Note Chapter13 SF027

8/3/2019 Note Chapter13 SF027

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PHYSICS CHAPTER 13

1

is defined as thecentral core of an

atom that ispositively

charged andcontains protons

and neutrons.

CHAPTER 13: Nucleus

(3 Hours)


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PHYSICS CHAPTER 13

2

At the end of this chapter, students should be able to:

State the properties of proton and neutron.

Define

proton number

nucleon number

isotopes

Use

to represent a nucleus. Explain the working principle and the use of mass

spectrometer to identify isotopes.

Learning Outcome:

www.kmph

.matrik.e

du.my/physic

s

13.1 Properties of nucleus (1 hour)

XAZ


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PHYSICS CHAPTER 13

3

13.1.1 Nuclear structure

A nucleus of an atom is made up of protons and neutrons thatknown as nucleons (is defined as the particles found insidethe nucleus) as shown in Figure 13.1.

13.1 Properties of nucleus

Proton

Neutron

Electron

Figure 13.1


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PHYSICS CHAPTER 13

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Proton and neutron are characterised by the following propertiesin Table 13.1.

For a neutral atom,

The number of protons inside the nucleus

= the number of electrons orbiting the nucleus

This is because the magnitude of an electron chargeequals to the magnitude of a proton charge but oppositein sign.

Proton (p) Neutron (n)

Charge (C)

Mass (kg)

+e 0

27

10672.1

27

10675.1

)1060.1(19

)uncharged(

Table 13.1


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PHYSICS CHAPTER 13

5

Nuclei are characterised by the number and type of nucleonsthey contain as shown in Table 13.2.

Any nucleus of elements in the periodic table called a nuclide is

characterised by its atomic numberZand its mass numberA.

The nuclide of an element can be represented as in the Figure13.2.

Table 13.2

Number Symbol Definition

Atomic number Z The number of protons in a nucleus

Neutron number N The number of neutrons in a nucleus

Mass (nucleon)number

A The number of nucleons in a nucleus

Relationship : (13.1)


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PHYSICS CHAPTER 13

6

The number of protonsZis not necessary equal to the number

of neutronsN.e.g. :

Figure 13.2

Atomic number

Mass number

Element X

Mg2412 S32

16; Pt195

78;

21Z12 ZAN


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PHYSICS CHAPTER 13

7

Since a nucleus can be modeled as tightly packed spherewhere each sphere is a nucleon, thus the average radius of

the nucleus is given by

31

0ARR (13.2)

where

nucleusofradiusaverage:R fm1.2ORm102.1constant: 150R

number(nucleon)mass:A

femtometre (fermi)

m101fm115


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PHYSICS CHAPTER 13

8

Based on the periodic table of element, Write down the symbol of

nuclide for following cases:

a.Z=20 ;A =40

b.Z=17 ;A =35

c. 50 nucleons ; 24 electrons

d. 106 nucleons ; 48 protonse. 214 nucleons ; 131 protons

Solution :

a. GivenZ=20 ;A =40

c. GivenA=50 andZ=number of protons = number of electrons =24

Example 1 :

XA

Z Ca40

20

XA

ZCr

50

24


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PHYSICS CHAPTER 13

9

What is meant by the following symbols?

State the mass number and sign of the charge for each entityabove.

Solution :

Example 2 :

n10 p1

1; e01;

n10

p11

e01

Neutron ;A=1

Charge : neutral (uncharged)

Proton ; A=1

Charge : positively charged

Electron ;A=0

Charge : negatively charged


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PHYSICS CHAPTER 13

10

Complete the Table 13.3.

Example 3 :

Table 13.3

Elementnuclide

Number ofprotons

Number ofneutrons

Total chargein nucleus

Number ofelectrons

H11

N147

Na2311

Co59

27

Be9

4

O168

S31

16

Cs13355

U

238

92

16 15 16e 16


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PHYSICS CHAPTER 13

11

is defined as the nuclides/elements/atoms that have the

same atomic numberZ but different in mass numberA. From the definition of isotope, thus the number of protons or

electrons are equal but different in the number of neutrons

Nfor two isotopes from the same element.

