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UNIT 17: NUCLEAR PHYSICSUNIT 17: NUCLEAR PHYSICS

is defined as the central is defined as the central

core of an atom that is core of an atom that is

positively charged and positively charged and

contains protons and contains protons and

neutrons.neutrons.

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� A nucleus of an atom is made up of protons and neutrons that known as nucleons(is defined as the particles found inside the nucleusthe particles found inside the nucleus) as shown in figure 17a.

� For a neutral atom :

�� The number of protons = The number of protons = the number of electrons orbiting nucleusthe number of electrons orbiting nucleus

the inside the nucleus the inside the nucleus

� This is because the magnitude of an electron charge equals to the magnitude of a proton charge but opposite sign.

17.1 Nucleus

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

Charge (Charge (CC))

Mass (Mass (kgkg))

+e+e 00

27106721 −×. 27106751 −×.

).( 1910601 −× )uncharged(

Table 17aTable 17a

� Proton and neutron are characterized by the following properties in table 15.1a.ProtonProton

NeutronNeutron

Electron Electron

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� Nuclei are characterized by the number and type of nucleons they contain as shown in table 15.1b.

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

� The nuclide of an element is represented as

DefinitionSymbolNumber

Atomic number/ Atomic number/

Proton numberProton number

Z The number of protons in a nucleus

Neutron numberNeutron number N The number of neutrons in a nucleus

Mass number/ Mass number/

Nucleon numberNucleon numberA The number of nucleons in a nucleus

Table 17bTable 17b

Relationship :Relationship :

Atomic numberAtomic number

Mass numberMass number

Element XElement X

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� The number of protons Z is not necessary equalnot necessary equal to the number of neutrons N.

e.g. :

� Example 1 :

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 (exercise)(exercise)

c. 50 nucleons ; 24 electrons

d. 106 nucleons ; 48 protons (exercise)(exercise)

e. 214 nucleons ; 131 protons (exercise)(exercise)

11Z =

Na23

11 S32

16; Pt195

78;

12ZAN =−=

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Solution:

a. Given Z=20 ; A=40

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

� Example 2 :

What is meant by the symbols below :

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

Solution:

XA

Z Ca40

20

XA

Z Cr50

24

n1

0 p1

1; e0

1−;

n1

0

p1

1

e0

1−

Neutron ;Neutron ; A=1A=1

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

Proton ;Proton ; A=1A=1

Charge :Charge : positively chargedpositively charged

Electron ;Electron ; A=0A=0

Charge :Charge : negatively chargednegatively charged

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� Example 3 : (exercise)

Complete the table below :

Number of

electrons

Total charge in

nucleus

Number of

neutrons

Number of

protons

Element

nuclide

H1

1

N14

7

Na23

11

Co59

27

Be9

4

O16

8

S31

16

Cs133

55

U238

92

1111 1212 11e11e 1111

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� Definition – the nuclides/elements/atoms that have the same atomic number Zbut different in mass number A.

� From the definition of isotope, thus the number of protons or electrons are equalnumber of protons or electrons are equal but

different in the number of neutronsdifferent in the number of neutronsN for two isotopes from the same element.For example :

� Hydrogen isotopes :

� Oxygen isotopes :

Isotope

H1

1

H2

1

H3

1

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

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

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

protonproton )( p1

1

deuteriumdeuterium )( D2

1

tritiumtritium )( T3

1

O16

8

O17

8

O18

8

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

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

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

equalequal

equalequal not equalnot equal

not equalnot equal

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17.2: Radioactivity17.2: Radioactivity

the spontaneous the spontaneous

disintegration of certain disintegration of certain

atomic nuclei accompanied by atomic nuclei accompanied by

the emission of alpha the emission of alpha

particles, beta particles or particles, beta particles or

gamma radiation.gamma radiation.

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�� The phenomenon in which an unstable nucleus disintegrates to acqThe phenomenon in which an unstable nucleus disintegrates to acquire a more uire a more

stable nucleus without absorb an external energystable nucleus without absorb an external energy.

� The radioactive decay is a spontaneousspontaneous reaction that is unplannedunplanned, cannot be cannot be

predictedpredicted and independentindependent of physical conditionsphysical conditions (such as pressure, temperature) and chemical changeschemical changes.

