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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?