9 – nuclear reaction ratesfaculty.tamuc.edu/cbertulani/ast/lectures/lec9.pdf9.1 – nuclear...

introduc)ontoAstrophysics,C.Bertulani,TexasA&M-Commerce 1

9 – Nuclear reaction rates



9.1 – Nuclear decays: Neutron decay By increasing the neutron number, the gain in binding energy of a nucleus due to the volume term is exceeded by the loss due to the growing asymmetry term. After that, no more neutrons can be bound, the neutron drip line is reached beyond which, neutron decay occurs:

Neutron drip line: Sn = 0

(Z, N) à (Z, N - 1) + n

Q-value: Qn = m(Z, N) - m(Z, N-1) - mn

Beyond the drip line Sn < 0

Neutron Separation Energy Sn Sn(Z, N) = m(Z, N-1) + mn - m(Z, N) = -Qn for n-decay

neutron unbound

(9.1)(9.2)

(9.3)

Gamma-decay •  When a reaction places a nucleus in an excited state, it drops to the lowest

energy through gamma emission •  Excited states and decays in nuclei are similar to atoms: (a) they conserve

angular momentum and parity, (b) the photon has spin = 1 and parity = -1 •  For orbital with angular momentum L, parity is given by P= (-1)L •  To first order, the photon carries away the energy from an electric dipole

transition in nuclei. But in nuclei it is easier to have higher order terms in than in atoms.


Proton decay

Proton drip line: Sp = 0

beyond the drip line: Sp < 0

Proton Separation Energy Sp: Sp(Z,N) = m(Z-1,N) + mp - m(Z,N)

the nuclei are proton unbound

α-decay is the emission of an a particle ( = 4He nucleus)

Qα = m(Z,A)−m(Z− 2,A− 4)−mα"# $%c

2The Q-value for a decay is given by

Nuclear lifetimes for decay through all the modes described so far depend not only on the Q-values, but also on the conservation of other physical quantities, such as angular momentum. Typical decay times for similar lifetimes are (a) 10-10 s for beta decay, (b) 10-16 s for gamma-decay and (c) up to many years for alpha-decay.

α-decay

Decay lifetimes

(9.4)

(9.5)


Fission Very heavy nuclei can fission into two parts (Q > 0) This happens because for large nuclei the surface energy is less important – large deformations are then possible. With a small amount of additional energy the nucleus can be deformed sufficiently so that Coulomb repulsion wins over nucleon-nucleon attraction and the nucleus fissions. This will be possible if

Qf = m(Z,A)− 2m(Z / 2,A / 2)"# $%c2 > 0

Using the LDM formula, Eq. (7.5) we can shown that Qα > 0 this happens for A > 90.

Nuc le a rC ha rg e Yie ld in Fissio n o f 234U

2 5 3 0 3 5 4 0 4 5 5 0 5 5 6 0 6 5

8 0 1 0 0 1 2 0 1 4 0 1 6 0 Ma ssNum b e rA

Pro to n Num b e rZ

0

5

1 0

1 5

2 0

YieldY(Z)

(%)

Example from Moller et al., Nature 409 (2001) 485

Symmetric fission (fission into two nearly equal size fragments) as well as asymmetric fission are likely to occur.

(9.6)


1.  Most nuclei found in nature are the stable and not too heavy to undergo alpha decay or fission)

2. There are many more unstable nuclei that can exist for short times after their creation. 3. Alpha particle is favored building blocks of nuclei because of their comparably low Coulomb barrier when fused with other nuclei high binding energy per nucleon 4. The binding energy per nucleon has a maximum in the Iron-Nickel region

This explains abundance peaks at nuclei composed of multiples of alpha particles in solar system abundance distribution

9.2 - Observations

The iron peak in solar abundance distribution implies that some fraction of matter was brought into equilibrium, probably during supernovae explosions.


