matriculation physics ( nucleus )

1

PHYSICS CHAPTER 13

is defined as the is defined as the central core of an central core of an

atom that is atom that is positively charged positively charged

and contains and contains protons and protons and

neutrons.neutrons.

CHAPTER 13: NucleusCHAPTER 13: Nucleus(3 Hours)(3 Hours)

PHYSICS CHAPTER 13

2

At the end of this chapter, students should be able to: At the end of this chapter, students should be able to: StateState the properties of proton and neutron. the properties of proton and neutron. Define Define

proton numberproton number nucleon numbernucleon number isotopesisotopes

UseUse

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

spectrometer to identify isotopes.spectrometer to identify isotopes.

Learning Outcome:

ww

w.k

mp

h.m

atri

k.ed

u.m

y/p

hys

ics

ww

w.k

mp

h.m

atri

k.ed

u.m

y/p

hys

ics

13.1 Properties of nucleus (1 hour)

XAZ

PHYSICS CHAPTER 13

3

13.1.1 Nuclear structure A nucleus of an atom is made up of protons and neutrons that

known as nucleonsnucleons (is defined as the particles found inside the particles found inside the nucleusthe nucleus) as shown in Figure 13.1.

13.1 Properties of nucleus

ProtonProton

NeutronNeutron

Electron Electron

Figure 13.1Figure 13.1

PHYSICS CHAPTER 13

4

Proton and neutron are characterised by the following properties in Table 13.1.

For a neutral atom, The number of protons inside the nucleus The number of protons inside the nucleus

= the number of electrons orbiting the nucleus= the number of electrons orbiting the nucleus This is because the magnitude of an electron charge magnitude of an electron charge

equals to the magnitude of a proton charge but opposite equals to the magnitude of a proton charge but opposite in signin sign.

Proton (Proton (pp)) Neutron (Neutron (nn))

Charge (C)

Mass (kg)

+e+e 00

2710672.1 2710675.1

)1060.1( 19 )uncharged(

Table 13.1Table 13.1

PHYSICS CHAPTER 13

5

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

Any nucleus of elements in the periodic table called a nuclidenuclide is characterised by its atomic number Z and its mass number A.

The nuclide of an element can be represented as in the Figure 13.2.


Number Symbol Definition

Atomic number ZZ The number of protons in a nucleusThe number of protons in a nucleus

Neutron number NN The number of neutrons in a nucleusThe number of neutrons in a nucleus

Mass (nucleon)

numberAA The number of nucleons in a nucleusThe number of nucleons in a nucleus

Relationship :Relationship : (13.1)(13.1)

PHYSICS CHAPTER 13

6

The number of protons Z is not necessary equalnot necessary equal to the number

of neutrons N.e.g. :


Atomic numberAtomic number

Mass numberMass number

Element XElement X

Mg2412 S32

16; Pt19578;

21Z12 ZAN

PHYSICS CHAPTER 13

7

Since a nucleus can be modeled as tightly packed spheretightly packed sphere where each sphere is a nucleoneach sphere is a nucleon, thus the average radius of the nucleus is given by

31

0 ARR (13.2)(13.2)

where nucleus of radius average :Rfm 1.2 OR m102.1constant : 15

0R

number (nucleon) mass :A

femtometre (fermi)femtometre (fermi)

m 101fm 1 15

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 electronsd. 106 nucleons ; 48 protons e. 214 nucleons ; 131 protons

Solution :Solution :

a. Given Z =20 ; A =40

c. Given A=50 and Z=number of protons = number of electrons =24

Example 1 :

XAZ

Ca4020

XAZ Cr50

24

PHYSICS CHAPTER 13

9

What is meant by the following symbols?

State the mass number and sign of the charge for each entity above.


Example 2 :

n10 p1

1; e01;

n10

p11

e01

Neutron ;Neutron ; AA=1=1

Charge :Charge : neutral (uncharged)neutral (uncharged)

Proton ;Proton ; AA=1=1

Charge :Charge : positively chargedpositively charged

Electron ;Electron ; AA=0=0

Charge :Charge : negatively chargednegatively charged

PHYSICS CHAPTER 13

10

Complete the Table 13.3.Example 3 :


Element nuclide

Number of protons

Number of neutrons

Total charge in nucleus

Number of electrons

H11

N147

Na2311

Co5927

Be94

O168

S3116

Cs13355

U23892

1616 1515 16e16e 1616

PHYSICS CHAPTER 13

11

is defined as the nuclides/elements/atoms that have the the nuclides/elements/atoms that have the same atomic number same atomic number ZZ but different in mass but different in mass

number number AA. From the definition of isotope, thus the number of protons or number of protons or

electrons are equalelectrons are equal but different in the number of neutronsdifferent in the number of neutrons N for two isotopes from the same element.

