NUCLEAR STRUCTURE AND PROPERTIES OF NUCLEI

ByA Saxena

Nucleus (of an atom), is a bound quantum system and can exist in different quantum states, depending upon energies, angular momenta etc.

Static properties of nuclei (in ground states) electric charge, mass, binding energy, size, shape, angular momentum, magnetic dipole moment, electric quadrupole moment, statistics, parity and iso-spin.

Dynamic characteristics of nuclei play role in processes of nuclear reactions, nuclear excitation and nuclear decay.

Nuclear mass and Binding energy  

Nucleons Atomic nuclei contains two different types elementary particles, protons and neutrons.

Proton nucleus of hydrogen atom, having positive charge of one electronic unit (+e) mass ~1836 times electronic mass (me)

Neutron electrically neutral, slightly heavier than proton

In the nucleus protons and neutrons held togetherby very strong short range attractive force, constitutes specifically nuclear interaction.  

+1e -e

+pn

-e

Numbers of neutrons (N) and protons (Z) inside nucleus is its mass number A = N + Z.

Z = atomic number of element in periodic table.

Nucleus of an atom X of atomic number Z and mass number A is written as

ZAX or AX.

e.g. 24He or 4He nucleus of helium

atom, atomic number 2 and mass number 4. This is α-particle.

Isotopes nuclei (of an atom) with same at. no. Z, but different mass no.A i.e. Nuclei contain same no. of protons, but different no. of neutrons. e.g. Lithium (Z = 3) has two stable isotopes 6Li and 7Li no. of protons in both nuclei are same Z =3. 6Li N = 3 7 Li N =4.  

Relative abundance Elements having more than one stable isotope in natural state are mixtures of these isotopes in fixed proportions (which remain more or less the same, Irrespective of their source).

Natural lithium is a mixture of two stable isotopes 6Li & 7Li with relative abundances 7.4% and 92.6%.  

Hydrogen has two stable isotopes 1H (99.99%), 2H (0.01%). 2H is called deuterium and its nucleus deuteron. Another unstable Isotope with A = 3 called tritium 3H.

Isobars Nuclei with same mass no. A, but different no. of proton Z

Isotones nuclei with same no. of neutrons but different mass no. 

Nuclear mass : Atomic mass M by subtracting the masses of Z orbital electrons (me is mass of electron)

Mnucl = M - Z me  

This expression is not exact, since it does not take into account binding energies of the electrons in atom.  

Binding energy of nucleus: Minimum amount of energy supplied to

nucleus to break up a nucleus Z protons and N neutrons completely, i.e. all are separated from each other.  

In formation of a nucleus, same amount of energy is required, which comes from disappearance of a fraction of its total mass(of Z protons and N neutrons), If quantity of mass disappear ∆M, then binding energy EB = ∆Mc2  

(e.g. 1g mass completely converted into energy = 9 x 1013 J)

So mass of the nucleus < sum of the masses of constituents neutrons and protons.

In hydrogen atom mass of hydrogen atom MH and neutron Mn

∆M = Z MH + N Mn - M(A, Z)  

where M(A, Z) mass of the atom (of mass no. A and at. no. Z).

Then binding energy of the nucleus EB= [Z MH + N Mn - M(A,Z)]c2

masses of the Z electrons (Zme) cancel out.

Hence ∆M for the nucleus ∆M = Z Mp + N Mn – Mnuc (A,Z)

Due to mass-energy equivalence masses can be written in energy units, then c2 can be omitted.