Observations – Solar abundances


9.3 - Nuclear reactions in stars The plasma within a star is a mix of fully ionized projectiles and target nuclei at a temperature T. Using the definition of cross section in Eq. (7.24), we get that for particles with a given relative velocity v and within a volume V with projectile number density ni, and target number density nj:

λ =σ nivR =σ nivn jV

so for reaction rate per second and cm3 is r = nin jσ v

This is proportional to the number of i-j pairs in the volume. However, when the species are identical, i.i., i =j, one has to divide by 2 to avoid double counting

r = 11+δij

nin jσ v

(9.7)

(9.8)

(9.9)


Reaction rates at temperature T Assuming Local Thermal Equilibrium LTE the velocity distribution of particles in a plasma follow a Maxwell-Boltzmann velocity distribution:

where Φ(v) is the probability to find a particle with a velocity between v and v + dv.

kT2v

22/3 2

e2

4)(m

vkTmv

−

⎟⎟⎠

⎞⎜⎜⎝

⎛=Φ

ππ ∫ =Φ 1)( dvvwith

max at E = kT

This distribution is easily deduced from the Boltzmann distribution f(E) = A exp(-E/kT) by replacing E = mv2/2 and adjusting the constant A to conform with the normalization condition given by Eq. (9.11).

(9.10) (9.11)


r = 11+δij

nin j σ (v)Φ(v)vdv∫

kT2v

22/3 2

e2

4)(µ

πµ

π−

⎟⎟⎠

⎞⎜⎜⎝

⎛=Φ v

kTv

In terms of the Maxwell-Boltzmann distribution, with the mass m replaced by the reduced mass m of the i and j particles, i.e.,

µ =mimj

mi +m j

In the stellar environments, the reaction rates given in Eq. (9.9) has to be averaged over velocities,

r = 11+δij

nin j <σ v >

using over the distribution Φ(v),

Reaction rates at temperature T

(9.12) (9.13)

(9.14)

(9.15)


λ =1

1+δijYiρNA <σ v >

r = 11+δij

YiYjρ2 NA

2 <σ v >

Sometimes, one uses another notation for reactions per second and target nucleus j, by dividing the above equation by YjρNA, i.e.,

One usually call this by the stellar reaction rate of a specific reaction

In terms of units of stellar reaction rate given by NA<σv> in cm3/s/g, the reaction rate per second and per cm3, is

Reaction rates in terms of abundances

(9.16)

(9.17)

As an application, lets assume that the species i and j are destructed by j capturing the projectile i in the form i + j à k. Then

dn jdt

= −n jλ = −n jYiρNA <σ v >

dnkdt

= nkλ

(9.18)


Since the dnj/dt is proportional to –nj, the solution of the equation is a decaying exponential. And since the species k is built from the destruction of of j, its time dependence should be proportional to [1 – nj(t)]. In other words,

n j(t) = n0 j e−λ t

nk (t) = n0k (1− e−λ t )

Reaction rates in terms of abundances

(9.19)

In terms of the abundances: Yj(t) =Y0j e

−λ t

Yk (t) =Y0k (1− e−λ t )

(9.20)

The lifetime for the destruction of j is

τ =1λ=

1YjρNA <σ v >

(9.21)

After a time equal to a few multiples of τ, the species j is almost completely destroyed while k is almost completely built up from i and j capture. One often uses the name “half-life”, τ1/2, instead of lifetime, where τ1/2 = τ ln2.


9.4 - Energy generation Lets consider the energy generation by nuclear reaction sin stars. As we know the energy released is given by the Q-values

⎟⎟⎠

⎞⎜⎜⎝

⎛−= ∑ ∑

i nuclei initial j nuclei final

2ji mmcQ

where m are the respective masses of the particles. The energy generated per gram and second by a reaction i + j is then

ε =rQρ=Q 11+δij

YiYjρNA2 <σ v >

We can now define the reaction flow as the abundance of nuclei converted in time T from species j to k via a specific reaction as

F = dtdYj

dt

!