For example :

Hydrogen isotopes:

Oxygen isotopes:

13.1.2 Isotope

H11

H21

H31

: Z=1,A=1,N=0

: Z=1,A=2,N=1

: Z=1,A=3,N=2

proton )p(1

1

deuterium )D(21

tritium )T(31

O16

8

O17

8

O18

8

: Z=8,A=16,N=8

: Z=8,A=17,N=9

:Z=8,A=18,N=10

equal

equal not equal

not equal


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PHYSICS CHAPTER 13

12

Mass spectrometer is a device that detect the presence of

isotopes and determines the mass of the isotope from knownmass of the common or stable isotope.

Figure 13.3 shows a schematic diagram of a Bainbridge massspectrometer.

13.1.3 Bainbridge mass spectrometer

Figure 13.3

+-

+++++

+

-----

-

2B

E

1

B

1r2r

S1S2

S3

Photographic plate

Plate P1 Plate P2

Ion source

Ions beam

Evacuatedchamber

1m 2m

d

Separationbetween isotopes


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PHYSICS CHAPTER 13

13

Plate P1 Plate P2

Working principle

Ions from an ion source such as a discharge tube are narrowed

to a fine beam by the slits S1 and S2. The ions beam then passes through a velocity selector (plates

P1 and P2) which uses a uniform magnetic fieldB1 and a uniform

electric fieldEthat are perpendicular to each other.

The beam with selected velocity v passes through the velocityselector without deflection and emerge from the slit S3. Hence,

the force on an ion due to the magnetic fieldB1 and the electric

fieldEare equal in magnitude but opposite in direction (Figure13.4).

v

EF

BF

Using Flemings

left hand rule.Figure 13.4


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PHYSICS CHAPTER 13

15

Since the magnetic fieldsB1 andB2 and the electric fieldEareconstants and every ion entering the spectrometer contains the

same amount of charge q, therefore

If ions of masses m1 and m2 strike the photographic plate with

radii r1 and r2 respectively as shown in Figure 13.3 then theratio of their masses is given by

kmr and constant21

qBB

Ek

mr

2

1

2

1

r

r

m

m (13.4)

qBB

mEr

21

(13.3)


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PHYSICS CHAPTER 13

16

A beam of singly charged ions of isotopes Ne-20 and Ne-22 travels

straight through the velocity selector of a Bainbridge massspectrometer. The mutually perpendicular electric and magneticfields in the velocity selector are 0.4 MV m1 and 0.7 T respectively.These ions then enter a chamber of uniform magnetic flux density1.0 T. Calculate

a. the selected velocity of the ions,b. the separation between two isotopes on the photographic plate.

(Given the mass of Ne-20 = 3.32 1026 kg; mass of Ne-22 =

3.65 1026 kg and charge of the beam is 1.60 1019 C)

Solution :

a. The selected velocity of the ions is

Example 4 :

T0.1T;7.0;mV104.0 2116

BBE

1B

Ev

7.0

104.0 6v

15sm1071.5

v


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PHYSICS CHAPTER 13

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Solution :

b. The radius of the circular path made by isotope Ne-20 is

and the radius of the circular path made by isotope Ne-22 is

Therefore the separation between the isotope of Ne is given by

T0.1T;7.0;mV104.0 2116 BBE

qBB

Emr

21

11

m119.0

19

626

11060.10.17.0

104.01032.3

r

m130.0

19

626

21060.10.17.0

104.01065.3

r

12 ddd 12 22 rrd 122 rr 119.0130.02

m102.2

2d

Figure 13.3


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PHYSICS CHAPTER 13

18

At the end of this chapter, students should be able to:

Define mass defect and binding energy.

Use formulae

Identify the average value of binding energy pernucleon of stable nuclei from the graph of binding

energy per nucleon against nucleon number.

Learning Outcome:

www.kmph.matrik.e

du.my/physics

13.2 Binding energy and mass defect (2 hours)

2mcE


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PHYSICS CHAPTER 13

20

Unit conversion of mass and energy

The electron-volt (eV)

is a unit of energy.

is defined as the kinetic energy gained by an electron inbeing accelerated by a potential difference (voltage) of 1volt.

The atomic mass unit (u)

is a unit of mass.

is defined as exactly the mass of a neutral carbon-12atom.