Atomic mass unit : 1 mole of 12C has mass 12g. 1 mole contains N0 atoms, Where N0 = 6.02205 x 1023 is Avogadro number,

12 X 10-3 or 12x 1.660566 x 10-27 kg No

Then mass of each 12C atom is 1 u = 1 x 12 X 10-3 = 1.660566 x 10-27 kg 12 No

Energy-equivalent of this mass is 1 u = 1.660566x 10-27 x c2 = 1.660566 X 10-27 x 8.98755 X 1016 = 14.924427x 10-11 Joule = 14.924427 x 10-11 = 931.502 MeV

1.60219 x 10-13

 

Importance of accurate determination of atomic masses

Atomic masses are determined with accuracy better than 1 ppm by mass spectroscopes. Required for determination of nuclear binding energies and calculation of nuclear disintegration energies.

e.g. α-disintegration of a heavy element like 226Ra (Z= 88) Nuclei spontaneously disintegrate by emission of α-particles (of few MeV kinetic energy) from conversion of a part of nucleus mass into energy (by mass-energy equivalence relation ). 226Ra 222Rn + 4He Product 222Rn (Z=86)

The masses of different atoms take part in processM (226Ra ) = 226.025436 u M(222Rn) = 222.017608 u M (4He) = 4.002603 u

then α-disintegration energy Qα = [ M ( 226Ra) – M( 222Rn) - M (4He)] c2 = (226.025436 - 222.017608 - 4.002603) x 931.502

= 0.005225 x 931.502 = 4.87 MeV   disintegration energy is less than one

part in 40,000 of the mass of disintegrating nucleus.

Study of nuclear disintegration energies provides direct experimental evidence for mass-energy equivalence principle.

Systematic of nuclear binding energy

Accurate determination of atomic masses, are very close to whole numbers, which are actually the mass no. of the atoms. .

Atoms with A < 20 and A > 180, atomic masses are slightly greater than the corresponding mass numbers. While for 20 < A < 180, slightly less.  

Mass defect ∆M departure of measured atomic mass M(A, Z) from the mass no. (A) is quite significant. ∆M = M(A, Z) - A

e.g. 4He at. mass 4.002603 u is slightly greater than the mass no. 4,∆M = + 0.002603 u.

& for 75As ∆M = - 0.078403 u. For very light and very heavy atoms, mass defect is

positive, while in the intermediate region it is negative. Packing fraction (f) mass defect of an atom

divided by its mass no. f = ∆M = M(A,Z) - A

A  A = M(A,Z) - 1

A  Having same sign of mass defect Measured atomic mass

M (A, Z) = A (1 + f )

Packing fraction curve

Packing fraction f varies in systematic manner with mass number A. It is explained from nuclear binding energy considerations.

A

Packin

g f

racti

on

20 60 100 140 180 220

-10

0

1

0

30

50

70

If the binding energy EB of a nucleus z

AX, then binding fraction  

fB = EB = Z MH + N Mn - M(A,Z)

A A measured atomic mass M(A, Z)

In mass unit (c2 not required ) It shows relative strengths of their

binding energy.

e.g. for deuteron ( Z = 1, N = 1) EB (2H) = MH + Mn - Md .

= (1.007825 + 1.008665 - 2.014102) X 931.5

=2.224 MeV

fB (2H) = EB =2.224 =1.112 MeV/u

A 2 For α-particle (4He) , since Z = 2, N= 2,

EB (4He) = Z MH + N Mn - MHe

EB (4He) = (2 X 1.007825 + 2 X 1.008665 - 4.002603 ) X 931.5

= 28.3 MeV fB (4He) = 28.3 = 7.075 MeV/u

4 Since fB (2H) << fB (4He), thus 2H is very weekly

bound than 4He.

Fig. Binding fraction curve .

A

EB/A

(

MeV

/u )

0 8 16 24 28 // 60 90 120 150 180

2

4

6

8

10----------------------------------------------------------------------

Variation of binding fraction fB

with A (a)fB for very light nuclei is very small

and rises rapidly with A, attaining a value of ~ 8 MeV/u for A ~ 20.

then rises slowly and attains a maximum of

~ 8.7 MeV/u at A ~ 56(b) For 20 < A < 180 fB is very slight, almost constant, mean value ~ 8.5 MeV/u.(c) For very heavy nuclei (A > 180) decreases with increase in A, fB ~ 7.5 MeV/u.