"##

$

%&&

0

T

∫via specific reaction

= λ(t)Yj(t)dt0

T

∫

(9.22)

(9.23)

(9.24)


Branching Consider possibilities as in the figure on the side in terms of several possible reaction rates λi

The total destruction rate for the nucleus is

∑=i

iλλ

And its total lifetime is given by

∑==

iiλλ

τ11

The reaction flow branching into reaction i, bi is the fraction of destructive flow through reaction i. (or the fraction of nuclei destroyed via reaction i)

∑=

jj

iib λ

λ

j

k

m (9.25)

(9.26)

i

(9.27)


As we san in Lecture 8, the lifetime corresponds to a width in the excitation energy of the state according to Heisenberg

!=Δ⋅Δ tETherefore, a lifetime τ = Δt corresponds to a width Γ = E :

τ!

=Γ

The lifetime for the individual decay modes, or decay “channels” for γ,p,n, and α emission are usually given as partial widths

Γγ, Γp, Γn, and Γα with ∑Γ=Γ i

Branching

(9.28)

(9.29)

(9.30)(9.31)


9.5 – Reaction rates – Theory x experiment Hence, we have seen that in order to determine the elemental abundance and energy generation in stars, we need the lifetime due to several nuclear decays and also the cross sections as a function of energy (or velocity) to integrate over the Maxwell-Boltzmann distribution. Since there are often several possible decay and reaction mechanisms the experiments needed to determine these quantities in the lab are many and laborious. The cross sections are also frequently very small at the relevant astrophysical energies. The particles have to be accelerated in the lab to energies of the order of E ~ kT ~ 10 Mi K ~ 1 keV for the Sun, or E ~ kT ~ 1 Bi K ~ 100 keV for a massive star. Although these numbers might look large, these are very small energies for accelerator based experiments. Typical nuclear physics laboratories use cyclotrons and accelerate particles (e.g., protons) to 100 MeV (CERN accelerates particles to TeVs !). Experimental nuclear physics is very expensive. Somehow, one needs to rely on theory. A combination of theory and experiment is often the best choice.


Nuclear structure notation As we discussed in our QM classes, the nucleons in the nucleus are confined and can only have discrete energies. Therefore, the nucleus, like an atom, can be excited into discrete energy levels, also called by excited states. The figure below illustrates a generic situation

Excitation energy

ground state

0

1st excited state

2nd excited state

3rd excited state

0

2.13

4.45

5.03

3/2-

1/2-

5/2+

3/2-

Excitation energy (MeV)

Spin

Parity (+ or - )

Nuclear states depend on •  energy (mass) •  spin •  parity •  lifetimes of γ,p,n, and α emission


Nuclear Reaction Types

1. Fusion & Radiative capture reactions A + C à B + γ 2. Elastic collision : entrance channel=exit channel A + B à A + B: Q = 0 3. Inelastic collision (Q ≠ 0) A + B à A* + B (A* = excited state) A + B à A + B*, etc.. 4. Breakup reactions A(= a + b) + C à a + b + C 5. Transfer reactions A + B à C + D


Ex: Radiative capture

I. Direct capture

Neutron capture A + n -> B + γ

A+n

B

En Sn

γdirect transition into bound states

II. Resonant capture

Step 1: capture to an unbound state)

A+n

B

En Sn

Γ

Step 2: decay to a bound state

B

Γ

γ


is the interaction matrix element, and Pl(E) is the penetrability factor, i.e., the probability that the projectile reaches the target nucleus and interacts with it. It depends on the projectile angular momentum l and its kinetic energy E. Using one gets

9.6 - Theory For radiative capture reactions of the type a + A -> B + γ, QM tell us that we need to know the wavefunctions of the relative motion of the initial set of particles, i.e., a + A, and the wavefunctions of the final particle, i.e., the nucleus B. The energy field of the photon, Vγ, is the responsible for the capture process. The equation obtained for the cross section is

σ ∝πλa2 B Vγ a +A

2Pl (E)

where πλ2 is a geometrical factor (deBroglie wave length of projectile - “size” of projectile).

