Noyaux Masse Et Energie
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اى ،ا وا Iا '' آ '' - (1أ : ﺃﺛﺒﺘﺖ ﺍﳌﻴﻜﺎﻧﻴﻚ ﺍﻟﻨﺴﺒﻮﻳﺔ ﺍﳋﺎﺻﺔ ﺍﻟﱵ ﺃﻧﺸﺄﻫﺎ ﺃﻟﺒﲑﺕ ﺃﻳﻨﺸﺘﺎﻳﻦ ﺳﻨﺔ 1905ﻡ ﺃﻥ ﻫﻨﺎﻙ ﺗﻜﺎﻓﺆﺍ ﺑﲔ ﺍﻟﻜﺘﻠﺔ ﻭﺍﻟﻄﺎﻗﺔ)ﺃﻱ ﺃﻥ ﺍﻟﻜﺘﻠﺔ ﺗﻌﺘـﱪ ﺷﻜﻼ ﻣﻦ ﺃﺷﻜﺎﻝ ﺍﻟﻄﺎﻗﺔ( ،ﲝﻴﺚ ﺃﻥ ﻛﻞ ﳎﻤﻮﻋﺔ ﻣﺎﺩﻳﺔ ﻛﺘﻠﺘﻬﺎ mﲤﺘﻠﻚ ﻃﺎﻗﺔ Eﺗﺴﻤﻰ ﺍﻟﻄﺎﻗﺔ ﺍﻟﻜﺘﻠﻴﺔ ﻭﻫﻲ ﺗﺴﺎﻭﻱ ﺣﺎﺻﻞ ﺿﺮﺏ ﺍﻟﻜﺘﻠﺔ ﻭﻣﺮﺑﻊ ﺳﺮﻋﺔ ﺍﻧﺘﺸﺎﺭ ﺍﻟﻀﻮﺀ ﰲ ﺍﻟﻔﺮﺍﻍ .
ﻭﻭﺣﺪﺓ ﺍﻟﻄﺎﻗﺔ ﺍﻟﻜﺘﻠﻴﺔ ﰲ ﺍﻟﻔﻴﺰﻳﺎﺀ ﺍﻟﻨﻮﻭﻳﺔ ﻫﻲ ﺍﻹﻟﻜﺘﺮﻭﻥ -ﻓﻮﻟﻂ ) (eVﺍﻟﺬﻱ ﺗﺮﺑﻄﻪ ﺑﺎﳉﻮﻝ ﺍﻟﻌﻼﻗﺔ ﺍﻟﺘﺎﻟﻴﺔ. 1eV = 1,6 × 10 −19 J : ﻭﻣﻦ ﻣﻀﺎﻋﻔﺎﺗﻪ ﺍﳌﻴﻐﺎ ﺇﻟﻜﺘﺮﻭﻥ ﻓﻮﻟﻂ 1MeV = 10 6 eV = 1,6 × 10 −13 J
ﻭﺗﺒﲔ ﻫﺬﻩ ﺍﻟﻌﻼﻗﺔ ﺃﻥ ﻛﻞ ﺗﻐﲑ ﻟﻜﺘﻠﺔ ﳎﻤﻮﻋﺔ ﻣﺎ ﺑﺎﳌﻘﺪﺍﺭ ∆mﻳﻮﺍﻓﻘﻪ ﺗﻐﲑ ﻟﻠﻄﺎﻗﺔ ﺍﻟﻜﺘﻠﻴﺔ ﳍﺬﻩ ﺍﻤﻮﻋﺔ ﺑﺎﳌﻘﺪﺍﺭ
∆E = ∆m.c 2
(2ﻭﺣﺪﺓ ﺍﻟﻜﺘﻠﺔ ﺍﻟﺬﺭﻳﺔ: ﻧﻈﺮﺍ ﻟﻜﻮﻥ ﻛﺘﻞ ﺍﻟﻨﻮﻯ ﻭﺍﻟﺪﻗﺎﺋﻖ ﺻﻐﲑﺓ ﺟﺪﺍ ،ﻳـﻌﱪ ﻋﻨﻬﺎ ﰲ ﺍﻟﻔﻴﺰﻳﺎﺀ ﺍﻟﻨﻮﻭﻳﺔ ﺑﻮﺣﺪﺓ ﻣﻼﺋﻤﺔ ﺗﺴـﻤـﻰ ﺏ :ﻭﺣﺪﺓ ﺍﻟﻜﺘﻠﺔ ﺍﻟﺬﺭﻳﺔ : Unité de masse atomiqueﺃﻱ u.m.a.
ﻭﺍﻟﱵ ﻳﺮﻣﺰ ﺇﻟﻴﻬﺎ ﺏ uﻓﻘﻂ ﻣﻦ ﺃﺟﻞ ﺍﻟﺘﺒﺴﻴﻂ.
