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PEMBAHASAN BUKU KENNETH KRANE
Kata Pengantar
Puji syukur kehadirat Tuhan Yang Maha Esa, yang telah memberikan nikmat kepada kita sehingga kita senantiasa diberkahi kesehatan, dan penulis bisa menyelesaikan pembahasan soal genap bab 9 dan bab 10 buku fisika modern oleh Kenneth Krane ini. Dalam pembahasan ini, diuraikan tentang langkah pengerjaan dan hasil hitung pada soal genap pada bab struktur inti, radioaktivitas, dan reaksi inti. Harapan yang paling besar dari penyusunan modul pembahasan ini ialah, mudahmudahan apa yang kami susun ini dapat bermanfaat sebagai pedoman dalam mengerjakan latihan soal fisika inti, baik untuk pribadi maupun untuk pembaca, khususnya mahasiswa fisika yang sedang menempuh mata kuliah fisika inti. Akhirnya, kami mengucapkan terima kasih kepada semua pihak yang telah memberikan semangat serta fasilitas kepada kami, hingga terselesaikannya modul pembahasan ini. Terimakasih kepada dosen pengampu mata kuliah, bapak Drs. Bambang Supriyadi, M.Sc, yang telah memberikan ilmunya kepada kami, mengenai mata kuliah Fisika Inti. Mudah-mudahan Allah Swt. Selalu memberikan rahmat dan bimbinganNya kepada kita semua. Amin.
Jember, 19 Mei 2016
Penyusun
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PEMBAHASAN BUKU KENNETH KRANE
DAFTAR ISI
HALAMAN JUDUL............................................ .................................................................. ............................................ ........................... ..... 1 KATA PENGANTAR .................................................... .......................................................................... ...................................... ................ 2 DAFTAR ISI ............................. .................................................... ............................................. ............................................. ............................... ........ 3 BAB 9 STRUKTUR INTI DAN RADIOAKTIFITAS SOAL GENAP BAB 9 9.1
Struktur Inti ............................................ .................................................................. .......................................... .................... 4
9.2
Radioaktifitas ............................................ .................................................................. ...................................... ................ 4
PEMBAHASAN BAB 9 SOAL GENAP 9.1
Struktur Inti ............................................ .................................................................. .......................................... .................... 6
9.2
Radioaktifitas ............................................ .................................................................. ...................................... ................ 10
BAB 10 REAKSI INTI SOAL GENAP BAB 10 10.1 Reaksi Inti. ............................................................ .................................................................................. ........................... ..... 16 PEMBAHASAN BAB 10 SOAL GENAP 10.1 Reaksi Inti ......................................... ............................................................... ............................................. ......................... 19 DAFTAR PUSTAKA .......................................... ................................................................ ............................................ ........................... ..... 31
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PEMBAHASAN BUKU KENNETH KRANE
2.
Timah memiliki isotop stabil yang lebih banyak dari unsure lainnya; nomor massa isotop-isotop itu adalah: 114, 115, 116, 117, 118, 119, 120, 122, 124. Berikan lambang isotop yang sesuai bagi semua isotop ini.
4.
Berapakah jari-jari inti (a)
6.
Hitunglah energi ikat total inti, dan energi ikat per nukleon, bagi (a) (b)
8.
; (c)
; (d)
; (b)
; (c)
.
Hitunglah Energi ikat total inti dari
3
He dan
3
H .
;
Terangkan perbedaan
yang ada dengan meninjau interaksi Coulomb dari proton lebih pada
3
He !
10.
Hitunglah energi pisah proton (lihat nomor 9) dari 4He, 12C, dan 40Ca !
12.
Interaksi lemah (gaya yang menyebabkan terjadinya peluruhan beta) diduga berasal dari partikel tukar dengan massa kurang lebih 75 GeV. Berapa jauhkah jangkauan gaya yang yang ini?
14.
Suatu cuplikan bahan radioaktif tertentu meluruh dengan laju 548 cacah per detik pada t = 0. Pada t = 48 menit, perhitungan lajunya menurun menjadi 213 cacah per detik. (a) berapakh usia paruh cuplikan tersebut? (b) berapakah tetapan luruhnya? (c) berapakah laju cacahnya pada t = 125 menit?
16.
