C 14 I 131 Mendeteksi fungsi kelenjar gondok. Na 24 Pb 210

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

Why we study this chapter?

Berbagai manfaat radioisotopes:

Co

60

_{Membunuh sel kanker}

C

14

Mendeteksi usia fosil

I

131

_{Mendeteksi fungsi kelenjar gondok}

Na

24

Mengetahui efektivitas kerja jantung

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

m

_p

= 1.007276 amu

m

_n

= 1.008665 amu

Mass of proton Mass of neutron Mass of electron

m

_e

= 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) 200_{79𝐴𝑢 and (b)} 83 209_𝐵𝑖 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

Nuclear Binding Energy

This mass difference between the individual nucleons and the fully formed nucleus of

any atom can be calculated by:

m

_inti

-( Z.m

_p

+ 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.

Example

Massa inti atom ₂₀40_{𝐶𝑎 adalah 40,078 sma.}

Jika massa proton = 1,0078 sma dan neutron = 1,0087 sma, defek massa pembentukan ₂₀40_{𝐶𝑎 adalah ….}

m

_inti

-( Z.m

_p

+ N.m

_n

)

∆m =

40.078 sma

-[

20(1.0078)

+

20 (1.0087)

]

∆m =

∆m =

20(2.0165) sma

-

40.078 sma

∆m =

0.252 sma

Nuclear Binding Energy

From Einstein mass – energy equation,

The “Energi Ikat Atom”

can be calculated:

c

2

∆m

=

E

_ikat

Untuk atom yang defek massanya adalah 1 sma, Maka energi ikat

atomnya adalah:

(1 sma)

=

E

_ikat

(3x10

8

₎

2

E

_ikat

= (1.66x10

-27

_kg)(3x10

8

_m/s)

2

E

_ikat

= 1.49x10

-10

_{J = 931 MeV}

Example (UN)

Massa inti atom ₁₂Mg24 _{adalah 23,993 sma. Massa nukleon bebas inti}

tersebut adalah massa proton 1,007 sma dan massa neutron 1,008 sma. Energi ikat inti ₁₂Mg24 _{sebesar …. (1 sma =931 MeV)}

m

_inti

-( Z.m

_p

+ N.m

_n

)

∆m =

23.993 sma

-[

12(1.007)

+

12 (1.008)

]

∆m =

∆m =

12(2.015) sma

-

23.993 sma

931

∆m

=

E

_ikat

=

0.187 sma

Example (UN)

Massa inti atom ₁₂Mg24 _{adalah 23,993 sma. Massa nukleon bebas inti}

tersebut adalah massa proton 1,007 sma dan massa neutron 1,008 sma. Energi ikat inti ₁₂Mg24 _{sebesar …. (1 sma =931 MeV)}

931

∆m

=

E

_ikat

∆m

=

E

_ikat

931 MeV

(0.187 sma)

=

E

_ikat

(931 MeV)

174,1 MeV

=

E

_ikat

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.

6

12

_𝐶

6

13

_𝐶

6

14

_𝐶

6

15

_𝐶

Radioactivity

Most elements have stable isotopes. However, the nuclei of some

isotopes are unstable.

therefore they

_emit

_{highly energetic particles}

decompose

and

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}

α –

alpha radiation

Alpha particles are identical to the nuclei of helium atoms

2 neutrons and 2 protons

α

2

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

2

4

_∝

₊

90

234

_𝑇ℎ

92

238

_𝑈

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.

0

1

_𝑛

1

1

_𝑝

₊

α and β – particle decay

6

14

_𝐶

−1

0

𝛽

+

14

₇

𝑁

Example,

11Na24 undergoes beta decay. What is the

resulting nucleus element?

11

24

_𝑁𝑎

−1

0

𝛽

+

₁₂

24

𝑋

X is Magnesium (Mg)

γ –

gamma rays

Gamma (γ) rays are a type of electromagnetic radiation of very high energy.

γ

0

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

γ –

gamma rays

94

248∗

_𝑃𝑢

0

0

_𝛾

₊

94

248

_𝑃𝑢

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 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

1 2

_𝐻

2 4

_𝐻𝑒

₊

0 1

_𝑛

+

₁3

_𝐻

0 1

_𝑛

55 140

_𝐶𝑠

+

3793

𝑅𝑏

+

235₉₂

_𝑈

+

3

₀1

𝑛

pereaksi

pereaksi

Hasil reaksi

Hasil reaksi

∆m = m

_pereaksi

- m

_{hasil reaksi}

E = ∆m 931 MeV

Energi pada reaksi nuklir:

+ E

Example (UN)

m_pereaksi - m_{hasil reaksi}

∆m =

( 1.0078 + 1.0078 ) - ( 2.01410 + 0.00055 )

∆m =

∆m =

2.0156 sma - 2.0146 sma

=

Nilai E (energi yang dihasilkan)

pada reaksi fusi tersebut

adalah…..

0.00095 sma ∆m

E =

(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

𝑁

0

Initial Number

of Nuclei

Activity

=

𝜆 𝑁

₀

of a radioactive material Large 𝝀 → decays quickly Small 𝝀 → decays slowly

The unit of Activity is Becquerel (Bq)

Nuclear Decay

𝑁

₀

Initial Number of Nuclei

𝑁

Undecayed Number of Nuclei

𝑁 = 𝑁

₀

_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 =}

____

𝑁

0

Nuclear Decay – Half-life

𝑁 = 𝑁

₀

_𝑒

−𝜆𝑡

𝑁

₀

____

2

= 𝑁

0

𝑒

−𝜆𝑡

1

____

2

=

𝑒

−𝜆𝑡

𝑙𝑛

𝑙𝑛

− ln 2

=

– 𝜆𝑡

ln 2

= 𝜆𝑡

ln 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

_ൗ

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 10

5

= 0.2 𝑥10

−5

Activity = 𝜆 N

₀

= ( 2 𝑥10

−6

) ( 2 𝑥 10

10

)

Activity = 4 𝑥 10

4

_𝐵𝑞

Half-life Example

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

𝑁 = 𝑁

₀

_𝑒

−𝜆𝑡

= (2 x 10

10

) 𝑒

−(2𝑥10

−6

)(3.5𝑥10

5

)

= (2 x 10

10

) 𝑒

−(0.7)

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 (226_{Ra) nucleus is 5.0 x 10}10

seconds. A sample contains 3.0 x 1016 _nuclei.

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

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?