NE 301 - Introduction to Nuclear ScienceSpring 2012

Classroom Session 3:

•Radioactive Decay Types•Radioactive Decay and Growth•Isotopes and Decay Diagrams•Nuclear Reactions

• Energy of nuclear reactions• Neutron Cross Sections• Activation Calculations

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ReminderLoad TurningPoint Reset slides Load List

Let’s do some accounting…Mass of Oxygen Atom:

Mp=1.007276 amuMn=1.008665 amuMe=5.48e-4 amu

3

168

16.131912 amu

8 1.007276 amu

( ) 8 1.008665 amu 15.994915 amu8 5.48 4 amu

p

n O

e

Zm

A Z m MZm e

Mass Defect = Binding

Energy (BE)

1 amu = 931.49 MeV

168O

16 O

931.49 MeVBE = (16.131912-15.994915 amu) 127.61 MeV1 amu

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Chart of the Nuclides

Z

N

IsobarsIsotopes

Isotones

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Notice radioactive decay stabilizes atoms:

Question:

Do fission products normally have - or + decay?

Reaction EnergeticsReaction reactants and products

If E is positive: reaction exothermic

releases energyIf E is negative, reaction endothermic

requires energyEndoergic and exoergic is sometimes used

A + B C + D + E

The Energy Released (or consumed), Q

Change in BE:

Or since BE is related to mass defect

Change in M:

A + B C + D + E

( )C D A BQ BE BE BE BE BE

( )A B C DQ M M M M M

Preferred!because we have table B.1.

Remember: The Equation Has to Be BALANCED!

Please remember…

BALANCE!

Before starting to work

Balancing Reactions

nucleons 1 +16 = 16+1Charges

01n 8

16O 716N1

1p

1 16 16 0 10 8 7 1 1

1 16 16 10 8 7 1

n O N e p or

n O N H

(+) 0 + 8 = 7 + 1(-) -0 -8 = -7 -0 e- missing

0 1So in reality the reaction is:

Calculating Q…

Q-value for the reaction is:

Using atomic mass tables:

1 16 16 10 8 7 1n O N H

( )1.008665 15.994915 16.006101 1.007825 0.010346 amu

931.494 MeV 0.010346 amu 9.637 MeV1 amu

A B C DM M M M MM

Endothermic reaction. Only a few fission neutrons can do it

A beryllium target is irradiated in a proton accelerator to produce 10B. What is Q of the reaction?

11 5.5 M

eV

4.5 MeV

3 MeV

6.5 MeV

85 MeV

14%

0%7%

79%

0%

1 9 101 4 5p Be B

1. 5.5 MeV2. 4.5 MeV3. 3 MeV4. 6.5 MeV5. 85 MeV

For clicker

1 9 101 4 5

(1.007825 9.012182 10.012937) 931.494 6.586H Be BQ MeV

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Excited NucleiMany reactions involve excited nucleiSometimes long lived states (isomers)Excitation energy has to be added to the mass of the excited nuclei when calculating Q

e.g. The mass of 22Ne* at 1274 MeV is:

M ZAX * M Z

AX E *

c 2

22 2210 10*

1amu* 21.991386 1274 MeV 23.3591 amu931.494MeVNe Ne

M M

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Decay SeriesThe radioactive minerals contain many nuclidesAll of them decay by either or decay A changes by 4, Z by 2 A does not change, A by 1

Th has one long lived isotope 232ThU has two long lived 235U, 238U

Series identified by relation Parent to Dauthers mass:

A in multiples of 4

There are 3 natural series

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16

NoticeBranching

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Series are:A = 4n --- Thorium Series

A = 4n+2 -- Uranium Series

A = 4n+3 – Actinium Series

Which one is missing?

A = 4n+1 – Neptunium Series (Artificial)

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It was there from the beginning… but notice: half life of 237Np is relatively low.

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Main Radioactive Decay Modes (Table 5.1 -page 89-Shultis)

Decay Type Description EmissionGamma ()

Decay of excited nucleusGamma photon

alpha ()Alpha particle is emitted

Alpha particle

negatron (-) np++e-+ Electron and anti-neutrino

positron (β+) p+n+e++ Positron and neutrino

Electron Capture (EC)

Orbital e- absorbed: p++e-n +

Neutrino

proton (p) Proton ejected Protonneutron (n) Neutron ejected Neutron

Internal Conversion (IC)

Electron (K-Shell) ejected*

Electron

Spontaneous Fission

(sf)

Fission fragments

*A AZ ZP P

1A AZ ZP D

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A AZ ZP D

1A AZ ZP D

*1

A AZ ZP e D

*A AZ ZP P e

1 2 nAZ P D D x

Comments:, +, - are common modes of decayLong T1/2 usually are -emittersn, p emission are rare (excess p+ atoms) is predominant for Z>83 (above Bismuth) and atoms away from the line of -stability.Some high Z atoms (Z>96) have dominant spontaneous fission mostly dominates again at Z>105

