the atomic mass
- it is clear that binding energy contributes a much more significant percentage and whilst we may be able to omit ionisation energy from atomic mass calculations, we must account for binding energy in nuclear mass calculations
Mass Defect
m = Eb / c²
Finding Binding Energy Experimentally
- found by measuring nuclear mass
- atomic masses may be measured by deflecting ions in electric and magnetic fields
- an ion in such a field follows a curved path with a particular radius of curvature
- |E gives re which is related to kinetic energy and |V give rmag which is related to momentum, from these we can determine mass
Mass Spectrometry
- particles pass through electric and magnetic fields on a curved path
- the radius of curvature is determined by energy for the electric field and momentum for the magnetic field
- the fields are tuned so that only particles of a particular mass will reach the detector
Atomic Mass Unit
-we measure atomic masses relative to the reference mass (12,6)C, carbon 12
-we define the atomic mass unit, where M is the mass of carbon 12, as:
1u = 1/12 M
1u = 1.66 x 10^(-27) kg
1u = 931.5 Me²/c²
When can’t mass spectrometry be used?
- for mass spectrometry to be useful, the nuclide must survive the journey through the instrument
- this method is unsuitable for very unstable short-lived nuclides
Q Value
Definition
-the change in binding energy before and after a collision event
-e.g. A + B -> C + D
Q = (mA+mB)c² - (mC+mD)c²
-this turns out to also be equivalent to the change in kinetic energy of the nuclides:
Q = EkC + EkD - EkA EkB
Q Values and Types of Reaction
-Q is defined such that:
Q>0 => exothermic reation
Q<0 => endothermic reaction
Q=0 => elastic collision
Nuclear Reactions and Calculating Mass
- if we know all the quantities in a reaction apart from one mass i.e. kinetic energies of all particles and mass of all except one, then we can calculate the unknown mass
- this is an alternative method to mass spectrometry
Q Value and Photon Energy
-as well as kinetic energy of the nuclides, the energy released can also be in the form of a photon
-Q is then a combination of photon energy and kinetic energy or recoil
-e.g. A -> B + γ
A + B -> C +γ
-in the centre of mass frame we have ptotal=0 so by conservation of momentum;
|mv| = |Eγ/c|
m²v² = Eγ²/c²
1/2mv² = Eγ²/2mc²
-where Eγ is the photon energy and m is the nuclide mass
-often Eγ<
Features of the Nuclear Binding Energy Graph
- plot nucleon number, A on the x axis and binding energy per nucleon Eb/A on the y axis
- gradual decrease in binding energy for large A
- peaks at regular intervals corresponding to sets of 4 nucleons (2xP & 2xN)
- low binding energy for small A
Liquid Drop Model
Description
- it is known from experiment that charge density of the nucleus is very uniform, adding more nucleons results in increased volume not density
- the forces inside the nucleus are very short range, SNF only effects closest neighbouring nucleons
- thus we can model the nucleus as a drop of liquid being pulled by its surface tension into a spherical shape
List the 5 contributions to binding energy in the liquid drop model
1) the volume term
2) the surface term
3) the coulomb term
4) the asymmetry term
5) the pairing term
The Liquid Drop Model
Volume Term
- each nucleon in the interior of the nucleus is surrounded on all sides by other nucleons with which it can interact by the SNF
- each nucleon therefore increases binding energy by an almost equal amount, av*c², or equivalently reduces the nuclear mass by av
- volume term: -av*A
The Liquid Drop Model
Surface Term
-it is immediately clear that nucleons at the surface of the nucleus have fewer neighbours and thus won’t reduce the mass by as much as those in the interior
-however with the volume term we have treated each nucleon equally as an interior nucleon
-thus we need a correction term proportional to the surface area
-assuming volume is proportional to A:
-surface term:
+as*A^(2/3)
The Liquid Drop Model
Coulomb Term
-the protons in the nucleus experience a mutual repulsion which lowers the binding energy (and thus increases the mass)
-we must calculate the energy required per proton
-this is a positive term
-Coulomb term:
+acZ²A^(-1/3)
The Liquid Drop Model
Asymmetry Term
-the asymmetry term does not strictly follow from the liquid-drop model as it is related to the quantum nature of the nucleus
-nucleons are Fermions and therefore subject to the Pauli exclusion principle
-lowest energy levels are filled first and subsequent nucleons forced to occupy higher states
-although some of this is accounted for by other terms, the difference in proton and neutron number is not
-the greater the asymmetry, the higher the energy levels that must be occupied
-a power series is calculated from analysis of Fermi energy of the nucleus, its first term is proportional to A so is absorbed into the volume term
-its second term is proportional to (N-Z)²/A
-asymmetry term:
aA*|N-Z|²/4A
The Liquid Drop Model
Pairing Term
- takes into account the relative stability of nuclides with even N and Z compared to those of odd N and Z
- experimentally we find that if A is odd ap=0 , if N & Z are both even ap<0 which decreases mass and if N & Z are both odd ap>0 which increases mass
The Semi-Empirical Mass Formula
Formula
M(A,Z) = Nmn + Zmp + Zme - avA + asA^(2/3) + acZ²/A^(1/3) + aa*(N-Z)²/4A + ap/A^(1/2)
The Semi-Empirical Mass Formula
Description
- semi empirical because not all of the terms are derivable
- constant are chosen to fit as many data points as possible
- as a result there are different sets of the constants that are used
The Semi-Empirical Mass Formula
Binding Energy Equation
binding energy =
[avA - asA^(2/3) - acZ²/A^(1/3) - aa(N-Z)²/4A - ap/A^(1/2)]*c²