(d) For very light nuclei rapid fluctuations in fB,

peaks are observed for the even-even nuclei 4He, 8Be, 12C,16O etc., for which A= 4n where n is an integer. But less prominent peaks are observed at values of Z or N = 20, 28,50, 82 and 126. These are known as magic numbers. Peaks show greater stability of corresponding nuclei relative to nuclei in their immediate neighborhood.

Nature of binding energy curve is complimentary to packing fraction curve.

 

MH =1 + fH and Mn = 1 + fn

(where fH = 0.007825 u and fn = 0.008665 u are constant)

then we have EB = Z ( 1 + fH ) + N (1 + fn ) - M (A, Z)

= ( Z + N ) + Z fH + N fn - A (1 + f )

= A + Z fH + N fn - A - ∆M  

where ∆M = A f , Hence we get EB = Z fH + N fn - ∆M

fB = EB = Z fH + N fn _ ∆M

A A A = Z fH + N fn _ f

A  

First term is almost constant, specially for lower A, when Z = N = AI2  

Thus fB increases or decreases as f decreases or increases.

Graphs variation f and fB are complementary with A.

Binding fraction curve explains ( qualitatively) α-disintegration of heavy nuclei and also energy release in nuclear fission and fusion processes.

Nuclear size Rutherford's theory of α-particle scattering gives

smallness of nuclear size.

Scattering experiments with relatively higher energy α-particles gives deviation from Rutherford scattering formula at large angles i.e. for small impact parameters b. When b becomes comparable to nuclear radius R, α-particle feels effect of nuclear force. Since Rutherford's scattering formula is deduced on the assumption that force acting on α-particle is purely electrostatic, deviation is expected, this is no longer true. Putting b = R, we get limiting angle of scattering θc above which ratio of measured scattering cross section (σ) to that Rutherford's formula (σR) will be different from unity. 

Above θc, anomalous scattering takes place. Rutherford estimated values of nuclear

radius R for a few light elements. e.g. magnesium. order of 10-15 m (not very accurate).

In case of nuclear radius, we assume that nucleus has spherical shape, because of short range nuclear force.

However, small deviations from the spherical shape of certain nuclei have been observed, due to electric quadrupole moment of nuclei. (zero for spherical nuclei).

It is assumed that nuclear charge is uniformly distributed. Experiments show that nuclear charge density ρc and distribution of nuclear matter (i.e., protons and neutrons) is nearly uniform.

then nuclear matter density ρm is also approximately constant. Since nuclear mass is almost linearly proportional to the mass number A, this means ρm ~ A/V = constant

i.e. nuclear volume V α A. (nucleus with radius R). then V = 4 πR3 α A 3 R α AI/3  R = r0 A 1/3  

where ro is a constant, known as nuclear radius parameter.  

Nuclear radius radius of

nuclear mass distribution or nuclear charge distribution. Since nuclear charge parameter (i.e. at. no.) Z is almost linearly proportional to mass no. A and nuclear charge density ρc is approximately same throughout nuclear volume,

distribution of nuclear charge +Ze follow pattern of nuclear mass distribution. Hence charge radius and mass radius are very nearly same (due to strong attractive forces within nucleus ). very nearly same for both types of nucleons, protons and neutrons and hence their distributions within nuclear volume follow the same pattern.

Potential energy of nucleous for a charged particle ( proton or α-particle )

Electrostatic repulsive force of nuclear charge +Ze acted from outside the nucleus (r > R), while inside the nucleus (r < R) a negative potential (due to short range nuclear force). Where r is the distance from the nuclear centre. We assume electrostatic force is not effective inside nucleus and nuclear force becomes zero at nuclear surface (r = R).  

Potential energy diagram of nucleus

.

