phy 310 chapter 7

is defined as the

central core of an

atom that is

positively

charged and

contains protons

and neutrons.

CHAPTER 7: Nucleus

(3 Hours)

Dr Ahmad Taufek Abdul RahmanSchool of Physics & Material Studies, Faculty of Applied Sciences, Universiti Teknologi MARA Malaysia, Campus of Negeri Sembilan

1

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:

7.1 Properties of nucleus (1 hour)

XAZ

2

DR.ATAR @ UiTM.NS PHY310 NUCLEUS

7.1.1 Nuclear structure

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

known as nucleons (is defined as the particles found inside

the nucleus) as shown in Figure 7.1.

7.1 Properties of nucleus

Proton

Neutron

Electron

Figure 7.1

3


Proton and neutron are characterised by the following properties

in Table 7.1.

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 charge

equals to the magnitude of a proton charge but opposite

in sign.

Proton (p) Neutron (n)

Charge (C)

Mass (kg)

+e 0

2710672.1 2710675.1

)1060.1( 19 )uncharged(

Table 7.1

4


Nuclei are characterised by the number and type of nucleons they contain as shown in Table 7.2.

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

characterised by its atomic number Z and its mass number A.

The nuclide of an element can be represented as in the Figure 7.2.

Table 7.2

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)

numberA The number of nucleons in a nucleus

Relationship : (7.1)

5


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

of neutrons N.e.g. :

Figure 7.2

Atomic number

Mass number

Element X

Mg24

12 S32

16; Pt195

78;

21Z12 ZAN 6


Since a nucleus can be modeled as tightly packed sphere

where each sphere is a nucleon, thus the average radius of

the nucleus is given by

31

0 ARR (7.2)

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

0

Rnumber (nucleon) mass :A

femtometre (fermi)

m 101fm 1 15

7


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 protons

e. 214 nucleons ; 131 protons

Solution :

a. Given Z =20 ; A =40

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

Example 1 :

XA

ZCa40

20

XA

ZCr50

24

8


What is meant by the following symbols?

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

Solution :

Example 2 :

n1

0 p1

1; e0

1;

n1

0

p1

1

e0

1

Neutron ; A=1

Charge : neutral (uncharged)

Proton ; A=1

Charge : positively charged

Electron ; A=0

Charge : negatively charged

9


Complete the Table 7.3.

Example 3 :

Table 7.3

Element

nuclide

Number of

protons

Number of

neutrons

Total charge

in nucleus

Number of

electrons

H1

1

N14

7

Na23

11

Co59

27

Be9

4

O16

8

S31

16

Cs133

55

U238

92

16 15 16e 16

10


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

same atomic number Z but different in mass number A.

From the definition of isotope, thus the number of protons or electrons are equal but different in the number of neutrons

N for two isotopes from the same element.

For example :

Hydrogen isotopes:

Oxygen isotopes:

7.1.2 Isotope

H1

1

H2

1

H3

1

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

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

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

proton )p(1

1

deuterium )D(2

1

tritium )T(3

1

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

11


Mass spectrometer is a device that detect the presence of

isotopes and determines the mass of the isotope from known

mass of the common or stable isotope.

Figure 7.3 shows a schematic diagram of a Bainbridge mass

spectrometer.

29.1.3 Bainbridge mass spectrometer

Figure 7.3

+-

++++

++

----

--

2B

E

1B

1r2r

S1S2

S3

Photographic plate

Plate P1 Plate P2

Ion source

Ions beam

Evacuated

chamber

1m 2m

d

Separation

between isotopes

12


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 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

7.4).

v

EF

BF

Using Fleming’s

left hand rule.Figure 7.4 13


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

14


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 7.3 then the ratio

of their masses is given by

kmr and constant21

qBB

Ek

mr

2

1

2

1

r

r

m

m (7.4)

qBB

mEr

21

(7.3)

15


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 :

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 v16


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

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 7.3

17


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 per

nucleon of stable nuclei from the graph of binding

energy per nucleon against nucleon number.

Learning Outcome:

7.2 Binding energy and mass defect (2 hours)

2mcE

18


7.2.1 Einstein mass-energy relation

From the theory of relativity leads to the idea that mass is a

form of energy.

