NUCLEAR

PHYSICS

OVERVIEW

• Properties of Nuclei

• Mass defect (mass difference)

• Nuclear binding energy

• Radioactivity

a) Alpha decay

b) Beta decay

c) Gamma decay

• Nuclear Reaction

• The Decay Processes

Why we study this chapter?

Berbagai manfaat radioisotopes:

Co60

Membunuh sel kanker

C14

Mendeteksi usia fosil

I131

Mendeteksi fungsi kelenjar gondok

Na24

Mengetahui efektivitas kerja jantung

Pb210

Mendeteksi pemalsuan keramik

Examples (UN)

Radiasi dari radioisotop Co-60 dimanfaatkan untuk...

A. Penghancuran batu ginjal

B. Detektor Asap

C. Menentukan umur fosil

D. Terapi pada kelenjar gondok

E. Membunuh sel kanker

Examples (UN)

Zat radioisotop C-14 dapat digunakan untuk...

A. Mendeteksi fungsi kelenjar gondok

B. Mengetahui efektifitas kerja jantung

C. Membunuh sel kanker

D. Mendeteksi pemalsuan keramik

E. Menentukan usia fosil

The Nucleus

Only 10-13

percent of atom's volume

It contains 99.9% of the total mass

of the atom.

Proton (p)

Neutron (n)

Nucleons

The Nucleus

Since the mass of nucleons is far too small to

measure in kg, other standard unit is used:

Atomic Mass Unit (amu)

1 amu = 1.66 x 10-27

kg

mp

= 1.007276 amu

mn

= 1.008665 amu

Mass of proton

Mass of neutron

Mass of electron

me

= 5.486 x 10-4

amu

Extremely small

1836 times smaller

than a proton

Atomic Number and Mass Number

The number of

protons in an

atom is called

The Atomic

Number

Z

The total

number of

protons and

neutrons in an

atom is called

The Atomic

Mass Number

A

Atomic Number and Mass Number

Li

7

3

Atomic Number = 3

Atomic Mass Number = 7

The difference

between Z and A

Will give us

The number of

neutrons (N) in

the nucleus

The number

of Neutrons

(N)

= A - Z

7 - 3=N 4=

Examples

Calculate the number of proton and neutron in an atom notated by:

(a) 79200𝐴𝑢 and (b) 83

209𝐵𝑖

Atomic Number = 79(a) (protons)

Jumlah Neutron = 200 - 79 = 121

Atomic Number = 83(b) (protons)

Jumlah Neutron = 209 - 83 = 126

Nuclear Force

There must be a massive repulsive force between the protons,

however they remain in the nucleus together.

There must be another force that attracts the nucleons.

This force is called Nuclear Attractive Force, or Nuclear Force or Strong

Force.

Dari mana datangnya gaya inti itu?

O.o?

Nuclear Binding Energy

Source of energy that holds protons and

neutrons together in a nucleus arises from

the transformation of mass to energy.

The amount of mass that

is transformed into energy

is called

Mass Defect.

For any atoms, the total mass

in measured experimentally

is always smaller than

the sum of the mass of protons

and neutrons.

Masa

inti< ( Z.m

p + N.m

n)

Jumlah

Proton

Jumlah

Neutron

Nuclear Binding Energy

This mass difference between the individual

nucleons and the fully formed nucleus of

any atom can be calculated by:

minti

-( Z.mp + N.m

n)∆m =

The energy of this ‘mass defect' is the same energy preventing the

nucleus from breaking up.

The energy which binds the protons and neutrons together in the nucleus

is called the nuclear binding energy.

Defek Massa

Example

Massa inti atom 2040𝐶𝑎 adalah 40,078 sma.

Jika massa proton = 1,0078 sma dan neutron = 1,0087 sma,

defek massa pembentukan 2040𝐶𝑎 adalah ….

minti-( Z.m

p + N.m

n)∆m =

40.078 sma-[ 20(1.0078) + 20 (1.0087) ]∆m =

∆m = 40.078 sma20(2.0165) sma -

∆m = 0.252 sma

Nuclear Binding Energy

From Einstein mass – energy equation,

The “Energi Ikat Atom” can be calculated:

c2

∆m=Eikat

Untuk atom yang defek

massanya adalah 1 sma,

Maka energi ikat

atomnya adalah:

(1 sma)=Eikat

(3x108

)2

Eikat

= (1.66x10-27

kg)(3x108

m/s)2

Eikat

= 1.49x10-10

J = 931 MeV

Example (UN)

Massa inti atom 12

Mg24

adalah 23,993 sma. Massa nukleon bebas inti

tersebut adalah massa proton 1,007 sma dan massa neutron 1,008 sma.

