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Selected Topics in Physicsa lecture course for 1st year students

by W.B. von SchlippeSpring Semester 2007

Lecture 111.) Determination of parameters of the SEMF2.) α decay3.) Nuclear energy levels4.) Spontaneous fission

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1.) Determination of parameters of the SEMF1.1) Determination of the Coulomb term from mirror nuclei1.2) Determination of the asymmetry term from line of maximum β stability

( ) ( )22

31 2 3 41

3

1( , )

Z Z N ZB A Z a A a A a a

AAδ

− −= − − − +

1.3) Estimate of the pairing term

In Lecture 9 we have derived a formula for the binding energy, based on theliquid drop model:

1.1) Determination of the Coulomb term from mirror nuclei

where the terms are, in that order: volume term, surface term, Coulomb term,asymmetry term and pairing term.

We can see that the volume and surface terms are identical for isobars.Therefore, if we compare the binding energies of two isobars, then theseterms cancel.

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If we choose these isobars to be odd-A mirror nuclides, then the asymmetryterm also cancels.

Two nuclides are called mirror nuclei if they have equal massnumbers A, and if the number of protons Z in one of them isequal to the number of neutrons N in the other.

Examples of mirror nuclei:11 115 6

13 136 7

boron-11 and carbon-11

carbon-13 and nitrogen-13

:

:

B C

C N

−

−

If we define the neutron excess of a nucleus by

T N Z= −

(also called isotopic number), then we have for odd-A mirror nuclei

and hence1, 1T Z N= ± = ±

Mirror nuclei:

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Thus these mirror nuclides lie next to the line N = Z in the Segre chart.

Now we know from the Segre chart that the band of stability moveswith increasing A away from the line N = Z towards neutron-rich nuclides,and unstable nuclides are never very far away from the band of stability.

Therefore we must expectthat mirror nuclides arefound only up to massnumbers A around 60.

All mirror nuclides withA >10 are shown in thefollowing table

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N - Z = +1 N - Z = -1

A Nuclide Z Delta Nuclide Z Delta

11 B 5 8.668 C 6 10.650

13 C 6 3.125 N 7 5.345

15 N 7 0.101 O 8 2.856

17 O 8 -0.809 F 9 1.952

19 F 9 -1.487 Ne 10 1.751

21 Ne 10 -5.732 Na 11 -2.184

23 Na 11 -9.530 Mg 12 -5.474

25 Mg 12 -13.193 Al 13 -8.916

27 Al 13 -17.197 Si 14 -12.384

29 Si 14 -21.895 P 15 -16.953

31 P 15 -24.441 S 16 -19.045

33 S 16 -26.586 Cl 17 -21.003

35 Cl 17 -29.014 Ar 18 -23.047

37 Ar 18 -30.948 K 19 -24.800

-46.90032Ge-56.54731Ga63

-47.09031Ga-56.35030Zn61

-47.26030Zn-56.35729Cu59

-47.31029Cu-56.08228Ni57

-45.34028Ni-54.02827Co55

-42.64027Co-50.94526Fe53

-40.22026Fe-48.24125Mn51

-37.62025Mn-45.33124Cr49

-34.56024Cr-42.00223V47

-31.88023V-39.00622Ti45

-29.32122Ti-36.18821Sc43

-28.64221Sc-35.13520Ca41

-27.27420Ca-33.80719K39

DeltaZNuclideDeltaZNuclideA

N - Z = -1N - Z = +1

Mirror nuclides from A = 11 to A = 63: there are no mirror nuclides with A > 63

Delta is the mass excess in MeV;the values are from the BNL Wallet Cards 2005, http://www.nndc.bnl.gov/wallet/

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Let us now compare the binding energies of mirror nuclides using the SEMF:

( ) ( )22

31 2 3 41

3

1( , )

Z Z N ZB A Z a A a A a a

AAδ

− −= − − − +

we have

or in terms of A and T = N – Z:

( )( ) 223

1 2 3 413

21( , )4

A T A T TB A T a A a A a aAA

δ− − −

= − − − +

( )( ) ( )( )

( )

3 13

13

3

( , 1) ( , 1)1 3 1 11

4

1

B B A T B A TA A A A

aA

a A A−

Δ ≡ = + − = −

− − − + −= −

= −

hence

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Thus, if we plot the measured differences of the binding energies asa function of (A-1)/A1/3, then we expect a straight line through the origin.This is shown in the next figure.

