Nuclei summary

- 1 -

R

( )rV

0V−

r

Nuclei: Nucleus… radius ~1fm = m1510− Isotopes … same Z, different A Isobars … same A, different Z Z = number of protons A = number of protons + neutrons Proton = uud Neutron = ddu (quarks) Deuteron:

(1) … inside nucleus… r < R (2) … outside nucleus… r > R

Can write the Schrödinger equation:

( ) ( )( ) ( ) 02

22 =−+∇ rurVE

mru r

h

Thus, if spherical symmetry:

( ) ( )( ) ( ) 02

22

2

=−+ rurVEm

rudrd r

h rm = reduced mass

Assume no relative angular momentum

(1) r < R ( )

BEEVrV

−=−= 0 (binding energy)

∴ ( ) 02

022

2

=−+ uEVm

drud

Br

h

(2) r > R ( )

BEErV

−== 0

∴ ( ) 02

22

2

=− uEm

drud

Br

h

Nuclei summary

- 2 -

Now, the reduced mass can be simplified:

2n

nn

nn

np

npr

mmm

mmmm

mmm =

+≈

+=

Now, a solution to the TISE is of the form: ( ) ikrikr BeAeru −+= Thus, for (1) & (2):

( )Bn EVmk −= 011h

BnEmi

kh

=2

Now, the boundary conditions: ( ) 00 =u ( ) 0lim =

∞→ru

r

Hence, applying to (1): ( ) 11111 00 BABAu −=⇒=+= ∴ ( ) ( )rikrik eeAru 21

11−−=

( )rkiA 11 sin2= And to (2): 0=A ∴ ( ) rikeBru 2

22−=

Also, the wavefunction must be continuous at r = R:

⇒ ( ) ( )RuRu 21 = RR dr

dudrdu 21 =

⇒ ( ) RikeBRkiA 2

211 sin2 −= ( ) RikeBikRkkiA 222111 cos2 −=

Dividing the two expressions results: ( ) 211 cot ikRkk = We can measure BE by firing a photon at a deuteron, and the minimum energy which will split the deuteron is the binding energy:

Nuclei summary

- 3 -

nHD 10

11

21 +→+γ

Liquid drop model:

( ) ( )2

,,c

AZBNmZmAZM nH −+=

If the system is bound: ( ) 0, >AZB Know the shape of the B/A – A graph… which is approx constant on 8MeV, with a max at A = 56. Semi-Empirical Mass formula: Volume term: AaB VV =

Coulomb term: ( )

31

1

A

ZZaB CC

−−=

As 31

AR ∝ … repulsive… Z(Z-1) proton pairs

Surface term: 32

AaB SS −=

Surface area ( )22 3

1

AR ∝∝

Asymmetry term: A

AZ

aB aa

2

2

−

−=

Keeps difference between number of protons & neutrons small

Stability term: 21−

−= Aa pδ Describes if odd/even for A/Z So, in total, the binding energy can be written:

( ) ( )δ+

−

−−−

−=A

AZ

aAaA

ZZaAaAZB aSCV

2

21, 3

2

31

Thus, the mass can be written:

( ) ( )δ−

−

++−

+−+=A

AZ

aAaA

ZZaAaNmZmAZM aSCVnH

2

21, 3

2

31

Stable nuclei: Most bound nuclei for a given A must have a max. binding energy:

Nuclei summary

- 4 -

0=

∂∂

AZB

Thus:

02212

31 =

−

−−

−=

∂∂

A

AZ

aA

Za

ZB

aCA

⇒ A

AZ

aA

Za aC

−

=−

− 2212

31

Can solve this, with ZZ 212 ≈− … to get the most stable Z for a given A… 0Z . Particles will decay down to 0Z : 0ZZ < −β decay epn υβ ++→ −

0ZZ > +β decay enp νβ ++→ + Probing nuclear structure: In Rutherford scattering, alpha particles are incident upon thin gold foil. Alpha interacts with both strong & coulomb forces.

reZZ

V AuC

0

2

4πεα= … coulomb barrier

2cot θ∝b … whereθ is the scattering angle & b the impact parameter: the distance from the centre of the nucleus in which the particle impacts. Hence can infer nuclear size from this.

