SPH4UI Physics

Nuclear Binding, Radioactivity

Modern understanding: the ``onion’’ picture

Let’s see what’s inside!

3

Nice Try

        

Introduction: Development of Nuclear Physics

1896  – the birth of nuclear physics Becquerel discovered radioactivity in uranium compounds Rutherford showed the radiation had three types    Alpha (He nucleus) Beta (electrons) Gamma (high-energy photons) 1911   Rutherford, Geiger and Marsden performed scattering experiments Established the point mass nature of the nucleus

Nuclear force

was a new type of force 1919 Rutherford and coworkers first observed nuclear reactions in which naturally occurring alpha particles bombarded nitrogen nuclei to produce oxygen 1932 Cockcroft and Walton first used artificially accelerated protons to produce nuclear reactions 1932 Chadwick discovered the neutron 1933 the Curies discovered artificial radioactivity 1938 Hahn and Strassman discovered nuclear fission 1942 Fermi achieved the first controlled nuclear fission reactor

   

Some Properties of Nuclei

All nuclei are composed of protons and neutrons  Exception is ordinary hydrogen with just a proton The

atomic number (charge)

, Z , equals the number of protons in the nucleus The

neutron number

, N , is the number of neutrons in the nucleus The

mass number

, A , is the number of nucleons in the nucleus  A = Z + N   Nucleon is a generic term used to refer to either a proton or a neutron The mass number is not the same as the mass   Notation

Z A X

where X is the chemical symbol of the element

Example: 27

Al

13    Mass number is 27 Atomic number is 13 The Z   Contains 13 protons Contains 14 (27 – 13) neutrons may be omitted since the element can be used to determine Z

Charge and mass

  

Charge:

The electron has a single negative charge, -e ( e = 1.60217733 x 10 -19 C ) The proton has a single positive charge, +e  Thus, charge of a nucleus is equal to Ze The neutron has no charge  Makes it difficult to detect  

Mass:

It is convenient to use

atomic mass units ,

u , to express masses  1 u = 1.660559 x 10 -27 kg  Based on definition that the mass of one atom of C-12 is exactly 12 u Mass can also be expressed in MeV/c 2   From E R = m c 2 1 u = 931.494 MeV/c 2

Summary of Masses

Masses

Particle

Proton Neutron Electron

kg

1.6726 x 10 -27 1.6750 x 10 -27 9.101 x 10 -31

u

1.007276

1.008665

5.486x10

-4

MeV/c 2

938.28

939.57

0.511

Quick problem: protons in your body

What is the order of magnitude of the number of protons in your body? Of the number of neutrons? Of the number of electrons? Take your mass approximately equal to 70 kg.

An iron nucleus (in hemoglobin) has a few more neutrons than protons, but in a typical water molecule there are eight neutrons and ten protons. So protons and neutrons are nearly equally numerous in your body, each contributing 35 kg out of a total body mass of 70 kg.

N

 35

kg

  1 nucleon 1.67 10  27

kg

   28 10 protons Same amount of neutrons and electrons.

The Size of the Nucleus

   First investigated by Rutherford in scattering experiments He found an expression for how close an alpha particle moving toward the nucleus can come before being turned around by the Coulomb force The KE of the particle must be completely converted to PE 1 2

mv

2 

k e q q

1 2

r



k e

or

d

 4

k Ze e

2

mv

2

d



For gold: d = 3.2 x 10 -14 m , for silver: d = 2 x 10 -14 Such small lengths are often expressed in 10 -15 m a fermi) femtometers m where 1 fm = (also called

Size of Nucleus

 Since the time of Rutherford, many other experiments have concluded the following   Most nuclei are approximately spherical Average radius is

r



r A o

1 3  r o = 1.2 x 10 -15 m Example:

Z A

27

Al

13

has radius

r

  15

m

Density of Nuclei



The volume of the nucleus (assumed to be spherical) is directly proportional to the total number of nucleons



This suggests that

all nuclei have nearly the same density



Nucleons combine to form a nucleus as though they were tightly packed spheres

Nuclear Stability

  There are very large repulsive electrostatic forces between protons  These forces should cause the nucleus to fly apart The nuclei are stable because of the presence of another, short range force, called the

nuclear (or strong) force

  This is an attractive force that acts between all nuclear particles The nuclear attractive force is stronger than the force at the short ranges within the nucleus Coulomb repulsive