Tritium, isotop hitfdrogen bernomor massa 3, memiliki usia paruh 12,3 tahun. Berapakah fraksi jumlah atom tritium yang tertinggal dalam suatu cuplikan bahan setelah 50,0 tahun?
18.
Andaikan kita mempunyai suatu cuplikan bahan yang mengandung inti radioaktif
// =8,04 ℎ (
) dengan aktivitas 2,00 mCi. (a) Berapa
banyak pekuruhan perdetikyang terjadi dalam cuplikan itu? (b) Berapa banyak peluruhan perdetik yang akan terjadi dalam cuplikan itu selama
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PEMBAHASAN BUKU KENNETH KRANE
20.
Sebuah detektor radiasi berbentuk piringan bundar berdiameter 3,0 cm, diletakkan sejauh 25 cm dari suatu sumber radioaktif. Detektor itu mencatat 1250 cacahan per detik. Dengan menganggap bahwa detektor mencatat tiap radiasi yang jatuh padanya, hitunglah aktivitas cuplikan (dalam curie).
22.
24.
Lengkapilah kelima peluruhan berikut: a)
27
b)
74
Si27 Al
As74Se
c)
288
d)
93
e)
131
U
Mo e
I 131Xe
Hitunglah energi kinetik partikel alfa yang dipancarkan dalam peluruhan 234
U!
26.
Inti
ℎ
meluruh alfa ke suatu keadaan eksitasi
, yang kemudian
meluruh ke keadaan dasarnya dengan memancarkan foton 217-ke V. Hitung energi kinetik partikel alfa. Massa 28.
ℎ
adalah 228,028726 u.
Hitunglah energi kinetik pental proton dalam peluruhan beta neutron (a) ketika elektron mencapai energi maksimumnya; (b) ketika neutrino mencapai energi maksimumnya.
30.
Inti
meluruh ke
dengan peluruhan beta positron. (a) Berapakah
nilai Q bagi peluruhan ini? (b) Berapakah energy kinetik maksimum positron? 32.
Dalam peluruhan beta dari
, sebuah elektron diamati memiliki energi
kinetik 2,15 MeV. Berapakah energi neutrino yang menyertainya? 34.
Sepotong kayu dari sebuah pohon yang baru ditebang memperlihatkan inti
Yang meluruh sebanyak 12,4 peluruhan per menit. Suatu cuplikan
dengan ukuran sama dari sebuah pohon yang ditebang ribuan tahun lalu memperlihatkan 3,5 peluruhan per menit. Berapakah usia cuplikan itu?
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PEMBAHASAN BUKU KENNETH KRANE
2.
Diketahui
: Timah memiliki nomor massa 114, 115, 116, 117, 118, 119, 120, 122, 124.
Ditanya
: Lambang isotop yang sesuai?
Jawab
:
Nomor Atom (Z) Timah adalah 50. 50. Jadi, lambang isotop yang sesuai adalah:
,
,
,
,
,
,
4.
(a) R (a) R = (1.2 fm) A1/3 = (1.2 fm)(197) 1/3 = 7.0 fm (b) R (b) R = (1.2 fm) A fm) A1/3 = (1.2 fm)(4)1/3 = 1.9 fm (c) R (c) R = (1.2 fm) A fm) A1/3 = (1.2 fm)(20) 1/3 = 3.3 fm
6.
B = [ Nm Nm + Zm( Zm( H) − m( A A X )]c )]c (a)
B
= [126(1.008665 u) +82(1.007825 u) − 207.976652
u](931.50MeV/u) = 1636.4MeV B / A = (1636.4MeV) / 208 = 7.868MeV per nucleon (b)
B = [78(1.008665 u) + 55(1.007825 u) −132.905452 u](931.50MeV/u) = 1118.5MeV
B / A
= (1118.5MeV) /133 = 8.410 MeV per nucleon
(c)
B = [50(1.008665 u) + 40(1.007825 u) −89.904704 u](931.50MeV/u) = 783.9MeV
B / A
= (783.9 MeV) / 90
,
,
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PEMBAHASAN BUKU KENNETH KRANE
(d)
B = [32(1.008665 u) + 27(1.007825 u) − 58.933195 u](931.50MeV/u) = 517.3MeV
B / A
= (517.3MeV) / 59 = 8.768MeV per nucleon
8.