)ﺍﻟﻜﺘﻠﺔ ﺍﳌﻮﻟﻴﺔ ﻝ 126 C :ﺗﺴﺎﻭﻱ (12 g / mol و ن آ ! $ل وا ! "#ذرات ا ن وي ، 12 gو ي 1ا ل NA=6,02.1023ذرة. ) m(126 C 12 g 12 × 10 −3 Kg = 1.u = = ﺃﻱ= 1,66 × 10 − 27 Kg : 23 12 12.N A 12 × 6,02 × 10
ﻭﺑﺎﻟﺘﺎﱄ 1u =1,66×10−27 Kg : eV/ c2
ﻛﻤﺎ ﻧﺴﺘﻌﻤﻞ ﻛﻮﺣﺪﺓ ﻟﻠﻜﺘﻠﺔ ﰲ ﺍﻟﻔﻴﺰﻳﺎﺀ ﺍﻟﻨﻮﻭﻳﺔ ﺍﻟﻮﺣﺪﺓ ﺍﻟﺘﺎﻟﻴﺔ :
1,6 × 10 −13 J 1,78 × 10 −30 u −30 = 1 , 78 × 10 Kg = ≈ 1,073 × 10 −3 u (3 × 10 8 ) 2 m 2 / s 2 1,66 × 10 − 27
= 1MeV / c 2
ﻭﻣﻨﻪ 1u ≈ 931 ,5 MeV / c 2 : ﻟﻨﺘﺤﻘﻖ ﻣﻦ ﲡﺎﻧﺲ ﺍﻟﻮﺣﺪﺍﺕ ﰲ ﺍﻟﻌﻼﻗﺔ ﺍﻟﺴﺎﺑﻘﺔ
J N ×m Kg × m.s −2 × m = = = Kg : 2 m 2 / s 2 m 2 × s −2 m2 × s−
ﻭﻫﻜﺬﺍ ﰲ ﺍﻟﻔﻴﺰﻳﺎﺀ ﺍﻟﻨﻮﻭﻳﺔ ﻧﺴﺘﻌﻤﻞ ﻛﻮﺣﺪﺓ ﻟﻠﻜﺘﻠﺔ ﺇﻣﺎ :ﺍﻝ:
uﺃﻭ ﺍﻝ:
MeV/ c2
Iا"! اة: (1ا $%ا :# ﺑﻴﻨﺖ ﻗﻴﺎﺳﺎﺕ ﺩﻗﻴﻘﺔ ﺃﳒﺰﺕ ﺑﻮﺍﺳﻄﺔ ﻣﻄﻴﺎﻑ ﺍﻟﻜﺘﻠﺔ ﺑﺎﻟﻨﺴﺒﺔ ﻟﻨﻮﺍﺓ ZA Xﺃﻥ ﻛﺘﻠﺔ ﺍﻟﻨﻮﺍﺓ ،ﻣﺮﺗﺒﻄﺔ ،ﺗﻜﻮﻥ ﺩﺍﺋﻤﺎ ﺃﺻﻐﺮ ﻣﻦ ﻛﺘﻞ ﺍﻟﺪﻗﺎﺋﻖ ﺍﳌﻜﻮﻧﺔ ﳍﺎ )ﻏﲑ ﻣﺮﺗﺒﻄﺔ( ﻭﻳﺴﻤﻰ ﻫﺬﺍ ﺍﻟﻔﺮﻕ ﺑﺎﻟﻨﻘﺺ ﺍﻟﻜﺘﻠﻲ ( Le.défaut.de.masse) .
1
ﻧﺴﻤﻲ ﺍﻟﻨﻘﺺ ﺍﻟﻜﺘﻠﻲ ∆mﻟﻨﻮﺍﺓ ZA Xﺍﻟﻔﺮﻕ ﺑﲔ ﳎﻤﻮﻉ ﻛﺘﻞ ﺍﻟﻨﻮﻳﺎﺕ ﻭﻛﺘﻠﺔ ﺍﻟﻨﻮﺍﺓ : ) ∆m = Zm p + ( A − Z )mn − m( ZAX
ﻭﻫﻮ ﻣﻘﺪﺍﺭ ﻣﻮﺟﺐ.
(2ا"! : ﻃﺎﻗﺔ ﺍﻟﺮﺑﻂ E lﻟﻨﻮﺍﺓ ZA Xﻫﻲ ﺍﻟﻄﺎﻗﺔ ﺍﻟﱵ ﳚﺐ ﺇﻋﻄﺎﺅﻫﺎ ﻟﻠﻨﻮﺍﺓ ﰲ ﺣﺎﻟﺔ ﺳﻜﻮﻥ ﻟﻔﺼﻞ ﻧﻮﻳﺎﺎ ﻭﺗﺒﻘﻰ ﰲ ﺣﺎﻟﺔ ﺳﻜﻮﻥ.