J1060.1eV119

J10601eV10Me1 136 .V

12

1

12

Cofmassu1

12

6

kg10661u127

.


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PHYSICS CHAPTER 13

21

1 atomic mass unit (u) can be converted into the unit ofenergy by using the mass-energy relation (eq. 13.5).

in joule,

in eV/c2orMeV/c2,

J1049.110E

2

mcE 2827 )1000.3)(1066.1(

J1049.1u1

10

26

19

10

eV/105.931

1060.1

1049.1 cE

26eV/105.931u1 c

2MeV/5.931u1 c

OR


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PHYSICS CHAPTER 13

22

The mass of a nucleus (MA) is always less than the total mass

of its constituent nucleons (Zmp+Nmn) i.e.

Thus the difference in this mass is given by

where m is called mass defect and is defined as the massdifference between the total mass of the constituent

nucleons and the mass of a nucleus. The reduction in mass arises because the act of combining

the nucleons to form the nucleus causes some of their massto be released as energy.

Any attempt to separate the nucleons would involve thembeing given this same amount of energy. This energy is calledthe binding energy of the nucleus.

13.2.2 Mass defect

np NmZmMA where

neutronaofmass:nmprotonaofmass:pm

AMNmZmm np (13.6)


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PHYSICS CHAPTER 13

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The binding energy of a nucleus is defined as the energy

required to separate completely all the nucleons in thenucleus.

The binding energy of the nucleus is equal to the energyequivalent of the mass defect. Hence

Since the nucleus is viewed as a closed packed of nucleons,thus its stability depends only on the forces exist inside it.

The forces involve inside the nucleus are

repulsive electrostatic (Coulomb) forces betweenprotons and

attractive forces that bind all nucleons together in thenucleus.

13.2.3 Binding energy

2

B cmE (13.7)Binding energyin jouleSpeed of light invacuum

Mass defect in kg13.2.4 Nucleus stability


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PHYSICS CHAPTER 13

24

These attractive force is called nuclear force and is responsible

for nucleus stability.

The general properties of the nuclear force are summarizedas follow :

The nuclear force is attractive and is the strongest force

in nature.

It is a short range force . It means that a nucleon is

attracted only to its nearest neighbours in the nucleus.

It does not depend on charge; neutrons as well as protons

are bound and the nuclear force is same for both.

e.g.

The nuclear force depends on the binding energy per

nucleon.

proton-proton (p-p)

neutron-neutron (n-n)proton-neutron (p-n)

The magnitude of nuclear

forces are same.


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PHYSICS CHAPTER 13

25

Note that a nucleus is stable if the nuclear force greater than

the Coulomb force and vice versa.

The binding energy per nucleon of a nucleus is a measure ofthe nucleus stability where

Figure 13.5 shows a graph of the binding energy per nucleon as

a function of mass (nucleon) numberA.

)number(Nucleon

)(energyBindingnucleonperenergyBinding B

A

E

A

mc2

nucleonperenergyBinding

(13.8)


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PHYSICS CHAPTER 13

26

Mass numberA

Bindingenergyper

nucleon(M

eV/nucleon)

Greatest stability

Figure 13.5


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PHYSICS CHAPTER 13

27

From Figure 13.5,

The value ofEB/A rises rapidly from 1 MeV/nucleon to 8

MeV/nucleon with increasing mass number A for light nuclei.

For the nuclei withA between 50 and 80, the value ofEB/Aranges between 8.0 and 8.9 Mev/nucleon. The nuclei in

these range are very stable. The maximum value of the

curve occurs in the vicinity of nickel, which has the moststable nucleus.

ForA > 62, the values ofEB/A decreases slowly, indicatingthat the nucleons are on average less tightly bound.

For heavy nuclei withA between 200 to 240, the binding

energy is between 7.5 and 8.0 MeV/nucleon. These nuclei

are unstable and radioactive.

Figure 13.6 shows a graph of neutron numberNagainst atomic

numberZfor a number of stable nuclei.


28/39

PHYSICS CHAPTER 13

28Atomic numberZ

Neutron

number,N

N=Z

Line ofstability

Figure 13.6


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PHYSICS CHAPTER 13

29

From Figure 13.6,

The stable nuclei are represented by the blue dots, which lie

in a narrow range called the line of stability.