Mass and energy can be related by the following relation:

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

7.2 Binding energy and mass defect

2mcE (7.5)

where

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

energy ofamount : E

J1000.9 16E

2mcE 28)1000.3)(1(

19


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

The atomic mass unit (u)

is a unit of mass.

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

J1060.1eV 1 19

J 10601eV 10Me 1 136 .V

12

1

12

Cof massu 1

126

kg10661u 1 27 .

20


1 atomic mass unit (u) can be converted into the unit of

energy by using the mass-energy relation (eq. 7.5).

in joule,

in eV/c2 or MeV/c2,

J 1049.1 10E

2mcE 2827 )1000.3)(1066.1(

J 1049.1u 1 10

26

19

10

eV/ 105.9311060.1

1049.1 cE

26 eV/ 105.931u 1 c

2MeV/ 5.931u 1 c

OR

21


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 mass difference 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 mass to be released as energy.

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

7.2.2 Mass defect

np NmZmM A where

neutron a of mass: nmproton a of mass: pm

AMNmZmm np(7.6)

22


The binding energy of a nucleus is defined as the energy

required to separate completely all the nucleons in the

nucleus.

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 it.

The forces involve inside the nucleus are

repulsive electrostatic (Coulomb) forces between protons and

attractive forces that bind all nucleons together in the nucleus.

7.2.3 Binding energy

2B cmE (7.7)Binding energy

in jouleSpeed of light in

vacuumMass defect in kg

7.2.4 Nucleus stability

23


These attractive force is called nuclear force and is responsible

for nucleus stability.

The general properties of the nuclear force are summarized

as 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.

24


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 of

the nucleus stability where

Figure 7.5 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

mc 2

nucleonper energy Binding

(7.8)

25


Mass number A

Bin

din

g e

ne

rgy p

er

nu

cle

on

(M

eV

/nu

cle

on

)

Greatest stability

Figure 7.526


From Figure 7.5,

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/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 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 7.6 shows a graph of neutron number N against atomic

number Z for a number of stable nuclei.

27


Atomic number Z

Ne

utr

on

nu

mb

er,

N

N=Z

Line of

stability

Figure 7.628


From Figure 7.6,

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 condition N=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 number

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

deviates upward from the line of N=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.

29


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 1327N

14N

u 2415.0m

98154.2600867.11400782.113

Alnp MNmZmm

30


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 m

kg 100089.4 28

kg 10661u 1 27 .

2828

B 1000.3100089.4 E

J 10608.3 11

B

E

2

B cmE

in kg

MeV 226B E

13

11

B1060.1

10608.3

E

J 10601MeV 1 13 .

31


Solution :

2nd method: 2

B cmE

MeV 225B E

22

u 1

MeV/ 5.931c

cm

2MeV/ 5.931u 1 c

in u

22

u 1

MeV/ 5.931u 2415.0 c

c

32


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)

Solution :

The mass defect of the boron nucleus is

Example 6 :

B105

B10

5 5Z and 510N5N

u 06951.0m

01294.1000867.1500782.15

Bnp MNmZmm

33


Solution :

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 11

B

E

2

B cmE

J/nucleon 1004.1 12B A

E10

1004.1 11

B

A

E

34


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 :

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

35


Solution :

From the aspect of binding energy per nucleon:

Carbon-12 :

The mass defect :

The binding energy per nucleon:

C12

66Z and 6N

Cnp MNmZmm

u 09894.0m

00000.1200867.1600782.16

A

cm

A

E 2

C

B

12

u 1

MeV/ 5.931u 09894.0 2

2

cc

nMeV/nucleo 68.7C

B

A

E

36


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.

U238

9292Z and 146N

u 93447.1m

05079.23800867.114600782.192 m

238

u 1

MeV/ 5.931u 93447.1 2

2

U

B

cc

A

E

nMeV/nucleo 57.7U

B

A

E

37


Exercise 7.1 :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. : 2.781013 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. : 8.03 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, 3rd edition, James S. Walker, Q39, p.1108)

ANS. : 15.7 MeV

Ne2010

O16

8

38


Next Chapter…CHAPTER 8 :

RADIOACTIVITY

39


phy 310 chapter 7

Technology

element number of number

number of protons z

atomic number z

number of neutrons n

nucleus table

properties of nucleus

nsphy310 nucleusexample

nsphy310 nucleus7