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7/31/2019 20120531130532Radiasi & Fizik Nuklear
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6/12/12
Radiasi & Fizik Nuklear
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Hasil Pembelajaran
Terangkan , +, dan mereput.
Mentakrifkan aktiviti, A pereputan
dan berterusan, .Terbitkan dan menggunakan
Mentakrifkan separuh hayat dan
N
dt
dN=
teNN = 0 teAA = 0
2ln2/1 =T
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Radioaktif
Radioaktif ditakrifkan sebagaifenomena di mana nukleus yangtidak stabil akan mereput untuk
memperoleh satu nukleus yang lebihstabil tanpa menyerap tenagaluaran.
Sinar radioaktif: dipancarkan apabilanukleus yang tidak stabil mereput.Sinaran zarah alfa, zarah beta dan
sinar gamma.
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Pereputan Alfa
Partikel alfa terdiri daripada duaproton dan dua neutron.
Ia adalah serupa dengan nukleushelium dan simbol adalah
Ia bercas positif zarah dan nilainyaadalah 2 e dengan jisim 4.002603 u.
Apabila nukleus mengalami
He42 42
XAZ +Y42
AZ + QHe42
OR
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Contoh Pereputan Alfa
Q++ HePbPo 4221482
21884
Q++ HeRaTh 422268823090
Q++ HeRnRa 42222
86226
88
Q++ HeThU 42234
9023892
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Pereputan Beta
Zarah beta elektron atau positron(kadang-kadang dipanggil beta-tolakdan zarah beta plus).
Simbol-simbol mewakili beta-tolakdan beta-plus (positron) adalahseperti berikut:
e01 OR
e01+OR
Beta-minus(electron) :
Beta-plus(positron) :
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Pereputan Beta
Beta-tolak zarah bercas negatif -1e dan jisim samadengan jisim elektron.Beta-postif (positron) bercas positif 1 e(antiparticle elektron) dan ia mempunyai jisim
sama seperti elektron.Dari pereputan beta-tolak, elektron dipancarkan,maka nombor jisim tidak berubah seperti yangditunjukkan di bawah:
(Parent) ( particle)(Daughter)
XA
Z +Y1
A
Z+ + Qe
0
1
Dari pereputan beta plus, positronyang dipancarkan, kali ini caj nukleusinduk berkurangan demi satu seperti
yang ditunjukkan di bawah: X
A
Z +Y1A
Z + Qe0
1
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Contoh-contoh pereputanBeta
Q++ ePaTh 012349123490
Q++ eUPa01
23492
23491
Q++ ePoBi 012148421483
Qv +++ enp 0110
11
Neutrino tidak dicas zarahdengan jisim yang bolehdiabaikan.
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Sinar Gamma
Sinar gamma adalah foton tenagayang tinggi (sinaran elektromagnet).
Pelepasan sinar gamma tidakmenukar nukleus induk menjadisebuah nuclid yang lain, kerana
bukan caj mahupun nombor nukleonyang akan ditukar.
Satu foton sinar gamma yangdipancarkan apabila satu nukleus
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Ia tidak bercas (neutral) ray dan sifarjisimnya.
Berbeza antara gamma dan sinar x-
ray panjang gelombang yang samahanya dengan cara di mana iadihasilkan, gamma-ray adalah hasil
proses nuklear, manakala x-rayberasal di luar nukleus.
++ HePbPo 42214
8221884
++
eUPa 01234
92234
91
+ TiTi 20881208
81Gamma ray
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Perbandingan
Alpha Beta GammaCas
Dibelokkan denganmedan magnet
Pengionan
Penembusan
Boleh memberi kesankepada kepingan
fotografik
Boleh menghasilkanfluoroscence
+2e 1e OR +1e 0 (uncharged)
Yes Yes No
Strong Moderate Weak
Weak Moderate Strong
YesYes Yes
Yes Yes Yes
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Kesan ke atas medan elektrik danmagnet B
+
+++++++
E
Radioactivesource
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Pemalar Pereputan
Hukum pereputan radioaktifmenyatakan:Bagi sumber radioaktif, kadar
pereputan berkadar langsung denganbilangan nukleus radioaktif Ntertinggal dalam sumber.
dt
dN
Ndt
dN
NdtdN =
Tanda negatif bermaksud bilangannuklei yang tinggal berkuranganterhadap masa
Pemalarpereputan
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Pereputan Radioaktif
Pemalar pereputan adalah ciri-ciri nukleus radioaktif.
