INTRODUCTION TO INERTIAL CONFINEMENT FUSION · INTRODUCTION TO . INERTIAL CONFINEMENT FUSION . R....

INTRODUCTION TO INERTIAL CONFINEMENT FUSION

R. Betti

Lecture 2

Binding Energy Cross sections Reaction rates

Binding Energy Curve

Binding energy curve

Why are nuclear reactions more easily initiated for heavy elements (fission) or light elements (fusion)?

BEA

A

More binding energy. Energy release

More binding Energy. Energy release

56

26Fe56

( ) 2B n p AE Nm Zm m c= + −Binding energy

Binding energy per nucleon /BE A

Shape of binding energy curve

The shape of the curve results from the competition between weak long-range Coulomb forces (repulsive) and strong short-range nuclear forces (attractive)

Long range (weak and repulsive) Coulomb force between two protons

Forces acting on nucleons

2( )

204

C eFrπε

=

Short range (strong and attractive) nuclear force between two nucleons

( )4

N NkFr

= −

Model: Calculate Net Force on a charged particle at the surface of a nucleus - Coulomb

Can be evaluated analytically. Here d is the distance from the point on the surface to an arbitrary point inside the nucleus

Sum i over all nucleons Approximate sum with volume integral

a.u.

A

A1/3

|FC|

Model: Calculate Net Force on a charged particle at the surface of a nucleus – Strong Force

A

a.u.

|FN|

Coulomb force

d4

As more nucleons are added, the nucleus becomes large enough that new nucleons do not feel the attractive force from distant nucleons because of the nuclear force’s short range, but still feel the repulsive force.

A

Ftot a.u.

tot N CTotal Attractive Force F = F - F

Conclusion: The shape of the binding energy curve is determined by the strong nuclear force which increases for small nuclei, but eventually saturates for large nuclei and cannot compete with the continually increasing repulsive coulomb force. As a result, the binding energy (per nucleon) is weak for very light and very heavy elements

Cross Sections and

Reaction Rates

Cross section

A=cross sectional area

dV= volume

Targets (particle 1) Cross section of each target ~ nuclear area

2Rσ πTarget particle density: 1n

particle 2 Total targets in volume:

1 1 1dN n dV n Adx= =

dx

Total target cross sectional area: 1dN σProbability P of reaction with Particle 2:

1 11

dN n A dxdP n dxA Aσ σ σ= = =

Probability per unit length

1dP ndx

σ=

Denote with Γ the flux of particles 2: Γ= number of particles 2 per unit area and per unit time

Decrease in Γ due to reactions with particles 1 is

1d dP n dxσΓ = −Γ = − ΓSolution: exponential decay of Γ

0 exp( / )mx λΓ = Γ − 11/m nλ σ=

λm is the mean free path for that reaction

Collision frequency is the time between reactions/collisions

1c

c

ντ

=1

1v vm

c nλτ

σ= = v velocity=

Reaction rate • R12= number of fusion collisions/reactions per unit volume, per unit time • Assume particles 1 are at rest: v1=0

• The number of particles 2 flowing through the area dA over the time dt is dN2 =n2v2dAdt

• The number of reactions over that time and that volume is

2 1 2 2 1 2 2v vdN dP n n dAdxdt n n dVdtσ σ= =

• In the time dt the flux of particles 2 has travelled the distance dx=v2dt

212 1 2 2vdN dPR n n

dVdtσ= =

• Generalize to moving particles 1

12 1 2 1 2v vR n n σ= −

• Generalize to particle populations with distribution function in velocity f(r,v,t)

1 1 1 1f ( , , )n d t⇒ ∫ v r v

2 2 2 2f ( , , )n d t⇒ ∫ v r v

( )1 2 1 1 2 2 1 2 1 2

1 2

f ( , , )f ( , , )v

d d t t

n n

σσ

− −≡ ∫ v v r v r v v v v v

• Note that I have assumed that the cross section depends only on relative velocity

