Lecture 8 Atomic Nuclear Physics

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



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:



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.



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).



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.



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



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



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?



• 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



• 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) /



• 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



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).



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.



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



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



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



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



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.



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.



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.



U235

92 +Ba141

56+ n1

03n

1

0 +Kr92

36

Example:



U23592n

10

A neutron travels at high speed towards a

uranium-235 nucleus.



U23592n

10

A neutron travels at high speed towards auranium-235 nucleus.



U23592n

10

The neutron strikes the nucleus which thencaptures the neutron.



U23692

The nucleus changes from being uranium-235 touranium-236 as it has captured a neutron.



• The uranium-236 nucleus formed is very unstable.

• It transforms into an elongated shape for a shorttime.



It then splits into 2 fission fragments and releasesneutrons.

14156Ba

9236Kr

n10

n10

n10




14156Ba

9236Kr

n10

n10

n10




141

56Ba

9236Kr

n10

n10

n10



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.



Fissionproduces

a chainreaction



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



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



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



• 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





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:



H2

1 +He4

2+ n1

0H

3

1 +Energy

Example of Fusion Process:



H21

H31

The Fusion Process



The Fusion Process

H21

H

3

1



The Fusion Process

H21

H31



The Fusion Process



The Fusion Process

He42

n

1

0



The Fusion Process

He42

n10



The Fusion Process

He42

n

1

0

Energy from Fusion



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



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.



See You NextSemester ISA inElectronics Course

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Lecture 8 Atomic Nuclear Physics

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