For example : Hydrogen isotopes:

Oxygen isotopes:

13.1.2 Isotope

H11

H21

H31

:: Z Z=1, =1, AA=1, =1, NN=0=0

:: Z Z=1, =1, AA=2, =2, NN=1=1

:: Z Z=1, =1, AA=3, =3, NN=2=2

protonproton )p(11

deuteriumdeuterium )D(21

tritiumtritium )T(31

O168

O178

O188

:: Z Z=8, =8, AA=16, =16, NN=8=8

:: Z Z=8, =8, AA=17, =17, NN=9=9

: : ZZ=8, =8, AA=18, =18, NN=10=10

equalequal

equalequal not equalnot equal

not equalnot equal

PHYSICS CHAPTER 13

12

Mass spectrometer is a device that detect the presence of detect the presence of isotopesisotopes and determines the mass of the isotopemass of the isotope from known mass of the common or stable isotope.

Figure 13.3 shows a schematic diagram of a Bainbridge mass spectrometer.

13.1.3 Bainbridge mass spectrometer


++--

++++++++++++

------------

2B

E

1B

1r2r

S1S2

S3

Photographic plate

Plate P1 Plate P2

Ion source

Ions beam

Evacuated chamber

1m 2m

d

Separation between isotopes

PHYSICS CHAPTER 13

13

Plate P1 Plate P2

Working principleWorking 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 field B1 and a uniform

electric field E that are perpendicular to each other. The beam with selected velocity v passes through the velocity

selector without deflection and emerge from the slit S3. Hence,

the force on an ion due to the magnetic field B1 and the electric

field E are equal in magnitude but opposite in direction (Figure 13.4).

v

EF

BF

Using Fleming’s left hand rule.


PHYSICS CHAPTER 13

14

Thus the selected velocity is

The ions beam emerging from the slit S3 enter an evacuated

chamber of uniform magnetic field B2 which is perpendicular to

the selected velocity v. The force due to the magnetic field B2

causes an ion to move in a semicircle path of radius r given by

EB FF qEqvB 90sin1

1B

Ev

cB FF

r

mvqvB

2

2 90sin

qB

mvr

2

and 1B

Ev

PHYSICS CHAPTER 13

15

Since the magnetic fields B1 and B2 and the electric field E are

constants 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 the

ratio of their masses is given by

kmr and constant21

qBB

Ek

mr

2

1

2

1

r

r

m

m (13.4)(13.4)

qBB

mEr

21

(13.3)(13.3)

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 mass spectrometer. The mutually perpendicular electric and magnetic fields in the velocity selector are 0.4 MV m1 and 0.7 T respectively. These ions then enter a chamber of uniform magnetic flux density 1.0 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 :Solution :a. The selected velocity of the ions is

Example 4 :

T 0.1 T; 7.0;m V 104.0 2116 BBE

1B

Ev

7.0

104.0 6v

15 s m 1071.5 v

PHYSICS CHAPTER 13

17


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

T 0.1 T; 7.0;m V 104.0 2116 BBE

qBB

Emr

21

11

m 119.0

19

626

11060.10.17.0

104.01032.3

r

m 130.0

19

626

21060.10.17.0

104.01065.3

r

12 ddd 12 22 rrd 122 rr 119.0130.02

m 102.2 2d

Figure 13.3

PHYSICS CHAPTER 13

18

At the end of this chapter, students should be able to: At the end of this chapter, students should be able to: DefineDefine mass defect and binding energy. mass defect and binding energy. UseUse formulae formulae

IdentifyIdentify the average value of binding energy per the average value of binding energy per nucleon of stable nuclei from the graph of binding nucleon of stable nuclei from the graph of binding energy per nucleon against nucleon number.energy per nucleon against nucleon number.

Learning Outcome:

ww

w.k

mp

h.m

atri

k.ed

u.m

y/p

hys

ics

ww

w.k

mp

h.m

atri

k.ed

u.m

y/p

hys

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13.2 Binding energy and mass defect (2 hours)

2mcE

PHYSICS CHAPTER 13

19

13.2.1 Einstein mass-energy relation From the theory of relativity leads to the idea that mass is a mass is a

form of energyform of energy. Mass and energy can be related by the following relation:

e.g. The energy for 1 kg of substance is

13.2 Binding energy and mass defect

2mcE (13.5)(13.5)

wheremassrest : m

)s m10003( in vacuumlight of speed: 18 .c

energy ofamount : E

J1000.9 16E

2mcE 28 )1000.3)(1(

PHYSICS CHAPTER 13

20

Unit conversion of mass and energyUnit conversion of mass and energy The electron-volt (electron-volt (eVeV))

is a unit of energyunit of energy. is defined as the kinetic energy gained by an electron in the kinetic energy gained by an electron in

being accelerated by a potential difference (voltage) of 1 being accelerated by a potential difference (voltage) of 1 voltvolt.

