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Studyadz 3as

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Page 1 sur 6 ‫ﺍﻟﺸﻌﺒﻴﺔ‬ ‫ﺍﻟﺪﻳﻤﻘﺮﺍﻃﻴﺔ‬ ‫ﺍﻟﺠﺰﺍﺋﺮﻳﺔ‬ ‫ﺍﻟﺠﻤﻬﻮﺭﻳﺔ‬ ‫ﻣﻌﻨﺼﺮ‬ ‫ﺛﺎﻧﻮﻳﺔ‬ ‫ﺍﻟﺒﻮﺍﻗﻲ‬ ‫ﺍﻡ‬ ‫ﻟﻮﻻﻳﺔ‬ ‫ﺍﻟﺘﺮﺑﻴﺔ‬ ‫ﻣﺪﻳﺮﻳﺔ‬ ‫ﻛﺮﺷﺔ‬ ‫ﻋﻴﻦ‬ ‫ﺃﻭﻧﻴﺲ‬ ‫ﺍﻟﺜﺎﻧﻮﻱ‬ ‫ﺍﻟﺘﻌﻠﻴﻢ‬ ‫ﻟﺒﻜﺎﻟﻮﺭﻳﺎ‬ ‫ﺗﺠﺮﻳﺒﻲ‬ ‫ﺍﻣﺘﺤﺎﻥ‬ . ‫ﻣﺎﻱ‬ 2019 ‫ﺍﻟﺸﻌﺒﺔ‬ : ‫ﺗﺠﺮﻳﺒﻴﺔ‬ ‫ﻋﻠﻮﻡ‬ ‫ﻣﺎﺩﺓ‬ ‫ﻓﻲ‬ ‫ﺍﺧﺘﺒﺎﺭ‬ : ‫ﺍﻟﻤﺪﺓ‬ ‫ﺍﻟﻔﻴﺰﻳﺎﺋﻴﺔ‬ ‫ﺍﻟﻌﻠﻮﻡ‬ : 03 ‫ﻭ‬ ‫ﺳﺎ‬ 30 ‫ﺩ‬ ‫ﺍﻵﺗﻴﻴﻦ‬ ‫ﺍﻟﻤﻮﺿﻮﻋﻴﻦ‬ ‫ﺃﺣﺪ‬ ‫ﻳﺨﺘﺎﺭ‬ ‫ﺃﻥ‬ ‫ﺍﻟﻤﺘﺮﺷﺢ‬ ‫ﻋﻠﻰ‬ : ‫ﺍﻷﻭﻝ‬ ‫ﺍﻟﻤﻮﺿﻮﻉ‬ ‫ﺍﻟﻤ‬ ‫ﻳﺤﺘﻮﻱ‬ ‫ﻋﻠﻰ‬ ‫ﺍﻷﻭﻝ‬ ‫ﻮﺿﻮﻉ‬ 03 ‫ﺻﻔﺤﺎﺕ‬ ) ‫ﺻﻔﺤﺔ‬ ‫ﻣﻦ‬ 1 ‫ﻣﻦ‬ 6 ‫ﺍﻟﺼﻔﺤﺔ‬ ‫ﺇﻟﻰ‬ 3 ‫ﻣﻦ‬ 6 ( ‫ﺍﻷﻭﻝ‬ ‫ﺍﻟﺠﺰﺀ‬ : ) 13 ‫ﻧﻘﻄﺔ‬ ( ‫ﺍﻷﻭﻝ‬ ‫ﺍﻟﺘﻤﺮﻳﻦ‬ : ) 06 ‫ﻧﻘﺎﻁ‬ ( ‫ﺃﺳﺎﺳﻲ‬ ‫ﺍﻟﻬﺮﻣﻮﻥ‬ ‫ﻭﻫﺬﺍ‬ ‫ﺍﻟﺪﻡ‬ ‫ﻣﻦ‬ ‫ﺍﻻﺧﻴﺮﺓ‬ ‫ﻫﺬﻩ‬ ‫ﺗﺴﺘﻨﺰﻓﻪ‬ ‫ﺣﻴﺚ‬ ‫ﺍﻟﺪﺭﻗﻴﺔ‬ ‫ﺍﻟﻐﺪﺓ‬ ‫ﻫﺮﻣﻮﻧﺎﺕ‬ ‫ﻟﺘﺮﻛﻴﺐ‬ ‫ﺍﻷﺳﺎﺳﻴﺔ‬ ‫ﺍﻟﻤﺎﺩﺓ‬ ‫ﺍﻟﻴﻮﺩ‬ ‫ﻳﻌﺘﺒﺮ‬ ‫ﻟﻠﻨﻤﻮ‬ . ‫ﺍﻟﺘﺮﻛﻴﺐ‬ ‫ﻟﻬﺬﺍ‬ ‫ﺍﻟﻼﺯﻣﺔ‬ ‫ﺍﻟﻤﺎﺩﺓ‬ ‫ﻛﻤﻴﺎﺕ‬ ‫ﻗﻠﻴﻠﺔ‬ . ‫ﻛﻞ‬ ‫ﻳﺤﺘﻮﻱ‬ ،‫ﻋﺎﺩﻱ‬ ‫ﻟﺸﺨﺺ‬ ‫ﺑﺎﻟﻨﺴﺒﺔ‬ 100mL ‫ﻋﻠﻰ‬ ‫ﺍﻟﺪﻡ‬ ‫ﻣﻦ‬ 10 ‫ﻣﻦ‬ ‫ﻫﺮﻣﻮﻧﺎﺕ‬ ‫ﺷﻜﻞ‬ ‫ﻋﻠﻰ‬ ‫ﺗﻔﺮﺯﻩ‬ ‫ﻭﺍﻟﺬﻱ‬ ‫ﻳﻮﺩﻭﺭ‬ ‫ﺷﻜﻞ‬ ‫ﻋﻠﻰ‬ ‫ﺍﻟﺪﻡ‬ ‫ﻣﻦ‬ ‫ﺍﻟﻜﻤﻴﺔ‬ ‫ﻧﻔﺲ‬ ‫ﺍﻟﺪﺭﻗﻴﺔ‬ ‫ﺍﻟﻐﺪﺓ‬ ‫ﺗﺄﺧﺬ‬ ‫ﻟﺤﻈﺔ‬ ‫ﻛﻞ‬ ‫ﻓﻲ‬ ،‫ﺍﻟﻴﻮﺩ‬ . ‫ﺛﺎﺑﺘﺔ‬ ‫ﺗﺒﻘﻰ‬ ‫ﺍﻟﺪﻡ‬ ‫ﻓﻲ‬ ‫ﺍﻟﻤﺘﻮﺍﺟﺪﺓ‬ ‫ﺍﻟﻴﻮﺩ‬ ‫ﻛﻤﻴﺎﺕ‬ . ‫ﺍﺳﺘﻘﻼﺏ‬ ‫ﺗﺘﺒﻊ‬ ‫ﺍﻟﻤﺸﻊ‬ ‫ﺍﻟﻴﻮﺩ‬ ‫ﺑﺈﺳﺘﻌﻤﺎﻝ‬ (métabolisme) ‫ﺍﻟﻴﻮ‬ ‫ﻭﻗﻴﺎﺱ‬ ‫ﺩ‬ ‫ﺍﻟﻤﺮﻛﺒﺔ‬ ‫ﺍﻟﻬﺮﻣﻮﻧﺎﺕ‬ . ‫ﻧﺠﺪ‬ ‫ﺍﻟﻴﻮﺩ‬ ‫ﻧﻈﺎﺋﺮ‬ ‫ﺑﻴﻦ‬ ‫ﻣﻦ‬ ‫ﻭ‬ ‫ﺍﻟﻨﺸﺎﻁ‬ ‫ﺇﺷﻌﺎﻋﻲ‬ ‫ﻭﺍﻵﺧﺮ‬ ‫ﻣﺴﺘﻘﺮ‬ ‫ﺃﺣﺪﻫﻤﺎ‬ β . 1 . ‫ﺍﻟﻨﻮﻭﻱ‬ ‫ﺍﻟﺘﺮﺍﺑﻂ‬ ‫ﻃﺎﻗﺔ‬ ‫ﻋﺮﻑ‬ . 2 . ‫ﺍﻟﻜﺘﺮﻭﻥ‬ ‫ﺑﺎﻟﻤﻴﻐﺎ‬ ‫ﻟﻠﻨﻮﺍﺓ‬ ‫ﺍﻟﺮﺑﻂ‬ ‫ﻃﺎﻗﺔ‬ ‫ﺃﺣﺴﺐ‬ - ‫ﺍﻟﻨﻈﻴﺮﻳﻦ‬ ‫ﻣﻦ‬ ‫ﻟﻜﻞ‬ ‫ﻓﻮﻟﻂ‬ . 3 . ‫ﺑﺎ‬ ‫ﺍﻟﺮﺑﻂ‬ ‫ﻃﺎﻗﺔ‬ ‫ﺃﺣﺴﺐ‬ ‫ﻟﻜﻞ‬ ‫ﻧﻮﻳﺔ‬ ‫ﺑﺎﻟﻨﺴﺒﺔ‬ ‫ﺍﻟﻨﻈﻴﺮ‬ ‫ﻣﻦ‬ ‫ﻟﻜﻞ‬ ‫ﺍﻟﻤﺸﻊ‬ ‫ﻭﺍﻟﻨﻈﻴﺮ‬ ‫ﺍﻟﻤﺴﺘﻘﺮ‬ ‫ﺍﻟﻨﻈﻴﺮ‬ ‫ﻭﺍﺳﺘﻨﺘﺞ‬ ، ‫ﻳﻦ‬ . 