Energi ikat inti12

Mg24

sebesar …. (1 sma =931 MeV)

minti-( Z.m

p + N.m

n)∆m =

23.993 sma-[ 12(1.007) + 12 (1.008) ]∆m =

∆m = 23.993 sma12(2.015) sma -

931∆m=Eikat

= 0.187 sma

Example (UN)

Massa inti atom 12

Mg24

adalah 23,993 sma. Massa nukleon bebas inti

tersebut adalah massa proton 1,007 sma dan massa neutron 1,008 sma.

Energi ikat inti12

Mg24

sebesar …. (1 sma =931 MeV)

931∆m=Eikat∆m=E

ikat931 MeV

(0.187 sma)=Eikat

(931 MeV)

174,1 MeV=Eikat

Isotopes

Isotopes merupakan kumpulan atom unsur yang memiliki sifat kimia

sama, tetapi berbeda sifat fisikanya.

Isotop yang tidak stabil dinamakan radioisotop.

Isotop yang tidak stabil ini cenderung memancarkan partikel partikel

radioaktif untuk mencapai kestabilannya.

612𝐶 6

13𝐶 614𝐶 6

15𝐶

Radioactivity

Radioactivity

Most elements have stable isotopes.

However, the nuclei of some

isotopes are unstable.

therefore they decompose and

emit highly energetic particles

Radioactive

Emitted nuclear radiation

Example: radium,

uranium,

plutonium

only a small number are found naturally,

most of them are artificially produced in laboratories.

Types of Radiation

The radioisotope is located in a chamber where

its radiation can pass through a magnetic field

to strike a photographic plate

3 different darkened

spots appear

α – alpha radiation

β – beta radiation

γ – gamma radiation

Penetration Power

α – alpha radiation

Alpha particles are identical to

the nuclei of helium atoms

2 neutrons and 2 protons α2

4

α - particle

(+2) Charged particle

When an element undergoing an alpha decay,

it emits 2 protons and 2 neutrons.

There is a decrease of 2 in the atomic number

and a decrease of 4 in the atomic mass number of the nucleus.

α – alpha decay

24∝ + 90

234𝑇ℎ92238𝑈

β– beta decay

β - beta particles are identical to electrons,

negatively charged (-1) and negligible mass.β-1

0

β - particle

When an element undergoing an beta decay,

it emits 1 electron (coming from the split of neutron)

There is an increase of 1 in the atomic number

and no change in the atomic mass number of the nucleus.

01𝑛 1

1𝑝 + −10𝛽

α and β – particle decay

614𝐶 −1

0𝛽 + 714𝑁

Example,

11Na

24undergoes beta decay. What is the

resulting nucleus element?

1124𝑁𝑎 −1

0𝛽 + 1224𝑋

X is Magnesium (Mg)

γ – gamma rays

Gamma (γ) rays are a

type of electromagnetic

radiation of very high

energy.

γ0

0

γ - rays

There is no change in the atomic number

and no change in the atomic mass number of the nucleus.

The mass and charge of gamma rays are both zero.

The emission of γ -rays

from a nucleus is similar to

the emission of photons

from an excited atom.

γ – gamma rays

94248∗𝑃𝑢 0

0𝛾 + 94248𝑃𝑢

The unstable nucleus (*) then

loses excess energy in the form

of gamma rays

to become a more stable

nucleus of lower energy.

Nuclear

Reaction

Nuclear Reaction

Tend to be unstable and split

into two or more nuclei

(FISSION)

Tend to be unstable

and combine with

other nuclei

(FUSION)

Any process that involves a change in the nucleus of an atom is called

nuclear reaction

Reaksi Fisi

Reaksi Fusi

Nuclear Reaction

Reaksi nuklir menghasilkankan energi yang sangat besar,

lebih dari jutaan kali lipat energi yang diperoleh dari reaksi kimia biasa.

Energi yang sangat besar

itu diperoleh karena

massa inti yang dihasilkan

lebih kecil daripada

jumlah total massa inti

yang bereaksi.

Jadi, sekali lagi,

persamaan Einstein E =mc2

menjelaskan bahwa massa

yang hilang telah dikonversi

menjadi energi pada hasil

reaksinya.