The atomic mass unit (atomic mass unit (uu)) is a unit of massunit of mass. is defined as exactly the mass of a neutral carbon-12 the mass of a neutral carbon-12

atom.atom.

J1060.1eV 1 19J 10601eV 10Me 1 136 .V

12

1

12

Cof massu 1

126

kg10661u 1 27 .

PHYSICS CHAPTER 13

21

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

in joule,

in eV/c2 or MeV/c2,

J 1049.1 10E

2mcE 2827 )1000.3)(1066.1(

J 1049.1u 1 10

2619

10

eV/ 105.9311060.1

1049.1 cE

26 eV/ 105.931u 1 c

2MeV/ 5.931u 1 cOR

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 defectmass defect and is defined as the mass mass difference between the total mass of the constituent difference between the total mass of the constituent nucleons and the mass of a nucleusnucleons and the mass of a nucleus.

The reduction in mass arises because the act of combining act of combining the nucleons to form the nucleusthe nucleons to form the nucleus causes some of their mass their mass to be released as energyto be released as energy.

Any attempt to separate the nucleonsseparate the nucleons would involve them being given this same amount of energysame amount of energy. This energy is called the binding energybinding energy of the nucleus.

13.2.2 Mass defect

np NmZmM A where

neutron a of mass: nmproton a of mass: pm

AMNmZmm np (13.6)(13.6)

PHYSICS CHAPTER 13

23

The binding energy of a nucleus is defined as the energy the energy required to separate completely all the nucleons in the required to separate completely all the nucleons in the nucleusnucleus.

The binding energy of the nucleus is equal to the energy equivalent 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 itstability depends only on the forces exist inside it.

The forces involve inside the nucleus are repulsive electrostatic (Coulomb) forces between repulsive electrostatic (Coulomb) forces between

protonsprotons and attractive forces that bind all nucleonsattractive forces that bind all nucleons together in the

nucleus.

13.2.3 Binding energy

2B cmE (13.7)(13.7)Binding energy Binding energy

in joulein jouleSpeed of light in Speed of light in vacuumvacuumMass defect in kgMass defect in kg

13.2.4 Nucleus stability

PHYSICS CHAPTER 13

24

These attractive force is called nuclear forcenuclear force and is responsible for nucleus stability.

The general properties of the nuclear forcegeneral properties of the nuclear force are summarized as follow : The nuclear force is attractive and is the strongest forceattractive and is the strongest force

in nature. It is a short range forceshort range force . It means that a nucleon is

attracted only to its nearest neighbours in the nucleus. It does not depend on chargedoes not depend on charge; neutrons as well as protons

are bound and the nuclear force is same for bothnuclear force is same for both.

e.g.

The nuclear force depends on the binding energy per depends on the binding energy per nucleonnucleon.

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

The magnitude of nuclear magnitude of nuclear forcesforces are samesame.

PHYSICS CHAPTER 13

25

Note that a nucleus nucleus is stablestable if the nuclear force greater thannuclear force greater than the Coulomb forceCoulomb force and vice versa.

The binding energy per nucleonbinding energy per nucleon of a nucleus is a measure of measure of the nucleus stabilitythe nucleus stability where

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

a function of mass (nucleon) number A.

)number(Nucleon

)(energy Bindingnucleonper energy Binding B

A

E

A

mc2

nucleonper energy Binding

(13.8)(13.8)

PHYSICS CHAPTER 13

26Mass number A

Bin

din

g e

ne

rgy

per

n

ucle

on

(M

eV

/nu

cle

on)

Greatest stabilityGreatest stability


PHYSICS CHAPTER 13

27

From Figure 13.5,

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

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

For the nuclei with A between 50 and 80, the value of EB/A

ranges 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 most stable nucleus.

For A > 62, the values of EB/A decreases slowly, indicating

that the nucleons are on average less tightly bound.

For heavy nuclei with A 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 number N against atomic number Z for a number of stable nuclei.

PHYSICS CHAPTER 13

28Atomic number Z

Ne

utr

on

nu

mb

er,

N

NN==ZZ

Line of Line of stabilitystability


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 stabilityline of stability.

