Basic Nuclear Physics

Roppon Picha

created: November 2005updated: April 8, 2010

Dalton’s atoms (1808)

J.J. Thomson’s Experiment

cathode rays = electrons (1897)

Rutherford, Geiger, Marsden

22688 Ra → 222

86 Rn +α → α + 21884 Po

rate of alpha scattering at angle θ from nucleus of chargeZ :

R(θ) ∝(

Ze2

mαv2α

)2 1sin4(θ/2)

Electron configuration

Rutherford model (1911): Electrons orbit the nucleus likeplanets orbit the Sun.

Bohr model of the atom (1913): Electrons stay in theatom on special orbits (orbitals).

Experimentally verified by James Franck and GustavLudwig Hertz in 1914. Atoms only absorb certain“chunks” of energy.

Electron configuration

principal quantum number: n = 1,2,3, . . .

e− most strongly bound at n = 1.

example: sodium (Na) has 11 electrons. In ground state,2 electrons are in n = 1 level, 8 in n = 2, and 1 in n = 3.

Hydrogen

V (r) = − e2

4πε0

1r

En = −13.6n2 eV

(Bohr formula, 1913)

hydrogenic (1 electron, Ze nuclear charge):

En = −13.6Z 2

n2

Sub configurations

Besides n, we have orbital angular momentumquantum number l .

l = 0,1,2, . . . ,n − 1letters: s, p, d, f, g, h, . . .

Then, there is spin quantum number s.

Quantum angular momentum

total angular momentum quantum number j :

j = s + l

values jump in integer steps:

|l − s| ≤ j ≤ l + s

Quantum angular momentum

example:for the electron, s = 1/2. if l = 1, what are possible valuesof j?

s = 1/2 and l = 3?

What are all possible j values for electron in n = 4 level?

Proton (1919) was discovered by Rutherford.

α+ N → H + O

Protos = first

Chadwick’s Neutron Discovery

• Existence suggested since 1920 by Rutherford.• Finally found via experiments in 1932.

94Be5 +4

2 He2+2 −→ 12

6 C +10 n1

or (α, n) reaction

mass: neutron 939.6 MeV/c2 ≈ proton 938.3 MeV/c2

Neutron energy

Fast neutrons = high-energy neutrons. E > 1 eV.

Thermal neutrons = those with average thermal energycorresponding to room temperature (T = 300 K).

Eth =32

kBT ≈ 140

eV

where kB = 1.38× 10−23 J/K.

Energy and Velocity

For a nucleon of kinetic energy 15 MeV, the velocity canbe calculated via

T =12

mv2

v =

√2Tm

≈ c

√2 · 15938

≈ 0.18c

de Broglie wavelength of this nucleon is

λ =h

mv=

4.1× 10−21 MeV s938MeV c−2 · 0.18c

≈ 7.3 fm

Accelerated Charge

EM radiation

Electric field far away does not know of particle’smovement.

The electric field form a wavefront consisting radial(Coulomb) and transverse components.

radiated power = P =q2a2

6πε0c3 Larmor’s equation

Electromagnetic Spectrum

p++

Region of Stability

Binding Energy per Nucleon

Binding energy

binding energy of most nuclei ∼ 8 MeV/nucleon

electrons are bound at ∼ 10 eV to atoms.

Separation Energy

removing a proton:

AZ XN −→ A−1

Z−1YN

removing a neutron:

AZ XN −→ A−1

Z YN−1

Separation energy (S) is the difference between bindingenergies (B) of initial nucleus and final nucleus.

Separation Energy

S > 0 when we change a stable nucleus (high B) into aless stable nucleus (low B).

B = (∑

mconstituents −matom)c2

S ≡ Bi − Bf

Sp = B(AZ XN)− B(A−1

Z−1YN)

Sn = B(AZ XN)− B(A−1

Z YN−1)

Ionization vs. Separation

Quantum behaviors

Subatomic particles can be described by quantummechanics.

States are represented by wave function ψ(x , t).

Particles = Wave packets = superpositions of waves.

Wave functions

Wave = non-localized state.

