6. Radioactivity

Question 1

Radioactivity

Definition

Answer 1

-spontaneous change of unstable nuclei to a more stable state

Question 2

Valley of Stability

Answer 2

• the stable nuclides lie on a stepped line on the nuclide map (in the N, Z plane), this is the valley of stability
• the further we go from the valley, the less stable the nuclides
• this related to the nuclides binding energy: less tightly bound nuclides are less stable and will tend to decay to nuclides with a higher binding energy

Question 3

Decay Constant

Answer 3

• since nuclear decay is governed by quantum mechanics, there is an element of randomness involved
• we cannot know in advance when a particular nucleus will decay
• we can only measure the probability of decay in a given time interval
• this is the decay constant λ

Question 4

Radioactivity Equations

Answer 4

dN/dt = -λN
=>
N(t) = N(0)*e^(-λt)

Question 5

Lifetime

Definition

Answer 5

-we define the lifetime of the nuclide as:
τ = 1/λ
-this is the time taken for N to drop by a factor of e

Question 6

Half Life

Answer 6

t1/2 = τ*ln2

-this is the time taken for N to drop by a factor of 2

Question 7

Spontaneous Deacy

Answer 7

-since the decay is spontaneous with no input of energy it can only occur if energetically favoured, i.e if energy is released during the process
-so we require the mass of the initial (parent) nuclide to be greater than the sum of the masses of the final (daughter) nuclides:
M(A,Z) > Σ M(Ai,Zi)

Question 8

Half Life and the Valley of Stability

Answer 8

-half-lives/lifetimes decrease the further we go from the valley of stability

Question 9

Beta Minus Deacy

Answer 9

n -> p + e- + ant ie- neutrino

-with a half-life of around 10 mins

Question 10

Beta Plus Decay

Answer 10

p -> n + e+ + (e- neutrino)

Question 11

How is beta plus decay possible spontaneously?

Answer 11

-for a free proton, beta plus decay will never happen, however in some cases it can be energetically favourable
-consider the semi-empirical mass formula, rewrite to eliminate N, N=A-Z
-collect like terms in Z:
M(A,Z) = c1A + c2Z + c3*Z² + c4
-since beta decay does not alter A, we can see that the relevant nuclide masses for this decay mode are quadratic in Z, there are two cases to consider

Question 12

Beta Decay

Semi-Empirical Mass Formula

Answer 12

-consider the semi-empirical mass formula, rewrite to eliminate N, N=A-Z
-collect like terms in Z:
M(A,Z) = c1A + c2Z + c3Z² + c4
c1 = mn - av + asA^(-1/3) + aa/4
c2 = mp + me - mn - aa
c3 = aa/A + ac/A^(1/3)
c4 = 𝛿/A^(1/2)
-with 𝛿=±11.2MeV/c²
-since beta decay doesn’t alter A, we can see that the relevant nuclide masses for this decay mode are quadratic in Z

Question 13

Beta Decay

Case 1) A is odd

Answer 13

-the nuclear masses of isobars fall on a parabola
-the minimum of this parabola lies at:
Z’stab = -c2/2c3
-since Z can only take integer values, the actual minimum Zstab, is the integer closest to Z’stab
-those nuclides with non-minimal mass for a given A will under go β-decay:
β- decay if Z < Zstab
β+ decay if Z > Zstab

Question 14

Beta Decay

A is even

Answer 14

• if A is even, then either N and Z are both even or they are both odd
• but β+ or β- decay in this case switches between the two possibilities
• so there are two parabolas to consider, the lower-mass parabola has N, Z even
• each decay jumps from one to the other which can lead to more than one stable nuclide on the same curve since it is each jump that must be individually energetically favoured rather than a whole chain of such jumps

Question 15

Discovery of the Neutrino

Answer 15

• during β decay the nucleus changes atomic number and a β± particle is emitted, these are the only two products detected
• two-body decays have a very well defined momentum Ek, e.g. for α decay, but β decay was found to have a much broader spectrum of possible momenta
• this is what would be expected from a three body decay and observations showed that β decay conserved mass number and charge but not spin or momentum
• there must be an unobserved third particle produced, neutral, with small mass and spin 1/2
• this is how the existence of the neutrino was predicted

Question 16

Electron Capture

Answer 16

X + e- => Y + νe

• where X has numbers A, Z
• and Y has numbers A, Z-1

Question 17

Mass Conditions for Beta Decay

Answer 17

-we have three possibilities, B-, B+ and electron capture
-all of these will only occur if energetically favoured
-in particular, a free electron can undergo B- decay since:
mn > mp + me + mν
-but a free proton is stable

Question 18

Mass Condition for Beta Minus Decay

Answer 18

-for beta minus decay we require:
M(A,Z) > M(A,Z+1) + me - me
-where +me is the emitted electron and -me is the missing electron in the daughter shell
-so:
M(A,Z) > M(A,Z+1)

Question 19

Mass Condition for Beta Plus Decay

Answer 19

-for beta plus decay we require:
M(A,Z) > M(A,Z-1) + me + me
-where +me is from the emitted positron and +me from the surplus electron in the daughter shell
-so:
M(A,Z) > M(A,Z-1) + 2me