� This reaction is randomrandom reaction because the probabilityprobability of a nucleus decaying at a given instant is the samesame for all the nuclei in the sample.

� Radioactive radiations are emitted when an unstable nucleus decays. The radiations are alpha particles, beta particles and gamma-rays.

a) Alpha particle (αααα)� An alpha particle consists of two protons and two neutrons.

� It is identical to a helium nucleus and its symbol is

� It is positively charged particle and its value is +2e with mass of 4.001502 u.

17.2.1 Radioactive Decay

He4

2 α4

2oror

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� When a nucleus undergoes alpha decay it loses four nucleons, two of which are protons, thus the reaction can be represented by general equation below :

� Examples of α decay :

b) Beta particle (ββββ)� Beta particles are electrons or positrons (sometimes is called beta-minus and

beta-plus particles).

� The symbols represent the beta-minus and beta-plus (positron) are shown below:

QHePbPo 4

2

214

82

218

84 ++→

(Parent)(Parent) ((αααααααα particle)particle)(Daughter)(Daughter)

XA

Z +Y4A

2Z

−−→ + QHe4

2

QHeRaTh 4

2

226

88

230

90 ++→QHeRnRa 4

2

222

86

226

88 ++→

QHeThU 4

2

234

90

238

92 ++→

e0

1−−βoror e0

1

+βororBetaBeta--minus minus

(electron) :(electron) :BetaBeta--plus plus

(positron) :(positron) :

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� Beta-minus particle is negatively charged of -1e and its mass equals to the mass of an electron.

� Beta-plus (positron) is positively charged of +1e (antiparticle of electron) and it has the same mass as the electron.

� In beta-minus decay, an electron is emitted, thus the mass number does not charge but the charge of the parent nucleus increases by one as shown below :

� Examples of β minus decay :

� In beta-plus decay, a positron is emitted, this time the charge of the parent nucleus decreases by one as shown below :

(Parent)(Parent) ((ββββββββ particle)particle)(Daughter)(Daughter)

XA

Z +YA

1Z +→ + Qe0

1−

QePaTh 0

1

234

91

234

90 ++→ −

QeUPa 0

1

234

92

234

91 ++→ −

QePoBi 0

1

214

84

214

83 ++→ −

(Parent)(Parent) (Positron)(Positron)(Daughter)(Daughter)

XA

Z +YA

1Z −→ + Qe0

1

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� For example of β plus decay is

c) Gamma ray(γγγγ)� Gamma rays are high energy photons (electromagnetic radiation).

� Emission of gamma ray does not change the parent nucleus into a different nuclide, since neither the charge nor the nucleon number is changed.

� A gamma ray photon is emitted when a nucleus in an excited state makes a transition to a ground state.

� Examples of γ decay are :

� It is uncharged (neutral) ray and zero mass.

� The differ between gamma-rays and x-rays of the same wavelength only in the manner in which they are produced; gamma-rays are a result of nuclear processes, whereas x-rays originate outside the nucleus.

Qvenp 0

1

1

0

1

1 +++→

γHePbPo 4

2

214

82

218

84 ++→∗

Neutrino is uncharged particle Neutrino is uncharged particle

with negligible mass.with negligible mass.

γeUPa 0

1

234

92

234

91 ++→ −∗

γTiTi 208

81

208

81 +→∗

Gamma rayGamma ray

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17.2.2 Comparison of the properties of the alpha particle, beta particle and gamma ray.

� Table 17c shows the comparison between the radioactive radiations.

Ability to produce

fluorescence

Ability to affect a

photographic plate

Penetration power

Ionization power

Deflection by electric and

magnetic fields

Charge

GammaBetaAlpha

+2e -1e or +1e 0 (uncharged)

Yes Yes No

Strong Moderate Weak

Weak Moderate Strong

Yes Yes Yes

Yes Yes Yes

Table 17cTable 17c

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Br

Fig. 17cFig. 17c

++++++++

−−−−−−−− E

r

Fig. 17bFig. 17b

� Figures 17b and 17c show a deflection of α, β and γ in electric and magnetic field.

