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Year 2 Nuclear and Particle Physics > 3. Nuclear Masses > Flashcards

3. Nuclear Masses Flashcards

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1
Q

Z

A
  • proton number / atomic number

- determines the place of an element in the periodic table

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2
Q

N

A

-neutron number

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3
Q

A

A

-mass number or nucleon number

A = Z + N

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4
Q

Nuclide

Definition

A

-a specific set of numbers Z, N, A determines a nuclide

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5
Q

Isobar

Definition

A

-nuclides with the same mass number, A, are isobars

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6
Q

Isotope

Definition

A

-nuclides with the same proton/atomic number, Z, are isotopes

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7
Q

Isotone

Definition

A

-nuclides with the same neutron number, N, are isotones

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8
Q

Why is this formula for nuclear mass incorrect?

M = Zmp + Nmn

A

-due to the strong force, we cannot simply calculate the nuclear mass in this way as the binding energy makes a large contribution

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9
Q

Nuclide Map

A
  • 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
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10
Q

Binding Energy

Definition

A

-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

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11
Q

Nuclear Mass

Equation

A

M = Zmp + Nmn - Eb/c²

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12
Q

Binding Energy vs Ionisation Energy

A
  • 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
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13
Q

Mass Defect

A

m = Eb / c²

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14
Q

Finding Binding Energy Experimentally

A
  • 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
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15
Q

Mass Spectrometry

A
  • 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
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16
Q

Atomic Mass Unit

A

-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

17
Q

When can’t mass spectrometry be used?

A
  • 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
18
Q

Q Value

Definition

A

-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

19
Q

Q Values and Types of Reaction

A

-Q is defined such that:
Q>0 => exothermic reation
Q<0 => endothermic reaction
Q=0 => elastic collision

20
Q

Nuclear Reactions and Calculating Mass

A
  • 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
21
Q

Q Value and Photon Energy

A

-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<

22
Q

Features of the Nuclear Binding Energy Graph

A
  • 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
23
Q

Liquid Drop Model

Description

A
  • 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
24
Q

List the 5 contributions to binding energy in the liquid drop model

A

1) the volume term
2) the surface term
3) the coulomb term
4) the asymmetry term
5) the pairing term

25
Q

The Liquid Drop Model

Volume Term

A
  • 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
26
Q

The Liquid Drop Model

Surface Term

A

-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)

27
Q

The Liquid Drop Model

Coulomb Term

A

-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)

28
Q

The Liquid Drop Model

Asymmetry Term

A

-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

29
Q

The Liquid Drop Model

Pairing Term

A
  • 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
30
Q

The Semi-Empirical Mass Formula

Formula

A

M(A,Z) = Nmn + Zmp + Zme - avA + asA^(2/3) + acZ²/A^(1/3) + aa*(N-Z)²/4A + ap/A^(1/2)

31
Q

The Semi-Empirical Mass Formula

Description

A
  • 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
32
Q

The Semi-Empirical Mass Formula

Binding Energy Equation

A

binding energy =

[avA - asA^(2/3) - acZ²/A^(1/3) - aa(N-Z)²/4A - ap/A^(1/2)]*c²

Year 2 Nuclear and Particle Physics (9 decks)

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