4 . ‫ﺍﻟﻤﺸﻊ‬ ‫ﻟﻠﻨﻈﻴﺮ‬ ‫ﺍﻟﺘﻔﻜﻚ‬ ‫ﻣﻌﺎﺩﻟﺔ‬ ‫ﺍﻛﺘﺐ‬ . ‫ﻋﻨﻪ‬ ‫ﻳﻨﺘﺞ‬ ‫ﺍﻟﺬﻱ‬ Xe ‫ﻣﺤﺪﺩﺍ‬ A ‫ﻭ‬ Z . ‫ﺍﻟﻌﻤﺮ‬ ‫ﻧﺼﻒ‬ ‫ﺯﻣﻦ‬ 131 ‫ﻫﻮ‬ ‫ﺍﻟﻤﺸﻊ‬ ‫ﻟﻠﻴﻮﺩ‬ / = 8 : 5 . ‫ﺍﺣﺴﺐ‬ ‫ﺯﻣﻦ‬ ‫ﺍﻟﻌﻤﺮ‬ ‫ﻧﺼﻒ‬ ‫ﻟﻨﻮﺍ‬ ‫ﺓ‬ ‫ﺛﻢ‬ ‫ﺍﻟﻤﺸﻌﺔ‬ ‫ﺍﻟﻴﻮﺩ‬ ‫ﺍﻟﺰﻣﻦ‬ ‫ﺛﺎﺑﺖ‬ ‫ﺍﺳﺘﻨﺘﺞ‬ τ . 6 . ‫ﺍﻟﻴﻮﺩ‬ ‫ﺍﺳﺘﻘﻼﺏ‬ ‫ﻟﺘﺘﺒﻊ‬ ‫ﺑﺎﻹﻳﻤﺎﺽ‬ ‫ﺍﻟﻔﺤﺺ‬ ‫ﺗﻘﻨﻴﺔ‬ ‫ﺗﻌﺘﻤﺪ‬ ‫ﺍﻟﻤﺮﻛﺒﺔ‬ ‫ﺍﻟﻬﺮﻣﻮﻧﺎﺕ‬ ‫ﻭﻗﻴﺎﺱ‬ (scintigraphie) ‫ﻟﻠﻐﺪﺓ‬ ‫ﺍﻟﻜﻈﺮﻳﺔ‬ (glandes surrénales) ‫ﻟﻠﻴﻮﺩ‬ ‫ﺑﻤﺤﻠﻮﻝ‬ ‫ﺍﻟﻤﺮﻳﺾ‬ ‫ﺣﻘﻦ‬ ‫ﻋﻠﻰ‬ ‫ﻧﺸﺎﻃﻪ‬ ‫ﺍﻹﺷﻌﺎﻋﻲ‬ 37MBq . ‫ﺃ‬ . ‫ﻧﺸﺎﻁ‬ ‫ﺍﺣﺴﺐ‬ 1g ‫ﺍﻟﻴﻮﺩ‬ ‫ﻣﻦ‬ . ‫ﺏ‬ . ‫ﺍﻟﻴﻮﺩ‬ ‫ﻣﺎﻛﺘﻠﺔ‬ ‫؟‬ ‫ﺣﻘﻨﻬﺎ‬ ‫ﺗﻢ‬ ‫ﺍﻟﺘﻲ‬ ‫ﺝ‬ . ‫ﺍﻋ‬ ‫ﺍﻟﻨﺸﺎﻁ‬ ‫ﻗﻴﺎﺱ‬ ‫ﻄﻰ‬ ‫ﺍﻹﺷﻌﺎﻋﻲ‬ ‫ﺍﻟﻠﺤﻈﺔ‬ ‫ﻋﻨﺪ‬ ‫ﺍﻟﻤﺪﺭﻭﺳﺔ‬ ‫ﻟﻠﻌﻴﻨﺔ‬ : = 24 . ‫ﺃﻥ‬ ‫ﺑﻴﻦ‬ t = τ ln . . . ‫ﺑـ‬ ‫ﻗﻴﻤﺘﻬﺎ‬ ‫ﺍﺣﺴﺐ‬ ‫ﺛﻢ‬ (jrs) . 7 . ‫ﺑﺠﺮﻋﺔ‬ ‫ﻣﺮﻳﺾ‬ ‫ﺣﻘﻦ‬ 1,00 ‫ﺍﻟﻴﻮﺩ‬ ‫ﻣﻦ‬ 131 ‫ﺍﻟﻤﺸﻊ‬ : ‫ﺃ‬ . ‫ﺍﻟﻴﻮﺩ‬ ‫ﻛﺘﻠﺔ‬ ‫ﺍﺣﺴﺐ‬ 131 ‫ﻓﻲ‬ ‫ﺍﻟﻤﺘﺒﻘﻲ‬ ‫ﻣﺮﻭﺭ‬ ‫ﺑﻌﺪ‬ ‫ﺍﻟﻤﺮﻳﺾ‬ ‫ﺟﺴﻢ‬ 4 ‫ﺑﻌﺪ‬ ‫ﺛﻢ‬ ، 8 ‫ﺑﻌﺪ‬ ‫ﺛﻢ‬، 16 . ‫ﺏ‬ . ‫ﺍﻟﻴﻮﺩ‬ ‫ﺃﻧﻮﻳﺔ‬ ‫ﻋﺪﺩ‬ ‫ﺍﺣﺴﺐ‬ 131 ‫ﻣﺮﻭﺭ‬ ‫ﺑﻌﺪ‬ ‫ﺍﻟﻤﺘﺒﻘﻴﺔ‬ 4 . ‫ﺝ‬ . ‫ﻟﺘﻔﻜﻚ‬ ‫ﺍﻟﻮﻗﺖ‬ ‫ﻣﻦ‬ ‫ﻳﻠﺰﻡ‬ ‫ﻛﻢ‬ 99% ‫ﺍﻟﻴﻮﺩ‬ ‫ﻣﻦ‬ 131 . ‫ﺍﻟﻤﻌﻄﻴﺎﺕ‬ : ‫ﺍﻟﺪﻗﻴﻘﺔ‬ ‫ﺃﻭ‬ ‫ﺍﻟﻨﻮﺍﺓ‬ 1 Xe U ‫ﺑـ‬ ‫ﻛﺘﻠﺘﻬﺎ‬ 130.9061 126.9044 1.00877 5.4858 × 10 1.00722 130.9050 ( ) = 131 . , = 6,02 × 10 , 1 = 1,6 × 10 , 1 = 931,5 e n c y - e d u c a t i o n . c o m / e x a m s 3as.ency-education.com
‫ﺍ‬ ‫ﺧ‬ ‫ﺘ‬ ‫ﺒ‬ ‫ﺎ‬ ‫ﺭ‬ ‫ﻓ‬ ‫ﻲ‬ ‫ﻣ‬ ‫ﺎ‬ ‫ﺩ‬ ‫ﺓ‬ ‫ﺍ‬ ‫ﻟ‬ ‫ﻌ‬ ‫ﻠ‬ ‫ﻮ‬ ‫ﻡ‬ ‫ﺍ‬ ‫ﻟ‬ ‫ﻔ‬ ‫ﻴ‬ ‫ﺰ‬ ‫ﻳ‬ ‫ﺎ‬ ‫ﺋ‬ ‫ﻴ‬ ‫ﺔ‬ / ‫ﺍ‬ ‫ﻟ‬ ‫ﺸ‬ ‫ﻌ‬ ‫ﺒ‬ ‫ﺔ‬ ، ‫ﻋ‬ ‫ﻠ‬ ‫ﻮ‬ ‫ﻡ‬ ‫ﺗ‬ ‫ﺠ‬ ‫ﺮ‬ ‫ﻳ‬ ‫ﺒ‬ ‫ﻴ‬ ‫ﺔ‬ / ‫ﺑ‬ ‫ﻜ‬ ‫ﺎ‬ ‫ﻟ‬ ‫ﻮ‬ ‫ﺭ‬ ‫ﻳ‬ ‫ﺎ‬ ‫ﺗ‬ ‫ﺠ‬ ‫ﺮ‬ ‫ﻳ‬ ‫ﺒ‬ ‫ﻲ‬ 2019 Page 2 sur 6 E L R K ‫اﻟﺸﻜﻞ‬ 1 2 ( ) 2 ( ) ‫اﻟﺸﻜﻞ‬ ) 2 ( 1 2 ‫ﺍﻟﺜﺎﻧﻲ‬ ‫ﺍﻟﺘﻤﺮﻳﻦ‬ : ) 07 ‫ﻧﻘﺎﻁ‬ ( ‫ﺍﻟﻤﻮﺿ‬ ‫ﺍﻟﺘﺮﻛﻴﺐ‬ ‫ﺑﺎﺳﺘﻌﻤﺎﻝ‬ ‫ﺍﻟﺘﺎﻟﻴﺔ‬ ‫ﺍﻟﺘﺠﺮﺑﺔ‬ ‫ﻧﻨﺠﺰ‬ ‫ﺍﻟﺸﻜﻞ‬ ‫ﻓﻲ‬ ‫ﺢ‬ ) 1 ( ‫ﻭﺍﻟﻤﻜﻮﻥ‬ ‫ﻣﻦ‬ : ‫ﻣﺜﺎﻟﻴﺔ‬ ‫ﻭﺷﻴﻌﺔ‬ (B) ‫ﺫﺍﺗﻴﺘﻬﺎ‬ L ‫ﺃﻭﻣﻲ‬ ‫ﻭﻧﺎﻗﻞ‬ (R) ‫ﻣﻘﺎﻭﻣﺘﻪ‬ = 200Ω ‫ﻭ‬ ، ‫ﺍﻟﻜﻬﺮﺑﺎﺋﻴﺔ‬ ‫ﻗﻮﺗﻪ‬ ‫ﺛﺎﺑﺖ‬ ‫ﻟﺘﻮﺗﺮ‬ ‫ﻣﻮﻟﺪ‬ E . ‫ﻭﻗﺎﻃﻌﺔ‬ K . ‫ﻋﻨﺪ‬ ‫ﺍﻟﻘﺎﻃﻌﺔ‬ ‫ﻧﻐﻠﻖ‬ ‫ﻟﻸﺯﻣﻨﺔ‬ ‫ﻣﺒﺪﺃ‬ ‫ﻧﻌﺘﺒﺮﻫﺎ‬ ‫ﻟﺤﻈﺔ‬ . 