Determination of the Coulomb coefficient from mirror nuclei

y = 0.6448x

0

2

4

6

8

10

12

0 5 10 15 20

(A-1)/A (1/3)

Del

ta E

_Cou

lom

b (M

eV)

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The slope of the line, which is constrained to pass through the origin,is the empirical coefficient a3 of the Coulomb term. The least squaresfit gives

3 MeV0.645a =

1.2) Determination of the asymmetry term from line of maximum β stability

( ),0

A const

M A ZZ =

∂=

∂

In the previous lecture we have defined the position of maximum β stability:

Now, the mass M(A,Z) is given by

( ) ( ) ( ) 2, ,H nM A Z Zm A Z m B A Z c= + − −

hence

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2 2 1 33 4

min 1 3 13 4

42 8

n Hm c m c a A aZa A a A

−

− −

− + +=

+

thus the position of the minimum of the mass parabola is determinedby the coefficients a3 (Coulomb term) and a4 (asymmetry term).

Solving for a4 we get

( )2 2 1 33 min

4 1min

1 28 4

n Hm c m c a Z Aa

Z A

−

−

− + −=

−

Empirically, fitting the masses of isobars with parabolas, we will getsomewhat different values of a4. Then these values are averaged.

From fits to only a few mass parabolas of mass numbers A = 15, 63,65, 97, 101, 107 and 135 I have got the following result:

4 MeV21.8 1.3a = ±

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1.3) Estimate of the pairing term

The pairing term δ is nonzero only for even-A nuclides.

In lecture 9 we have seen in the discussion of stability rules that of the 174stable even-A nuclides, 166 are even-even and only 8 are odd-odd nuclides.

Four of the odd-odd stable nuclides are light nuclides:deuterium, lithium, boron and nitrogen;they are absolutely stable and symmetric, i.e. N = Z.

The remaining four odd-odd nuclides are unstable but have lifetimes greater then 109 years, i.e. greater than the age of the earth:

40 919

50 1723

138 1157

180 1573

1.3 10

1.4 10

1.1 10

1.2 10m

K years

V years

La years

Ta years

×

×

×

> ×

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We will be guided by these stability rules to estimate the pairing term δ

To do this we rewrite the expression for the minimum of the mass parabolain terms of the neutron excess T:

( )1 33

min min 1 3 13 4

( 1)2

4n Ha A A m m

T A Za A a A

−

− −

− − −= − =

+

and then use this to get the following result:

( ) ( )

( )

2 2min

21 3 13 4 min

, ,

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U M A T c M A T c

a A a A T T− −

Δ ≡ −

⎛ ⎞= + −⎜ ⎟⎝ ⎠

Then we apply this to even-A nuclides. The stability rules suggest thatthe mass parabolas of the odd-odd nuclides are shifted upwards fromthe mass parabolas of the even-even nuclides. We denote the separationbetween these parabolas by δ.

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Mass parabolas for even-A nuclides: the energy U=Mc2 is plotted vs T(Figures from Blatt and Weisskopf; their notation differs from ours: they writeδ/A instead of our δ, also 4uC = a3 and 4uτ = a4)

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Shown on the left is the situation where there is one stable odd-odd nuclideat T = Tmin and no stable even-even nuclides. Here

1 3 13 4

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a A a Aδ − −⎛ ⎞< +⎜ ⎟⎝ ⎠

From the stability rules we know that this does not happen in nature.Therefore we conclude that

1 3 13 4

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a A a Aδ − −⎛ ⎞> +⎜ ⎟⎝ ⎠

On the right-hand figure there are three stable even-even nuclides,the middle one at T = Tmin. Their stable even-even neighbors are separatedby 2 units in T, and hence their energy difference is

1 3 13 4

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U a A a Aδ − −⎛ ⎞= +⎜ ⎟⎝ ⎠

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The unstable odd-odd nuclides lie above the minimum of their mass parabolaby ΔU with T - Tmin = 1, and therefore above the minimum of the mass parabolaof the even-even nuclides by δ + ΔU, thus

1 3 1 1 3 13 4 3 4

1 144 4

a A a A a A a Aδ − − − −⎛ ⎞ ⎛ ⎞+ + > +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

1 3 13 4

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a A a Aδ − −⎛ ⎞> +⎜ ⎟⎝ ⎠

hence

1 3 1 1 3 13 4 3 4

1 134 4

a A a A a A a Aδ− − − −⎛ ⎞+ < ≤ +⎜ ⎟⎝ ⎠

This situation occurs rarely in nature; we must therefore conclude, togetherwith the previous result, that

Now, we had δ = a5/A, and that gives us an estimate of a5:

2 3 2 33 4 5 3 4

1 134 4

a A a a a A a⎛ ⎞+ < ≤ +⎜ ⎟⎝ ⎠

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For a numerical estimate let us take an intermediate value of A = 100;then, with our values for a3 and a4 we get

532 96a< ≤

We will continue using the SEMF next to discuss the (in)stability ofnuclei against α decay.

but for small values of A the limits get tighter: if we put A = 1, then

522 65a< ≤

We must conclude that this parameter can be determined onlysemi-quantitatively. But then we must remember that the basis of theSEMF, the liquid drop model of nuclear matter, is only the simplestmodel that gives at least reasonable results. More sophisticated modelsrequire quantum mechanics. These are outside the scope of the presentlecture course.

