the united states nuclear regulatory commission and...

The United States Nuclear Regulatory Commission and Duke University

Present: Regulatory and Radiation Protection Issues in Radionuclide Therapy

Copyright 2008 Duke Radiation Safety and Duke University. All Rights Reserved.

Welcome!

� This is the ninth of a series of training

modules on radiation physics.

� These modules provide a basic introduction

to interaction of neutrons with matter.

� Sponsored by the United States Nuclear

Regulatory Commission and Duke University

� Author: Dr. Rathnayaka Gunasingha, PhD

Your Instructor

� Dr. Rathnayaka Gunasingha is an

Accelerator Physicist with

background in High Energy physics.

� Dr. Gunasingha is a physicist in the

Radiation Safety division and

member of the Faculty of the Duke

Medical Physics Graduate Program.

� Contact:

[email protected]

Goals of the Course

� Upon completing these instructional

modules, you should be able to:

� understand the Basic Interactions of Radiation

with Matter

� apply the knowledge in various calculations in

Medical and Health Physics

� understand the basic principles behind various

instrumentation used in Medical and Health

Physics

This Module Will Cover

� Elastic scattering of neutrons with nuclei

� Energy loss spectrum for neutron and proton

scattering

� Threshold energy for an inelastic scattering of

neutron

� Fission and neutron activation and,

� Criticality

Interaction of Neutrons with Matter

• Introduction:

1932: Neutron was discovered by Chadwick

1939: Nuclear fission, induced by the capture

of a slow neutron in 235U discovered

by Hahn and Strassman

1942: First man-made nuclear reactor : by

Fermi

Neutron Classification

• Neutrons are classified according to their energies:

1. Thermal: <E> = 0.025 eV at 20º C; upper limit ~0.4

eV (Cadmium cutoff)

2. Epithermal : just above 0.4 eV

3. Intermediate: 0.4eV – 10 keV ( Secondary

protons do not cause enough ionization)

4. Fast: above: 10 keV

5. High energy : above 20 MeV

Neutron Sources

• The major neutron sources are:

• Nuclear reactors:

neutrons from a few keV to several MeV

• Radio-isotopic sources:

(α, X) or (γ, X), such as Am-Be, Ra-Be, Pu-Be, Pu-

Li, Pu-B, Pu-F, Sb-Be

• Particle Accelerators:

Planned productions ( d-D, d-T, ..) or

unwanted background from beam losses

• Cosmic rays generate secondary neutrons

Interaction of Neutrons with Matter

• Neutrons are uncharged and hence can travel appreciable

distance in matter without interacting

• Like photons, neutrons do not interact with orbital electrons (

i.e. there is no Coulomb’s interaction )

Interaction Mechanism

• Neutrons can interact with atomic nucleus with several

mechanisms depending on their energy.

• Those are:

1. Elastic scattering

2. Inelastic scattering

3. Non-elastic scattering

4. Neutron Capture

5. Neutron Spallation

Elastic Scattering of neutrons

• This is the most important process for slowing down neutrons.

• Total kinetic energy is conserved in elastic scattering.

• In this process, energy lost by the neutron is transferred to the recoiling nucleus

• Maximum energy transfer is occurred with an head-on collision

• Energy of the recoiled nucleus depends on the recoiled angle φ of nucleus.


• Using the conservation laws of energy and linear momentum

• Energy transferred to the nucleus or Q value is given by

•

• En = neutron energy

• M = nucleus mass m = neutron mass

φ)cos(m)(M

mME4 Q

2

2

n

+=


• Maximum energy transfer to the recoiled nucleus occurs when

θ = 1800 and φ = 00

In neutron mass units, i.e

2

n

maxm)(M

mME4 Q

+=

m

M A =

2

n

max

A)1(

AE4 Q

+=


• Since the energy transfer to the nucleus is Qmax

The energy transferred to the neutron =

• Scattered Neutron energy

2

n

nmaxn A)1(

AE4 E - QE

+−=

2

n

2

A)1(

EA)1(

+−

=


• Following table gives the

maximum fraction of energy

lost ( ) by a neutron in

a single elastic collision

• Neutron looses all of energy

in a head-on collision with

H

Nucleus

H 1.0

2H 0.889

4He 0.640

9Be 0.360

12C 0.284

16O 0.221

n

max

E

Q

n

max

E

Q

Energy-Loss spectrum for neutron and

proton elastic scattering• proton and neutron, masses are

nearly equal m=M

In a CM system, the approaching velocity of each is

2

V

After the collision, they move back with the same velocity

In lab frame consider a neutron with velocity v, incident on a

nucleus of mass M at rest

mM m M

v

O′

uv - uu = ½ v

Q

P

P

Oωφ

θ

φ

Lab CM System


proton elastic scattering

• In the center of mass system the scattering is isotropic

• Since masses are equal the speed of CM system is

• If O is the collision point of p and n, center of mass is

O’ and proton is p, neutron is N

• The scattering angle of proton is φ in the lab system

and ω in the CM system.

• Using conservation of momentum, we can show that

each particle speed v/2 after the collision.