Nuclear Stability chart

   Light nuclei are most stable if N = Z Heavy nuclei are most stable when N > Z  As the number of protons increase, the Coulomb force increases and so more nucleons are needed to keep the nucleus stable No nuclei are stable when Z > 83

Isotopes

  The nuclei of all atoms of a particular element must contain the same number of protons They may contain varying numbers of neutrons 

Isotopes

of an element have the same Z but differing N and A values

Z A X

atomic number (charge) , Z neutron number , N nucleon number , A ,

 Example : 11

C

6 12

C

6 13

C

6 14

C

6

Binding Energy

 The total energy of the bound system (the nucleus) is less than the combined energy of the separated nucleons  This difference in energy is called the

binding energy

of the nucleus  It can be thought of as the amount of energy you need to add to the nucleus to break it apart into separated protons and neutrons Binding Energy per Nucleon

Binding Energy Plot

Iron (Fe) is most binding energy/nucleon. Lighter have too few nucleons, heavier have too many.

Fission Fission = Breaking large atoms into small Fusion = Combining small atoms into large

Binding Energy Notes

   Except for light nuclei, the binding energy is about 8 MeV nucleon per The curve peaks in the vicinity of A = 60  Nuclei with mass numbers greater than or less than 60 are not as strongly bound as those near the middle of the periodic table The curve is slowly varying at A > 40  This suggests that the nuclear force saturates  A particular nucleon can interact with only a limited number of other nucleons

Question

Where does the energy released in the nuclear reactions of the sun come from?

(1) covalent bonds between atoms (2) binding energy of electrons to the nucleus (3) binding energy of nucleons

Question

Which element has the highest binding energy/nucleon?

• Neon (Z=10) • Iron (Z=26) • Iodine (Z=53)

Question

Which of the following is most correct for the total binding energy of an Iron atom (Z=26)?

9 MeV 234 MeV 270 MeV 504 Mev For Fe, B.E./nucleon



9MeV

56

Fe

26

has 56 nucleons Total B.E



56x9=504 MeV

Binding Energy

Einstein’s famous equation

E = m c

2 Proton: mc 2 = 938.3MeV

Neutron: mc 2 = 939.5MeV

Adding these, get 1877.8MeV

Deuteron: mc 2 =1875.6MeV

Difference is Binding energy, 2.2MeV

M

Deuteron

= M

Proton

+ M

Neutron

– |Binding Energy|

Big Problem: binding energy

41

Nb

Using atomic mass units

Calculate the average binding energy per nucleon of

93 41

Nb

Given: m p = 1.007276u

m n = 1.008665u

1u=931.5 MeV Find: E b = ?

E b



A

In order to compute binding energy, let’s first find the mass difference between the total mass of all protons and neutrons in Nb and subtract mass of the Nb :

Number of protons:

N p

 41 Number of neutrons:

Mass difference:

 41

m



p

 52

m n



u



m Nb



N n

 

u

92.9063768

u

  0.8425192

u

2

Thus, binding energy is

  0.842519

u

 931.5

93

MeV u

  8.44

MeV

nucleon

Radioactivity

  

Radioactivity

is the spontaneous emission of radiation Experiments suggested that radioactivity was the result of the decay, or disintegration, of unstable nuclei Three types of radiation can be emitted   Alpha particles  The particles are 4 He nuclei Beta particles  The particles are either electrons or positrons   A positron is the

antiparticle

of the electron  It is similar to the electron except its charge is +e Gamma rays  The “rays” are high energy photons

B field into screen

Types of Radioactivity

Radioactive sources detector

a b g 

He

particles: electrons rays) Barely penetrate a piece of paper Can penetrate a few mm of aluminum photons (more energetic than x Can penetrate several cm of lead

The Decay Processes – General Rules

    When one element changes into another element, the process is called

spontaneous decay

or

transmutation

The sum of the mass numbers, A , must be the same on both sides of the equation The sum of the atomic numbers, Z , must be the same on both sides of the equation Conservation of mass-energy and conservation of momentum must hold

Alpha Decay

  When a nucleus emits an alpha particle it loses two protons and two neutrons  N decreases by 2  Z decreases by 2  A decreases by 4 Symbolically