[
]
ﻭﺗﻌﻄﻴﻬﺎ ﺍﻟﻌﻼﻗﺔ ﺍﻟﺘﺎﻟﻴﺔE l = ∆m.c 2 = Z .m p + ( A − N )mn − m( ZAX ) .c 2 :
(3ا"! !)( : El ﻧﺴﺘﻌﻤﻞ ﺃﺣﻴﺎﻧﺎ ﻃﺎﻗﺔ ﺍﻟﺮﺑﻂ ﺑﺎﻟﻨﺴﺒﺔ ﻟﻨﻮﻳﺔ ﻭﺗﻌﻄﻴﻬﺎ ﺍﻟﻌﻼﻗﺔ ﺍﻟﺘﺎﻟﻴﺔ : A ﻭﻭﺣﺪﺎ . MeV / nucléon :
= ξﺣﻴﺚ E l :ﻫﻲ ﻃﺎﻗﺔ ﺍﻟﺮﺑﻂ ﻟﻠﻨﻮﺍﺓ ﻭ Aﻋﺪﺩ ﺍﻟﻨﻮﻳﺎﺕ.
ﻛﻠﻤﺎ ﻛﺎﻧﺖ ﻃﺎﻗﺔ ﺍﻟﺮﺑﻂ ﺑﺎﻟﻨﺴﺒﺔ ﻟﻨﻮﻳﺔ ﻛﺒﲑﺓ ﻛﻠﻤﺎ ﻛﺎﻧﺖ ﺍﻟﻨﻮﺍﺓ ﺃﻛﺜﺮ ﺍﺳﺘﻘﺮﺍﺭﺍ.
:+,(- (4 238 ، m( 238ﻛﺘﻠﺔ ﺍﻟﱪﻭﺗﻮﻥ m p = 1,0073u :ﻭ ﻛﺘﻠﺔ ﺍﻟﻨﻮﺗﺮﺭﻭﻥ = 1,0087u ﻋﻠﻤﺎ ﺃﻥ ﻛﺘﻠﺔﻧﻮﺍﺓ ﺍﻷﻭﺭﺍﻨﻴﻭﻡ 92 Uﻫﻲ 92 U ) = 238,003u
. mn
. 238 ﺃ( ﺍﺣﺴﺐ ﻃﺎﻗﺔ ﺍﻟﺮﺑﻂ ﻟﻨﻮﺍﺓ ﺍﻷﻭﺭﺍﻧﻴﻮﻡ 92 U . 238 ﺏ( ﺍﺣﺴﺐ ﻃﺎﻗﺔ ﺍﻟﺮﺑﻂ ﺑﺎﻟﻨﺴﺒﺔ ﻟﻨﻮﻳﺔ ﻟﻨﻮﺍﺓ ﺍﻻﻭﺭﺍﻧﻴﻮﻡ 92 U ﻧﻌﻄﻲ1u = 931,5MeV / c 2 : --------------------------------------------
ﺃ( ﻟﺪﻳﻨﺎ :
]
[
= E l = ∆m.c = Z .m p + ( A − N )mn − m( U ) .c = [92 × 1,0073 + (238 − 92) × 1,0087 − 238,0003]u × c 2 2
238 92
2
............................................................................ = 1,9415u × c 2 = (1,945 × 931,5MeV / c 2 ) × c 2 = 1808,5MeV E 1808,5 = ξ= l ﺏ( ≈ 7,59MeV / nucléon A 138
123 (5أ0ن : El ﻝﻤﻘﺎﺭﻨﺔ ﺍﺴﺘﻘﺭﺍﺭ ﺍﻝﻨﻭﻯ ،ﻨﺨﻁ ﺍﻝﻤﻨﺤﻨﻰ ﺍﻝﻤﻤﺜل ﻝﺘﻐﻴﺭﺍﺕ A
−ﺒﺩﻻﻝﺔ ﻋﺩﺩ ﺍﻝﻨﻭﻴﺎﺕ .ﻴﺴﻤﻰ ﻫﺫﺍ ﺍﻝﻤﻨﺤﻨﻰ :ﻤﻨﺤﻨﻰ ﺃﺴﻁﻭﻥ .
ا$ى ا(آ' إ ارا "#$%أ ا ،ا ة إذا آ *+وإذا آ ."% -و$ % 01ل إ $ى أآ' إ ارا.
