lecture 8 atomic nuclear physics
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7/23/2019 Lecture 8 Atomic Nuclear Physics
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Dr. Ahmed Said Eltrass
Electrical Engineering Department
Alexandria University, Alexandria, Egypt
Fall 2015Office hours: Sunday (10:00 to 12:00 )
4th floor, Electrical Engineering Building
Modern Physics: Lecture 8
Introduction to Nuclear Physics
7/23/2019 Lecture 8 Atomic Nuclear Physics
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Review of Decay processes
4 4
2 2
A A
Z Z X Y energy
Alpha Decay:
0
1 1
A A
Z Z X Y energy
Beta-minus Decay:
0
1 1
A A
Z Z X Y energy
Beta-plus Decay:
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Review: Conservation Laws
Conservation of Charge: The total charge of asystem can neither be increased nor decreased.
Conservation of Nucleons: The total number ofnucleons in a reaction must be unchanged.
Conservation of Mass Energy: The total mass-
energy of a system must not change in anuclear reaction.
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Kinetic Energy of particles
QY X A
Z
A
Z
4
2
4
2
• The kinetic energy K.E. of the emitted particle is
related to Q and the mass number A of the originalnucleus
Q A
A E K
4..
• The kinetic energy of the new nucleus is given by
Q
A
E K 4..
Q is the energy released in the decay process(called the disintegration energy).
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Rate of decay
dN/dt = -l*N
where
l is the decay constantN is the number of nuclei present in a sample ofradioactive nuclide at a certain time ( number ofundecayed nuclei)
The minus sign comes from the fact that dN/dt isDECREASING rather than growing.
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dN/dt = -l*N
• We can solve this differential equation for N(t):
dN/N = -l dt , or ln (N) = -l t + ln (No)
ln (N/No) = -l t , or N(t) = No e-lt
• The rate of decay for radioactive substances can beexpressed in terms of the activity A :
A = lN = lNo
e-lt = A o
e-lt
A = A oe-lt
This means that the activity decreases exponentially
with time also
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Half LifeN(t) = No e
-lt
• The number of radioactive atoms decreases withtime.
• The measure of the time it takes for half of the
radioactive nuclei in a sample to decay intoanother element is called a half-life.
N(T=half life) = No /2 = Noe
-lT
, or 1/2 = e
-lT
or -lT = ln(1/2) = ln(1) – ln(2) = 0 - ln(2), or
T(half life) = ln(2) / l
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Half Life
NO. After a second half life, we only have half asmuch as we did after the first half life:
N(t = 0) = No N(t = 1 half life) = ½ No
N(t = 2 half lives) = ½ (½ No) = (½)2 No
…
N(t = n half lives) = (½)n No
After two half lives, is the number now down to
zero?
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• After one half life there is 1/2 of original sample left.
• After two half-lives, there will be 1/2 of the 1/2 = 1/4the original sample.
Graph of Amount of Remaining Nuclei vs Time
A=A oe-lt
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• We can find T (half life) if we can wait for N (or A )to decrease by half.
• We can find by measuring N and A .
• If the half life is large, l is small. This means that
if the radioactive isotope will last a long time(half life is large) , its activity will be small;
if the half life is small, the activity will be large
but only for a short time!
Review: N(t) = No e- t
A = lN = Aoe-lt
T(half life) = ln(2) /
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• Table N provides uswith a list of variousnuclides, their decay
modes, and their half-lifes.
• Using Table N, what is
the decay mode andhalf-life for Radium-226?
Radium-226 undergoes
alpha decay.
RadonRadium
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Using Half-life
• Table N also tells us that Radium-226 has a
half-life of 1600 years.
• Starting with a100g sample, after1 half-life (or 1600years), 50g remain.
•
After another1600 years, half ofthe 50g will remain(25g).
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Example 1: A 10 gram of sample of Iodine-131 undergoes
decay, what will be the mass of iodine remainingafter 24 days?
From Table N, the ½ life of iodine is determinedto be approximately 8 days.
That means that 24 days is equivalent to 3 half-lifes.