Diketahui: mn = 1,008665 u m ( 1 H ) = 1,007825 u c2 = 931,5 MeV/u Ditanya: Energi ikat total inti dari
3
3
He dan H ?
perbedaan dengan meninjau
interaksi Coulomb dari proton lebih pada
3
He ?
Jawab: 3
He :
B Nmn
Zm1 H m A X c 2
B 1(1,008665u) 2(1,007825u) 3,016029u(931 931,50 MeV MeV / u) 7,718 718MeV 3
H :
B Nmn
Zm1 H m A X c 2
B 2(1,008665u) 1(1,007825u) 3,016049u(931 931,50 MeV MeV / u) 8,482 482 Me Me / V 13
Karena A Karena A = 3 sehingga R (1,2 fm) A
(1,2 fm)31 3
1,7 fm dan energi
tolakan dari dua proton dengan jari-jari inti sebesar 1,2 fm adalah
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PEMBAHASAN BUKU KENNETH KRANE
10.
Diketahui : S p (Energi pisah proton) = [m[ 1H] + m( m(1H)
= 1,007825 u
− − – m m (
c2
m(4He) = 4,002603 u m(12C) = 12,000000 u m(40Ca) = 39,96259 39,96259 u Ditanya : Energi pisah proton dari : a.
4
b.
12
c.
40
He C
Ca
Jawab : a. Energi pisah proton dari 4He
− −
= 3H
m(3H)
= 3,016049 u
S p (4He)
=
[m(1H) + m(3H) – H) – m( m(4He)]c2
= [1,007825 u + 3,016049 u – 4,002603 4,002603 u](931,50 MeV/u) = [4,023874 u – u – 4,002603 4,002603 u] (931,50 MeV/u) = [0,021271 u](931,50 MeV/c) = 19,8139365 MeV Jadi energi pisah proton dari 4He adalah 19,8139365 MeV b. Energi pisah proton dari
− −
m(11B)
12
C
= 11B = 11,009305 u
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PEMBAHASAN BUKU KENNETH KRANE
m(39K)
= 38,963707 u
S p (40Ca)
=
[m(1H) + m(39K) – K) – m( m(40Ca)]c2
= [1,007825u + 38,963707u – 38,963707u – 39,962591u] 39,962591u] (931,50 MeV/u) = [39,971532 u – u – 39,962591 39,962591 u](931,50 MeV/c) = [0,008941 u](931,50 MeV/u) = 8,3285415 MeV Jadi energi pisah proton dari
12.
Diket :
12
C adalah 8,3285415 MeV
∆=m = 75 GeV = 75 × 10
Ditanya : x = ...? Jawab :
Berdasarkan Asas Ketidakpastian
∆ ℏ∆ ~ ℏℏ ∆= ∆ = ∆ = = ∆ℏ = ℎℏ2 = 1=214204. 0 .. ℏ= ℏ = 12402 .. 3, 3 , 14 1 4 ℏ=197,45 .
Untuk mencari besarnya jangkauan gaya inti adalah
Dimana nilai,
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PEMBAHASAN BUKU KENNETH KRANE
14.
Diketahui
Ditanya
=213= 54ℎ 8 ℎ → = 0 ,=48 =2880 :
: (a) t1/2 = ? (b) tetapan luruh =? (c)
Penyelesaian
:
= ? → = 125 = 7500
Untuk mencari usia paruh cuplikan, terlebih dahulu mencari tetapan luruhnya (b) tetapan luruh;
(a)
waktu paruh;
= −−. 212133 = 548 . −. = 548 0,ln0,3388=l 88=n−. −. 0, 9 46=. 2 880 = 0,2880946 = 3,28 . 10− = ln2 0,693
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PEMBAHASAN BUKU KENNETH KRANE
16.
Diketahui
: Tritium
, dengan t1/2 = 12,3 tahun
t = 50 tahun
Ditanya
: Fraksi atom tritium yang tertinggal setelah 50 tahun ( ) ?
Jawab
:
=
= =
, , ,, .. x
x
x
x
= 1,81 x 10 -9 s-1
N
−−
= N0 = = = =
−,−,/ /, −,
= 0,0598
Jadi, fraksi jumlah atom tritium yang tertinggal dalam cuplikan bahan setelah 50 tahun, adalah 0,0598 bagian.
18.