(6ا ,52ا7 8 9 ,وي: ﺃ( ﺘﻌﻤﻴﻡ :
7 9- "( <7و <3د : :
X 2 → Y1 + Y 2 A2 Z2
A1 Z1
2
A2 Z2
X1+
A1 Z1
]) ∆E = [E l ( X 1 ) + E l ( X 2 ) − E l (Y1 ) − E l (Y2
ﺗﻜﺘﺐ ﺍﳊﺼﻴﻠﺔ ﺍﻟﻄﺎﻗﻴﺔ ﺍﳌﻘﺮﻭﻧﺔ ﺬﺍ ﺍﻟﺘﻔﺎﻋﻞ ﻛﻤﺎ ﻳﻠﻲ:
(2ﺍﳊﺼﻴﻠﺔ ﺍﻟﻄﺎﻗﻴﺔ ﻟﻠﺘﺤﻮﻻﺕ ﺍﻟﻨﻮﻭﻳﺔ ﺍﻟﺘﻠﻠﻘﺎﺋﻴﺔ:
* ا ا"2 Cرة Aل اط ا>=< : α #
X →AZ−−42Y +24He
<3د ا : α D 9 ا ا"2 Cرة Aل اط ا>=< α #ه:#
A Z
* ا ا"2 Cرة Aل اط ا>=< : β − # <3د ا : β − D 9 ا ا"2 Cرة Aل اط ا>=< β − #ه:#
X →Z +A1Y +−10e
A Z
* ا ا"2 Cرة Aل اط ا>=< : β + # <3د ا : β + D 9 ا ا"2 Cرة Aل اط ا>=< β + #ه:#
X →Z −A1Y ++10e
A Z
( IIIﺍﻟﺘﺄﺛﲑﺍﺕ ﺍﻟﺒﻴﻮﻟﻮﺟﻴﺔ ﻟﻠﻨﺸﺎﻁ ﺍﻹﺷﻌﺎﻋﻲ: 3%ا 5 678ت ا$و; :# 5ا 8ن < و.%
ا"? >= ا $<% : αن د? أ $ Bاة ا$م He q = +2e = +2 × 1,6 × 10 −19 c E#$
و HE%
4 2
ا $ %ي 5و B $%و و Bو > 7
5د? >= أ BGا * B ?Fا دة ا6+ا6+
10
1
5ا$Jء و < Bآ KG Eا Eو 7اF6K; E#$ن ?"ر 5 %ا-Nاق O*% F J L أن L%ق ا Q;RKت ا(و "#ا Nن أو Pآ ب .
ا : β − ,! +FGه اNآو ت e
0 −1
16000km / s 5 ";R% 5 5 % 5 ? U
و 5 K
ا 6و KGا Vن "? >= ?$ة ا-اق أ$? B 5ة ا-اق ا"? >= أ O*% .أن L%ق :#ا Nن ا Pة و 1>"5 WE%ا وق K 6 6% .ا ;FLا. U
ا(γ X 67
ه ' أ #$ X 67ت آو (=> ?" ) U
5ه
5ا +را$Jء
ذات ?$ة ا-اق 5و "ة <+آR; ;( ، Eم "5ة ات Bا Pص ،و 6% ^L+%ا(اض $Vر.و _ا (ن أ 7 0% N X 67و Nآ ن ا * ; N ?Fا"6د ا< أو ا"6د ا1ري 1رة.
3
:+,(- (7 (1 )
( 2) ( 3)
226 88
Ra →
60 27 13 7
222 86
:ﻨﻌﺘﺒﺭ ﺍﻝﺘﺤﻭﻻﺕ ﺍﻝﺘﺎﻝﻴﺔ Rn + He 4 2
60 Co→ 28 Ni + −10 e
N →136 C + 10 e
m( 24 He) = 4,0015u ، m( 222 ، m( 226 86 Rn) = 221,9702u 88 Ra ) = 225,9770u : ﻧﻌﻄﻲ m(e) = 5,49 × 10 −4 u
60 ، m( 28 Ni ) = 59,915u
60 ، m( 28 Co) = 59,9190u