2

v

Energy-Loss spectrum for neutron

and proton elastic scattering

• From geometry, OO` = O`P =O`N

The scattering angle of proton is in the lab system and, in the CM

system. φ ω

φφω d2 dω 2 =⇒=

diagram, From

O′O

ωφ

P

N

v/2

v/2

v/2

dA

dω

C

RRsinω

ω

Energy-Loss spectrum for neutron

and proton elastic scattering

ω RdωsinR2dA

dωωω

R4

dAdA

2

) and (between

areaan scatter toproton ay that probabilit

π

π

=

+

=

dωωπR

RdωωπRdP sin

2

1

4

sin2)(

2==ωω

φφφ d+andbetweenangleantosctteringofyprobabilit

ωωφφ d)(Pd)(P =



φφφ

φφφφ

ωωφφ

d)cos()sin(2

d2)2sin(2

1d)(P

d)(Pd)(P

=

×=

=

dQQQ +→Probability that neutron loss energy

nn

n

2

n

1

E

dQdQ

sincosE2

cossin2dQ)Q(P

sincosE2d

dQcosEQ

dQd

dQdcossin2d)(PdQ)Q(P

==

==

==

−

Let

φφφφ

φφφ

φ

φφφφφφ



nEFigure shows the normalized spectrum for scattering of neutrons of energy by

protons.

since spectrum is flat. max

1 1

2 2avg n

Q Q E= =

and probability that a neutron losses an amount of energy n

E

Q∆=

En

1/En

Q



Example: What is the maximum energy that a 4 MeV neutron can transfer to a

nucleus in an elastic collision? B10

5

( ) ( )

66.0Q2

1Q

32.1121

160

11

4104E

A1

A4E

m4M

mM4Q

max

2n2n2max

==

==××

=+

==

average



( ) n2E

A1

A4

+

( )

2

nn2n

A1

A1EE

A1

A4EE

+−

=+

−=′

•Since energy transfer to the nucleus is

The energy of the scattered neutron is

• For scattering of protons nEE0 <<

Elastic Scattering in heavy nuclei

Example: For Lead, M=207 m=1

( ) nn2maxE019.0E

mM

Mm4Q =

+=

Only 1.9% energy is lost in each collision

• Lead is not good for neutron shielding (except at high energy)

Elastic Scattering in heavy nuclei

• For a light material :

Example: polyethylene

1m)(M

Mm4H

1 m1 M H12M

2=

+

==→=

for

•In a single collision, all neutron energy can be lost.

•Therefore, Hydrogenous material is best for neutron shielding

Inelastic Scattering

• The neutron is absorbed and then re-emitted

• Some energy is absorbed to the nucleus and left nucleus in the excited state

• Nucleus de-excited by emission of gamma rays or other particles

• In general, A(n, n’)A*, A(n, 2n’)B

• Example: 14N(n, n’ ) 14N* , Eg = 10 MeV

• Inelastic scattering occurs above a threshold energy. ( Eth ~ 1 MeV ). Only elastic scattering allowed below Eth

Threshold Energy for the Reaction

• Consider a particle of mass m1 collides with a particle of mass m2, initially at rest. After the reaction, the identity of the particle changes and their masses are m3 and m4

as shown

Q value of the reaction

Q = (m1 + m2) – (m3 + m4)

m1 m2

θ

φ

m4

m3

Threshold Energy

Total energy conservation:

E1 = E3 + E4 – Q (1)

Linear momentum conservation:

p1 = p3 cos θ + p4 cos φ (2)

P3 sin θ = p4 sin φ (3)

(2) & (3) Gives

(4)

→

↑

2

331

2

1

2

4 pθ cospp2pp +−=

Threshold Energy

(5)

Using , and (4) in (5) 111Em2p =

4

2

4

4

m2

pE =

333Em2p =

4

33113311

4m

EmEmθcos2EmEmE

−+=

Threshold Energy

0cEb2E33

=+−

0)m(m

Qm)Em(mE

)m(m

θcosmEm2E

43

4141

3

43

311

3=

+−−

++

−

3EinequationquadraticaisThis

)m(m

Qm)Em(mcand

)m(m

θcosmEmb

43

4141

43

311

+−−

=+

=

(1)inEforExpression 4

QEEE413

+−=

Threshold Energy

4331

2

441

2

31

344

1

mmmmmmmθcosmm

)Qm(mmE

+−+−+−

≥

0c4b42 ≥−solutions realFor

for i.e. collision on head aFor 0θ =

314

34

1

mmm

)Qm(mE

+−+−

≥

Threshold Energy

• The smallest possible for E1 is the threshold energy of the

incident particle.

• When m1 is small compared to the (m3+m4) this value is equal

to the Q value of the reaction

314

34

thmmm

)Qm(mE

+−+−

=

Neutron Capture

• Non-Elastic Scattering ( Spallation)

In this process, secondary particle is not a neutron after the

capture of initial neutron

• Neutron Capture:

In this process, neutron disappears in the capture process.