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a) Decay constant

� Law of radioactive decay states:

““For a radioactive source, the decay rate is direcFor a radioactive source, the decay rate is directly proportional to the tly proportional to the

number of radioactive nuclei N present in the source.number of radioactive nuclei N present in the source.

i.e.

� Rearranging eq. (17.2a) :

Hence the decay constantdecay constant is defined as the probability that a radioactive nucleus will the probability that a radioactive nucleus will

decay in one seconddecay in one second. Its unit is ss--11.

17.2.3 Decay constant and Half-life

−dt

dN

Ndt

dN∝

−

(17.2a)(17.2a)

Negative sign means the number of nuclei Negative sign means the number of nuclei

present decreases with timepresent decreases with time

Decay constantDecay constant

N

dt

dN

−=λpresent nuclei eradioactiv ofnumber

decay of rate=λ

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� The decay constant is a characteristic of the radioactive nuclei.

� Rearrange eq. (17.2a), we get

At time t=0, N=N0 (initial number of radioactive nuclei in the sample) and after a time t, the number of radioactive nuclei present is N. Integration of eq. (17.2b) from t=0 to time t:

� From the eq. (17.2c), thus the graph of N, the number of radioactive nuclei present in a sample, against the time t is shown in figure 17.2a.

dtN

dNλ−= (17.2b)(17.2b)

∫∫ λ−=t

0

N

Ndt

N

dN

0

(17.2c)(17.2c)

[ ] [ ]t

0

N

N tN0

λ−=ln

λtN

N

0

−=ln

Exponential law of Exponential law of

radioactive decayradioactive decay

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From the graph above, the life of any radioactive nuclide is infinite, therefore to

talk about the life of radioactive nuclide, we refer to its half-life.

t

0eNN λ−=

2

N 0

0N

8

N 04

N 0

016

N 0

2

1T

2

1T22

1T32

1T4ttime ,

2

1T5

N

lifehalfT2

1 −:

Fig. 17b Fig. 17b Simulation

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b) Half-life

� Definition –the time taken for a sample of radioactive nuclides disintegrate to half of

the initial number of nuclei.

� From the eq. (17.2c) : and the definition of half-life, when

� thus

Taking ln,

� The half-life of any given radioactive nuclide is constant, it does not depend on the number of nuclei present.

t

0eNN λ−=

2

1Tt =2

NN 0=;

2

1T

00 eN

2

N λ−

=

2

1T

e2λ

=

2

1T

e2

1 λ−

=

2

1T

e2λ

= lnln

(17.2d)(17.2d)HalfHalf--lifelife

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� The units of the half-life are second (s), minute (min), hour (hr), day and year (y). Its unit depend on the unit of decay constant.

� Table 17.2a shows the value of half-life for several isotopes.

� Example 1:

Initially, a radioactive sample contains of 1.0 x 106 nuclei. The half-life of the sample is T1/2. Calculate the number of nuclei present after 0.5T1/2.

Solution: N0=1.0 x 106 nuclei, t=0.5T1/2

By rearranging the equation of half-life, we get

20 minutes

3.8 days

24 days

4.5 x 109 years

Half-lifeIsotope

U238

92

Bi214

83

Rn222

86

Th234

90

Table 17.2aTable 17.2a

(1)(1)

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From the exponential law:

By substituting eq. (1) and t =0.5T1/2 into eq. (2), hence

� Example 2:

A radioactive source contains 1.0 x 10-6 g of Pu-239. If the source emits 2300 alpha particles per second, calculate

a. the decay constant.

b. the half-life.

(Given Avogadro constant, NA=6.02 x 1023 mol-1)

Solution:

a. 239 g of Pu-239 contains 6.02 x 1023 nuclei

1.0 x 10-6 g of Pu-239 contains

(2)(2)

1- s2300dt

dN−=

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Hence N0 = 2.52 x 1015 nuclei

By applying the equation for law of radioactive decay, thus

b. The half-life of Pu-239 is

c) Activity of radioactive sample (A)

� Definition –

� Its unit is number of decays per second.

� Other units for activity are curie (Ci) and becquerel (Bq) – S.I. unit.