. m(136C ) = 13,000062u
، m(137N ) = 13,001898u
. ﺃ( ﺍﺣﺴﺐ ﺗﻐﲑ ﺍﻟﻜﺘﻠﺔ ﺍﳌﺼﺎﺣﺐ ﻟﻜﻞ ﲢﻮﻝ ﻧﻮﻭﻱ .ﺏ( ﺍﺣﺴﺐ ﻃﺎﻗﺔ ﻛﻞ ﲢﻮﻝ ﻣﻦ ﺍﻟﺘﺤﻮﻻﺕ ﺍﻟﺴﺎﺑﻘﺔ 1u = 931,5MeV / c :ﻧﻌﻄﻲ 2
--------------------------------------------
∆m1 = m( Rn) + m( He) − m( Ra ) = −5,3 × 10 −3 u ≈ −4,94 Mev / c 2 :
226 88
∆m2 = m( Ni ) + m(e) − m(Co) = −3,45 × 10 −3 u ≈ −3,21Mev / c 2 ∆m3 = m(C ) + m(e) − m( N ) = −1,287 × 10 −3 u ≈ −1,2 Mev / c 2
4 Ra → 222 86 Rn + 2 He ﺃ( ﺑﺎﻟﻨﺴﺒﺔ ﻟﻠﺘﺤﻮﻝ ﺍﻻﻭﻝ
:
60 27
60 Co→ 28 Ni + −10 e ﺑﺎﻟﻨﺴﺒﺔ ﻟﻠﺘﺤﻮﻝ ﺍﻟﺜﺎﱐ
:
13 7
N →136 C + 10 e ﺑﺎﻟﻨﺴﺒﺔ ﻟﻠﺘﺤﻮﻝ ﺍﻟﺜﺎﻟﺚ
E1 = ∆m1 × c 2 = (−4,94 MeV / c 2 ) × c 2 = −4,94 MeV -:ﺏ( ﺍﻟﻄﺎﻗﺔ ﺍﳌﺘﺤﺮﺭﺓ ﺧﻼﻝ ﺍﻟﺘﺤﻮﻝ ﺍﻷﻭﻝ E 2 = ∆m2 × c 2 = (−3,21MeV / c 2 ) × c 2 = −3,21MeV -:ﺍﻟﻄﺎﻗﺔ ﺍﳌﺘﺤﺮﺭﺓ ﺧﻼﻝ ﺍﻟﺘﺤﻮﻝ ﺍﻟﺜﺎﱐ E 3 = ∆m3 × c 2 = (−1,2 MeV / c 2 ) × c 2 = −1,2 MeV -:ﺍﻟﻄﺎﻗﺔ ﺍﳌﺘﺤﺮﺭﺓ ﺧﻼﻝ ﺍﻟﺘﺤﻮﻝ ﺍﻟﺜﺎﻟﺚ
SBIRO Abdelkrim Lycée Agricole et Lycée Abdellah Cheffchouni Oulad –Taima région d’Agadir Maroc Mai : [email protected] Pour toutes vos observations contacter mon émail
4
ﻭﻭﺣﺪﺓ ﺍﻟﻄﺎﻗﺔ ﺍﻟﻜﺘﻠﻴﺔ ﰲ ﺍﻟﻔﻴﺰﻳﺎﺀ ﺍﻟﻨﻮﻭﻳﺔ ﻫﻲ ﺍﻹﻟﻜﺘﺮﻭﻥ -ﻓﻮﻟﻂ ) (eVﺍﻟﺬﻱ ﺗﺮﺑﻄﻪ ﺑﺎﳉﻮﻝ ﺍﻟﻌﻼﻗﺔ ﺍﻟﺘﺎﻟﻴﺔ. 1eV = 1,6 × 10 −19 J : ﻭﻣﻦ ﻣﻀﺎﻋﻔﺎﺗﻪ ﺍﳌﻴﻐﺎ ﺇﻟﻜﺘﺮﻭﻥ ﻓﻮﻟﻂ 1MeV = 10 6 eV = 1,6 × 10 −13 J
ﻭﺗﺒﲔ ﻫﺬﻩ ﺍﻟﻌﻼﻗﺔ ﺃﻥ ﻛﻞ ﺗﻐﲑ ﻟﻜﺘﻠﺔ ﳎﻤﻮﻋﺔ ﻣﺎ ﺑﺎﳌﻘﺪﺍﺭ ∆mﻳﻮﺍﻓﻘﻪ ﺗﻐﲑ ﻟﻠﻄﺎﻗﺔ ﺍﻟﻜﺘﻠﻴﺔ ﳍﺬﻩ ﺍﻤﻮﻋﺔ ﺑﺎﳌﻘﺪﺍﺭ
∆E = ∆m.c 2
(2ﻭﺣﺪﺓ ﺍﻟﻜﺘﻠﺔ ﺍﻟﺬﺭﻳﺔ: ﻧﻈﺮﺍ ﻟﻜﻮﻥ ﻛﺘﻞ ﺍﻟﻨﻮﻯ ﻭﺍﻟﺪﻗﺎﺋﻖ ﺻﻐﲑﺓ ﺟﺪﺍ ،ﻳـﻌﱪ ﻋﻨﻬﺎ ﰲ ﺍﻟﻔﻴﺰﻳﺎﺀ ﺍﻟﻨﻮﻭﻳﺔ ﺑﻮﺣﺪﺓ ﻣﻼﺋﻤﺔ ﺗﺴـﻤـﻰ ﺏ :ﻭﺣﺪﺓ ﺍﻟﻜﺘﻠﺔ ﺍﻟﺬﺭﻳﺔ : Unité de masse atomiqueﺃﻱ u.m.a.
ﻭﺍﻟﱵ ﻳﺮﻣﺰ ﺇﻟﻴﻬﺎ ﺏ uﻓﻘﻂ ﻣﻦ ﺃﺟﻞ ﺍﻟﺘﺒﺴﻴﻂ.