1 . ‫ﻭﻛﻴﻔﻴﺔ‬ ‫ﻓﻴﻬﺎ‬ ‫ﺍﻟﻤﺎﺭ‬ ‫ﻭﺍﻟﺘﻴﺎﺭ‬ ‫ﺍﻟﺘﻮﺗﺮﺍﺕ‬ ‫ﺟﻬﺔ‬ ‫ﻋﻠﻴﻬﺎ‬ ‫ﻣﻮﺿﺤﺎ‬ ‫ﺍﻟﺪﺍﺭﺓ‬ ‫ﺭﺳﻢ‬ ‫ﺍﻋﺪ‬ ‫ﺭﺍﺳﻢ‬ ‫ﺗﻮﺻﻴﻞ‬ ‫ﺍﻟﺘﻮﺗﺮﻳﻦ‬ ‫ﻟﻤﺸﺎﻫﺪﺓ‬ ‫ﺍﻟﻤﻬﺒﻄﺔ‬ ‫ﺍﻻﻫﺘﺰﺍﺯ‬ ‫ﻭ‬ . 2 . ‫ﻗﻴﻤﺔ‬ ‫ﺃﻥ‬ ‫ﺑﻴﻦ‬ ‫ﺍﻟﻤﻨﺤﻨﻰ‬ ‫ﻋﻠﻰ‬ ‫ﻭﺍﻋﺘﻤﺎﺩﺍ‬ ، ‫ﺍﻟﺘﻮﺗﺮﺍﺕ‬ ‫ﺟﻤﻊ‬ ‫ﻗﺎﻧﻮﻥ‬ ‫ﺑﺘﻄﻴﻖ‬ ‫ﻫﻲ‬ 10 . 3 . ‫ﻳ‬ ‫ﺍﻟﺘﻲ‬ ‫ﺍﻟﺘﻘﺎﺿﻠﻴﺔ‬ ‫ﺍﻟﻤﻌﺎﺩﻟﺔ‬ ‫ﺍﻥ‬ ‫ﺑﻴﻦ‬ ‫ﺤ‬ ‫ﺍﻟﺘﻮﺗﺮ‬ ‫ﻘﻘﻬﺎ‬ ‫ﺍﻷﻭﻣﻲ‬ ‫ﺍﻟﻨﺎﻗﻞ‬ ‫ﻃﺮﻓﻲ‬ ‫ﺑﻴﻦ‬ ‫ﺍﻟﺸﻜﻞ‬ ‫ﻋﻠﻰ‬ ‫ﺗﻜﺘﺐ‬ : + − = 0 4 . ‫ﺍ‬ ‫ﺍﻟﻤﻌﺎﺩﻟﺔ‬ ‫ﺣﻞ‬ ‫ﺍﻥ‬ ‫ﻋﻠﻤﺎ‬ ‫ﺍﻟﺸﻜﻞ‬ ‫ﻋﻠﻰ‬ ‫ﻳﻜﺘﺐ‬ ‫ﺍﻟﺴﺎﺑﻘﺔ‬ ‫ﻟﺘﻔﺎﺿﻠﻴﺔ‬ : = (1 − ) ‫ﻋﺒﺎﺭﺓ‬ ‫ﺣﺪﺩ‬ ، α ‫ﺛﻮﺍﺑﺖ‬ ‫ﺑﺪﻻﻟﺔ‬ ‫ﺍﻟﺪﺍﺭﺓ‬ . 5 . ‫ﺍﻟﻮﺷﻌﺔ‬ ‫ﻃﺮﻓﻲ‬ ‫ﺑﻴﻦ‬ ‫ﺍﻟﺘﻮﺗﺮ‬ ‫ﻋﺒﺎﺭﺓ‬ ‫ﺍﺳﺘﻨﺘﺞ‬ . 6 . ‫ﺍﻟﺸﻜﻞ‬ ‫ﻳﻤﺜﻞ‬ ) 2 ( ‫ﺍﻟﻮﺷﻴﻌﺔ‬ ‫ﻃﺮﻓﻲ‬ ‫ﺑﻴﻦ‬ ‫ﺍﻟﺘﻮﺗﺮ‬ ‫ﺗﻐﻴﺮﺍﺕ‬ ‫ﺍﻟﻤﻘﺎﺑﻞ‬ ‫ﺍﻷﻭﻣﻲ‬ ‫ﺍﻟﻨﺎﻗﻞ‬ ‫ﻭ‬ ‫ﺑﺪﻻﻟﺔ‬ ‫ﺍﻟﺰﻣﻦ‬ t . ‫ﺃ‬ . ‫ﻣﻦ‬ ‫ﻟﻜﻞ‬ ‫ﺍﻟﻤﻮﺍﻓﻖ‬ ‫ﺍﻟﻤﻨﺤﻨﻰ‬ ‫ﺣﺪﺩ‬ ‫ﻭ‬ ‫ﺍﻟﺘﻌﻠﻴﻞ‬ ‫ﻣﻊ‬ . ‫ﺏ‬ . ‫ﺑﻴﻦ‬ ‫ﺃﻥ‬ ‫ﺍﻟﺘﻮﺗﺮﻳﻦ‬ ‫ﺗﻘﺎﻃﻊ‬ ‫ﻟﺤﻈﺔ‬ ( ) ‫ﻭ‬ ( ) ‫ﺗﻜﺘﺐ‬ ‫ﺍﻟﺸﻜﻞ‬ ‫ﻋﻠﻰ‬ : = . ln (2) . ‫ﺝ‬ . ‫ﻗﻴﻤﺔ‬ ‫ﺍﺳﺘﻨﺘﺞ‬ ‫ﻟﻠﺪﺍﺭﺓ‬ ‫ﺍﻟﺰﻣﻦ‬ ‫ﺛﺎﺑﺖ‬ . ‫ﺩ‬ . ‫ﺍﻟﺬﺍﺗﻴﺔ‬ ‫ﻗﻴﻤﺔ‬ ‫ﺍﺳﺘﻨﺘﺞ‬ L ‫ﻟﻠﻮﺷﻴﻌﺔ‬ . 7 . ‫ﻋﺒﺎﺭﺓ‬ ‫ﺃﻭﺟﺪ‬ ‫ﻓﻲ‬ ‫ﺍﻟﻤﺎﺭ‬ ‫ﺍﻟﺘﻴﺎﺭ‬ ‫ﺷﺪﺓ‬ ، ‫ﺍﻟﺪﺍﺋﻢ‬ ‫ﺍﻟﻨﻈﺎﻡ‬ ‫ﻓﻲ‬ ‫ﺍﻟﺪﺍﺭﺓ‬ ‫ﻗﻴﻤﺘﻪ‬ ‫ﺍﺣﺴﺐ‬ ‫ﺛﻢ‬ . 8 . ‫ﺍﻟﺸﻜﻞ‬ ‫ﻋﻠﻰ‬ ‫ﺗﻜﺘﺐ‬ ‫ﺍﻟﺰﻣﻦ‬ ‫ﺑﺪﻻﻟﺔ‬ ‫ﺍﻟﻮﺷﻴﻌﺔ‬ ‫ﻓﻲ‬ ‫ﺍﻟﻤﺨﺰﻧﺔ‬ ‫ﺍﻟﻄﺎﻗﺔ‬ ‫ﻋﺒﺎﺭﺓ‬ ‫ﺃﻥ‬ ‫ﺑﻴﻦ‬ : = 1 − ‫ﺛﻢ‬ ، ‫ﺍﻟﻠﺤﻈﺔ‬ ‫ﻋﻨﺪ‬ ‫ﻗﻴﻤﺘﻬﺎ‬ ‫ﺃﺣﺴﺐ‬ = . ‫ﺍﻟﺜﺎﻧﻲ‬ ‫ﺍﻟﺠﺰﺀ‬ : ) 07 ‫ﻧﻘﺎﻁ‬ ( ‫ﺍﻟﺘﺠﺮﻳﺒﻲ‬ ‫ﺍﻟﺘﻤﺮﻳﻦ‬ : ‫ﻓﻲ‬ ‫ﺣﺼﺔ‬ ‫ﺍﻷﻋﻤﺎﻝ‬ ‫ﺍﻟﺘﻄﺒﻴ‬ ‫ﻘﻴﺔ‬ ‫ﺃﺭﺍﺩ‬ ‫ﻓﻮﺟﺎﻥ‬ ‫ﻣﻦ‬ ‫ﺍﻟﺘﻼﻣﻴﺬ‬ ‫ﺗﺤﺪﻳﺪ‬ ‫ﺍﻟﺘﺮﻛﻴﺰ‬ ‫ﺍﻟﻜﺘﻠﻲ‬   m C ‫ﻟﻤﺤﻠﻮﻝ‬ ‫ﺣﻤﺾ‬ ‫ﺍﻷﺳﻜﻮﺭﺑﻴﻚ‬   6 8 6 C H O ‫ﺑﻄﺮﻳﻘﺘﻴﻦ‬ . ‫ﻳﻤﻠﻚ‬ ‫ﺣﻤﺾ‬ ‫ﺍﻷﺳﻜﻮﺭﺑﻴﻚ‬ ‫ﺧﺎﺻﻴﺔ‬ ‫ﺣﻤﻀﻴﺔ‬ ‫ﻭﺧﺎﺻﻴﺔ‬ ‫ﻣﺮﺟﻌﺔ‬ . ‫ﺍﻟﺜﻨﺎﺋﻴﺎﺕ‬ ) ‫ﻣﺮ‬ / ‫ﻣﺆ‬ : (       2 2 6 6 6 6 8 6 2 4 6 2 3 C H O / C H O , I / I , S O / S O    . ‫ﺍﻟﺜﻨﺎﺋﻴﺎﺕ‬ ) ‫ﺃﺳﺎﺱ‬ / ‫ﺣﻤﺾ‬ : (     6 8 6 6 7 6 2 C H O / C H O , H O / HO   . ‫ﺍﻷﻭﻝ‬ ‫ﺍﻟﻔﻮﺝ‬ : ‫ﻗﺎﻡ‬ ‫ﺍﻟﺘﻼﻣﻴﺬ‬ ‫ﺑﺄﻛﺴﺪﺓ‬ ‫ﺣﻤﺾ‬ ‫ﺍﻷﺳﻜﻮﺭﺑﻴﻚ‬ ، ‫ﻭﺫﻟﻚ‬ ‫ﺑﺈﺿﺎﻓﺔ‬ ‫ﺔ‬ ‫ﻴ‬‫ﻛﻤ‬ ‫ﺯﺍﺋﺪﺓ‬ ‫ﻣﻦ‬ ‫ﻣﺤﻠﻮﻝ‬ ‫ﺛﻨﺎﺋﻲ‬ ‫ﺍﻟﻴﻮﺩ‬ 2 I ‫ﺇﻟﻰ‬ ‫ﺑﻴﺸﺮ‬ ‫ﻳﺤﺘﻮﻱ‬ ‫ﻋﻠﻰ‬ ‫ﺣﺠﻢ‬ 1 10 V mL  ‫ﻣﻦ‬ ‫ﺣﻤﺾ‬ ‫ﺍﻷﺳﻜﻮﺭﺑﻴﻚ‬ . ‫ﺣﺠﻢ‬ ‫ﺛﻨﺎﺋﻲ‬ ‫ﺍﻟﻴﻮﺩ‬ ‫ﺍﻟﻤﻀﺎﻑ‬ ‫ﻫﻮ‬ 2 20 V mL  ‫ﻭﺗﺮﻛﻴﺰﻩ‬ ‫ﺍﻟﻤﻮﻟﻲ‬ 2 2 3 5 10 C , mol / L    . e n c y - e d u c a t i o n . c o m / e x a m s 3as.ency-education.com
‫ﺍ‬ ‫ﺧ‬ ‫ﺘ‬ ‫ﺒ‬ ‫ﺎ‬ ‫ﺭ‬ ‫ﻓ‬ ‫ﻲ‬ ‫ﻣ‬ ‫ﺎ‬ ‫ﺩ‬ ‫ﺓ‬ ‫ﺍ‬ ‫ﻟ‬ ‫ﻌ‬ ‫ﻠ‬ ‫ﻮ‬ ‫ﻡ‬ ‫ﺍ‬ ‫ﻟ‬ ‫ﻔ‬ ‫ﻴ‬ ‫ﺰ‬ ‫ﻳ‬ ‫ﺎ‬ ‫ﺋ‬ ‫ﻴ‬ ‫ﺔ‬ / ‫ﺍ‬ ‫ﻟ‬ ‫ﺸ‬ ‫ﻌ‬ ‫ﺒ‬ ‫ﺔ‬ ، ‫ﻋ‬ ‫ﻠ‬ ‫ﻮ‬ ‫ﻡ‬ ‫ﺗ‬ ‫ﺠ‬ ‫ﺮ‬ ‫ﻳ‬ ‫ﺒ‬ ‫ﻴ‬ ‫ﺔ‬ / ‫ﺑ‬ ‫ﻜ‬ ‫ﺎ‬ ‫ﻟ‬ ‫ﻮ‬ ‫ﺭ‬ ‫ﻳ‬ ‫ﺎ‬ ‫ﺗ‬ ‫ﺠ‬ ‫ﺮ‬ ‫ﻳ‬ ‫ﺒ‬ ‫ﻲ‬ 2019 Page 3 sur 6 2 2 ( ) ‫ﻭﻓﻲ‬ ‫ﻧﻬﺎﻳﺔ‬ ‫ﺍﻟﺘﻔﺎﻋﻞ‬ ‫ﻗﺎﻡ‬ ‫ﺍﻟﺘﻼﻣﻴﺬ‬ ‫ﺑﻤﻌﺎﻳﺮﺓ‬ ‫ﺛﻨﺎﺋﻲ‬ ‫ﺍﻟﻴﻮﺩ‬ ‫ﻓﻲ‬ ‫ﺍﻟﺒﻴﺸﺮ‬ ‫ﺑﻮﺍﺳﻄﺔ‬ ‫ﻣﺤﻠﻮﻝ‬ ‫ﻣﺎﺋﻲ‬ ‫ﻟﺜﻴﻮﻛﺒﺮﻳﺘﺎﺕ‬ ‫ﺍﻟﺼﻮﺩﻳﻮﻡ‬   2 2 3 2Na ,S O   ‫ﺍﻟﻤﻮﻟﻲ‬ ‫ﺗﺮﻛﻴﺰﻩ‬ 2 3 2 5 10 C , mol / L    ، ‫ﻓﺎﺣﺘﺎﺟﻮﺍ‬ ‫ﺇﻟﻰ‬ ‫ﺣﺠﻢ‬ ‫ﻣﻨﻪ‬ 20 E V mL  ‫ﻻﺳﺘﻬﻼﻙ‬ ‫ﺁﻝ‬ ‫ﺛﻨﺎﺋﻲ‬ ‫ﺍﻟﻴﻮﺩ‬ ‫ﺍﻟﻤﻮﺟﻮﺩ‬ ‫ﻓﻲ‬ ‫ﺍﻟﺒﻴﺸﺮ‬ . 1 - ‫ﺍﻛﺘﺐ‬ ‫ﻣﻌﺎﺩﻟﺔ‬ ‫ﺍﻟﺘﻔﺎﻋﻞ‬ ‫ﺑﻴﻦ‬ ‫ﺣﻤﺾ‬ ‫ﺍﻷﺳﻜﻮﺭﺑﻴﻚ‬ ‫ﻭﺛﻨﺎﺋﻲ‬ ‫ﺍﻟﻴﻮﺩ‬ . 2 - ‫ﺃﻧﺸﺊ‬ ‫ﺟﺪﻭﻝ‬ ‫ﺍﻟﺘﻘﺪﻡ‬ ‫ﻟﻬﺬﺍ‬ ‫ﺍﻟﺘﻔﺎﻋﻞ‬ . 3 - ‫ﺍﺫﻛﺮ‬ ‫ﺍﻟﺸﺮﻭﻁ‬ ‫ﺍﻟﺘﻲ‬ ‫ﺗﺘﻮﻓﺮ‬ ‫ﻓﻲ‬ ‫ﻣﺤﻠﻮﻝ‬ ‫ﺛﻴﻮﻛﺒﺮﻳﺘﺎﺕ‬ ‫ﺍﻟﺼﻮﺩﻳﻮﻡ‬ ‫ﻻﺳﺘﻌﻤﺎﻟﻪ‬ ‫ﻓﻲ‬ ‫ﻫﺬﻩ‬ ‫ﺍﻟﻤﻌﺎﻳﺮﺓ‬ . 4 - ‫ﺍﻛﺘﺐ‬ ‫ﻣﻌﺎﺩﻟﺔ‬ ‫ﺗﻔﺎﻋﻞ‬ ‫ﻣﻌﺎﻳﺮﺓ‬ ‫ﺛﻨﺎﺋﻲ‬ ‫ﺍﻟﻴﻮﺩ‬ ‫ﺑﺜﻴﻮﻛﺒﺮﻳﺘﺎﺕ‬ ‫ﺍﻟﺼﻮﺩﻳﻮﻡ‬ . 5 - ‫ﺍﺣﺴﺐ‬ ‫ﻛﻤﻴﺔ‬ ‫ﻣﺎﺩﺓ‬ ‫ﺛﻨﺎﺋﻲ‬ ‫ﺍﻟﻴﻮﺩ‬ ‫ﻏﻴﺮ‬ ‫ﺍﻟﻤﺘﻔﺎﻋﻞ‬ ‫ﻣﻊ‬ ‫ﺣﻤﺾ‬ ‫ﺍﻷﺳﻜﻮﺭﺑﻴﻚ‬ . 6 - ‫ﺍﺣﺴﺐ‬ ‫ﺍﻟﺘﺮﻛﻴﺰ‬ ‫ﺍﻟﻜﺘﻠﻲ‬   m C ‫ﻟﺤﻤﺾ‬ ‫ﺍﻷﺳﻜﻮﺭﺑﻴﻚ‬ . ‫ﻗﺎﺭﻥ‬ ‫ﻧﺘﻴﺠﺘﻲ‬ ‫ﺍﻟﻔﻮﺟﻴﻦ‬ .     2 5 2 5 4 9 12 1 16 a pK C H COOH / C H COO , , C ,H O g / mol      ‫ﺍﻟﺜﺎﻧﻲ‬ ‫ﺍﻟﻔﻮﺝ‬ : ‫ﻗﺎﻡ‬ ‫ﺑﺎﻟﻤﻌﺎﻳﺮﺓ‬ ‫ﺍﻟـ‬ pH ‫ﻣﺘﺮﻳﺔ‬ ‫ﻟﺤﻤﺾ‬ ‫ﺍﻷﺳﻜﻮﺭﺑﻴﻚ‬ ، ‫ﺣﻴﺚ‬ ‫ﺃﺧﺬ‬ ‫ﺍﻟﺘﻼﻣﻴﺬ‬ ‫ﻓﻲ‬ ‫ﺑﻴﺸﺮ‬ ‫ﺣﺠﻤﺎ‬ 0 V ‫ﻣﻦ‬ ‫ﺍﻟﺤﻤﺾ‬ ‫ﻭﺃﺿﺎﻓﻮﺍ‬ ‫ﻟﻪ‬ ‫ﻧﻔﺲ‬ ‫ﺍﻟﺤﺠﻢ‬ ‫ﻣﻦ‬ ‫ﺍﻟﻤﺎﺀ‬ ،‫ﺍﻟﻤﻘﻄﺮ‬ ‫ﺛﻢ‬ ‫ﺃﺧﺬﻭﺍ‬ ‫ﻣﻦ‬ ‫ﺍﻟﻤﺤﻠﻮﻝ‬ ‫ﺍﻟﺠﺪﻳﺪ‬ ‫ﺣﺠﻤﺎ‬ 20 a V mL  ، ‫ﻭﻣﻠﺌﻮﺍ‬ ‫ﺳﺤﺎﺣﺔ‬ ‫ﺟﺔ‬ ‫ﺭ‬‫ﻣﺪ‬ ‫ﺑﻤﺤﻠﻮﻝ‬ ‫ﻣﺎﺋﻲ‬ ‫ﻟﻬﻴﺪﺭﻭﻛﺴﻴﺪ‬ ‫ﺍﻟﺒﻮﺗﺎﺳﻴﻮﻡ‬   K ,OH   ‫ﺗﺮﻛﻴﺰﻩ‬ ‫ﺍﻟﻤﻮﻟﻲ‬ 2 5 10 B C mol / L    ، ‫ﻭﺑﻌﺪ‬ ‫ﺍﻟﺤﺼﻮﻝ‬ ‫ﻋﻠﻰ‬ ‫ﺍﻟﻘﻴﺎﺳﺎﺕ‬ ‫ﻗﺎﻣﻮﺍ‬ ‫ﺑﺘﻤﺜﻴﻞ‬ ‫ﺍﻟﺒﻴﺎﻥ‬ B pH f (V )  . 