ne 301 - introduction to nuclear science spring 2012 classroom session 3: radioactive decay types...
TRANSCRIPT
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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Reminder
Load 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 amu
8 5.48 4 amu
p
n O
e
Zm
A Z m M
Zm e
Mass Defect = Binding
Energy (BE)
1 amu = 931.49 MeV
168O
16 O
931.49 MeVBE = (16.131912-15.994915 amu) 127.61 MeV
1 amu
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Chart of the Nuclides
Z
N
Isobars
Isotopes
Isotones
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Notice radioactive decay stabilizes atoms:
Question:
Do fission products normally have - or + decay?
Reaction Energetics
Reaction 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 MeV
1 amu
A B C DM M M M M
M
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?
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5.5 MeV
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 6.586
H Be B
Q MeV
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Excited Nuclei
Many 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 amu
931.494MeVNe NeM M
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Decay Series
The 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 Emission
Gamma () Decay of excited nucleus
Gamma 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 Proton
neutron (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 -emitters
n, 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 m
KE 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 Products
Parent 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
m1
m1
m2
Original atom that will split in 2 pieces
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Kinematics of radioactive decay…
2 21 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 v
2 2m v
v = replacing...m
m v1 1m ( ) m v
2 m 2
m v1 1 1m v and replacing m v by KE
2 m 2 2
m solving for KE
m
Q
Q
Q KE KE
KE Q
1 21
1 2 1 2
similarly: m m
KE 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 m
KE 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:
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3.638 MeV
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)
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0.09 MeV
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 241Am
There 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. 241Am
Taken 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 Q
m m
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3.6
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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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’s
’s
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Main Radioactive Decay Modes (Table 5.1 -page 89-Shultis)
Decay Type Description Emission
Gamma () Decay of excited nucleus
Gamma 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 Proton
neutron (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