)ﺍﻟﻜﺘﻠﺔ ﺍﳌﻮﻟﻴﺔ ﻝ 126 C :ﺗﺴﺎﻭﻱ (12 g / mol و ن آ ! $ل وا ! "#ذرات ا ن وي ، 12 gو ي 1ا ل NA=6,02.1023ذرة. ) m(126 C 12 g 12 × 10 −3 Kg = 1.u = = ﺃﻱ= 1,66 × 10 − 27 Kg : 23 12 12.N A 12 × 6,02 × 10
ﻭﺑﺎﻟﺘﺎﱄ 1u =1,66×10−27 Kg : eV/ c2
ﻛﻤﺎ ﻧﺴﺘﻌﻤﻞ ﻛﻮﺣﺪﺓ ﻟﻠﻜﺘﻠﺔ ﰲ ﺍﻟﻔﻴﺰﻳﺎﺀ ﺍﻟﻨﻮﻭﻳﺔ ﺍﻟﻮﺣﺪﺓ ﺍﻟﺘﺎﻟﻴﺔ :
1,6 × 10 −13 J 1,78 × 10 −30 u −30 = 1 , 78 × 10 Kg = ≈ 1,073 × 10 −3 u (3 × 10 8 ) 2 m 2 / s 2 1,66 × 10 − 27
= 1MeV / c 2
ﻭﻣﻨﻪ 1u ≈ 931 ,5 MeV / c 2 : ﻟﻨﺘﺤﻘﻖ ﻣﻦ ﲡﺎﻧﺲ ﺍﻟﻮﺣﺪﺍﺕ ﰲ ﺍﻟﻌﻼﻗﺔ ﺍﻟﺴﺎﺑﻘﺔ
J N ×m Kg × m.s −2 × m = = = Kg : 2 m 2 / s 2 m 2 × s −2 m2 × s−
ﻭﻫﻜﺬﺍ ﰲ ﺍﻟﻔﻴﺰﻳﺎﺀ ﺍﻟﻨﻮﻭﻳﺔ ﻧﺴﺘﻌﻤﻞ ﻛﻮﺣﺪﺓ ﻟﻠﻜﺘﻠﺔ ﺇﻣﺎ :ﺍﻝ:
uﺃﻭ ﺍﻝ:
MeV/ c2
Iا"! اة: (1ا $%ا :# ﺑﻴﻨﺖ ﻗﻴﺎﺳﺎﺕ ﺩﻗﻴﻘﺔ ﺃﳒﺰﺕ ﺑﻮﺍﺳﻄﺔ ﻣﻄﻴﺎﻑ ﺍﻟﻜﺘﻠﺔ ﺑﺎﻟﻨﺴﺒﺔ ﻟﻨﻮﺍﺓ ZA Xﺃﻥ ﻛﺘﻠﺔ ﺍﻟﻨﻮﺍﺓ ،ﻣﺮﺗﺒﻄﺔ ،ﺗﻜﻮﻥ ﺩﺍﺋﻤﺎ ﺃﺻﻐﺮ ﻣﻦ ﻛﺘﻞ ﺍﻟﺪﻗﺎﺋﻖ ﺍﳌﻜﻮﻧﺔ ﳍﺎ )ﻏﲑ ﻣﺮﺗﺒﻄﺔ( ﻭﻳﺴﻤﻰ ﻫﺬﺍ ﺍﻟﻔﺮﻕ ﺑﺎﻟﻨﻘﺺ ﺍﻟﻜﺘﻠﻲ ( Le.défaut.de.masse) .
1
ﻧﺴﻤﻲ ﺍﻟﻨﻘﺺ ﺍﻟﻜﺘﻠﻲ ∆mﻟﻨﻮﺍﺓ ZA Xﺍﻟﻔﺮﻕ ﺑﲔ ﳎﻤﻮﻉ ﻛﺘﻞ ﺍﻟﻨﻮﻳﺎﺕ ﻭﻛﺘﻠﺔ ﺍﻟﻨﻮﺍﺓ : ) ∆m = Zm p + ( A − Z )mn − m( ZAX
ﻭﻫﻮ ﻣﻘﺪﺍﺭ ﻣﻮﺟﺐ.
(2ا"! : ﻃﺎﻗﺔ ﺍﻟﺮﺑﻂ E lﻟﻨﻮﺍﺓ ZA Xﻫﻲ ﺍﻟﻄﺎﻗﺔ ﺍﻟﱵ ﳚﺐ ﺇﻋﻄﺎﺅﻫﺎ ﻟﻠﻨﻮﺍﺓ ﰲ ﺣﺎﻟﺔ ﺳﻜﻮﻥ ﻟﻔﺼﻞ ﻧﻮﻳﺎﺎ ﻭﺗﺒﻘﻰ ﰲ ﺣﺎﻟﺔ ﺳﻜﻮﻥ.
[
]
ﻭﺗﻌﻄﻴﻬﺎ ﺍﻟﻌﻼﻗﺔ ﺍﻟﺘﺎﻟﻴﺔE l = ∆m.c 2 = Z .m p + ( A − N )mn − m( ZAX ) .c 2 :
(3ا"! !)( : El ﻧﺴﺘﻌﻤﻞ ﺃﺣﻴﺎﻧﺎ ﻃﺎﻗﺔ ﺍﻟﺮﺑﻂ ﺑﺎﻟﻨﺴﺒﺔ ﻟﻨﻮﻳﺔ ﻭﺗﻌﻄﻴﻬﺎ ﺍﻟﻌﻼﻗﺔ ﺍﻟﺘﺎﻟﻴﺔ : A ﻭﻭﺣﺪﺎ . MeV / nucléon :
= ξﺣﻴﺚ E l :ﻫﻲ ﻃﺎﻗﺔ ﺍﻟﺮﺑﻂ ﻟﻠﻨﻮﺍﺓ ﻭ Aﻋﺪﺩ ﺍﻟﻨﻮﻳﺎﺕ.