1 - ‫ﺍﻛﺘﺐ‬ ‫ﻣﻌﺎﺩﻟﺔ‬ ‫ﺗﻔﺎﻋﻞ‬ ‫ﺍﻟﻤﻌﺎﻳﺮﺓ‬ . 2 - ‫ﻑ‬ ‫ﺮ‬‫ﻋ‬ ‫ﺍﻟﺘﻜﺎﻓﺆ‬ ‫ﺣﻤﺾ‬ – ‫ﺃﺳﺎﺱ‬ ، ‫ﺛﻢ‬ ‫ﺩ‬ ‫ﺪ‬‫ﺣ‬ ‫ﺇﺣﺪﺍﺛﻲ‬ ‫ﻧﻘﻄﺔ‬ ‫ﺍﻟﺘﻜﺎﻓﺆ‬ ‫ﺣﻤﺾ‬ – ‫ﺃﺳﺎﺱ‬ . 3 - ‫ﻋﻴﻦ‬ a pK ‫ﺍﻟﺜﻨﺎﺋﻴﺔ‬   6 8 6 6 7 6 C H O / C H O  . 4 - ‫ﺍﺣﺴﺐ‬ ‫ﺍﻟﺘﺮﻛﻴﺰ‬ ‫ﺍﻟﻜﺘﻠﻲ‬   m C ‫ﻟﺤﻤﺾ‬ ‫ﺍﻷﺳﻜﻮﺭﺑﻴﻚ‬ . 5 - ‫ﻦ‬ ‫ﻴ‬‫ﺑ‬ ‫ﺑﻄﺮﻳﻘﺘﻴﻦ‬ ‫ﺃﻥ‬ ‫ﺣﻤﺾ‬ ‫ﺍﻷﺳﻜﻮﺭﺑﻴﻚ‬ ‫ﺿﻌﻴﻒ‬ ‫ﺍﻟﻤﺎﺀ‬ ‫ﻓﻲ‬ . 6 - ‫ﺍﺣﺴﺐ‬ ‫ﺍﻟﺘﺮﻛﻴﺰ‬ ‫ﺍﻟﻤﻮﻟﻲ‬ ‫ﻟﺤﻤﺾ‬ ‫ﺍﻷﺳﻜﻮﺭﺑﻴﻚ‬ ‫ﻓﻲ‬ ‫ﺍﻟﺒﻴﺸﺮ‬ ‫ﻋﻨﺪ‬ ‫ﺍﻟﺘﻜﺎﻓﺆ‬ ، ‫ﺛﻢ‬ ‫ﺍﺳﺘﻨﺘﺞ‬ ‫ﺃﻧﻪ‬ ‫ﻳﻤﻜﻦ‬ ‫ﺗﻔﺎﻋﻞ‬ ‫ﺍﻋﺘﺒﺎﺭ‬ ‫ﺍﻟﻤﻌﺎﻳﺮﺓ‬ ‫ﺗﺎﻣﺎ‬ . 7 - ‫ﻗﺎﺭﻥ‬ ‫ﻗﻮﺓ‬ ‫ﺣﻤﺾ‬ ‫ﺍﻷﺳﻜﻮﺭﺑﻴﻚ‬ ‫ﻣﻊ‬ ‫ﺣﻤﺾ‬ ‫ﺍﻟﺒﺮﻭﺑﺎﻧﻮﻳﻚ‬   2 5 C H COOH . 8 - ‫ﻓﻲ‬ ‫ﺣﺎﻟﺔ‬ ‫ﺍﺳﺘﻌﻤﺎﻝ‬ ‫ﻛﺎﺷﻒ‬ ‫ﻥ‬ ‫ﻮ‬‫ﻣﻠ‬ ‫ﻟﺘﺤﺪﻳﺪ‬ ‫ﻧﻘﻄﺔ‬ ‫ﺍﻟﺘﻜﺎﻓﺆ‬ ، ‫ﻣﺎ‬ ‫ﻫﻮ‬ ‫ﺍﻟﻜﺎﺷﻒ‬ ‫ﺍﻷﻧﺴﺐ‬ ‫ﻣﻦ‬ ‫ﺑﻴﻦ‬ ‫ﺍﻟﻜﻮﺍﺷﻒ‬ ‫ﺍﻟﺘﺎﻟﻴﺔ‬ ‫ﻟﻬﺬﻩ‬ ‫ﺍﻟﻤﻌﺎﻳﺮﺓ‬ ‫؟‬ ‫ﺍﻟﻬﻠﻴﺎﻧﺘﻴﻦ‬ : ‫ﻣﺠﺎﻝ‬ ‫ﺗﻐﻴﺮ‬ ‫ﺍﻟﻠﻮﻥ‬   3 1 4 4 , ,  . ‫ﺍﻟﻔﻴﻨ‬ ‫ﻮﻝ‬ ‫ﻓﺘﺎﻟﻴﻦ‬ : ‫ﻣﺠﺎﻝ‬ ‫ﺗﻐﻴﺮ‬ ‫ﺍﻟﻠﻮﻥ‬   8 2 10 ,  . ‫ﺃﺯﺭﻕ‬ ‫ﺍﻟﺒﺮﻭﻣﻮﺗﻴﻤﻮﻝ‬ : ‫ﻣﺠﺎﻝ‬ ‫ﺗﻐﻴﺮ‬ ‫ﺍﻟﻠﻮﻥ‬   6 7 6 ,  . ‫ﺍ‬ ‫ﻧ‬ ‫ﺘ‬ ‫ﻬ‬ ‫ﻰ‬ ‫ﺍ‬ ‫ﻟ‬ ‫ﻤ‬ ‫ﻮ‬ ‫ﺿ‬ ‫ﻮ‬ ‫ﻉ‬ ‫ﺍ‬ ‫ﻷ‬ ‫ﻭﻝ‬ e n c y - e d u c a t i o n . c o m / e x a m s 3as.ency-education.com
‫ﺍ‬ ‫ﺧ‬ ‫ﺘ‬ ‫ﺒ‬ ‫ﺎ‬ ‫ﺭ‬ ‫ﻓ‬ ‫ﻲ‬ ‫ﻣ‬ ‫ﺎ‬ ‫ﺩ‬ ‫ﺓ‬ ‫ﺍ‬ ‫ﻟ‬ ‫ﻌ‬ ‫ﻠ‬ ‫ﻮ‬ ‫ﻡ‬ ‫ﺍ‬ ‫ﻟ‬ ‫ﻔ‬ ‫ﻴ‬ ‫ﺰ‬ ‫ﻳ‬ ‫ﺎ‬ ‫ﺋ‬ ‫ﻴ‬ ‫ﺔ‬ / ‫ﺍ‬ ‫ﻟ‬ ‫ﺸ‬ ‫ﻌ‬ ‫ﺒ‬ ‫ﺔ‬ ، ‫ﻋ‬ ‫ﻠ‬ ‫ﻮ‬ ‫ﻡ‬ ‫ﺗ‬ ‫ﺠ‬ ‫ﺮ‬ ‫ﻳ‬ ‫ﺒ‬ ‫ﻴ‬ ‫ﺔ‬ / ‫ﺑ‬ ‫ﻜ‬ ‫ﺎ‬ ‫ﻟ‬ ‫ﻮ‬ ‫ﺭ‬ ‫ﻳ‬ ‫ﺎ‬ ‫ﺗ‬ ‫ﺠ‬ ‫ﺮ‬ ‫ﻳ‬ ‫ﺒ‬ ‫ﻲ‬ 2019 Page 4 sur 6 ‫ﺍﻟﺜﺎﻧﻲ‬ ‫ﺍﻟﻤﻮﺿﻮﻉ‬ ‫ﺍﻟ‬ ‫ﺍﻟﻤﻮﺿﻮﻉ‬ ‫ﻳﺤﺘﻮﻱ‬ ‫ﺜﺎﻧﻲ‬ ‫ﻋﻠﻰ‬ 3 ‫ﺻﻔﺤﺎﺕ‬ ) ‫ﺻﻔﺤﺔ‬ ‫ﻣﻦ‬ 4 ‫ﻣﻦ‬ 6 ‫ﺍﻟﺼﻔﺤﺔ‬ ‫ﺇﻟﻰ‬ 6 ‫ﻣﻦ‬ 6 ( ‫ﺍﻷﻭﻝ‬ ‫ﺍﻟﺠﺰﺀ‬ ) 13 ‫ﻥ‬ ( ‫ﺍﻷﻭﻝ‬ ‫ﺍﻟﺘﻤﺮﻳﻦ‬ : ) 06 ‫ﻧﻘﺎﻁ‬ ( ‫ﻧﻌ‬ ‫ﺘﺒﺮ‬ ‫ﺍﻟﺪﺍﺭﺓ‬ ‫ﺍﻟﻜﻬﺮﺑﺎﺋﻴﺔ‬ ‫ﺍﻟﻤﻤﺜﻠﺔ‬ ‫ﻓﻲ‬ ‫ﺍﻟﺸﻜﻞ‬ (1) ، ‫ﺍﻟﻤﻜﻮﻧﺔ‬ ‫ﻣﻦ‬ :  ‫ﻣﻮﻟﺪ‬ ‫ﺗﻮﺗﺮ‬ ‫ﻗﻮﺗﻪ‬ ‫ﺍﻟﻤﺤﺮﻛﺔ‬ ‫ﺍﻟﻜﻬﺮﺑﺎﺋﻴﺔ‬ .  ‫ﻧﺎﻗﻼﻥ‬ ‫ﺃﻭﻣﻴﺎﻥ‬ ‫ﻣﻘﺎﻭﻣﺘﻴﻬﻤﺎ‬ 1 =75 Ω ‫ﻭ‬ 2 ‫ﻣﺠﻬﻮﻟﺔ‬ .  ‫ﻣﻜﺜﻔﺔ‬ ‫ﺳﻌﺘﻬﺎ‬ ، ‫ﻏﻴﺮ‬ ‫ﻣﺸﺤﻮﻧﺔ‬ .  ‫ﺑﺎﺩﻟﺔ‬ 1 . ‫ﻋﻨﺪ‬ ‫ﺍﻟﻠﺤﻈﺔ‬ =0 ، ‫ﻧﻀﻊ‬ ‫ﺍﻟﺒﺎﺩﻟﺔ‬ ‫ﺍﻟﻮﺿﻊ‬ ‫ﻋﻠﻰ‬ 1 . ‫ﺃﻋﺪ‬ ‫ﺍﻟﻜﻬﺮﺑﺎ‬ ‫ﺍﻟﺪﺍﺭﺓ‬ ‫ﻧﻘﻞ‬ ‫ﺋﻴﺔ‬ ‫ﺟﻬﺔ‬ ‫ﻭ‬ ،‫ﺍﻟﺘﻴﺎﺭ‬ ‫ﺟﻬﺔ‬ ‫ﻋﻠﻴﻬﺎ‬ ‫ﻭﻣﺜﻞ‬ ‫ﺍﻟﺘﻮﺗﺮﺍﺕ‬ . ‫ﺃ‬ - ‫ﺃﻭﺟﺪ‬ ‫ﺍﻟﻤﻌﺎﺩﻟﺔ‬ ‫ﺍﻟﺘﻔﺎﺿﻠﻴﺔ‬ ‫ﺍﻟﺘﻲ‬ ‫ﺷﺪﺓ‬ ‫ﺗﺤﻘﻘﻬﺎ‬ ‫ﺍﻟﻜﻬﺮﺑﺎﺋﻲ‬ ‫ﺍﻟﺘﻴﺎﺭ‬ . ‫ﻋﻦ‬ ‫ﺍﻟﻤﻌﺒﺮﺓ‬ ‫ﺗﻠﻚ‬ ‫ﻣﻨﻬﺎ‬ ‫ﻭﺍﺳﺘﻨﺘﺞ‬ ‫ﻃﺮﻓﻲ‬ ‫ﺑﻴﻦ‬ ‫ﺍﻻﻭﻣﻲ‬ ‫ﺍﻟﻨﺎﻗﻞ‬ .R2 ‫ﺏ‬ - ‫ﺍﻟﺘﻔﺎﺿﻠﻴﺔ‬ ‫ﺍﻟﻤﻌﺎﺩﻟﺔ‬ ‫ﺣﻞ‬ ‫ﺍﻟﺸﻜﻞ‬ ‫ﻣﻦ‬ ‫ﻛﺘﺎﺑﺘﻬﺎ‬ ‫ﻳﻤﻜﻦ‬ = . ‫ﻋﺒ‬ ‫ﻋﻦ‬ ‫ﺮ‬ ‫ﻭ‬ ‫ﻋﻨﺎﺻﺮ‬ ‫ﻣﻤﻴﺰﺍﺕ‬ ‫ﺑﺪﻻﻟﺔ‬ ‫ﺍﻟﺪﺍﺭﺓ‬ . ‫ﺝ‬ - ‫ﺍﻟﻜﻬﺮﺑﺎﺋ‬ ‫ﺍﻟﺘﻮﺗﺮ‬ ‫ﻋﺒﺎﺭﺓ‬ ‫ﺍﺳﺘﻨﺘﺞ‬ ‫ﺍﻟﻤﻜﺜﻔﺔ‬ ‫ﻃﺮﻓﻲ‬ ‫ﺑﻴﻦ‬ ‫ﻲ‬ . ( ) 2 . ‫ﺫﺍﻛﺮﺓ‬ ‫ﺫﻭ‬ ‫ﺍﻟﻤﻬﺒﻄﻲ‬ ‫ﺍﻻﻫﺘﺰﺍﺯ‬ ‫ﺭﺍﺳﻢ‬ ‫ﻳﺴﻤﺢ‬ ‫ﺍﻟﺴﺎﺑﻘﻴﻦ‬ ‫ﺍﻟﺘﻮﺗﺮﻳﻦ‬ ‫ﺑﻤﻌﺎﻳﻨﺔ‬ ‫ﻭ‬ ). ‫ﺍﻟﺸﻜﻞ‬ - 2 - ( ‫ﺃ‬ - ‫ﺍﻟﺠﻬﺎﺯ‬ ‫ﺭﺑﻂ‬ ‫ﻛﻴﻔﻴﺔ‬ ‫ﻛﻴﻔﻲ‬ ‫ﺑﺮﺳﻢ‬ ‫ﻭﺿﺢ‬ ‫ﻟﻤﻌﺎﻳﻨﺔ‬ ‫ﺍﻟﻤﺪﺧﻞ‬ ‫ﻋﻞ‬ y1 ‫ﻭ‬ ‫ﻋﻠﻰ‬ y1 ‫ﺏ‬ - ‫ﺍﻟﻤﻨﺎﺳﺐ‬ ‫ﺍﻟﺘﻮﺗﺮ‬ ‫ﻣﺪﺧﻞ‬ ‫ﻟﻜﻞ‬ ‫ﺃﻧﺴﺐ‬ . ‫ﺝ‬ - ‫ﻣﻦ‬ ‫ﻛﻞ‬ ‫ﻗﻴﻢ‬ ‫ﺣﺪﺩ‬ ‫ﺍﻟﺸﻜﻞ‬ ‫ﻋﻠﻰ‬ ‫ﺍﻋﺘﻤﺎﺩﺍ‬ : E ، R2 ‫ﻭ‬ C . 3 . ‫ﺍﻟﺒﺎﺩﻟﺔ‬ ‫ﻧﻨﻘﻞ‬ ‫ﻣﺸﺤﻮﻧﺔ‬ ‫ﺍﻟﻤﻜﺜﻔﺔ‬ ‫ﺗﺼﺒﺢ‬ ‫ﻋﻨﺪﻣﺎ‬ ‫ﺍﻟﻮﺿﻊ‬ ‫ﺇﻟﻰ‬ ) 2 ( ‫ﻣﺒﺪﺍ‬ ‫ﻧﻌﺘﺒﺮﻫﺎ‬ ‫ﻟﺤﻈﺔ‬ ‫ﻓﻲ‬ ‫ﺍﻟﻌﺒﺎﺭﺓ‬ ‫ﺗﺼﺒﺢ‬ ، ‫ﻟﻠﺰﻣﻦ‬ ‫ﺟﺪﻳﺪ‬ ‫ﺍﻟﻠﺤﻈﻴﺔ‬ : = − . ‫ﺃ‬ - ‫ﺗ‬ ‫ﻛﻴﻒ‬ ‫ﺍﻟﺘﻮﺗﺮ‬ ‫ﺇﺷﺎﺭﺓ‬ ‫ﻔﺴﺮ‬ . ‫ﺏ‬ - ‫ﻛﻞ‬ ‫ﺗﻮﺟﻴﻪ‬ ‫ﺍﻟﺸﻜﻞ‬ ‫ﻋﻠﻰ‬ ‫ﻭﺿﺢ‬ ‫ﺍﻟﺤﺎﻟﺔ‬ ‫ﻫﺬﻩ‬ ‫ﻓﻲ‬ ‫ﺍﻟﻜﻬﺮﺑﺎﺋﻲ‬ ‫ﻭﺍﻟﺘﻮﺗﺮ‬ ‫ﺍﻟﺘﻴﺎﺭ‬ ‫ﺷﺪﺓ‬ ‫ﻣﻦ‬ . ‫ﺝ‬ - ‫ﺍﻟﻠﺤﻈﺔ‬ ‫ﺣﺪﺩ‬ t1 ‫ﺍ‬ ‫ﺍﻟﻄﺎﻗﺔ‬ ‫ﻓﻴﻬﺎ‬ ‫ﺗﺼﺒﺢ‬ ‫ﻟﺘﻲ‬ ‫ﺍﻟﻨﺎﻗﻞ‬ ‫ﻓﻲ‬ ‫ﺟﻮﻝ‬ ‫ﺑﻔﻌﻞ‬ ‫ﺍﻟﻤﺤﻮﻟﺔ‬ ‫ﺍﻻﻭﻣﻲ‬ R2 ‫ﻫﻲ‬ : We =0.32J ‫ﺍﻟﺜﺎﻧﻲ‬ ‫ﺍﻟﺘﻤﺮﻳﻦ‬ ) 7 ‫ﻧﻘﺎﻁ‬ :( ‫ﺍﻟﻤﻴﺜﺎﻧﻮﻳﻚ‬ ‫ﺣﻤﺾ‬ ‫ﻳﻌﺘﺒﺮ‬ ) ‫ﺣﻤﺾ‬ ‫ﺍﻟﻨﻤﻞ‬ ( ‫ﻟﻠﻨﻤﻞ‬ ‫ﺍﻟﺪﻓﺎﻉ‬ ‫ﻭﺳﺎﺋﻞ‬ ‫ﻣﻦ‬ . ‫ﻣﺤﻠﻮﻟﻬﺎﻟﻤـﺎﺋﻲ‬ ‫ﺧﻮﺍﺹ‬ ‫ﺑﻌﺾ‬ ‫ﺩﺭﺍﺳﺔ‬ ‫ﻧﺮﻳﺪ‬ . I - ‫ﺣﺠﻤﺎ‬ ‫ﻧﻀﻊ‬ = 2 ‫ﺍﻟﻤﻮﻟﻲ‬ ‫ﺍﻟﺘﺮﻛﻴﺰ‬ ‫ﺫﻱ‬ ‫ﺍﻟﻨﻤﻞ‬ ‫ﺣﻤﺾ‬ ‫ﻣﻦ‬ ‫ﺳﻌﺔ‬ ‫ﺫﺍﺕ‬ ‫ﻋﻴﺎﺭﻳﺔ‬ ‫ﺣﻮﺟﻠـﺔ‬ ‫ﻓـــﻲ‬ = 100 ‫ﺍﻟﺤﺠﻢ‬ ‫ﺛﻢ‬ ‫ﺍﻟﻌﻴﺎﺭ‬ ‫ﺧﻂ‬ ‫ﺇﻟﻰ‬ ‫ﺍﻟﻤﻘﻄﺮ‬ ‫ﺑﺎﻟﻤﺎﺀ‬ . ‫ﻣﺤﻠﻮﻝ‬ ‫ﻋﻠﻰ‬ ‫ﻓﻨﺤﺼﻞ‬ ‫ﺟﻴﺪﺍ‬ ‫ﺍﻟﻤﺤﻠﻮﻝ‬ ‫ﻧﺮﺝ‬ ( ) ‫ﺍﻟﻤﻮ‬ ‫ﺗﺮﻛﻴﺰ‬ ‫ﺫﻱ‬ ‫ﻟﻲ‬ ‫ﻗﻴﺎﺱ‬ ‫ﻋﻨﺪ‬ ‫ﻧﺎﻗﻠﻴﺘﻬﺎﻟﻨﻮﻋﻴ‬ ‫ﻧﺠﺪ‬ ‫ﺔ‬ = 0,25 / . ‫ﻳﻌﻄﻰ‬ : 3 2 5,46 10 . / HCOO S m mol      ، 3 3 2 35,00 10 . / H O S m mol      e n c y - e d u c a t i o n . c o m / e x a m s 3as.ency-education.com
‫ﺍ‬ ‫ﺧ‬ ‫ﺘ‬ ‫ﺒ‬ ‫ﺎ‬ ‫ﺭ‬ ‫ﻓ‬ ‫ﻲ‬ ‫ﻣ‬ ‫ﺎ‬ ‫ﺩ‬ ‫ﺓ‬ ‫ﺍ‬ ‫ﻟ‬ ‫ﻌ‬ ‫ﻠ‬ ‫ﻮ‬ ‫ﻡ‬ ‫ﺍ‬ ‫ﻟ‬ ‫ﻔ‬ ‫ﻴ‬ ‫ﺰ‬ ‫ﻳ‬ ‫ﺎ‬ ‫ﺋ‬ ‫ﻴ‬ ‫ﺔ‬ / ‫ﺍ‬ ‫ﻟ‬ ‫ﺸ‬ ‫ﻌ‬ ‫ﺒ‬ ‫ﺔ‬ ، ‫ﻋ‬ ‫ﻠ‬ ‫ﻮ‬ ‫ﻡ‬ ‫ﺗ‬ ‫ﺠ‬ ‫ﺮ‬ ‫ﻳ‬ ‫ﺒ‬ ‫ﻴ‬ ‫ﺔ‬ / ‫ﺑ‬ ‫ﻜ‬ ‫ﺎ‬ ‫ﻟ‬ ‫ﻮ‬ ‫ﺭ‬ ‫ﻳ‬ ‫ﺎ‬ ‫ﺗ‬ ‫ﺠ‬ ‫ﺮ‬ ‫ﻳ‬ ‫ﺒ‬ ‫ﻲ‬ 2019 Page 5 sur 6 1 - ‫ﺍﻟﻤﺎﺀ‬ ‫ﻓﻲ‬ ‫ﺍﻟﻤﻴﺜﺎﻧﻮﻳﻚ‬ ‫ﺣﻤﺾ‬ ‫ﺍﻧﺤﻼﻝ‬ ‫ﻣﻌﺎﺩﻟﺔ‬ ‫ﺃﻛﺘﺐ‬ . 2 - ‫ﺃﻭﺟﺪ‬ ‫ﺟﺪ‬ ‫ﺑﻴﻦ‬ ‫ﺍﻟﻌﻼﻗﺔ‬ ‫ﻭ‬ . 1 - ‫ﻗﻴﻤﺔ‬ ‫ﺃﺣﺴﺐ‬ ‫ﺍﻟﻤﺤﻠﻮﻝ‬ ( ) . 2 - ‫ﻋﺒﺎﺭﺓ‬ ‫ﺃﻛﺘﺐ‬ ‫ﺍﻟﻨﻬﺎﺋﻲ‬ ‫ﺍﻟﺘﻘﺪﻡ‬ ‫ﻧﺴﺒﺔ‬ f  ‫ﺍﻟﻤﺤﻠﻮﻝ‬ ‫ﻓﻲ‬ ‫ﺍﻟﻤﺎﺀ‬ ‫ﻣﻊ‬ ‫ﺍﻟﻨﻤﻞ‬ ‫ﻟﺤﻤﺾ‬ ‫ﺍﻟﺤﺎﺻﻞ‬ ‫ﻟﻠﺘﺤﻮﻝ‬ ( ) ‫ﺑﺪﻻﻟﺔ‬ . -II ‫ﺍﻟﻤﻴﺜﺎﻧﻮﻳﻚ‬ ‫ﺣﻤﺾ‬ ‫ﺑﻴﻦ‬ ‫ﻳﺤﺪﺙ‬ ‫ﺍﻟﺬﻱ‬ ‫ﺍﻟﻜﻴﻤﻴﺎﺋﻲ‬ ‫ﺍﻟﺘﻔﺎﻋﻞ‬ ‫ﺩﺭﺍﺳﺔ‬ ‫ﻧﺮﻳﺪ‬ ‫ﺍﻟﻤﺠﻤﻠﺔ‬ ‫ﺍﻟﺠﺰﻳﺌﻴﺔ‬ ‫ﺻﻴﻐﺘﻪ‬ ‫ﻛﺤﻮﻝ‬ ‫ﻭ‬ . ‫ﺃ‬ ‫ﺛﻤﺎﻧﻴﺔ‬ ‫ﻓﻲ‬ ‫ﻧﻀﻊ‬ ‫ﻣﻦ‬ ‫ﻣﺮﻗـــﻤﺔ‬ ‫ﺍﺧـﺘﺒﺎﺭ‬ ‫ﻧﺎﺑـﻴﺐ‬ 01 ‫ﺇﻟﻰ‬ 08 ‫ﻣﻦ‬ ‫ﺍﻟﻤﺘﻜﻮﻥ‬ ‫ﺍﻟﻤﺰﻳﺞ‬ ‫ﻧﻔﺲ‬ 0,2 ‫ﺍﻟﺤـﻤﻀﻮ‬ ‫ﻣﻦ‬ 0,2 ‫ﺣﺮﺍﺭﺗﻪ‬ ‫ﺩﺭﺟﺔ‬ ‫ﻣﺎﺋﻲ‬ ‫ﺣﻤﺎﻡ‬ ‫ﻓﻲ‬ ‫ﺍﻷﻧﺎﺑﻴﺐ‬ ‫ﻫﺬﻩ‬ ‫ﻧﺪﺧﻞ‬ ‫ﺛﻢ‬ ‫ﺍﻟﻜﺤﻮﻝ‬ ‫ﻣﻦ‬ (180° ) ‫ﺳﺎﻋﺔ‬ ‫ﻛﻞ‬ ‫ﺑﻌﺪ‬ ‫ﻭ‬ ‫ﻣﻦ‬ ‫ﺑﺎﻟﺘﺮﺗﻴﺐ‬ ‫ﺍﻷﻧﺎﺑﻴﺐ‬ ‫ﻫﺬﻩ‬ ‫ﺃﺣﺪ‬ ‫ﻧﺨﺮﺝ‬ 01 ‫ﺇﻟﻰ‬ 08 ‫ﻣ‬ ‫ﺑﻮﺍﺳﻄﺔ‬ ‫ﻓﻴﻪ‬ ‫ﺍﻟﻤﺘﺒﻘﻲ‬ ‫ﺍﻟﺤﻤﺾ‬ ‫ﻧﻌﺎﻳﺮ‬ ‫ﺛﻢ‬ ، ‫ﺍﻟﺼﻮﺩﻳﻮﻡ‬ ‫ﻫﻴﺪﺭﻭﻛﺴﻴﺪ‬ ‫ﺤﻠﻮﻝ‬ . ( ( ) + ( )) ‫ﺍﻟﻨﺘﺎﺋﺞ‬ ‫ﻋﻠﻴﻬﺎ‬ ‫ﺍﻟﻤﺘﺤﺼﻞ‬ ‫ﺍﻟﺘﺎﻟﻲ‬ ‫ﺍﻟﺠﺪﻭﻝ‬ ‫ﻓﻲ‬ ‫ﻣﺪﻭﻧﺔ‬ : 08 07 06 05 04 03 02 01 ‫ﺍﻷﻧﺒﻮﺏ‬ ‫ﺭﻗﻢ‬ 7 6 5 4 3 2 1 0 (ℎ ) 0,067 0,067 0,067 0,068 0,074 0,084 0,114 0,200 (‫)ﺣﻤﺾ‬ (‫)ﺃﺳﺘﺮ‬ 1 - ‫ﺃﻋﻼﻩ‬ ‫ﺍﻟﺠﺪﻭﻝ‬ ‫ﺃﻛﻤﻞ‬ . 2 - ‫ﺍﻟﺒﻴﺎﻧﻲ‬ ‫ﺍﻟﻤﻨﺤﻨﻰ‬ ‫ﺃﺭﺳﻢ‬ (‫)ﺃﺳﺘﺮ‬ = ( ) . ‫ﻭﻓﻖ‬ ‫ﺍﻟﺴﻠﻢ‬ : ( 1 → 1ℎ  ‫ﻭ‬ 1 → 0,01 ) 3 - ‫ﺍﻟﺤﻤﺾ‬ ‫ﺑﻴﻦ‬ ‫ﺍﻟﺘﻔﺎﻋﻞ‬ ‫ﺗﻘﺪﻡ‬ ‫ﺟﺪﻭﻝ‬ ‫ﺃﻧﺸﺊ‬ ‫ﺍﻟﻜﺤﻮﻝ‬ ‫ﻭ‬ . 