λ =hp=

h2mE

σ ∝1Ef H a +A

2Pl (E)

B Vγ a +A = ΨB*(r)V∫ γ

(r)Ψa+A(r)d3r

(9.32)

(9.33)

(9.34)

(9.35)


Angular momentum Incident particles can have orbital angular momentum L. Classically it is given by L = pb, where p is the momentum and b the impact parameter

p b

In quantum mechanics the angular momentum is quantized and can only have discrete values

!)1( += llLWhere l = 0

l = 1 l = 2

s-wave p-wave d-wave

…

And the parity of the relative motion wave function is given by (-1)l

The particle kinetic energies in stars are pretty low. Let us take E ~ 10 keV for a proton, then its momentum is about p/ħ = (2mE)1/2/ħ ~ 0.02 fm-1. The protons can only react with other protons if they come close to each within a distance of b ~ 1 fm. The angular momentum under these conditions is thus l ~ pb/ħ ~ 0.02 << 1. Thus, at astrophysical energies only very low values of l (mostly s-waves) are of relevance.

(9.36)


Ex: s-wave neutron capture For a neutron incident on a nucleus, there is not Coulomb repulsion and the potential looks like in the figure below

Vl=0

The change in potential from a free neutron outside the nucleus to the nuclear attraction as it penetrates the nucleus still causes a quantum mechanical reflection. So, the penetrability is not constant (or just one), but depends on the energy as The cross section is thus or

Incoming wave transmitted wave

Reflected wave

Potential

P ~ E

E1

∝σv1

∝σ

(9.37)

(9.38) (9.39)


Example: 7Li(n,γ)

~ 1/v Deviation from 1/v due to resonant contribution

Because then v1

∝σ >=<= vconst σσv (9.40)


9.6.1 - Penetrability for charged particles

ReZZVc2

21=

V

r R

Coulomb Barrier Vc

Pote

ntia

l

When two nuclei with charge Z1 e and Z2e approach each other they repel due to the Coulomb force. When tehy come within the range of nuclear forces the repulsion turns into a stron attraction. In terms of the potential energy V (recal that force F = - dV/dr).

)(2.1

fm][44.1MeV][ 3/1

23/1

1

2121

AAZZ

RZZVc +

≈=

or

For example, if we consider the reaction 12C(p,γ) (relevant in the CN cycle), then VC = 3 MeV. In start the typical particle energies when this reaction occurs is kT = 10 keV. This is much smaller than the Coulomb barrier, meaning that the particle has to tunnel (a pure QM effect) to get close together and fuse together.

At the top of the Coulomb barrier the potential is

(9.41)

(9.42)


T E( ) = exp − 2! dr 2µ V(r)−E"# $%0

D

∫"

#'

$

%(

Each transmission probability Ti through each barrier can be obtained using Eq. (B.53) of slides on QM. This way of calculating the total transmission probability is called the WKB approximation.

Tunneling through a generic barrier

Ti E( ) = exp − 2! Δ 2µ(V−E)#

$%

&

'(

T (E) =i∏ Ti E( )

(9.43)

(9.44)

(9.45)


Penetrability for charged particles

V

r R

Coulomb Barrier Vc

Pote

ntia

l

The penetrability through the Coulomb barrier can be calculated from QM by dividing the barrier into small step barriers (much in the way we do integrals) and multiplying the transmission probabilities obtained through each barrier using Eq. (B.53). One gets

where

This implies that Where S(E) contains the information of what happens when the particles are inside the barriers, i.e., when they interact via the strong force.

(9.46)

(9.47)

πη2e)( −∝EPl

!