Menyusun semula persamaan. (15.1), kita akan mendapat
Pada masa t = 0, N =, N0 (nombor awal nukleusradioaktif dalam sampel) dan selepas masa t, bilangannukleus baki N.
dtN
dN=
=tN
N dtN
dN
00 [ ] [ ]tN
N tN 00ln =
tN
N=
0
lnt
eNN= 0 Hukum eksponenpereputan radioaktif
Kamirkan persamaan ini daripada t=0
sehingga masat
:
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1515
From the eq. (15.3), thus the graph ofN, the number of remaining radioactivenuclei in a sample, against the time tis shown in Figure 15.3.
teNN
= 0
2
0N
0N
4
0N
16
0N8
0N
2/1T 2/12T 2/13T 2/14T 2/15T0t,time
N
lifehalf:2/1 T
Figure 15.3
Stimulation 15.1
Note:
From the graph (decay curve),
the life of any radioactivenuclide is infinity, therefore totalk about the life of radioactivenuclide, we refer to its half-life.
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1616
is defined as the time taken for a
sample of radioactive nuclidesdisintegrate to half of the initialnumber of nuclei.
From the eq. (15.3), andthe definition of half-life,
when , thus
31.1.6 Half-life (T1/2)
teNN = 0
2/1Tt=2
; 0N
N=
2/1
00
2
TeN
N =2/12
Te
=
2/1
2
1 Te
=
2/1ln2lnT
e=
T
693.02ln
2/1 ==
Half-life
(15.4)
The half-life of any given radioactivenuclide is constant, it does not dependon the number of remaining nuclei.
The units of the half-life are second(s), minute (min), hour(hr), day (d) andyear(y). Its unit depend on the unit ofdecay constant.
Table 15.2 shows the value of half-lifefor several isotopes.
Isotope Half-life
4.5 109 years
1.6 103 years
138 days
24 days
3.8 days
20 minutes
U23892
Po210884Ra
226
88
Bi21483
Rn22286
Th23490
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Activity at time t
Relation between activity (A) ofradioactive sample and time t:
l From the law of radioactivedecay :
and definition of activity :
is defined as the decay rateof a radioactive sample.
Its unit is number of decays persecond.
Other units for activity are curie(Ci) and becquerel (Bq) S.I.unit.
Unit conversion:
1717
secondperdecays1073C i1 10.
=
31.1.7 Activity OfRadioactive
Sample (A)
dt
dN
secondperdecay1Bq1 =
Ndt
dN=
dt
dNA =
00 NA =
NA = and teNN = 0( )teNA = 0
teAA= 0
Activity at time, t=0
and
(15.5)
( ) teN = 0
l Thus
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Solution :
1818
A radioactive nuclide Adisintegrates into a stablenuclide B. The half-life of Ais 5.0 days. If the initialnumber of nuclide A is1.0 1020, calculate the
number of nuclide B after 20days.
Example 1 : QBA +
0.5
2ln=
days20;101.0days;0.520
02/1=== tNT
2/1
2ln
T= 1days139.0 =
teNN = 0 ( ) ( ) ( )20139.020100.1 = eN
nuclei102.6 18=
1820102.6100.1 =
nuclei1038.9 19=
19
9.38 10 nuclei=
Therefore the number of nuclide B formed is
The decay constant is given by
The number of remaining nuclide A is
The number of nuclide A that have decayed is
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Solution :
1919
A radioactive nuclide Adisintegrates into a stablenuclide B. The half-life of Ais 5.0 days. If the initialnumber of nuclide A is1.0 1020, calculate the
number of nuclide B after 20days.
Example 1 : QBA +
0.5
2ln=
days20;101.0days;0.520
02/1=== tNT
2/1
2ln
T= 1days139.0 =
teNN = 0 ( ) ( ) ( )20139.020100.1 = eN
nuclei102.6 18=
1820102.6100.1 =
nuclei1038.9 19=
19
9.38 10 nuclei=
Therefore the number of nuclide B formed is
The decay constant is given by
The number of remaining nuclide A is
The number of nuclide A that have decayed is
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2020
a. Radioactive decay is arandom and spontaneousnuclear Reaction. Explainthe terms random andspontaneous.
a. 80% of a radioactivesubstance decays in 4.0days. Determine
i. the decay constant,
ii.the half-life of thesubstance.
