### Fundamental Concepts end & nuclear enegetics - radiochem

```NE 105 - Introduction to Nuclear Engineering
Spring 2011
Classroom Session 4 - Fundamental Concepts End
Nuclear Energetics Intro
•Classic
and Relativistic Calculations
•Photon Interactions with Matter
•Nuclear Energetics
Electron Volt
Work done by one electron accelerated
through a potential difference of one volt
1 eV = 1.60217646x10-19 J
Example:
What is the speed (m/s) of a 12 eV
(from the chart of the nuclides:
134Xe
134Xe
ion?
Weights = 133.905394 AMU)
Use classic concept of KE for now
amu in table 1.5
Joule = Energy, Work = Force (N) x d =kg m2/s2
2
Correction of the book… REMEMBER!
Book: Page 6
c2. It is confusing
3
4156.4 m/s
~9,300 m.p.h
i.e. even very low energy ions are moving pretty fast
Please remember this is ONLY for
classical calculations.
At energies close to “c”, need to use
relativistic calculations
4
What is the speed of a 100.00 MeV proton:
m
/s
/s
8
3e
0
84
13
e8
38
1.
m
m
/s
g/
s
7
46
5,
5.
m
/s
4.
54
0
3.
20% 20% 20% 20% 20%
2,
2.
102,540 m/s
5,467 g/s
1.38e8 m/s
13840 m/s
3e8 m/s
10
1.
5
What is the speed of a 100.00 MeV proton:
100MeV proton = 0.46 c :close to the speed
of light.
i.e. classic equations do NOT hold
i.e. 0.46 is likely wrong
6
Newton Laws
For over 200 years, Newton’s laws worked


Accurately described many physical behaviors
Unifying the earth and the skies
Previously:


Sub-lunar sphere: impure and imperfect
Skies: perfect and immutable (circle, ether)
7
Special Theory of Relativity - Effects
“Mass Increase” with increasing velocity
m(v) 
m0
2
1 v c
2
Increase quantified by Lorentz factor ():

  1 always
  1 v c
2
2
 v<<<c   1 classic limit

 v~c   0 effect is max
8
Special Theory of Relativity - Effects
Length and time are also modified
relative to an object’s speed
L(v)  L0 1 v c
2
t (v) 

2
t0
1 v c
2
2
For example: To find speed…
9
Special Theory of Relativity - Effects
E  mc  m0c  KE
2
2
What is the kinetic energy of a 100.00 MeV
proton?
Hint: Relativistic speeds, i.e. use this equation:
m(v) 
m0
2
1 v c
2
10
Reminder: simple error is
Accepted Value - Obtained Value
100  % Error
Accepted Value
The error grows as v  c
11
Remember
Relativistic calculation required when:
kinetic energy ~ rest energy



What is the rest mass of an electron?
What is the rest mass of a p+ or n0?
What is the rest mass of heavy ions?
(Table 1.5 book)
Use:
eV
keV
MeV
12
What is the kinetic energy of a 1 MeV electron?
Rest mass of the electron, me=0.511MeV
20% 20% 20% 20% 20%
M
eV
0
M
eV
1
M
eV
0.
99
9
M
eV
9
48
0.
5.
M
eV
4.
1
3.
51
2.
0.511 MeV
0.489 MeV
0.999 MeV
1 MeV
0 MeV
0.
1.
13
What is the speed of a 1 MeV electron?
Rest mass of the electron, me=0.511MeV
20%
0.
99
3c
5.
20%
0.
94
c
4.
20%
0.
86
c
3.
20%
0.
81
c
2.
0.58c
0.81c
0.86c
0.94c
0.993c
0.
58
c
1.
20%
14
Solution:
mc 2  m0c 2  KE  0.511MeV  1MeV  1.511MeV
2
and solving for v, from relativistic equation mc =
m0 c2

:
2
 0.511 
v  1 
 c  0.94 c
 1.511 
15
Special Theory of Relativity - Effects
In Nuclear Engineering we rarely work with
neutrons of more than 10MeV.
We stick to classic calculations for KE of p,
n, , ions, and fission fragments
Homework 2.3. What is the error in computing speed of a
10 MeV neutron classically instead of relativistically?
16
Photon Interactions
Energy
Low
Photoelectric
Effect
Intermediate
Compton
Scattering
High
Pair
Production
19
Pair Production
20
Compton Scattering
21
The Photoelectric Effect
22
Compton Scattering – The Experiment
In 1922, Compton
obtained this data
an increase in
wavelength
Can you explain why?
E’
E
23
Compton Scattering – Light has p!
E’
E
Why the equation written for the
photon angle?
h
 ' 
(1  cos s )
me c
1 1
1
 
(1  cos  s )
E ' E me c
If light is a wave, then radiation scattered by an
electron should have no change in wavelength
In 1922, Compton demonstrated that that x-rays
scattered from electrons had a decrease in
wavelength.
This is only possible if light is treated as a particle with
linear momentum equal to p=h/
24
But pay attention to units
1 kg m 2 1e9 nm
h
6.63e-34 J . s



[] nm
2
me c 9.11e-31 kg  3e8 m / s 1 J . s
1m
25
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