Neutron capture occurs only at low energy, mostly for thermal

neutrons ( below Cd absorption edge 0.4 eV )

B)A(n,α

Neutron Capture

• The capture of a neutron with a proton in the tissue is the

major contributor to dose in tissue from thermal neutrons

MeV2.2EMeV2.2QH),n(p 2 == γγ

• Probability of neutron capture is proportional to E

1

v

1or

Neutron Capture

• Some significant capture processes are

( )( )

) (0.765

4.8

2.3

in tissue 0.63

shieldingin

6

10

14

MeVpHnHe

MeVHeHnLi

MeVHeLinB

MeVpCnN

MeV2.2EHnp

33

43

47

14

2

+→+

+→+

+→+

+→+

=+→+ γγ

• Cross sections are very high for last three processes.

Neutron Activation

Induction of radioactivity in materials, by bombardment

with neutrons is called "neutron activation"

This is a useful property of neutrons. In practice, it is used

to determine the concentratio

•

•

n of elements in a large amount

of materials ( Neutron Activation Analysis (NAA))

Capture of free neutrons by the target can create a new isotope.•

Neutron Activation

Na23

11

e

24

12

24

11

23

11eMgNaNan υ++→→+ −

Na24

11

•For example, becomes β− active after the absorption of a n.

If the depletion of the target is negligible, rate of production p of is constant

•As soon as, is produced, it is subject to decay, if is the decay constant

and N is the number of atoms at any time t,

Na24

11 λ

Npdt

dNλ−=

Neutron Activation

T

-2 1

Assume we have a target with N atoms and it is exposed to

a beam of neutrons with a fluence rate ( . sec ).

The capture cross section is

New material ( daughter ) is present

The ra

n cmφ

σ

−

•

•

•

�

Tte of production of daughter = N

Decay constant of daughter is , and if is the number of

daughter atoms at time

n

t

φσ

λ•

�

Neutron Activation

T

1 2

Decay constant of daughter = n

Net rate of change of daughter, (1)

Assume N and are constants.

General solutions,

T

t

dnN n

dt

n x x e λ

λ

φσ λ

φ

−

•

• = −

•

• = +

�

�

Neutron Activation

1

2

2

Substitute in (1)

Then,

at 0, 0

T

tT

T

Nx

Nn x e

Nt n x

λ

φσλ

φσλ

φσλ

−

• ⇒ =

= +

= = ⇒ = −

�

�

�

Neutron Activation

( )

( )

activity" saturation" thecalled is This

daughter theofActivity

Solution

T

t

T

tT

Nnt

e1Nn

e1N

n

σφλ

σφλλ

σφ

λ

λ

�

�

�

=∞→•

−=•

−=•

−

−

Neutron Activation

Induced activity vs. time t of irradiation is shown in

the plot below.

Nλ

t

λN

φσNT

0

.

Fission• Break up of heavy nucleus by the absorption of

neutrons is called Fission.

• Nuclei with odd numbered nucleons fissions more rapidly, since binding energy of those is less than that of even numbered nuclei.

• Fissionable nuclei can break up in a number of different modes. Some are charged fragments, neutrons, gammas, neutrinos.

Fission

• An example for a fission is an absorption of a thermal

neutron by

• Total energy released per fission in this reaction is

195 MeV. Major share of this energy is carried out by

charged fragments La and Br.

n 2BrLa U n 1

0

87

35

147

57

235

92

1

0++→+

U235

92

Fission

U235

• Distribution of energy

Average number of neutrons produced by a single fission of

is 2.5

•Activity of collective fission products is

daysin fission

curies tt

10A

2.1

16−

≈

Criticality

i

1i

effN

Nk +=

• An assembly of fissionable material is said to be critical, on average, if exactly

one of the several neutrons released in the fission, causes another fission process.

• Criticality depends on : geometrical factors and the materials

• Effective multiplication factor

• Ni = number of thermal neutrons in one generation and

Ni+1 is the number of next generation neutron

increases)(output calsupercriti

lsubcritica

critical

1K

1K

1K

eff

eff

eff

>

<

=

Criticality

∞K

∞= LKKeff

• Supercritical� more than one fission neutron produces fission of another nucleus

Assume = infinite multiplication factor

(independent of shape and size of assembly

Where L= probability that a neutron will not escape

Assume Ni = total thermal neutrons in its generation

f = fraction absorbed in the fissionable fuel ( )

h = average number of fission neutrons per

thermal neutron capture

U235

Criticality

ηεpfK =∞

neutrons thermalofnumber

neutronsfission ofnumber total=ε

∞= LKKeff

•Then, for all disappeared neutrons Ni, the number of fission neutrons produced = Nifηε

•If p is the probability that a fast neutron will slow down to a thermal neutron

without a capture

Then, Ni+1 = Nipfηε

Since, and , i

1i

effN

NK +=

Credits and References

� Glenn F. Knoll, Radiation Detection and

Measurement, 3rd ed. John Wiley, New York,

NY

� J. E. Turner, Atom, Radiation and Radiation

Protection, 3rd ed. ,Wiley-Vch(2007)

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