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2.) α decay

The reaction equation of α decay is

4 42 2

A AZ ZX Y Qα−

−→ + +

where Q is the kinetic energy released in the process.

In the rest frame of the mother nucleus X the momenta of thedaughter nucleus Y and the α particle are equal in magnitude.

From the previous lecture we have the following formula for the momentum

( ) ( ){ }1 22 2* 12 X Y X Y

X

p M M m M M mM α α

⎡ ⎤ ⎡ ⎤= − − − +⎣ ⎦ ⎣ ⎦

Thus for example in the α decay of radium-226

226 222 488 86 2Ra Rn Qα→ + +

we find p* = 189 MeV/c, and hence kinetic energies …

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of the radon nucleus and the α particle:

*2 2

*2 2

MeV,

MeV

4.78

0.09Rn Rn Rn

KE p m m

KE p M M

α α α= + − =

= + − =

as a general rule the kinetic energy of the daughter nucleus is smallcompared with the kinetic energy of the α particle.

For the decay to take place the Q value must be greater than zero

Thus the condition for α decay is

( ) ( ) ( )2 2 4 2, 4, 2 0Q M A Z c M A Z c m He c= − − − − >

or in terms of the mass excess Δ = M(A,Z) - A

( ) ( ) ( )4, 4, 2 0A Z A Z HeΔ −Δ − − −Δ >

and hence, taking the values of Δ from tables of nuclides, …

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we have( )( )( )

226

222

4

MeV

MeV

MeV

23.669 ,

16.374 ,

2.425

Ra

Rn

He

Δ =

Δ =

Δ =

and hence

( )226 222 MeV4.78Q Ra Rn α→ + =

Empirically one knows that spontaneous α decay of naturally occurringnuclides takes place only for heavy nuclides. The lightest α unstablenuclides are shown in the next table

A Z Element T1/2

144 60 Nd 2.3E18147 62 Sm 1.06E11148 62 Sm 7E18

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We can see that these nuclides are barely unstable: they much prefernot to decay.

Now let us see whether the SEMF is qualitatively in agreement with theempirical evidence.

From the SEMF we find the following expression for the Q value:

( ) ( ) ( )

( )

2 2 4 2

24 2

1 3 41 13 3

, 4, 2

84 4 1 4 1 23 3

Q M A Z c M A Z c m He c

a Z Z ZB He a a aA AA A

= − − − −

⎛ ⎞ ⎛ ⎞= − + + − − −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

and if we want to apply the condition to naturally occurring nuclides,then we must put Z equal to its value on the line of β stability:

( )23

3 42 2Z A a A a+

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This is not the kind of formula that allows us to see at a glance what’s going on,so we better write a little computer program in our preferred language (whichin my case is FORTRAN) to compute Q as a function of A. The results willdepend somewhat on the values of the constants we use. With the valuesof Lecture 9 we get the following result:

0 1460 147

Q for AQ for A< ≤> ≥

which is in better than just qualitative agreement with the empiricalevidence.

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3.) Nuclear energy levels

Since nuclei consist of nucleons we may expect by analogy with atomsthat there can be excited nuclear states.

To make a transition from its ground state into an excited state the nucleusmust absorb energy.

Excitation energy can be transferred to a nucleus by a collision withanother nucleus or by exposure to electromagnetic radiation.

An experiment to demonstrate the existence of excited nuclear statesis by collisions with protons.

A beam of protons is directed at a target of some pure substance.Measured are the energies of the scattered protons at some fixedscattering angle θ

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This experiment is a slight modification of the elastic scattering experimentdiscussed in a previous lecture, so I did not produce a new figure.

We denote the 4-momenta of the beam particle by p and of the targetparticle by P; the 4-momenta of the scattered and recoil particles arep’ and P’.