Question 20

Beta Decay Spectrum

Answer 20

-recall that we must have
Q = Mparent - Σ Mdaughters > 0
-for decay to occur and
Q = Σ Ek
-but how this kinetic energy is distributed can vary, this gives the beta decay spectrum
-a bell curve, Beta momentum on the x axis and relative intensity on the y axis
-typical beta decay pmax depends on the nuclide
-p=0 corresponds to the nucleus and neutrino carrying all kinetic energy away in opposite directions
-p=pmax corresponds to the nucleus and neutrino travelling in the same direction opposite to the beta particle

Question 21

Electron Capture at Nucleon Level

Answer 21

p + e- => n + ve

Question 22

Mass Condition for Electron Capture

Answer 22

-since there is no electron emitted, it would appear that we have:
M(A,Z) > M(A,Z-1)
-however the removal of the electron leaves a gap in the inner shell, an electron from a higher energy level is then able to transition down emitting an x-ray in the process
-this is the most likely outcome however it is also possible for the energy to cause ionisation of a third electron, this energy must be taken into account:
M(A,Z) > M(A,Z-1) + E/c²

Question 23

The Auger Effect

Answer 23

• when electron capture occurs and a gap is left in the inner shell an electron from a higher energy level can transition down to fill the gap and emit an x-ray photon
• OR more occasionally the energy will instead cause the ionisation of a third electron, this is known as the Auger effect and the emitted electron is an Auger electron
• since Auger electrons have a well defined energy, we see them as an additional peak in the electron spectrum
• beta decay gives a continuum of states whilst the Auger effect gives a set of discrete emission lines each corresponding to Auger emission from a particular shell

Question 24

Comparing mass conditions for beta plus decay and electron capture

Answer 24

• electron capture can occur whenever beta plus decay can occur
• however beta plus is more likely if possible as it requires no initial-state-electron
• this makes electron capture the dominant decay mode only when Q is insufficient for beta plus decay

Question 25

Neutron Drip and Proton Drip

Answer 25

- neutron drip is the emission of a n | - proton drip is the emission of a p

Question 26

Driplines

Answer 26

- the furthest we can go from the valley of stability is to the proton an neutron driplines - the lower line, the neutron dripline is where the asymmetry between N and Z is so great that neutrons are in sufficiently high-energy states that it is energetically favourable for them to simply drip off the nucleus - likewise for the upper line and proton drip - there is a third line, the alpha dripline above which alpha decay is energetically favourable

Question 27

Lifetimes of Proton Drip, Neutron Drip and Alpha Decay

Answer 27

- the lifetime for alpha decay can be very long due to the potential barrier involved - however proton and neutron drip can have lifetimes of a fraction of a second (<<0.1s)

Question 28

Q Value Conditions for Alpha Decay

Answer 28

-above the alpha dripline, alpha emission is energetically favoured, that is the Q value is positive: Q(A,Z) = B(A-4,Z-2) + 28.3MeV - B(A,Z) -where B(A-4,Z-2_ is the binding energy of the daughter and B(A,Z) is the binding energy of the parent -the 28.3MeV is the binding energy of the alpha particle

Question 29

Alpha Decay | General Equation

Answer 29

(A,Z)X - > | (A-4,Z-2)Y + (4,2)α

Question 30

Alpha Decay | Potential Barrier

Answer 30

- despite a positive Q, α decay can take a long time due to the potential barrier the α particle must overcome to exit the nucleus - outside Rnuc the potential is due almost entirely to Coulomb repulsion, inside the strong force creates a constant negative potential - classically we would expect α particles to have ~30MeV of kinetic energy to overcome the barrier but experiments show that they do it with just 5-8MeV - classically this would be impossible but quantum mechanically as the barrier is of finite thickness the α particle has non-zero transmission probability so can tunnel out of the nucleus

Question 31

Alpha Decay | Transmission Coefficient

Answer 31

-transmission probability is given by: Tα = exp( -2/ℏ ∫√[2m(V-Ek)]dx ) -where the integral is taken from 0 to D and D is the thickness of the barrier

Question 32

Alpha Decay | Gamow Factor

Answer 32

``` Tα = e^(-G) -where G is the Gamow factor given by: G = 2√[2m]/ℏ * ∫√[Z1*Z2*e²/4πεo - Ek] dr -and R1 is the nuclear radius and R2 is the distance from the centre that the α particle must be in order to escape ```

Question 33

How does changing G affect Tα?

Answer 33

- a relatively small change in G can make a huge difference to Tα and therefore t1/2 because of the exponential factor - Tα is by far the dominant factor in determining t1/2

Question 34

Alpha Decay and Beta Decay

Answer 34

- α emission leaves odd A odd and also leaves odd-odd N, Z odd-odd and even N, Z even-even - this explains why beta decays come in pairs before an α decay - e.g. beta decay from 234Th to 234Pa leave the nuclide odd-odd and therefore beta unstable

Question 35

Spontaneous Fission

Answer 35

- the binding energy per nucleon is greatest for Fe, Co, Ni with N=26, 27, 28 - larger nuclei have lower binding energies so in principal reactions that split these larger nuclei down in to smaller more stable nuclei should be able to occur spontaneously as they have positive Q values - in fact such a process is extremely unlikely to occur

Question 36

Why is spontaneous fission so unlikely?

Answer 36

-consider a nucleus being deformed from a spherical shape to am ellipsoid on its way to fissioning -increasing the deformation increases Es since there is more surface area and decreases Ec as we have increased average separation of charges -deformation is energetically favoured if : ΔE = Es + Ec < 0 i.e. if Z²/A > 47 -for nuclides with Z≥26 but Z²/A≤47, fission itself is energetically favourable but the deformation leading to fission is not