- - - - - - - - - - - - - - - - - - - - - - -Vc =E

- Vc

V

E

o

-Vo

R r

Nucleus is surrounded by a Coulomb potential barrier Vc = ZZ’ e2

4π ε0 r  

form incident particle of charge Z’e (for r > R) . At the nuclear surface barrier height is given by

VR = ZZ’ e2

4π ε0 R

Radius R is known as potential radius, slightly greater than charge or mass radius.

 e.g. For uranium nucleus with Z = 92 and

R = 8 X 10-15 m , VR = 16.5 MeV for a proton,

while VR = 33.1 MeV for an α-particle taking

ro = 1.3 X 10-15 m.  

Classically, a charged particle of energy E < VR cannot escape (or enter) from the nucleus.

But quantum mechanically (due to uncertainty principle) position of the particle within the nucleus is not so well-defined, i.e. a finite probability of particle to penetrate barrier, If E < VR .

If particle with an initial energy +E outside the nucleus reaches the point r = b where Vc = E. then it will be repelled by electrostatic force of positive charge of residual nucleus and goaway.  

Nuclear radius is expressed in units of 10-15 m i.e. femtometer (fm) or Fermi.

Mean square radius of nuclear charge distribution is defined as

<r2 > = oʃ∞ r2. 4πr2 ρ(r) dr

oʃ∞ 4πr2 ρ(r) dr where ρ(r) is nuclear charge density. For a uniformly charged sphere (ρ = constant) of radius R, this gives (since ρ = 0 for r> R)

<r2 > = oʃR r4 dr

oʃR r2 dr

R2 = 5 <r2 >3

Measurement of the charge radius by Electron scattering experiment

In scattering of high energy electrons by nuclei, no nuclear force acting on the electrons, only Coulomb attractive force acts due to nuclear charge. If de Broglie wavelength of the electrons is small compared to the nuclear radius, then this experiment gives many details of nuclear charge distribution.

According to de Broglie's theory of wave-corpuscular dualism,

the wavelength of a relativistic electron of rest mass mo and

total energy E > moc2 is given by

λ = ch

e [ V(V+2moc2/e)]1/2 

where eV = Ek is kinetic energy of the electron, e is charge.

Substituting values of c, h, e and mo we get

λ = 12.4Xl03 Ao

[V (V + 1.02 x 106) ] 1/2

For electrons Ek = 200 MeV,

V = 200 x 106 volts

which gives λ = 6.19 X 10-5 Ao

ƛ = λ/2π =10-15m =1fm 

It is smaller than radius of most nuclei.  

It shows that electrons of a few hundred MeV energy can give many details of nuclear charge distribution.

High energy electron scattering experiment

.

collimator

Accelerator AcceleratorBeamstopp

er /////////////////////////////////////////////////////////////

shielding/////////////////////////////////////////////////////////////////////////////////////////

/////////////////////

/////////

//////

/

spectrom

eter

-----------------------------

--------------

-

Scattering chamber

Deflecting magnet T

S

High energy electron beam from linear accelerator A is deflected by means of magnet and collimated by slit, Deflecting magnet directs beam to target inside scattering chamber. The elastically scattered beam of electrons is then analysed by the large magnetic spectrometer .

Quantum mechanical expression for differential scattering cross-section of a relativistic electron from a spin-less target at centre of mass angle θ is

σ(θ) =σM (θ) {F(q)}2  

σM (θ) is Mott cross section of elastic scattering from a point charge +Ze

σM (θ) = ( Ze2 )2 cos2θ/2

( 8πεoE) sin4θ/2

where E energy of the electrons in the C.M. system.

F(q) form factor which gives the ratio by which scattering cross-section is reduced, when the charge +Ze is spread out over finite volume. Due to destructive interference between electron waves scattered from different parts of the target nucleus, F(q) < 1.  

Using Born approximation method of quantum mechanics

F(q) = 1 ʃ ρ (r) exp (i.q.r) dτ Ze = 4π ʃ ρ (r) exp (sin qr)r dr Zeq where q = k - k' = 1/ ħ (p - p' )

measure of momentum transfer p - p' in elastic scattering. | q | depends on the angle of scattering

|q | = 2p sin θ ħ 2

ρ(r) is charge density within nucleus and exponential is a phase factor over the volume.