ﻛﻠﻤﺎ ﻛﺎﻧﺖ ﻃﺎﻗﺔ ﺍﻟﺮﺑﻂ ﺑﺎﻟﻨﺴﺒﺔ ﻟﻨﻮﻳﺔ ﻛﺒﲑﺓ ﻛﻠﻤﺎ ﻛﺎﻧﺖ ﺍﻟﻨﻮﺍﺓ ﺃﻛﺜﺮ ﺍﺳﺘﻘﺮﺍﺭﺍ.
:+,(- (4 238 ، m( 238ﻛﺘﻠﺔ ﺍﻟﱪﻭﺗﻮﻥ m p = 1,0073u :ﻭ ﻛﺘﻠﺔ ﺍﻟﻨﻮﺗﺮﺭﻭﻥ = 1,0087u ﻋﻠﻤﺎ ﺃﻥ ﻛﺘﻠﺔﻧﻮﺍﺓ ﺍﻷﻭﺭﺍﻨﻴﻭﻡ 92 Uﻫﻲ 92 U ) = 238,003u
. mn
. 238 ﺃ( ﺍﺣﺴﺐ ﻃﺎﻗﺔ ﺍﻟﺮﺑﻂ ﻟﻨﻮﺍﺓ ﺍﻷﻭﺭﺍﻧﻴﻮﻡ 92 U . 238 ﺏ( ﺍﺣﺴﺐ ﻃﺎﻗﺔ ﺍﻟﺮﺑﻂ ﺑﺎﻟﻨﺴﺒﺔ ﻟﻨﻮﻳﺔ ﻟﻨﻮﺍﺓ ﺍﻻﻭﺭﺍﻧﻴﻮﻡ 92 U ﻧﻌﻄﻲ1u = 931,5MeV / c 2 : --------------------------------------------
ﺃ( ﻟﺪﻳﻨﺎ :
]
[
= E l = ∆m.c = Z .m p + ( A − N )mn − m( U ) .c = [92 × 1,0073 + (238 − 92) × 1,0087 − 238,0003]u × c 2 2
238 92
2
............................................................................ = 1,9415u × c 2 = (1,945 × 931,5MeV / c 2 ) × c 2 = 1808,5MeV E 1808,5 = ξ= l ﺏ( ≈ 7,59MeV / nucléon A 138
123 (5أ0ن : El ﻝﻤﻘﺎﺭﻨﺔ ﺍﺴﺘﻘﺭﺍﺭ ﺍﻝﻨﻭﻯ ،ﻨﺨﻁ ﺍﻝﻤﻨﺤﻨﻰ ﺍﻝﻤﻤﺜل ﻝﺘﻐﻴﺭﺍﺕ A
−ﺒﺩﻻﻝﺔ ﻋﺩﺩ ﺍﻝﻨﻭﻴﺎﺕ .ﻴﺴﻤﻰ ﻫﺫﺍ ﺍﻝﻤﻨﺤﻨﻰ :ﻤﻨﺤﻨﻰ ﺃﺴﻁﻭﻥ .
ا$ى ا(آ' إ ارا "#$%أ ا ،ا ة إذا آ *+وإذا آ ."% -و$ % 01ل إ $ى أآ' إ ارا.
(6ا ,52ا7 8 9 ,وي: ﺃ( ﺘﻌﻤﻴﻡ :
7 9- "( <7و <3د : :
X 2 → Y1 + Y 2 A2 Z2
A1 Z1
2
A2 Z2
X1+
A1 Z1
]) ∆E = [E l ( X 1 ) + E l ( X 2 ) − E l (Y1 ) − E l (Y2
ﺗﻜﺘﺐ ﺍﳊﺼﻴﻠﺔ ﺍﻟﻄﺎﻗﻴﺔ ﺍﳌﻘﺮﻭﻧﺔ ﺬﺍ ﺍﻟﺘﻔﺎﻋﻞ ﻛﻤﺎ ﻳﻠﻲ:
(2ﺍﳊﺼﻴﻠﺔ ﺍﻟﻄﺎﻗﻴﺔ ﻟﻠﺘﺤﻮﻻﺕ ﺍﻟﻨﻮﻭﻳﺔ ﺍﻟﺘﻠﻠﻘﺎﺋﻴﺔ:
* ا ا"2 Cرة Aل اط ا>=< : α #
X →AZ−−42Y +24He
<3د ا : α D 9 ا ا"2 Cرة Aل اط ا>=< α #ه:#
A Z
* ا ا"2 Cرة Aل اط ا>=< : β − # <3د ا : β − D 9 ا ا"2 Cرة Aل اط ا>=< β − #ه:#
X →Z +A1Y +−10e
A Z
* ا ا"2 Cرة Aل اط ا>=< : β + # <3د ا : β + D 9 ا ا"2 Cرة Aل اط ا>=< β + #ه:#
X →Z −A1Y ++10e
A Z
( IIIﺍﻟﺘﺄﺛﲑﺍﺕ ﺍﻟﺒﻴﻮﻟﻮﺟﻴﺔ ﻟﻠﻨﺸﺎﻁ ﺍﻹﺷﻌﺎﻋﻲ: 3%ا 5 678ت ا$و; :# 5ا 8ن < و.%
ا"? >= ا $<% : αن د? أ $ Bاة ا$م He q = +2e = +2 × 1,6 × 10 −19 c E#$
و HE%
4 2
ا $ %ي 5و B $%و و Bو > 7
5د? >= أ BGا * B ?Fا دة ا6+ا6+
10
1
5ا$Jء و < Bآ KG Eا Eو 7اF6K; E#$ن ?"ر 5 %ا-Nاق O*% F J L أن L%ق ا Q;RKت ا(و "#ا Nن أو Pآ ب .
