NE 105 - Introduction to Nuclear Engineering Spring 2011 Classroom Session 4 - Fundamental Concepts End Nuclear Energetics Intro Classic and Relativistic.

NE 105 - Introduction to Nuclear EngineeringSpring 2011

Classroom Session 4 - Fundamental Concepts End

Nuclear Energetics Intro

•Classic and Relativistic Calculations•Photon Interactions with Matter•Nuclear Energetics

2

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 134Xe

ion?(from the chart of the nuclides: 134Xe Weights = 133.905394

AMU)Use classic concept of KE for nowamu in table 1.5Joule = Energy, Work = Force (N) x d =kg m2/s2

3

Correction of the book… REMEMBER!

Please ignore the c2. It is confusing

Book: Page 6

4

4156.4 m/s ~9,300 m.p.hi.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

5

What is the speed of a 100.00 MeV proton:

102

,540

m/s

5,4

67 g

/s

1.3

8e8

m/s

138

40 m

/s

3e8

m/s

20% 20% 20%20%20%1. 102,540 m/s2. 5,467 g/s3. 1.38e8 m/s4. 13840 m/s5. 3e8 m/s

6

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

What is the speed of a 100.00 MeV proton:

7

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)

8

Special Theory of Relativity - Effects

“Mass Increase” with increasing velocity

Increase quantified by Lorentz factor ():

m(v) m0

1 v 2 c 2

2 2 1

v<<<c 1 classic limit1 always

v~c 0 effect is max

v c

9

Length and time are also modified relative to an object’s speed

For example: To find speed…

L(v) L0 1 v 2 c 2

22

0

1)(

cv

tvt

Special Theory of Relativity - Effects

10

What is the kinetic energy of a 100.00 MeV proton?

Hint: Relativistic speeds, i.e. use this equation:

Special Theory of Relativity - Effects

2 20E mc m c KE

m(v) m0

1 v 2 c 2

11

The error grows as v c

Reminder: simple error is

Accepted Value - Obtained Value100 % Error

Accepted Value

12

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:eVkeVMeV

13

What is the kinetic energy of a 1 MeV electron? Rest mass of the electron, me=0.511MeV

0.5

11 M

eV

0.4

89 M

eV

0.9

99 M

eV

1 M

eV

0 M

eV

20% 20% 20%20%20%

1. 0.511 MeV2. 0.489 MeV3. 0.999 MeV4. 1 MeV5. 0 MeV

14

What is the speed of a 1 MeV electron? Rest mass of the electron, me=0.511MeV

0.58c 0.81c

0.86c 0.94c

0.993c

20% 20% 20%20%20%

1. 0.58c2. 0.81c3. 0.86c4. 0.94c5. 0.993c

15

Solution:2 2

0

22 0

2

0.511 1 1.511

m cand solving for v, from relativistic equation mc = :

0.5111 0.94

1.511

mc m c KE MeV MeV MeV

v c c

16

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?

Radiation Interaction with Matter

Ionizing Radiation

19

Photon Interactions

EnergyHighIntermediateLow

Pair Production

Compton Scattering

Photoelectric Effect

20

Pair Production

21

Compton Scattering

22

The Photoelectric Effect

23

Compton Scattering – The Experiment

E

E’

In 1922, Compton obtained this dataScattered X-Rays had an increase in wavelengthCan you explain why?

24

Compton Scattering – Light has p!

If light is a wave, then radiation scattered by an electron should have no change in wavelengthIn 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/

)cos1(' secm

h

Why the equation written for the photon angle?

1 1 1(1 cos )

' seE E m c

EE’

25

Follow equations

But pay attention to unitsFor wavelength please use nm

6.63 -34

e

h e J

m c

. s

9.11e-31 kg 3e8 m / s

1 kg

2m

1 J 2. s

1 9

1

e nm

m [ ] nm