The energy of the incident alpha is αE . Thus, if CVE >α , the coulomb barrier is overcome and the alpha is absorbed. This energy can be found, and thus the radius of the nucleus found. This also implies however, that we cannot useα -particles to probe nuclear structure.

VC

Nuclei summary

- 5 -

Electrons are not affected by the strong nuclear force… which is the force which “sucks” the alpha particles in… so can use −e to probe nuclear structure.

pcEe = eE

hcph

==λ

Thus, the minimum distance which an electron can resolve is given by its de Broglie wavelength. To increase the resolution (thus decrease minimum distance scales) need to increase the incident electron-energy. Now, if this data is plotted, a straight line is found, and:

31

0 Arr = fmr 2.10 = Radioactive decays: Alpha… beta… fission… gamma Decay law:

Ndt

dNλ=− N = #radioactive nuclei

λ =decay constant Thus, solving gives: ( ) teNtN λ−= 0

Half life… time when ( )2

0NtN = :

λ

2ln

21 =t

Average time nuclei survives before decay:

λ

τ1

0

0 ==

∫

∫∞

∞

dtdtdN

dtdtdN

t

Activity… decays per second:

NdtdN

A λ=−= ⇒ NA λ=

Nuclei summary

- 6 -

3 main sources of natural activity: Primordial… before earth; Cosmogenic… cosmic ray interactions; Human origin… fires/fission/reactors etc… Dating: If a decay happens: DP → Then: ( ) ( ) ( )tNtNtN DPP +== 0

∴ ( ) ( ) ( ) ( )[ ] tDP

tpP etNtNetNtN λλ −− +=== 0

Thus, as every quantity can be measured, solve for t to date things! Carbon dating: C12 … stable C14 … unstable Thus, can say: ( ) ( ) teCNCN λ−= 14

014

( ) ( )CNCN 120

12 = Hence, the ratio can be measured:

( )( )

( )( )

teCNCN

CNCN λ−

Π

=43421

120

140

12

14

Where Π is assumed to be constant. This is only true up until ~1900, as fossil fuels & such like have messed up this ratio. Alpha decay: 321

α≡

++−− +→ HeYX A

ZAZ

42

42

Now, if X initially at rest, by energy conservation: αα EcMEcMcM YYX +++= 222 Thus, rearranging, and defining the Q-value: ( ) QcMMMTT YXY ≡−−=+ 2

αα

Nuclei summary

- 7 -

Now,α -decay is only possible for 0>Q . Also, note that: XY BBBQ −+= α binding energies… To find out how much KE the α has… i.e. αT ; consider momentum conservation: Ypp =α ⇒ 22

Ypp =α Kinetic energies are:

Y

YY m

pT

mp

T

2

22

2

=

=α

αα

⇒ YYY mTp

mTp

2

22

2

=

= ααα

Thus:

YY mY

mT αα=1 ⇒ αα T

mm

TY

Y =

α

ααα Tm

mTTTQ

Yy +=+=

Hence, rearranging:

Ymm

QT

αα

+=

1 the kinetic energy of the resulting α -particles.

MeVE 50≈α Fission:

YXA +→

Symmetric fission is of the form: XA 2→ Thus, the energy released from fission:

( )

−=

2,

22,2

ZAMZAM

cEF

Nuclei summary

- 8 -

0>FE for fission to happen, and if the SEMF is looked at, all but the surface & coulomb terms cancel; leaving:

( )

31

32 1

37.026.0A

ZZaAaE CSF

−+−=

Which corresponds to:

02

≥A

Z should spontaneously fissure

But it does not…

… this is due to the activation energy which is needed to be overcome as atom deforms from being spherical, to more ellipsoidal.

MeVEF 200≈