∆x ·∆p > ~

(Heisenberg uncertainty relation)

To get the wave function and its evolution, solveSchrodinger’s equation:

i~∂ψ

∂t=

(− ~2

2m+ V

)ψ

Wave function

Normalization: ∫ ∞

−∞|ψ(x , t)|2 dx = 1

At any given time, the particle has to be somewhere.

expectation values:

〈x〉 =

∫ψ∗(x)ψ dx

〈p〉 =

∫ψ∗(p)ψ dx

Wave properties

de Broglie wavelength of a (non-zero mass) particle ofmomentum p

λ =hp

Experimental verification: Davisson and Germer (1954).

Davisson and Germer used 54-eV electron beam toscatter of a nickel crystal. An interference peak wasobserved, similar to Bragg peak in x-ray diffraction.

Photons

∼ 1900: Blackbody radiation study led Planck to thinkabout nature of electromagnetic energy.

1905: Einstein proposed that light consists of photons,each possessing a certain lump of energy.

Total energy = multiples of this number.

Energy

Planck-Einstein relation gives energy of a photon:

E = hν = ~ω =hcλ

ν and ω are frequency and angular frequency,respectively.

Energy

h = 6.63× 10−34 J s = 4.14 eV s

for λ given in angstrom:

E =12.4λ

keV

Characteristic radiation of atoms which has only certainvalues are due to the fact that the atoms only exist incertain stable states of discrete energies.

Photon interactions

excitation (and de-excitation)

hν + Am ↔ An

ionization (and recombination)

hν + A ↔ A+ + e−

Fermions and Bosons

Protons, neutrons, and electrons belong to the fermionfamily.

Quarks and leptons are also fermions.

They have odd half-integer spins: s = 1/2,3/2,5/2, . . ..

Bosons have integer spin: s = 0,1,2, . . ..examples: photons (s = ±1) and 4He atoms (s = 0)

Periodic table

Electrons are identical fermions. At a given orbital(n, l ,m), only two electrons can occupy the same state(one spin-up, one spin-down)

For each l , there are 2l + 1 values of ml . For each (l ,ml ,there is two spin states (ms = ±1

2 ).

Exercise: What are maximum number of electrons forl = 0,1,2,3?

Periodic table shows an integer increase of protons andelectrons. Shells are filled, from low to high energies.

Ground-state configs:

• H: (1s1)

• He: (1s2)

• Li: (He)(2s1)

• Be: (He)(2s2)

• B: (He)(2s2)(2p1)

• . . .

information about a radioisotope.

Decay Law

dN(t)dt

= −λN(t)

t is time. N(t) is number of nuclei. λ is decay constant.solution:

N(t) = N0e−λt

N0 = number of nuclei at the starting time.decay constant is inversely proportional to the half-life:

λ =ln 2t1/2

A parent nuclide decays and yields a daughter nuclide.

increase in number of daughter (D) = decrease in numberof parents (P)

Df − Di = Pi − Pf

Decay constant

Decays aren’t always 1-to-1:

A → B (55% of the time)→ C (40%)→ D (5%)

For branched decays, the total decay constant is just thesum of each mode constant:

λtot = λ1 + λ2 + λ3 + . . .

Lifetime

For a given decay constant λ, the lifetime of the state is

τ =1λ

It is the time taken the state to drop from N0 toN0/e ≈ 0.37N0.

branched decays:

τ =1

λ1 + λ2 + . . .

Activity

A ≡ −dNdt

= λN = −λN0e−λt = A0e−λt

A is also called “decay rate” or “disintegration rate.”

units: becquerel (1 s−1) or curie (3.7× 1010 s−1)

Mysterious rays

Henri becquerel discovered radioactivity from uranium orein 1896.

At Cambridge, Rutherford studied these unknown raysand published results in 1899.

Those that got absorbed by a sheet of paper or a few cmof air was named alpha rays.

The more penetrating ones were called beta rays.

Alpha Decay

Alpha (α) = 2p&2n bound state

Process:AZ XN −→ A−4

Z−2YN−2 + 42He2

Examples:

22688 Ra138 → 222

86 Rn136 + α23892 U146 → 234

90 Th144 + α

mX c2 = (mY c2 + TY ) + (mαc2 + Tα)

Q ≡ (mi −mf )c2 = (mX −mY −mα)c2

Alpha emitters with large Q tend to have short half-lives.

lnλ(E) = a− bZ√E

Geiger-Nuttall law. λ is the decay constant; a and b areconstants; Z is the atomic number; E is the decay energy.