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Modes of Decay, +, - are common modes of decayLong T1/2 usually are -emittersn, p emission are rare (excess p+ atoms) is predominant for Z>83 (above Bismuth) and atoms away from the line of -stability.Some high Z atoms (Z>96) have dominant spontaneous fission mostly dominates again at Z>105

Solving momentum and KE equations

2 11 2

1 2 1 2

m mKE Q KE Qm m m m

Remember the conditions:1. Parent nucleus at rest (usually the case)2. Binary products only (not -decay, but OK to

Emax)3. Calculate the correct Q (excited states are

prevalent, and balance)4. Finally, there usually reaction paths with

many outcomes, therefore multiple Q-values

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Kinetic Energy of Radioactive Decay ProductsParent nucleus is at rest (Eth~ 0.025 eV~17 oC)Conservation of Linear Momentum and Kinetic Energy requires products to travel in opposite directions (2 product).

m1v1=m2v2

Q=½ m1v12

+ ½ m2v22

What is the energy of emitted particle? (it is what we measure)

v1

m2

v2

m1m1

m2

Original atom that will split in 2 pieces

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Kinematics of radioactive decay…2 2

1 1 2 2 1 1 2 2

2 21

1

2 22 21 2 2

1

2 22 22 2

2 2 2 2 21

22 2 2

1

2

1 1m v =m v Q= m v m v2 2

m vv = replacing...m

m v1 1m ( ) m v2 m 2

m v1 1 1m v and replacing m v by KE2 m 2 2m solving for KEm

Q

Q

Q KE KE

KE Q

1 21

1 2 1 2

similarly: m mKE Qm m m m

Notice 2:1

Warm up:What % of the energy should go to the -particle?

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98% 2%50%

10% 1%

20% 20% 20%20%20%

HeThU 42

23490

23892

1 22 1

1 2 1 2

m mKE Q KE Qm m m m

1. 98%2. 2%3. 50%4. 10%5. 1%

Example of -spectroscopy?

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237Pa 237U

237Np 237Pu

237Am 237Cm

0% 0% 0%0%0%

100%241 ?Am

1. 237Pa2. 237U3. 237Np4. 237Pu5. 237Am6. 237Cm

Find Q for:

28 3.638 M

eV

4.638 MeV

5.638 MeV

6.638 MeV

7.638 MeV

20% 20% 20%20%20%

241 237 495 93 2Am Np He

1. 3.638 MeV2. 4.638 MeV3. 5.638 MeV4. 6.638 MeV5. 7.638 MeV

For Clicker slide:Q=(241.056823-237.048167-

4.002603)*931.494=5.638MeV

What is the KE of the particle in the radioactive decay of 241Am? (3 min)

30 0.09 M

eV

0.98 MeV

5.54 MeV

5.64 MeV

25% 25%25%25%

1. 0.09 MeV2. 0.98 MeV3. 5.54 MeV4. 5.64 MeV

For Clicker slide:

KE=5.638*237/(237+4)=5.545 MeV

Notice:If alpha particle ALWAYS leaves with exactly the same energy.We would expect to detect a monoenergetic beam of ’s.

In reality…

The real alpha spectrum of 241Am is:

At least 5 different energies…

Why?

Excited Nuclei!

The real decay path of 241AmThere are actually 6 alpha peaksLast two peaks are too close to be resolvedNotice frequencies (%’s)Every decay path happens all the time but not with equal probabilityLook in your book:

Page 578. 241AmTaken from J. K. Beling, et al. Phys. Rev. 87 (1952) 670-671

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Diagram means:

Energy of the -particle?

Same old same old

But Q is different each time

24195

*170 KeV

24195

*114 KeV

24195

*71 KeV

24195

*43 KeV

24195

*11 KeV

24195

237 * 493 2

237 * 493 2

237 * 493 2

237 * 493 2

237 * 493 2

237 493 2

Am

Am

Am

Am

Am

Am

Np He

Np He

Np He

Np He

Np He

Np He

2

mKE Qm m

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3.6

37

38

4.0

By the wayNotice also

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4.0

There are a lot more hard to see peaks

So how is the “real” diagram?For that we need the

TABLE OF ISOTOPES

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Diagram 241Am - 1 of 2

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Diagram 241Am - 2 of 2

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The Table also includes a more complete list of particles emitted during decay

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44

45

’s

’s

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Main Radioactive Decay Modes (Table 5.1 -page 89-Shultis)

Decay Type Description EmissionGamma ()

Decay of excited nucleusGamma photon

alpha ()Alpha particle is emitted

Alpha particle

negatron (-) np++e-+ Electron and anti-neutrino

positron (β+) p+n+e++ Positron and neutrino

Electron Capture (EC)

Orbital e- absorbed: p++e-n +

Neutrino

proton (p) Proton ejected Protonneutron (n) Neutron ejected Neutron

Internal Conversion (IC)

Electron (K-Shell) ejected*

Electron

Spontaneous Fission

(sf)

Fission fragments

*A AZ ZP P

1A AZ ZP D

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A AZ ZP D

1A AZ ZP D

*1

A AZ ZP e D

*A AZ ZP P e

1 2 nAZ P D D x