ا : β − ,! +FGه اNآو ت e
0 −1
16000km / s 5 ";R% 5 5 % 5 ? U
و 5 K
ا 6و KGا Vن "? >= ?$ة ا-اق أ$? B 5ة ا-اق ا"? >= أ O*% .أن L%ق :#ا Nن ا Pة و 1>"5 WE%ا وق K 6 6% .ا ;FLا. U
ا(γ X 67
ه ' أ #$ X 67ت آو (=> ?" ) U
5ه
5ا +را$Jء
ذات ?$ة ا-اق 5و "ة <+آR; ;( ، Eم "5ة ات Bا Pص ،و 6% ^L+%ا(اض $Vر.و _ا (ن أ 7 0% N X 67و Nآ ن ا * ; N ?Fا"6د ا< أو ا"6د ا1ري 1رة.
3
:+,(- (7 (1 )
( 2) ( 3)
226 88
Ra →
60 27 13 7
222 86
:ﻨﻌﺘﺒﺭ ﺍﻝﺘﺤﻭﻻﺕ ﺍﻝﺘﺎﻝﻴﺔ Rn + He 4 2
60 Co→ 28 Ni + −10 e
N →136 C + 10 e
m( 24 He) = 4,0015u ، m( 222 ، m( 226 86 Rn) = 221,9702u 88 Ra ) = 225,9770u : ﻧﻌﻄﻲ m(e) = 5,49 × 10 −4 u
60 ، m( 28 Ni ) = 59,915u
60 ، m( 28 Co) = 59,9190u
. m(136C ) = 13,000062u
، m(137N ) = 13,001898u
. ﺃ( ﺍﺣﺴﺐ ﺗﻐﲑ ﺍﻟﻜﺘﻠﺔ ﺍﳌﺼﺎﺣﺐ ﻟﻜﻞ ﲢﻮﻝ ﻧﻮﻭﻱ .ﺏ( ﺍﺣﺴﺐ ﻃﺎﻗﺔ ﻛﻞ ﲢﻮﻝ ﻣﻦ ﺍﻟﺘﺤﻮﻻﺕ ﺍﻟﺴﺎﺑﻘﺔ 1u = 931,5MeV / c :ﻧﻌﻄﻲ 2
--------------------------------------------
∆m1 = m( Rn) + m( He) − m( Ra ) = −5,3 × 10 −3 u ≈ −4,94 Mev / c 2 :
226 88
∆m2 = m( Ni ) + m(e) − m(Co) = −3,45 × 10 −3 u ≈ −3,21Mev / c 2 ∆m3 = m(C ) + m(e) − m( N ) = −1,287 × 10 −3 u ≈ −1,2 Mev / c 2
4 Ra → 222 86 Rn + 2 He ﺃ( ﺑﺎﻟﻨﺴﺒﺔ ﻟﻠﺘﺤﻮﻝ ﺍﻻﻭﻝ
:
60 27
60 Co→ 28 Ni + −10 e ﺑﺎﻟﻨﺴﺒﺔ ﻟﻠﺘﺤﻮﻝ ﺍﻟﺜﺎﱐ
:
13 7
N →136 C + 10 e ﺑﺎﻟﻨﺴﺒﺔ ﻟﻠﺘﺤﻮﻝ ﺍﻟﺜﺎﻟﺚ
E1 = ∆m1 × c 2 = (−4,94 MeV / c 2 ) × c 2 = −4,94 MeV -:ﺏ( ﺍﻟﻄﺎﻗﺔ ﺍﳌﺘﺤﺮﺭﺓ ﺧﻼﻝ ﺍﻟﺘﺤﻮﻝ ﺍﻷﻭﻝ E 2 = ∆m2 × c 2 = (−3,21MeV / c 2 ) × c 2 = −3,21MeV -:ﺍﻟﻄﺎﻗﺔ ﺍﳌﺘﺤﺮﺭﺓ ﺧﻼﻝ ﺍﻟﺘﺤﻮﻝ ﺍﻟﺜﺎﱐ E 3 = ∆m3 × c 2 = (−1,2 MeV / c 2 ) × c 2 = −1,2 MeV -:ﺍﻟﻄﺎﻗﺔ ﺍﳌﺘﺤﺮﺭﺓ ﺧﻼﻝ ﺍﻟﺘﺤﻮﻝ ﺍﻟﺜﺎﻟﺚ
SBIRO Abdelkrim Lycée Agricole et Lycée Abdellah Cheffchouni Oulad –Taima région d’Agadir Maroc Mai : [email protected] Pour toutes vos observations contacter mon émail
4
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