Nuclear Reaction

Reaksi Fusi

Reaksi Fisi

12𝐻 2

4𝐻𝑒 + 01𝑛+ 1

3𝐻

01𝑛 55

140𝐶𝑠 + 3793𝑅𝑏+ 92

235𝑈 + 3 01𝑛

pereaksi

pereaksi

Hasil reaksi

Hasil reaksi

∆m = mpereaksi - m

hasil reaksi

E = ∆m 931 MeV

Energi pada reaksi nuklir:

+ E

+ E

Example (UN)

mpereaksi

- mhasil reaksi∆m =

( 1.0078 + 1.0078 ) - ( 2.01410 + 0.00055 )∆m =

∆m = 2.0146 sma2.0156 sma - =

Nilai E (energi yang dihasilkan)

pada reaksi fusi tersebut

adalah…..

0.00095 sma

∆mE = (931 MeV)

(0.00095 sma)E = (931 MeV)

= 0.88 MeV

Nuclear

Decay

Nuclear Decay

Imagine that you are studying a sample of radioactive material.

You know that the atoms in the material are decaying into other types of atoms.

How many of the unstable parent atoms remain after a certain amount of time?

𝜆 Decay constant 𝑁0Initial Number

of Nuclei

Activity = 𝜆 𝑁0

of a radioactive material

Large 𝝀 → decays quickly

Small 𝝀 → decays slowly

The unit of Activity is Becquerel (Bq)

1 Curie (Ci) = 3.7 x 1010 Bq

Nuclear Decay

𝑁0 Initial Number of Nuclei

𝑁 Undecayed Number of Nuclei

𝑁 = 𝑁0 𝑒−𝜆𝑡

The rate of radioactive decay is generally expressed in terms of half-life.

Half-life is the period of time over which the number of radioactive nuclei

decreases by half.

Half-Life 𝑁 =𝑁0____2

Nuclear Decay – Half-life

𝑁 = 𝑁0 𝑒−𝜆𝑡

𝑁0____2

= 𝑁0 𝑒−𝜆𝑡

1____2

= 𝑒−𝜆𝑡𝑙𝑛 𝑙𝑛

− ln 2 = – 𝜆𝑡

ln 2 = 𝜆𝑡ln 2 is equal to 0.693

We can calculate the decay constant

based on the information of half-life

of a radioactive material by:

𝜆 =0.693_________𝑡 ൗ1 2

Half-life Example

A radioactive isotope, Radon-222 (86222

𝑅𝑛) has a half-life of 4 days. An

initial number of 2 x 1010 nuclei is released during radioactive decay.

a) What is the activity of the sample?

b) How many nuclei remain undecayed at the end of this period?

ANSWER

a) First, convert the unit of days into second!

4 days = 4 x 24 x 60 x 60 = 350000 seconds = 3.5 x 105 s

Then, before we find the activity, we have to calculate the decay

constant of Radon-222

Half-life Example

4 days = 4 x 24 x 60 x 60 = 350000 seconds = 3.5 x 105 s

Then, before we find the activity, we have to calculate the decay

constant of Radon-222

𝜆 =0.693_________𝑡 ൗ1 2

=0.693_________3.5 𝑥 105

= 0.2 𝑥10−5

Activity = 𝜆 N0 = ( 2 𝑥10−6 ) ( 2 𝑥 1010 )

Activity = 4 𝑥 104 𝐵𝑞

Half-life Example

b) How many nuclei remain undecayed at the end of this period?

𝑁 = 𝑁0 𝑒−𝜆𝑡

= (2 x 1010) 𝑒−(2𝑥10−6)(3.5𝑥105)

= (2 x 1010) 𝑒−(0.7)

= (2 x 1010) 0.5 = 1010 Nuclei

Half-life Example 2

How long will it take a sample of polonium-210 with a half-life of 140

days to decay to one-sixteenth its original strength?

Half-life Example 3

The half-life of the radioactive Radium (226Ra) nucleus is 5.0 x 1010

seconds. A sample contains 3.0 x 1016 nuclei.

(a) What is the decay constant for this radioactive?

(b) How many radium nuclei, in curies, will decay per second?

Half-life Example 3

A radioactive sample consists of 5.3 x 105 nuclei. There is one decay

every 4.2 hours.

(a) What is the decay constant for this sample?

(b) What is the half-life for the sample?