The dashed line corresponds to the conditionN=Z.

The light stable nuclei contain an equal number of

protons and neutrons (N=Z) but in heavy stable nuclei

the number of neutrons always greater than the numberof protons (aboveZ =20) hence the line of stability

deviates upward from the line ofN=Z.

This means as the number of protons increase, the

strength of repulsive coulomb force increases which

tends to break the nucleus apart.

As a result, more neutrons are needed to keep the

nucleus stable because neutrons experience only the

attractive nuclear force.


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PHYSICS CHAPTER 13

30

Calculate the binding energy of an aluminum nucleus in MeV.

(Given mass of neutron, mn=1.00867 u ; mass of proton,

mp=1.00782 u ; speed of light in vacuum, c=3.00108 m s1 and

atomic mass of aluminum,MAl=26.98154 u)

Solution :

The mass defect of the aluminum nucleus is

Example 5 :

Al2713

Al2713 13Z and 1327N14N

u2415.0m 98154.2600867.11400782.113

Alnp MNmZmm


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PHYSICS CHAPTER 13

31

Solution :

The binding energy of the aluminum nucleus can be calculated by

using two method.1st method:

Thus the binding energy in MeV is

271066.12415.0 mkg100089.4 28

kg10661u127

.

2828B 1000.3100089.4 EJ10608.3 11B

E

2B

cmE

in kg

MeV226B

E

13

11

B1060.1

10608.3

E

J10601MeV113

.


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PHYSICS CHAPTER 13

32

Solution :

2nd method: 2B

cmE

MeV225B E

22

u1

MeV/5.931c

cm

2MeV/5.931u1 c

in u

22

u1

MeV/5.931u2415.0 c

c


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PHYSICS CHAPTER 13

34

Solution :

The binding energy of the boron nucleus is given by

Hence the binding energy per nucleon is

2827 1000.31066.106951.0

J1004.1 11BE

2B cmE

J/nucleon1004.1 12B A

E

10

1004.1 11B

A

E

S CS C


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PHYSICS CHAPTER 13

35

Why is the uranium-238 nucleus less stable than carbon-12

nucleus ? Give an explanation by referring to the repulsivecoulomb force and the binding energy per nucleon.

(Given mass of neutron, mn=1.00867 u ; mass of proton,

mp=1.00782 u ; speed of light in vacuum, c=3.00108 m s1; atomic

mass of carbon-12,MC=12.00000 u and atomic mass of uranium-238,MU=238.05079 u )

Solution :

From the aspect of repulsive coulomb force :

Uranium-238 nucleus has 92 protons but the carbon-12nucleus has only 6 protons.

Therefore the coulomb force inside uranium-238 nucleus

is or 15.3 times the coulomb force inside carbon-12

nucleus.

Example 7 :

C126 U23892

6

92

PHYSICS CHAPTER 3


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PHYSICS CHAPTER 13

36

Solution :

From the aspect of binding energy per nucleon:

Carbon-12 :The mass defect :

The binding energy per nucleon:

C12

6 6Z and 6N

Cnp MNmZmm

u09894.0m

00000.1200867.1600782.16

A

cm

A

E2

C

B

12

u1MeV/5.931u09894.0 2

2

cc

nMeV/nucleo68.7C

B

A

E

PHYSICS CHAPTER 13


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PHYSICS CHAPTER 13

37

Uranium-238 :

The mass defect :

The binding energy per nucleon:

Since the binding energy of uranium-238 nucleus less than thebinding energy of carbon-12 and the coulomb force inside uranium-238 nucleus greater than the coulomb force inside carbon-12nucleus therefore uranium-238 nucleus less stable than carbon-12nucleus.

U23892 92Z and 146N

u93447.1m 05079.23800867.114600782.192 m

238u1

MeV/5.931u93447.1 2

2

U

B

cc

A

E

nMeV/nucleo57.7

U

B

A

E


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PHYSICS CHAPTER 13


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PHYSICS CHAPTER 13

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Next ChapterCHAPTER 14 :Nuclear reaction

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Note Chapter13 SF027