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� Unit conversion :

� Relation between activity (A) of radioactive sample and time t :

� From the law of radioactive decay :

and definition of activity :

thus,

secondper decay 1Bq1 =secondper decays 107.3 Ci1 10×=

Ndt

dNλ−=

dt

dNA =

00 NA λ−=

NA λ−= andand

andand

t

0eNN λ−=

( )t

0eNA λλ −−=

(17.2e)(17.2e)

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� Example 3:

Thorium-234 has a half-life of 24 days. The initial activity of this isotope is

10 µCi. Calculatea. the activity of the isotope after 72 days.

b. the time taken for the activity to fall to 2.5 µCi.

Solution: T1/2= 24 days, A0= 10 µCi

a. Given t=72 days

The decay constant of the thorium-234 is

The activity of the isotope after 72 days is

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b. Given A=2.5 µCi

By using the equation of activity for radioactive sample, thus

taking ln,

� Example 4:

A uranium-238 isotope which has a half-life of 4.47 x 109 years decays by

emitting alpha particle into thorium-234 nucleus. Calculate

a. the decay constant.

b. the mass of uranium-238 required to decay with activity of

6.00 µCi. c. the number of alpha particles per second for the decay of

30.0 g uranium-238.

(Given Avogadro constant, NA=6.02 x 1023 mol-1)

( )Th234

90

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Solution: T1/2= 4.47x109 years

a. The decay constant of the uranium-238 is

b. Using unit conversion ( Ci⇒ decay/second )

secondper decays 107.3 Ci1 10×=

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If 6.02 x 1023 nuclei of mass of 238 g uranium-238

4.52 x 1022 nuclei of mass of

Therefore the mass of the uranium-238 is given by

c. 238 g of uranium-238 contains 6.02 x 1023 nuclei

30 g of uranium-239 contains

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� Example 5:

The half-life of radium-226 nuclide is 1600 years. Assuming that only

one decay per atom, calculate

a. the percentage of atoms remaining after 100 years.

b. the number of decays per second for 1 mg of radium-226.

(Given Avogadro constant, NA=6.02 x 1023 mol-1)

Solution: T1/2= 1600 years

a. Given t =100 years, Initially, N0=100%=1

The decay constant of radium-226 is

By applying the exponential law :

Therefore the percentage of atoms remaining after 100 years

( )Ra226

88

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b. 226 g of radium-226 contains 6.02 x 1023 atoms

1.0 x 10-3 g of radium-226 contains

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� Example 6:

Radioactive can be used for radioactive dating, i.e. a method to determine the age of an

artifact based on decay rate and half-life of carbon-14. The half-life of carbon-14 is known

to be 5600 years. If a 10 gram of carbon sample from a live tree gives the decay rate of

500 per hour and a 10 gram sample from an artifact gives decay rate of 100 per hour,

calculate the age of the artifact.

Solution: T1/2= 5600 years;A0=500 decays/hour;

A=100 decays/hour

The decay constant of carbon-14 is

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Radioisotope as tracers

� Since radioisotope has the same chemical properties as the stable isotopes then they can be used to trace the path made by the stable isotopes.

� Its method :

� A small amount of suitable radioisotope is either swallowed by the patient or injected into the body of the patient.

� After a while certain part of the body will have absorbed either a normal amount, or an amount which is larger than normal or less than normal of the radioisotope. A detector (such as Geiger counter ,gamma camera, etc..) then measures the count rate at the part of the body concerned.

� It is used to investigate organs in human body such as kidney, thyroid gland, heart, brain, and etc..

Other uses of Radioisotope

� In medicine:

� To destroy cancer cells by gamma-ray from a high-activity source of Co-60.

� To treat deep-lying tumors by planting radium-226 or caesium-137 inside the body close to the tumors.

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� In agriculture:

� To enable scientists to formulate fertilizers that will increase the production of food.

� To develop new strains of food crops that are resistant to diseases, give high yield and are of high quality.

� In industry :

� To measure the wear and tear of machine part and the effectiveness of lubricants.

� To detect flaws in underground pipes e.g. pipes use to carry natural petroleum gas.

� To monitor the thickness of metal sheet during manufacture by passing it between gamma-ray and a suitable detector.

� In archaeology and geology:

� To estimate the age of an archaeological object found by referring to carbon-14 dating. (carbon dating)

� To estimate the geological age of a rock by referring to potassium-40 dating.