( )1 2σ σ= −v v

( )12 1 2 1 1 2 2 1 2 1 2f ( , , )f ( , , )R d d t t σ= − −∫ v v r v r v v v v v

• Reactivity (definition)

• In a collisional plasma with time scales t>>τcollision (Coulomb collision time) the distribution function is Maxwellian

( )23/2 v

2, 2

j

j

mTj

j jj

mf n e

Tπ

− =

r v

• Maximum reaction rate

12 1 2 vR n n σ=

1 in kn= 2 (1 ) in k n= −1 2 ionsn n n+ =

212 (1 ) viR k k n σ= − maximum for k=1/2

2max

12 v4inR σ=

( )1/2 3/2

2 0

8 1Dm T

DT D TD

R n n e dT m

µεµ σ ε ε επ

−∞ = ∫

D T

D T

m mm m

µ =+

( )σ ε DT fusion cross section using D as bombarding particles on T target

ε is the deuteron energy used in the cross section measurements

reduced mass

Simplified form DT reaction rate (homework)

• Assumes each particle acts like a billiard ball with radius R

• R∼A1/3 r0 where r0 is the radius of a nucleon

• Use r0=1.7fm (1fm=10-15m) and R ≈3r0≈5 ×10-15m (example) leading to σ≈πR2≈0.8×10-28m2≈0.8 barn (1barn= 10-28m2) • The unit 1 barn is the typical hard-sphere cross section of nuclei

Hard sphere cross section

Classical cross section

r

Coulomb force dominate. Force ~ 1/r2 Coulomb potential ~ 1/r

Nuclear well Nuclear force dominate

Coulomb barrier Vb

R 1/3

0R A r

Necessary conditions for DT fusion based on classical physics: Kinetic energy in center of mass frame> Coulomb barrier

22 2

0

1 12 2 4 ( )CM D rD T rT

D T

eK m v m vR Rπε

= + ≥+

v vTrD

D T

mm m

≡+

v vDrT

D T

mm m

≡+

v D T= −v v

22

0

1 v2 4CM

eKR

µπε

≡ ≥ D T

D T

m mm m

µ ≡+

Substitute R≈5×10-15m 288CMK keV≥

( )1/3 1/303 2R r= +

120

19

8.85 10 /

1.6 10

F me Cε −

−

= ×

= ×

Classical physics fails at nuclear scales. Need to include quantum effects

• Wavelike behavior: At the nuclear scale, nuclear particles exhibit both particle and wavelike behavior. Waves can penetrate – for a short distance- even a perfect reflector.

• Particles can penetrate the Coulomb barrier even if their kinetic energy is below 288keV. The further below the barrier, the lower the probability of tunneling through barrier cross section decreases.

• Short interactions at high energies: as the particle kinetic energy increases, two particles have less time to interact. If the relative velocity is very large (KCM>>288keV), the time available for a reaction is short thus decreasing the probability of interaction cross section decreases.

• Resonances: wavelike behavior can lead to resonances for certain

conditions of geometry and velocity. The net result is an increase in reaction probability corresponding to an increase in cross section

Interesting facts on fusion cross sections

• Fusion cross sections depends on particle center-of-mass energy E

• Fusion reactivity depends on plasma temperature (average over fMaxw ) • DT fusion cross section has a maximum of about 5 b at 120keV

• For comparison, fission cross section of U235 is 600 b at 0.025eV • at 20keV due to a resonance in DT

• DT reactivity is maximum at T≈64 keV

• At T=10keV

( ) 100 ( )DT DDE Eσ σ

( )DD Eσ

( )v Tσ

( )DT Eσ

100DT DD

v vσ σ

Fusion cross sections

DT Max at 120keV

Fusion reactivities

( )1/2 3/2

2 0

8 1Dm T

DT D TD

R n n e dT m

µεµ σ ε ε επ

−∞ = ∫

Plot for T=10keV

Reactivity for DT fusion

Max at 64keV