− = 2,0−0,⁄/3,7×10 −/ −=7, −,40×10 ⁄, =6, 6 2× = = 7, 4 0×10 10− (a)
(b)
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PEMBAHASAN BUKU KENNETH KRANE
4 425 − = =1250 32 = 1,39 1010− − 1, 3 9 10 = 3,7 10 = 0,375375 1010− = 37,5 1010− = 37,5 22.
24.
a)
27 14
27 Si13 13 Al 14
b)
74 33
c)
288 92
d)
93 42
e)
131 131 53 78 54
74 As41 34 Se40
e
v
e _ v
U 136 24 He2 224 90Th134
Mo51 I
e
93 41
Xe77
Nb52
e
v
v
Diketahui : Reaksi yang terjadi
: 234U
m(234U) = 234,040952 u
;
230
Th + α
m( 230Th) = 230,033134 u
Ditanya : Besar Energi kinetik partikel alfa yang dipancarkan ? Jawab :
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PEMBAHASAN BUKU KENNETH KRANE
Untuk mencari Energi kinetik partikel alfa kita perlu mencari energi ikat terlebih dahulu ;
= = 228, 228,02872 287266 224, 224,02044 204455 4,00260 026033 931,5 = 5,303303 = − = − 5,303 == 5,210210 Sehingga besar Energi Kinetik partikel alfa adalah :
28.
Diketahui
: elektron mencapai energi maksimumnya Neutrino mencapai energi maksimumnya
Ditanya
: Hitung energi kinetik pental proton ?
Penyelesaian
:
Elektron mencapai energi maksimumnya Energi kinetik pental proton adalah
→=[ =[− ̅ ] ℎ
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PEMBAHASAN BUKU KENNETH KRANE
30.
Diketahui
:
+
+ v
massa neutrino dapat diabaikan. Ditanya
: a) Q? b) Ek maksimum maksimum positron?
Jawab
:
a) Pada peluruhan beta positron, Q
= [m [ = [m [
]-m[
] – 2 2 me] c2
]-m[
] – ] – 2 2 me] c2
= [15,003065 u – u – 15,000109 15,000109 u – u – (2 (2 x 5,485803 x 10 -4 u)] (931,5 MeV/u) = 1,732 MeV Jadi, nilai Q bagi peluruhan ini adalah 1,732 MeV
b) Q
= K N + K e + K v. v. Energi Kinetik
sangat kecil, sehingga dapat
diabaikan. Elektron memiliki energy kinetic maksimum ketika energy neutrino sama dengan nol. Maka,
= Q = 1,732 MeV
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PEMBAHASAN BUKU KENNETH KRANE
ℎ = 12,4 ℎ ln− 3,5 ℎ /12,4 ℎ = ln 3,5 1,26 = 1 693 = 57300,1,62ℎ = 1,0404 1010ℎ
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PEMBAHASAN BUKU KENNETH KRANE
2.
Untuk tiap reaksi berikut, berikan inti gabungannya dan tiga modus peluruhan berbeda dari inti gabung itu: a)
4.
b)
3
c)
12
d)
115
In
H 74Ge C 209Bi 27
Al
(a) Daftarkan empat reaksi berbeda, dalam mana partikel-partikel ringan ditembakkan pada suatu sasaran stabil, yang menghasilkan pembentukan inti gabung
14
N*. (b) Daftarkan empat cara berbeda [tidak mengandung
keempat reaksi dari bagian (a)] yang memungkinkan inti gabung meluruh. meluruh. 6.
Seberkas partikel alfa ditembakan pada sebuah sasaran menghasilkan reaksi: α+
, yang
+ n. Anggaplah penampang reaksi bagi suatu energi
alfa tertentu adalah 1,25 b. Sasarannya berbentuk lembaran setebal 2,5 µA,
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PEMBAHASAN BUKU KENNETH KRANE
12.
Secuplik lukisan cat disinari dengan berkas neutron yang memiliki fluks
× 10 5,8menit)
neutron/cm2 /s selama 2,5 menit. Pada akhir penyinaran didapati
3,0
bahwa aktivitas cuplikan mengandung 105 peluruhan/s dari , dan 12 peluruhan/s dari
(/ =
(/ = 5,27 tahun) hun)
. Hitunglah
jumlah gram titanium dan kobal dalam cuplikan semula. Gunakan data berikut kobal adalah murni adalah 5,25 persen 14.
dengan penampang lintang 19 b; titanium
, penampang lintangnya 0,14 b.