The dashed line corresponds to the condition N=Z. The light stable nuclei contain an equal number of light stable nuclei contain an equal number of

protons and neutronsprotons and neutrons (N=Z) but in heavy stable nucleiheavy stable nuclei the number of neutrons always greater than the number number of neutrons always greater than the number

of protons (above of protons (above Z Z =20)=20) hence the line of stability

deviates upwarddeviates upward from the line of N=Z. This means as the number of protons increasenumber of protons increase, the

strength of repulsive coulomb force increasesstrength of repulsive coulomb force increases which tends to break the nucleus apart.

As a result, more neutrons are needed to keep the more neutrons are needed to keep the nucleus stablenucleus stable because neutrons experience only the neutrons experience only the attractive nuclear forceattractive nuclear force.

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)


The mass defect of the aluminum nucleus is

Example 5 : Al27

13

Al2713 13Z and 1327 N

14N

u 2415.0m 98154.2600867.11400782.113 Alnp MNmZmm

PHYSICS CHAPTER 13

31


The binding energy of the aluminum nucleus can be calculated by using two method.

11stst method: method:

Thus the binding energy in MeV is

271066.12415.0 mkg 100089.4 28

kg 10661u 1 27 .

2828B 1000.3100089.4 E

J 10608.3 11B

E

2B cmE

in kgin kg

MeV 226B E

13

11

B 1060.1

10608.3

EJ 10601MeV 1 13 .

PHYSICS CHAPTER 13

32


22ndnd method: method: 2B cmE

MeV 225B E

22

u 1

MeV/ 5.931c

cm

2MeV/ 5.931u 1 cin uin u

22

u 1

MeV/ 5.931u 2415.0 c

c

PHYSICS CHAPTER 13

33

Calculate the binding energy per nucleon of a boron nucleus in J/nucleon.


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

atomic mass of boron, MB=10.01294 u)


The mass defect of the boron nucleus is

Example 6 : B10

5

B105 5Z and 510 N

5N

u 06951.0m

01294.1000867.1500782.15

Bnp MNmZmm

PHYSICS CHAPTER 13

34


The binding energy of the boron nucleus is given by

Hence the binding energy per nucleon is

2827 1000.31066.106951.0

J 1004.1 11B

E

2B cmE

J/nucleon 1004.1 12B A

E10

1004.1 11B

A

E

PHYSICS CHAPTER 13

35

Why is the uranium-238 nucleus less stable than carbon-12 nucleus ? Give an explanation by referring to the repulsive coulomb force and the binding energy per nucleon.


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 :Solution :From the aspect of repulsive coulomb force :

Uranium-238 nucleus has 92 protons but the carbon-12 nucleus 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 13

36

Solution :Solution :From the aspect of binding energy per nucleon:

Carbon-12 :The mass defect :

The binding energy per nucleon:

C126

6Z and 6N

Cnp MNmZmm

u 09894.0m 00000.1200867.1600782.16

A

cm

A

E 2

C

B

12

u 1MeV/ 5.931

u 09894.0 22

cc

nMeV/nucleo 68.7C

B

A

E

PHYSICS CHAPTER 13

37

Uranium-238 :The mass defect :

The binding energy per nucleon:

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

U23892

92Z and 146N

u 93447.1m

05079.23800867.114600782.192 m

238

u 1MeV/ 5.931

u 93447.1 22

U

B

cc

A

E

nMeV/nucleo 57.7U

B

A

E

PHYSICS CHAPTER 13

38

Exercise 13.1 :Given c =3.00108 m s1, mn=1.00867 u, mp=1.00782 u

1. Calculate the binding energy in joule of a deuterium nucleus. The mass of a deuterium nucleus is 3.34428 1027 kg.

ANS. :ANS. : 2.782.7810101313 J J

2. The mass of neon-20 nucleus is 19.99244 u. Calculate the binding energy per nucleon of neon-20 nucleus in MeV per nucleon.

ANS. :ANS. : 8.03 MeV/nucleon8.03 MeV/nucleon

3. Determine the energy required to remove one neutron from an oxygen-16 . The atomic mass for oxygen-16 is 15.994915 u

(Physics, 3(Physics, 3rdrd edition, James S. Walker, Q39, p.1108) edition, James S. Walker, Q39, p.1108)

ANS. :ANS. : 15.7 MeV15.7 MeV

Ne2010

O168

39

PHYSICS CHAPTER 13

Next Chapter…CHAPTER 14 :Nuclear reaction

matriculation physics ( nucleus )

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