221 e

2ZZ

Eµ

η =

σ ∝1Ef H a +A

2Pl (E) ~

1Ee−2πη (E)S(E) (9.48)


The penetrability Pl(E) for charged particles decrease very fast as a function of E. And so does the cross section. An example is shown in the figure for the 12C(p,γ) cross section. The dots are experimental data. Very small cross sections cannot be measured because the number of events (reactions) occurring per unit time is very small requiring a very long time for data taking (expensive, or even impossible).

Penetrability for charged particles

We need the cross section here !

introduc)ontoAstrophysics,C.Bertulani,TexasA&M-Commerce 2727 27 27 27

9.6.2 - Astrophysical S-factor Coulomb barrier for charged particles

Cross section has a steep energy dependence

Extrapolation of data to low astrophysics energies better done with astrophysical S-factor S(E)

Astrophysical S-factor S(E)

The S-factor can be •  easier graphed •  easier fitted and tabulated •  easier extrapolated •  and contains all the essential nuclear physics

σ ∝=1ES(E)exp −2π

Z1Z2e2

!v

#

$%%

&

'(((9.49)


Astrophysical S-factor - example Astrophysical S-factor for the reaction 4He + 3He à 7Li + γ.


Adding up the cross sections for all angular momenta (partial wave expansion), In stars we have to average over velocities:

9.6.3 - Gamow window

σ v ~ σ E( )E exp − EkT

"

#$%

&'dE∫

σ (E) = σ (E)

∑

kT

Note that the relevant cross section is in tail of MB distribution, which is much larger than kT. This is very different than n-capture which occurs at energies ~ kT. Nonetheless, one has always E0 << Ecoul (coulomb barrier) and only a small number of partial waves (l = 0 mainly) contribute.

(9.50)

(9.51)


Reaction rates - examples

Reaction T (109 K) E0 (MeV) Ecoul (MeV) σ(E0)/σ(Ecoul) d + p 0.015 0.006 0.3 10-4

3He + 3He 0.015 0.021 1.2 10-13

α + 12C 0.2 0.3 3 10-11

12C + 12C 1 2.4 7 10-10

E0 ~ 0.122 Z2i Z2

j A( )1/3

T92/3 MeV

A=AiA j

Ai +A j

ECoul ~ 1.2Zi Z j

Ai1/3 + Aj

1/3 MeV

1 MeV ~ 1010 K

T9 = 109 K

σ v ~ σ E( )E exp − EkT

"

#$%

&'dE∫

(9.52)

(9.53) (9.54)


By approximating the Gamow window by a Gaussian function centered at the energy E0, one can calculate the reaction rate in an approximate form where in cm3/(s.mole). One still needs to know (measure, calculate) the cross section, or S-factor, at the Gamow energy E0.

Reaction rate estimates

<σ v >= 169 3

1µ

12παZ1Z2

S(E0)3E0kT

!

"#

$

%&

2

E0 =bkT2

!

"#

$

%&

3/2

= 0.12204 Z12Z2

2A( )1/3T92/3 MeV

NA <σ v >= 7.83⋅109 Z1Z2

AT92

"

#$$

%

&''

1/3

S(E0 )[MeV barn] e−4.2487

Z12Z2

2AT9

"

#$$

%

&''

1/3

(9.55)

(9.56)

(9.57)


9.6.4 - Resonant Reactions

B

Er ΓSa

Step 1

B

γ

Step 2

In some reactions, the energy of the formed nucleus B might coincide with a quantum state that would be in the system if the nuclear confining well would be infinitely high (impenetrable walls). These are like excited states above the particle a separation energy from B, which we call Sa. Such states serve as resonances.


If the reaction occurs close to a resonance, the cross section contribution due to the resonance is given by the Breit-Wigner formula

σ (E) = π!2ωΓaΓb

(E−Er )2 + (Γ / 2)2

)12)(12(12

21 ++

+=

JJJrω

where ω is a statistical factor given by

Γa is the partial width for decay of resonance by emission of particle a = rate for formation of B nucleus. Γb is the partial width for decay of resonance by emission of particle b.