By 4-momentumconservation we have

p P p P′ ′+ = +

hence

( )22P p P p′ ′= + −

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Let us denote the masses of the incident particle, target particle, scatteredparticle and struck particle by

, , ,p t pm M m M

hence

( )2 2 22 2 2t t pM M M E E m p p′ ′= + − + − ⋅

Observe the scattered protons at 90 degrees to the beam direction, then

0p p′⋅ =

and if we express the energy in terms of the kinetic energy:

pE T m= +then

( )2 2 212

t t p

t p

M M T M mT

M T m− + −

′ =+ +

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In the particular case of elastic scattering the formula simplifies:

t pelastic

t p

M mT T

M T m−

′ =+ +

and then we can write the formula for the general case as

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telastic

t p

M MT TM T m

−′ ′= −+ +

Thus we expect to see protons with kinetic energies less than thekinetic energy of elastically scattered protons.

In a particular experiment, protons of kinetic energy 10.02 MeV wereincident on a target containing boron-10 whose atomic mass is10.01 u. Thus, remembering that 1 u = 931.494 MeV, we get

MeV8.18elasticT ′ =

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In the figure the proton KE is plotted along the horizontal axis, and verticallythe number of protons in arbitrary units. The discrete set of lines is a clear demonstration of nuclear energy levels.

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Another way of producing excitation is by decay of anucleus. Then it can happenthat the daughter nucleus isin an excited state.

In the figure this is shown for theexample of α decays into thedifferent energy levels of thedaughter nucleus.

Observed is a discrete set ofα energies and associatedemission of γ rays

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4.) Spontaneous fission

Nuclear fission is the process of separation of a nucleus into fragmentsof similar size; this is usually accompanied by emission of one or severalneutrons.

Fission can take place spontaneously or by induction. The usualmechanism of induced fission is the bombardment of the nucleus withneutrons.

Consider the spontaneous fission of nucleus X into fragments A and B:

X A B Q→ + +

The masses of the fragments are usually about equal. Let us forsimplicity put

( ) ( ) ( )12

m A m B m X= =

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then the released kinetic energy is

( ) ( ) ( ) ( ) ( )2Q B A B B B X B A B X= + − = −

Now, the binding fraction at A = 240 is about 7.6 MeV per nucleon,and at A = 120 about 8.5 MeV per nucleon. Therefore

( ) MeV8.5 7.6 200Q A− ∼

The fragments have an excess ofneutrons. This can be seen byreference to the Segre chart

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Mother nucleus

daughter nuclei

By dividing the heavy nucleus into two fragments of about equalmass, the daughter nuclei end up further from the line of stabilityinto the region of greater neutron excess.

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The excess of neutrons can be reduced by emission of neutrons and/orby β decay. Usually both processes take place.

The fragments are initially in excited states; de-excitation takes place byemission of gamma rays.

Some of the kinetic energy is carried by the neutrinos which are alwaysemitted together with electrons in beta decay. These neutrinos escapewithout depositing their energy in the surrounding material. All other particlesare absorbed. The absorbed energy is ultimately converted into heat.

To produce 1 Joule of thermal energy requires of the order of 1010 fissions

1 Joule =0.623x1013 MeV

and we get about 200 MeV per fission:

13 JouleMeV Joule

200 1200 100.623

fission fissionN fissions

−= =

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hence

13 10 Joule0.62310 3 10200

N fissions per= ×

Bohr-Wheeler Theory of Nuclear FissionIn the liquid drop model, a deformation of the nucleus takes place withoutchange of the volume.As a result of the deformation the energy of the nucleus changes.The energy difference between the deformed and undeformed nucleiis the deformation energy.

Fission can occur if the deformation energy is less than zero.

Estimate of the deformation energy: in the SEMF only the surface energyand the Coulomb energy change.

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Consider a small deformation of a sphere into an ellipsoid:

Rc

b

b

343

V Rπ= 24

3V b cπ=

Let

( )1c R ε= + hence ( )1b R ε= +(ε = deformation parameter, |ε| << 1)

Surface energy:2 3 2 3 2

2 2215

a A a A ε⎛→ ⎞+⎜ ⎟⎝ ⎠

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Coulomb energy:

( )2 1 3 2 1 3 23 3

115

a Z A a Z A ε− ⎛ ⎞−⎜ ⎟⎝ ⎠

→

(Exercise!!!)

hence net increase of E as a result of the deformation (deformationenergy):

( ) ( )2 2 3 2 1 3 22 3 definition of

1 25

E a A a Z A k kε εΔ = − =

i.e.

( )2 3 2 1 32 3

1 25

k a A a Z A= −

Condition for stability: 0k >

Condition for instability 0k <

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Critical value: 0k =

hence2 3 2 1 3

2 32 0a A a Z A− =

or2

2 32Z A a a=

and with

2 3MeV MeV13 , 0.6a a= =

2 43.5Z Awe get

and hence with Z on the line of maximum β stability we get

0 110k for Z≤ ≥