Z A X



A Z

  4 2

Y



2 4

He

  X is called the

parent nucleus

Y is called the

daughter nucleus

Alpha Decay -- Example

 Decay of 226 Ra 226

Ra

88  222

Rn

86  2 4

He

   Half life for this decay is 1600 years Excess mass is converted into kinetic energy Momentum of the two particles is equal and opposite

Beta Decay

   During beta decay, the daughter nucleus has the same number of nucleons as the parent, but the atomic number is one less In addition, an electron (positron) was observed The emission of the electron is from the nucleus    The nucleus contains protons and neutrons The process occurs when a neutron is transformed into a proton and an electron Energy must be conserved

Beta Decay – Electron Energy

     The energy released in the decay process should almost all go to kinetic energy of the electron Experiments showed that kinetic energy few electrons had this amount of To account for this “missing” energy, in 1930 Pauli proposed the existence of another particle Enrico Fermi later named this particle the

neutrino

Properties of the neutrino  Zero electrical charge    Mass much smaller than the electron, probably not zero Spin of ½ Very weak interaction with matter

Beta Decay

 Symbolically  

Z A X



Z



A

1

Y



e

  

Z A X



Z



A

1

Y



e

     is the symbol for the neutrino is the symbol for the antineutrino  To summarize, in beta decay, the following pairs of particles are emitted  An electron and an antineutrino  A positron and a neutrino

Gamma Decay

    Gamma rays are given off when an excited nucleus “falls” to a lower energy state  Similar to the process of electron “jumps” to lower energy states and giving off photons The excited nuclear states result from “jumps” made by a proton or neutron The excited nuclear states may be the result of violent collision or more likely of an alpha or beta emission Example of a decay sequence  The first decay is a beta emission  The second step is a gamma emission   12

B

5  12

C

* 6  12

C

6 * 

e

 12

C

6  g   The C* indicates the Carbon nucleus is in an excited state Gamma emission doesn’t change either A or Z

Decay Rules

1) Nucleon Number is conserved.

2) Atomic Number (charge) is conserved.

3) Energy and momentum are conserved.

a

: example

238

U

92  234 90

Th

 a

1) 238 = 234 + 4 2) 92 = 90 + 2 recall

2 4

He

 a

Nucleon number conserved Charge conserved

b

: example

0 1 n  1 1 p   0 1

e

  0 0  g

: example

Z A P

* 

Z A P

 0 0 g

Neutrino needed to conserve energy and momentum.

Practice

A nucleus undergoes

a

FALSE?

decay. Which of the following is 1. Nucleon number decreases by 4 2. Neutron number decreases by 2 3. Charge on nucleus increases by 2

a

decay is the emission of

4 2

He

 a

A decreases by 4 Ex .

238 U 92  234 90 Th  2 4

He

Z decreases by 2 (charge decreases!)

Practice

Th

b 

Which of the following is true?

decay . 1. The number of protons in the daughter nucleus increases by one. 2. The number of neutrons in the daughter nucleus increases by one.

b 

decay is accompanied by the emission of an electron: creation of a charge -e.

  0 1 

e

0 1 

e

   0 0 0  0 

n e

neutrino “escape.”

 

Decay

Which of the following decays is NOT allowed?

1

238

U

92  234 90

Th

 a

2

214 Po 84  210 82 Pb  4 2 He

3

14 6 C  14 7 N  g

4

40 19 K  40 20 p   0 1

e

  0 0 

238 = 234 + 4 92 = 90 + 2 214 = 210 + 4 84 = 82 + 2 14 = 14+0 6 <> 7+0 40 = 40+0+0 19 = 20-1+0

Decay

Which of the following are possible reactions?

(a) and (b). Reactions (a) and (b) both conserve total charge and total mass number as required. Reaction (c) violates conservation of mass number with the sum of the mass numbers being 240 before reaction and being only 223 after reaction.

The Decay Constant

 The number of particles that decay in a given time is proportional to the total number of particles in a radioactive sample 

N

  λ is called the

decay constant

material will decay and determines the rate at which the The

decay rate

or

activity

, R, of a sample is defined as the number of decays per second

R

 

N



t

 

N

Radioactivity

Decays per second, or “activity”



N



t

  

N

No. of nuclei present decay constant Start with 16 14 C atoms.

After 6000 years, there are only 8 left.

How many will be left after another 6000 years?