Anggaplah penampang total bagi neutron yang dijatuhkan pada tembaga adalah 5,0 b. Berapa bagiankah intensitas neutron yang hilang setelah melintasi tembaga setebal (a) 1,0 mm, (b) 1,0 cm, (c) 1,0 m?
16.
Carilah Nilai Q reaksi-reaksi berikut: a)
6
Li n3 H 4 He
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PEMBAHASAN BUKU KENNETH KRANE
b.
Untuk neutron 3 MeV, berapa banyak energi yang hilang
c.
Berapa banyak tumbukan yang dilakukan sebuah neutron 3 MeV sebelum energinya berkurang menjadi 0,03 eV (energi thermal).
28.
Perlihatkan bahwa reaksi fusi-D-T membebaskan energi 17,6 MeV!
30.
Andaikanlah kita mempunyai 100,0 cm 3 air, yang mengandung 0,015 persen D2O. (a) Hitunglah energi yang dapat diperoleh jika semua deuterium 2
2
3
terpakai dalam reaksi H H H . (b) Sebagai pilihan lain, hitunglah energi yang dibebaskan dalam reaksi D-T, jika dua pertiga deuterium berfusi membentuk
3
H , yang kemudian bergabung dengan sepertiga
deuterium yang sisa. 32.
Hitunglah energi yang dibebaskan apabila tiga partikel alfa bergabung membentuk
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PEMBAHASAN BUKU KENNETH KRANE
2.
a)
115
, Modus Peluruhannya: In119 51 Sb
117 50
Sn n p
116 50
Sn 2n p
115 50
Sn 3n p
74 77 b) 23 He 32 Ge 34 Se , Modus Peluruhannya: 76 34
Se n p
75 34
Se 2n p
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PEMBAHASAN BUKU KENNETH KRANE
14 7
N
6.
Diket
8 3
Li 24 He 2 p
: ρ Cu = 8,96 g/cm 3 σ = 1,25 1,25 b = 1,25 x 10 10 -24 t (tebal)
d (diameter) Arus Ditanya
= 2,5 µA = 2,5 x 10-4 cm = 0,50 cm = 7,5 µA = 7,5 . 10 -6 A
: I = ...?
Jawab : Mencari volume penampang yang sasaran yang ditembaki partikel :
=
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PEMBAHASAN BUKU KENNETH KRANE
, / , /, . = , ., = ,, 10
= 6,2626 10 / 8.
Diketahui
: Pers 10.4 Pers 10.3
Ditanya
= (1−) =
: Buktikan persamaan 10.4 merupakan pemecahan bagi persamaan 10.3 ?
Penyelesaian:
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PEMBAHASAN BUKU KENNETH KRANE
ln ln= ln = − = =−
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PEMBAHASAN BUKU KENNETH KRANE
= 51,11 x 10 18 partikel
Arus datang
= 12 x 10-6 A = 12 x 10 -6 C/s.
Muatan
= 2 x 1,6 x 10 -19 C = 3,2 x 10 -19 C.
Maka, Intensitas berkas datang: I0
=
C / x , x −
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PEMBAHASAN BUKU KENNETH KRANE
12.
Titanium
atom dari
Titanium 5.25% ×1016 .
−) − = −,, =(1 , =407− = ∅ = ,×///,,× =9,68×10 50
Ti, nomor atom dari titanium 9.68×1014 / 0.0525 = 1.84
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PEMBAHASAN BUKU KENNETH KRANE
Jadi , sebesar 34,6% bagian hilang
(c)
= 1,0 = 100 ; = − = −, = 3,8585 1010−
Semua bagian diserap. 16.
a) 6 Li n3 H 4 He
Q m( 6 Li) m(n) m( 3 H ) m( 4 He) c 2
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PEMBAHASAN BUKU KENNETH KRANE
b.
Energi ambang proton kepada 4He yang diam X
= 1H dan X = 4He
K th th
= -Q
1 1 11 ,,
= -(-18,353MeV)
= 18,353 MeV (1 + 0,25179)
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PEMBAHASAN BUKU KENNETH KRANE
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PEMBAHASAN BUKU KENNETH KRANE