Example 2 :Solution :
a. Random means that the time ofdecay for each nucleus cannot bepredicted. The probability of decay foreach nucleus is the same.
Spontaneous means it happen by itselfwithout external stimuli. The decay isnot affected by the physical conditionsand chemical changes.
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2121
a. Radioactive decay is a
random and spontaneousnuclear Reaction. Explainthe terms random andspontaneous.
a. 80% of a radioactivesubstance decays in 4.0days. Determinei. the decay constant,
ii. the half-life of thesubstance.
.
Example 2 : Solution :b. At time
The number of remainingnuclei is:
days,0.4=t=
00100
80NNN
nuclei2.0 0N=
teNN = 0 ( )0.4002.0= eNN
( )0.42.0 = e( )0.4ln2.0ln = e
1day402.0 =( ) eln0.42.0ln =
2ln
2/1 =T 402.02ln
2/1 =T days72.12/1 =T
i. By applying the exponential law ofradioactive decay, thus the decay
constant is
ii. The half-life of the substance is
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2222
Phosphorus-32 is a betaemitter with a decay constantof 5.6 107 s1. For aparticular application, thephosphorus-32 emits 4.0 107 beta particles every
second. Determinea. the half-life of thephosphorus-32,
b. the mass of purephosphorus-32 will givethis decay rate.
(Given the Avogadro constant,NA =6.02 1023 mol1)
Example 3 :
2ln2/1 =T
7 1
7 1
5.6 10 s ;
4.0 10 s
dN
dt
=
=
7106.5
2ln
=
s1024.1 62/1 =T
Solution :
a. The half-life of the phosphorus-32is given by
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2323
Phosphorus-32 is a beta
emitter with a decayconstant of 5.6 107 s1.For a particularapplication, thephosphorus-32 emits 4.0
107 beta particles every
second. Determinea. the half-life of thephosphorus-32,b. the mass of purephosphorus-32 will give
this decay rate.(Given the Avogadroconstant, NA =6.021023 mol1)
Example 3 :Solution :
b. By using the radioactive decay law,thus
6.02 1023 nuclei of P-32 has amass
of 32 g
7.14 1013 nuclei of P-32 has amass
of
0Ndt
dN=
( ) 077 106.5100.4 N=nuclei1014.7 130 =N
13
23
7.14 1032
6.02 10
=
g1080.3
9
=
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2424
A thorium-228 isotope which has a
half-life of 1.913 years decays byemitting alpha particle into radium-224nucleus. Calculatea. the decay constant.b. the mass of thorium-228 required to
decay with activity of12.0 Ci.c. the number of alpha particles persecond for the decay of 15.0 g
thorium-228.
(Given the Avogadro constant, NA=6.02 1023 mol1)
Example 4 :
2ln2/1 =T
( )
1/2 1.913 y
1.913 365 24 60 60
T =
=
7ln 2 6.03 10 =
18 s1015.1 =
s1003.67
=
Solution :
a. The decay constant is given by
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2525
A thorium-228 isotope which has a
half-life of 1.913 years decays byemitting alpha particle into radium-224nucleus. Calculatea. the decay constant.b. the mass of thorium-228 required to
decay with activity of12.0 Ci.c. the number of alpha particles persecond for the decay of 15.0 g
thorium-228.
(Given the Avogadro constant, NA=6.02 1023 mol1)
Example 4 : Solution :Solution : ( Ci decay/second ),b. By using the unit conversion
the activity is
Since then
If 6.02 1023 nuclei of Th-228 has a
mass of 228 g thus 3 86 1019n clei of Th 228 has a mass of
( )
10
12.0 Ci
12.0 3.7 10
A =
= decays/s1044.4 11=
secondperdecays1073Ci110. =
NA =
AN =
( )8
11
1015.1
1044.4
=N
nuclei1086.3 19=
22810026
1086.323
19
g1046.12
=