20120531130532radiasi & fizik nuklear

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.
http://var/www/apps/conversion/current/tmp/scratch31463/AF_4409.htmlhttp://var/www/apps/conversion/current/tmp/scratch31463/AF_4409.html


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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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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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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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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

=

20120531130532radiasi & fizik nuklear

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