The decay of 10 grams of I-131 would produce:
1.25 grams of I-131 would remain after 24 days.
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The percent of C-14 is found to be 25% of what the original
C-14 concentration was. What is the age of the sample?
First, let’s analyze how many half -lives have taken place.
Two half-lives have gone by while the sample decayed fromthe original C-14 concentration to 25% of that concentration.
Based on Table N, the half-life of C-14 is 5730 years, so…
Example 2:
The age of the sample= 2* 5730= 11460 years
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Nuclear Reactions
It is possible to alter the structure of a nucleusby bombarding it with small particles. Suchevents are called nuclear reactions:
General Reaction: x + X
Y + y
For example, if an alpha particle bombardsa nitrogen-14 nucleus it produces a
hydrogen atom and oxygen-17:4 14 1 17
2 7 1 8 N H O
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Example 3: Use conservation criteria to determinethe unknown element in the following nuclearreaction and to calculate the energy released :
1 7 41 3 2
A Z H Li He X energy
Charge before = +1 + 3 = +4
Charge after = +2 + Z = +4
Z = 4 – 2 = 2
Nucleons before = 1 + 7 = 8
Nucleons after = 4 + A = 8
(Helium has Z = 2)
(Thus, A = 4)
Q is the energy released in the reaction.
1 7 4 4
1 3 2 2 H Li He He Q
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The energy released or absorbed is called the Q-value and can be found if the atomic masses areknown before and after.
1 7 4 4
1 3 2 2 H Li He He Q
1 7 4 4
1 3 2 2Q H Li He He
7
3 7.016003 u Li
4
2 4.002603 u He
1
1 1.007825 u H
Substitution of these masses gives:
Q = 0.018622 u*(931.5 MeV/u) Q =17.3 MeV
4
2 4.002603 u He
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Types of Nuclear Reactions
1- Nuclear Fission• The heaviest nuclei are less stable than the nuclei near
A=60. This suggests that energy can be released if heavy
nuclei split into smaller nuclei having masses nearer
A=60.
• The process of splitting a nucleus into smaller nuclei- is
called nuclear f ission.
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There are 2 types of fission that exist:
1. Spontaneous Fission
2. Induced Fission
• Heavy nuclei are highly unstable and decayspontaneously by splitting into 2 smaller nuclei.
• Such spontaneous decays are accompanied by the
release of neutrons.
• Nuclear fission can be induced by bombarding
atoms with neutrons.
• The nuclei of the atoms then split into 2 equal parts
• Induced fission decays are also accompanied by the
release of neutrons.
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What nuclei can split during nuclearfission?
• Only large nuclei like U orplutonium can split apartduring nuclear fission.
• U-236 is so unstable thatwhen U-236 is bombardedwith a neutron itimmediately splits into
barium & krypton nuclei,several neutrons & a largeamount of energy.
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U235
92 +Ba141
56+ n1
03n
1
0 +Kr92
36
Example:
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U23592n
10
A neutron travels at high speed towards a
uranium-235 nucleus.
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U23592n
10
A neutron travels at high speed towards auranium-235 nucleus.
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U23592n
10
A neutron travels at high speed towards auranium-235 nucleus.
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U23592n
10
The neutron strikes the nucleus which thencaptures the neutron.
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U23692
The nucleus changes from being uranium-235 touranium-236 as it has captured a neutron.
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• The uranium-236 nucleus formed is very unstable.
• It transforms into an elongated shape for a shorttime.
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• The uranium-236 nucleus formed is very unstable.
• It transforms into an elongated shape for a shorttime.
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• The uranium-236 nucleus formed is very unstable.
• It transforms into an elongated shape for a shorttime.
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It then splits into 2 fission fragments and releasesneutrons.
14156Ba
9236Kr
n10
n10
n10
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It then splits into 2 fission fragments and releasesneutrons.
14156Ba
9236Kr
n10
n10
n10
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It then splits into 2 fission fragments and releasesneutrons.
141
56Ba
9236Kr
n10
n10
n10
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What is a chain reaction?