Form factor F (q) is equal to Fourier transform of charge density. It is determined directly by scattering experiment, from the ratio σ(θ)/σM(θ). And  using the inverse Fourier transformation, ρ(r) is determined ( If large number of measurements for angles θ from expt.)

If not so possible, we have to assume a form of density distribution, and best fit with expt. data obtained by suitably adjusting parameters R1/2 and a, in the expression. A suitable particular form is

ρ(r) = ρo ____________________

I + exp {(r - R1/2)/ a)

This is called Fermi distribution, shown in Fig.

Fermi distribution of nuclear charge density

.

90% ----

<-----------R1/2-------->

10 % ----

<---- 4.4 a ------>

Nuclear Radius

Nucl

ear

Ch

arg

e

Densi

tySkin thickness

For r = R1/2 , ρ =ρo /2

where ρo is charge density at centre (r = 0).

Thus R1/2 is half radius.

Parameter a determines the skin-thickness of nucleus, in which ρ(r) falls from 0.9 to 0.1 ρo of nuclear surface. Hence t = 4.4 a. If we approximate above Eqn. for uniform charge distribution, then equivalentR = roA1/3

where ro = 1.32 x 10-15 m for A < 50

and ro = 1.21 x 10-15 m for A > 50

This confirms that nuclear matter is almost uniformly distributed within nuclear volume, if we assume mass and charge radii are equal for all nuclei a= 0.5 x 10-15 m = 0.5 fm.

Experimental data shows that for spherical nuclei mass with A > 15, charge distribution has a core of uniform density, surrounded by a skin thickness 2.3 fm. The radius of half the maximum density R1I2 = 1.07 A1/3 fm.

For 4 < A < 15 , no uniform core and density decreases steadily with increasing r. Also the charge density in the core region decreases as Z increases.

Nuclear Spin Protons & neutrons have intrinsic spin

angular momentum 1/2 ( unit of ħ ) each, like electrons.

In addition, nucleons also possess quantized orbital angular momenta about C.M. of nucleus, like electrons in atom.

Resultant angular momentum (I) of nucleus is vector sum of orbital angular momentum (L) and spin angular momentum (S) of nucleus :

I = L + S

Quantum mechanically, total orbital and spin angular momenta of the nucleus are given by pI

2 = I (I + 1) ħ2

pL2 = L (L + 1 ) ħ2

pS2 = S (S+ 1) ) ħ2

During measurement, it is largest component of angular momentum along the direction of applied electric or magnetic field which is determined. For the three cases mentioned above, these have magnitudes I, L and S in units of ħ2 .

Resultant spin angular momentum of nucleuis vector addition of spins of individual nucleons : S = Ʃ si

Similarly, resultant orbital angular momentum is L = Ʃ li

Since si = 1/2 , S can be either integral or half-integral, depending on whether number of nucleons A in the nucleus is even or odd.

Since Ij is integral (0, I, 2, etc.), L is integral or zero.

Thus total angular momentum I of nucleus can be either integral (for A even) or half odd integral (for A odd).

Total nuclear angular momentum I is called nuclear spin.

For ground state Nuclear spin I = 0 (for even Z, even N nuclei).

This shows the tendency of nucleons inside the nucleus, to form pairs with equal and oppositely aligned angular momenta, which cancel out in pairs for like nucleons.

Measured values of ground state spins of the nuclei are small integers or half odd integers, highest measured value being 9/2 which is small compared to sum of lj and si of all individual nucleons in the nucleus.

Majority of the nucleons of either type seems to form a core in which (even numbers of protons and neutrons are grouped in pairs of zero spin and orbital angular momenta so that the core itself has zero total angular momentum).

The few remaining nucleons outside the core determine the nuclear spin which is small number, Integral or half odd integral. .

Spins of excited states of nuclei are deduced from nuclear disintegration and nuclear reaction data.