Γ = Γa + Γb is the total width is in the denominator as a large total width reduces the relative probabilities for formation and decay into specific channels.

Resonant Reactions

(9.58)

(9.59)


Assuming a narrow resonance, then one can carry out the integral for the reaction rate can be done analytically and one finds, using

NA <σ v >=1.54 ⋅1011(AT9 )−3/2ωγ [MeV]e

−11.605 Er [MeV]

T9 cm3

s mole

Γ

ΓΓ

++

+= 21

1 )12)(12(12

T

r

JJJ

ωγ

that the contribution of a single narrow resonance to the stellar reaction rate is given by

The rate is entirely determined by the “resonance strength” and this has to be determined experimentally for the angular momenta and partial widths for the decay of nucleus B by emission of particles a and b.

rr

EEE

dE <<ΓΓ

ΓΓ=

Γ+−

ΓΓ∫∞

for;)2/()(

210 22

21 π

Resonant Reactions

(9.60)

(9.61)

(9.62)


9.7 - Complications in stellar environment Beyond certain temperature and density, there are additional effects related to the extreme stellar environments that affect reaction rates. In particular, experimental laboratory reaction rates need a (theoretical) correction to obtain the stellar reaction rates.

The most important two effects are:

1. Thermally excited target

2. Electron screening

At high stellar temperatures photons can be absorbed and excite the target. Reactions on excited target nuclei can have different angular momentum and parity selection rules and have a somewhat different Q-value. In the laboratory it is nearly impossible to prepare targets under such conditions.

In a stellar environment, atoms are fully ionized, but the electron gas still shields the nucleus and affects the effective Coulomb barrier. In the laboratory, the reactions are also screened by the atomic electrons, but the screening effect is by bound electrons.


9.7.1 - Electron screening (in stars) In stars the particles are surrounded by free electrons. Once can use the Debye-Hueckel screening (Salpeter 1959). This approximation works only if

(20 % effect)

More positive ions

More electrons

Ion under consideration

RD Debye Radius

σ vplasma

= f (E) σ vbare

Veff =Ze2

re−r/RD

n RD3 >>1

7Be(p,γ )8B (T ~107K) : f (E) ~1.2

f (E) = eUe /kT ,

RD ~ kT / ρ ~ 0.218 A0

(Sun)

One then finds a correction of the form

(9.63)

(9.64)(9.65)

(9.66)


Debyedynamicsvv thermal /,/ 12 ==α

But there are problems: Gamow energy >> thermal energy (dynamical screening) à Debye sphere is not really spherical

Also, the number of particles within the Debye spehere is not so large to justify the mean-field approximation à Fluctuations in the Debye sphere.

•  One cannot derive the screening from thermodynamics but one has to resort to kinetic equations.

•  Deviations from Debye-Hueckel can be large.

n RD3 ~ 3− 5

Carraro, Schaefer, Koonin, ApJ 1988

Electron screening (in stars)

(9.67)


In the adiabatic model one assumes that the projectile acquires atomic binding energy as it penetrates the electron cloud of the target. The extra amount of energy is due to the different atomic bindings for A and B which can be taken from an atomic table. One then finds the extra ΔE

With this one can calculate the change in the cross section:

9.7.2 - Electron screening (in the laboratory)

€

ΔE = E '−E

σ labfusion ~σ bare(E +ΔE)

~ exp πη(E)ΔEE

"

#$%

&'σ bare(E)

(9.68)

(9.69)


Also, very enhanced electron screening has been found in solids, and liquids.

Czerski, Kasagi, 2010

Rolfs, 1995 But the amount of screening needed to explain the experimental data has been more than either the adiabatic model, or a more complex theoretical calculation (dynamic screeening) can explain.

Electron screening (in the laboratory)

9 – nuclear reaction ratesfaculty.tamuc.edu/cbertulani/ast/lectures/lec9.pdf9.1 – nuclear...

Documents