• Free neutrons produced by fission can hitother nuclei emitting more neutronsrepeating the reaction over and over.
• A series of fission reactions is called a chainreaction.
• An uncontrolled chain reaction releases a
huge amount of energy in a short time &requires a critical mass of starting materialto produce more reactions.
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Fissionproduces
a chainreaction
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Energy from Fission
• Both the fission fragments and neutrons travel at
high speed.
• The kinetic energy of the products of fission arefar greater than that of the bombarding neutron
and target atom.
E K before fission << E K after fission
Energy is being released as a result of the fissionreaction.
235
138
1
1
96
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U235
92 +Cs138
55+ n1
02n1
0 +Rb96
37
Element Atomic Mass (kg)235
92U 3.9014 x 10-25
13855Cs 2.2895 x 10-25
96
37Rb 1.5925 x 10-25
10n 1.6750 x 10-27
The total mass before fission (LHS of the equation):
3.9014 x 10-25
+ 1.6750 x 10-27
= 3.91815 x 10-25
kg
The total mass after fission (RHS of the equation):
2.2895 x 10-25 + 1.5925 x 10-25 + (2 x 1.6750 x 10-27)
= 3.9155 x 10-25
kg
total mass before fission > total mass after fission
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total mass before fission > total mass after fission
This reduction in mass results in the release of energy.
mass difference m = 3.91815 x 10-25 – 3.91550 x 10-25
= 2.65 x 10-28 kg
E = mc2 =2.65 x 10-28 x (3 x 108)2
E = 2.385 x 10-11 J
• The energy released from this fission reactiondoes not seem a lot because it is produced fromthe fission of a single nucleus.
• Large amounts of energy are released when a
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• Large amounts of energy are released when alarge number of nuclei undergo fission reactions.
Each uranium-235 atom has a mass of 3.9014 x 10-25 kg.
No. of atoms in 1 kg of uranium-235 = 2.56 x 1024 atoms
Energy per fission , E = 2.385 x 10-11
J
• The amount of energy released by 1 kg ofuranium-235 can be calculated as follows:
Total energy = energy per fission x number of atoms
Total energy = 2.385 x 10-11 x 2.56 x 1024
Total energy = 6.1056 x 1013 J
Types of Nuclear Reactions
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Types of Nuclear Reactions
2- Nuclear Fusion
• In a nuclear fusion reaction, two small, light
nuclei combine to form one larger, heavier nucleus.
Fission Fusion
Both
Reactionsproduceenergy
Process of splittinga nucleus intosmaller nuclei
2 small, lightnuclei combine toform one larger,heavier nucleus
Example of Fusion Process:
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H2
1 +He4
2+ n1
0H
3
1 +Energy
Example of Fusion Process:
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H21
H31
The Fusion Process
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The Fusion Process
H21
H
3
1
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The Fusion Process
H21
H31
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The Fusion Process
H21
H31
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The Fusion Process
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The Fusion Process
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The Fusion Process
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The Fusion Process
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The Fusion Process
He42
n
1
0
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The Fusion Process
He42
n10
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The Fusion Process
He42
n10
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The Fusion Process
He42
n
1
0
Energy from Fusion
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Energy from Fusion
Element Atomic Mass (kg)21H 3.345 x 10-27
31H 5.008 x 10-27 42He 6.647 x 10-27 10n 1.6750 x 10-27
H2
1 +He
4
2+n
1
0H
3
1 +Energy
The total mass before fusion (LHS of the equation):
The total mass after fusion (RHS of the equation):
3.345 x 10-27 + 5.008 x 10-27 = 8.353 x 10-27 kg
6.647 x 10-27
+ 1.675 x 10-27
= 8.322 x 10-27
kg
m total mass before fusion total mass after fusion
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m = total mass before fusion – total mass after fusion
m = 8.353 x 10-27 – 8.322 x 10-27
m = 3.1 x 10-29 kg
E = mc2
E = 3.1 x 10-29 x (3 x 108)2
E = 2.79 x 10-12 J
The energy released per fusion is 2.79 x 10 -12
J.
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