1) 0 2) 4 3) 8 Every 6000 years ½ of atoms decay

Decay Curve

 The decay curve follows the equation

N



N e

0  

t

  The

half-life

is also a useful parameter The half-life is defined as the time it takes for half of any given number of radioactive nuclei to decay T 1 2  ln  2  0 .

693 

Units



The unit of activity, R , is the

Curie, Ci



1 Ci = 3.7 x 10

10

decays/second



The SI unit of activity is the

Becquerel, Bq



1 Bq = 1 decay / second

 Therefore, 1 Ci = 3.7 x 10 10 Bq 

The most commonly used units of activity are the mCi and the µCi

Decay Function



N e

0  

t



N

0

t T

1/2

time

Practice

The half-life for beta-decay of 14 C is ~6,000 years. You test a fossil and find that only 25% of its 14 C is un decayed. How old is the fossil?

3,000 years At 0 years: 100% remains 6,000 years 12,000 years At 6,000 years: 50% remains At 12,000 years: 25% remains

Radioactivity Quantitatively

Decays per second, or “activity”



N



t

  

N

No. of nuclei present Survival:



decay constant

N e

0  

t

No. of nuclei present at time t No. we started with at t=0 Instead of base e 1/2 : we can use base

t e

 

t T

1/2

where

T

1/ 2  0.693



Half life Then we can write



N e

0  

t



N

0

t T

1/2

Radioactivity Example

The half-life for beta-decay of around in 22920 years? 14 C is 5730 years. If you start with 1000 carbon-14 nuclei, how many will be

  

t



N

0

t T

1/2

T

1/ 2  0.693

 ) 

N

0

t T

1/2  1000 1    62 .5

22920 573 0

T

1/ 2  0.693

   0.693

5730   4 

N e

0  

t

 1000

e

   62.5

 4  22920) 

Uses of Radioactivity

   Carbon Dating   Beta decay of 14 C is used to date organic samples The ratio of 14 C to 12 C is used Smoke detectors    Ionization type smoke detectors use a radioactive source to ionize the air in a chamber A voltage and current are maintained When smoke enters the chamber, the current is decreased and the alarm sounds Radon pollution    Radon is an inert, gaseous element associated with the decay of radium It is present in uranium mines and in certain types of rocks, bricks, etc that may be used in home building May also come from the ground itself

Binding Energy

Which system “weighs” more?

1) Two balls attached by a relaxed spring.

2) Two balls attached by a stretched spring.

3) They have the same weight.

M 1 M 2 M 2 = M balls = M balls – M 1 + M + M spring spring = E spring /c 2 + E spring ≈ 10 -16 /c Kg 2

Strong Nuclear Force



Acts on Protons and Neutrons



Strong enough to overcome Coulomb repulsion



Acts over very short distances

Two atoms don’t feel force

Q Values (nice to know)

  Energy must also be conserved in nuclear reactions The energy required to balance a nuclear reaction is called the

Q value

of the reaction   An 

exothermic reaction

There is a mass “loss” in the reaction  There is a release of energy  Q is positive An 

endothermic reaction

There is a “gain” of mass in the reaction  Energy is needed, in the form of kinetic energy of the incoming particles  Q is negative

Problem: nuclear reactions

Determine the product of the reaction What is the Q value of the reaction?

7 3

Li

 2 4

He n

Determine the product of the reaction What is the Q value of the reaction?

7 3

Li

 2 4

He



Y X

?

 0 1

n

Given: reaction Find: Q = ?

It is easier to use atomic mass units rather than kg.

In order to balance the reaction, the total amount of nucleons (sum of must be the same on both sides. Same for the Z -number. A )

7

: A -numbers)

X

 1 

X

Number of protons ( Z ) :

3 2

Y

0  10 5

Thus, it is B , i.e.

3 7

Li

 2 4

He

 10

B

5  0 1

n

The Q-value is then

Q

2  

m

7

Li



m

4

He

  7.016005

u

 4.002603

u



m

10

B



n

 2  10.012938

u

 1.008665

u

 931.5

  2.79

MeV



Summary

 Nuclear Reactions  Nucleon number conserved  Charge conserved     Energy/Momentum conserved a particles = nucleii 2

He

b

-

particles = electrons g particles = high-energy photons

Survival :



N e

0  

t T

1/ 2  0.693

  Decays  Half Life is time for ½ of atoms to decay