Beta Decay

W. Pauli: There must be a neutrino. (1930)Cowan and Reines observed it. (1956)

Beta Decay

Processes:

n → p + e− + ν̄e β− decayp → n + e+ + νe β+ decay (rare)

p + e− → n + νe e capture (ε)

Examples:

23490 Th144 → 234

91 Pa143 + e− + ν̄e53m27 Co → 53

26Fe + e+ + νe15O + e− → 15N + νe

X-ray

Charged particles that decelerate create electromagneticradiation. This process is known as bremsstrahlung.

Photons can excite or ionize atoms.

Subsequent atomic transitions can produce additionalX-ray photons. This process is called X-rayfluorescence.

If an atomic electron absorbs such X-ray photon, it can beejected. These electrons are called Auger (oh-zhay)electrons.

Gamma Decay

A year after Rutherford discovered α and β rays, PaulVillard discovered a more penetrating radiation fromradium. This is the gamma (γ) ray.

Excited nuclear states can decay via γ emission. Typicalenergies ∼ 0.1− 10 MeV.

Examples:

99m43 Tc → 99

43Tc + γ isomeric transition6027Co → 60

28Ni + e− + ν̄e + γ with β−

Internal conversion

An excited nucleus can interact with an orbital electron,transferring energy Eex .

The electron gets ejected with energy

Ee = Eex − Eb

where Eb is the binding energy of the electron.

The gamma decay and internal conversion decaycontribute to total decay probability:

λ = λγ + λe

Radiation Units

quantity description units

activity (A) decay rate curie (Ci), becquerel (Bq)exposure (X ) air ionization roentgen (R), coulomb/kgabsorbed dose (D) absorbed energy rad, gray (Gy)dose equivalent (DE) bio. effects rem, sievert (Sv)

Quiz

1. What kind of radiation does not come from anucleus? [choices: α, β, x-ray, γ]

2. Be-7 decays by capturing an electron. What is theresulting nuclide?

3. 15.1% of natural samarium is 147Sm, which decays byemitting α. 10 grams of natural samarium gives 120 αper second. Calculate activity per gram of 147Sm.

Reaction Cross Section

for reactiona + X −→ Y + b

σ =reaction rate

fluxincident · densitytarget

=rate of detecting b

(flux of a) · (X areal density)

Nuclear Reactions: First reaction in lab

Creating new nuclides

making light radionuclides:

14N + n →14 C +1 H55Mn +2 H →55 Fe + 2n

59Co + n →60 Co + γ

making Np-239 (transuranic)

238U + n →239 U239U →239 Np + e− + ν̄e

Balancing nuclear equations

What is x in each of these nuclear reactions?

19779 Au +12

6 C → 20685 At + x

3216S +4 He → x + γ

2713Al + p → x + n

4He +177 N → x +1 H

EM interactions

Main processes:

Photoelectric absorption

Compton scattering

Pair production

Intensity attenuation:

I(x) = I(0)e−µx

half-value layer = thickness that reduces intensity by 50%.

Producing radionuclides

Ways to do it:

• Reactors• Accelerators• Generators

Reactors

AX + n → →

Longer irradiation time → higher specific activity.

Examples:

13051 Te + n → →

63Li + n → α+ t

as fission products:

8536Kr, 133

54 Xe, 9038Sr, 99

42Mo, 13755 Cs

Accelerators

Usual projectiles: p, d , α

Examples:

2010Ne(d , α) 18

9 F7634Se(p,n) 76

35Br3517Cl(α, n) 38

19K

Generators

Suppose you want to use a short-lived nuclide producedfrom a reactor. But you are far away from the reactor.What can you do?

Prepare the parent nuclide which has longer half-life, in adevice that can separate the daughter from the parent.

Examples:4422Ti (t1/2 = 6 y) ⇒ 44

21Sc (t1/2 = 3.9 h)8337Rb (86 d) ⇒ 83m

36 Kr (1.8 h)9942Mo (66 h) ⇒ 99m

43 Tc (6 h)

the End