4 - ‫ﺍﻟﺒﻴﺎﻥ‬ ‫ﻣﻦ‬ ‫ﺍﺳﺘﻨﺘﺞ‬ : ‫ﺃ‬ - ‫ﺍﻟﻠﺤﻈﺔ‬ ‫ﻋﻨﺪ‬ ‫ﺍﻟﺘﻔﺎﻋﻞ‬ ‫ﺳﺮﻋﺔ‬ = 2ℎ . ‫ﺏ‬ - ‫؟‬ ‫ﻫﺬﺍﺍﻟﺘﺤﻮﻝ‬ ‫ﻟﻨﻬﺎﻳﺔ‬ ‫ﺍﻟﻤﻮﺍﻓﻘﺔ‬ ‫ﺣﺪﺩﺍﻟﻠﺤﻈﺔ‬ ‫ﺟـ‬ - ‫ﺍﻷﺳﺘﺮﺓ‬ ‫ﻣﺮﺩﻭﺩ‬ . - ‫ﺍﻟﻤﻤﻜﻨﺔ‬ ‫ﺍﻟﻤﻔﺼﻠﺔ‬ ‫ﻧﺼﻒ‬ ‫ﺻﻴﻐﻪ‬ ‫ﻭ‬ ‫ﺍﻟﻤﺴﺘﻌﻤﻞ‬ ‫ﺍﻟﻜﺤﻮﻝ‬ ‫ﺻﻨﻒ‬ ‫ﺍﺳﺘﻨﺘﺞ‬ . 5 - ‫ﺍﻟﻤﺘﻔﺮﻋﺔ‬ ‫ﺍﻟﺼﻴﻐﺔ‬ ‫ﺫﻱ‬ ‫ﺍﻟﻜﺤﻮﻝ‬ ‫ﻭ‬ ‫ﺍﻟﺤﻤﺾ‬ ‫ﺑﻴﻦ‬ ‫ﺍﻟﺤﺎﺻﻞ‬ ‫ﻟﻠﺘﺤﻮﻝ‬ ‫ﺍﻟﻤﻨﻤﺬﺝ‬ ‫ﺍﻟﺘﻔﺎﻋﻞ‬ ‫ﻣﻌﺎﺩﻟﺔ‬ ‫ﺃﻛﺘﺐ‬ . ‫ﺍﻷﺳﺘﺮ‬ ‫ﺗﺴﻤﻴﺔ‬ ‫ﻣﻊ‬ ‫ﺍﻟﻨﺎﺗﺞ‬ 6 - ‫ﺍﻷ‬ ‫ﻧﺨﺮﺝ‬ ‫ﺭﻗﻢ‬ ‫ﻧﺒﻮﺏ‬ 07 ‫ﺍﻟﻠﺤﻈﺔ‬ ‫ﻋﻨﺪ‬ = 6 ℎ ‫ﻣﺒﺎﺷﺮﺓ‬ ‫ﻟﻪ‬ ‫ﻧﻀﻴﻒ‬ ‫ﺛﻢ‬ 0,2 ‫ﺍﻷﺳﺘﺮ‬ ‫ﻣﻦ‬ .  ‫ﻋﻠﻞ‬ ‫؟‬ ‫ﺍﻟﻜﻴﻤﻴﺎﺋﻴﺔ‬ ‫ﺍﻟﺠﻤﻠﺔ‬ ‫ﺗﻄﻮﺭ‬ ‫ﺗﺘﻮﻗﻊ‬ ‫ﺟﻬﺔ‬ ‫ﺃﻱ‬ ‫ﻓﻲ‬ . ‫ﺍﻟﺜــﺎﻧﻲ‬ ‫ﺍﻟﺠﺰﺀ‬ ) 7 ‫ﻥ‬ ( ‫ﺍﻟﺘﺠﺮﻳﺒﻲ‬ ‫ﺍﻟﺘﻤﺮﻳﻦ‬ : ) 7 ‫ﻥ‬ ( ‫ﻛﺮﻳﺔ‬ ( ) ‫ﻛﺘﻠﺘﻬﺎ‬ ‫ﻧﻘﺘﺮﺡ‬ ‫ﻗﻴﻤﺘﻬﺎ‬ ‫ﻟﺘﺤﺪﻳﺪ‬ ‫ﻣﺠﻬﻮﻟﺔ‬ . I - ‫ﺍﻷﻭﻟﻰ‬ ‫ﺍﻟﻄﺮﻳﻘﺔ‬ : ‫ﺍﻟﺴﻘ‬ ‫ﺣﺮﻛﺔ‬ ‫ﺩﺭﺍﺳﺔ‬ ‫ﺍﻟﻬﻮﺍﺀ‬ ‫ﻓﻲ‬ ‫ﻟﻠﻜﺮﻳﺔ‬ ‫ﺍﻟﺸﺎﻗﻮﻟﻲ‬ ‫ﻮﻁ‬ : ‫ﺍﻟﻨﻘﻄﺔ‬ ‫ﻣﻦ‬ ‫ﺍﺑﺘﺪﺍﺀ‬ ‫ﺍﻟﻬﻮﺍﺀ‬ ‫ﺍﺑﺘﺪﺍﺋﻴﺔﻓﻲ‬ ‫ﺳﺮﻋﺔ‬ ‫ﺩﻭﻥ‬ ‫ﺍﻟﻜﺮﻳﺔ‬ ‫ﺗﺴﻘﻂ‬ ‫ﻗﻮﺓ‬ ‫ﺣﺮﻛﺘﻬﺎ‬ ‫ﺗﻌﻴﻖ‬ ، ‫ﺍﻟﺪﺭﺍﺳﺔ‬ ‫ﻣﻌﻠﻢ‬ ‫ﺍﺣﺪﺍﺛﻴﺎﺕ‬ ‫ﻣﺒﺪﺃ‬ ‫ﻣﻦ‬ ‫ﻋﺒﺎﺭﺗﻬﺎ‬ ‫ﺍﺣﺘﻜﺎﻙ‬ ‫ﺍﻟﺸﻜﻞ‬ : f Kv  ). ‫ﺃﺭﺧﻤﻴﺪﺱ‬ ‫ﺩﺍﻓﻌﺔ‬ ‫ﻧﻬﻤﻞ‬ ( ‫ﺍﻟﺸﻜﻞ‬ ‫ﺍﻟﺒﻴﺎﻥ‬ ‫ﻳﻤﺜﻞ‬ - 3 ‫ﻋﻄﺎﻟ‬ ‫ﻣﺮﻛﺰ‬ ‫ﺳﺮﻋﺔ‬ ‫ﺗﻐﻴﺮﺍﺕ‬ ‫ﺍﻟﺰﻣﻦ‬ ‫ﺑﺪﻻﻟﺔ‬ ‫ﺍﻟﻜﺮﻳﺔ‬ ‫ﺔ‬ . e n c y - e d u c a t i o n . c o m / e x a m s 3as.ency-education.com
‫ﺍ‬ ‫ﺧ‬ ‫ﺘ‬ ‫ﺒ‬ ‫ﺎ‬ ‫ﺭ‬ ‫ﻓ‬ ‫ﻲ‬ ‫ﻣ‬ ‫ﺎ‬ ‫ﺩ‬ ‫ﺓ‬ ‫ﺍ‬ ‫ﻟ‬ ‫ﻌ‬ ‫ﻠ‬ ‫ﻮ‬ ‫ﻡ‬ ‫ﺍ‬ ‫ﻟ‬ ‫ﻔ‬ ‫ﻴ‬ ‫ﺰ‬ ‫ﻳ‬ ‫ﺎ‬ ‫ﺋ‬ ‫ﻴ‬ ‫ﺔ‬ / ‫ﺍ‬ ‫ﻟ‬ ‫ﺸ‬ ‫ﻌ‬ ‫ﺒ‬ ‫ﺔ‬ ، ‫ﻋ‬ ‫ﻠ‬ ‫ﻮ‬ ‫ﻡ‬ ‫ﺗ‬ ‫ﺠ‬ ‫ﺮ‬ ‫ﻳ‬ ‫ﺒ‬ ‫ﻴ‬ ‫ﺔ‬ / ‫ﺑ‬ ‫ﻜ‬ ‫ﺎ‬ ‫ﻟ‬ ‫ﻮ‬ ‫ﺭ‬ ‫ﻳ‬ ‫ﺎ‬ ‫ﺗ‬ ‫ﺠ‬ ‫ﺮ‬ ‫ﻳ‬ ‫ﺒ‬ ‫ﻲ‬ 2019 Page 6 sur 6 ‫اﻟﺷﻛل‬ – 04 0 0,1 1,88 ( ) ( / ) ‫اﻟﺷﻛ‬ ‫ل‬ - 5 - ‫ﻳﻌﻄﻰ‬ : 2 3.57 10 / K Kg s    ، 2 10 / g m s  . 1 . ‫؟‬ ‫ﻫﺬﻫﺎﻟﺤﺮﻛﺔ‬ ‫ﻟﺪﺭﺍﺳﺔ‬ ‫ﺍﻟﻤﻨﺎﺳﺐ‬ ‫ﺍﻟﻤﺮﺟﻊ‬ ‫ﻫﻮ‬ ‫ﻣﺎ‬ - ‫ﺗﺴﻤﺢ‬ ‫ﺍﻟﺘﻲ‬ ‫ﻭ‬ ‫ﺍﻟﻤﺮﺟﻊ‬ ‫ﺑﻬﺬﺍ‬ ‫ﺍﻟﻤﺘﻌﻠﻘﺔ‬ ‫ﺍﻟﻔﺮﺿﻴﺔ‬ ‫ﻫﻲ‬ ‫ﻣﺎ‬ ‫ﻟﻨﻴﻮﺗﻦ؟‬ ‫ﺍﻟﺜﺎﻧﻲ‬ ‫ﺍﻟﻘﺎﻧﻮﻥ‬ ‫ﺑﺘﻄﺒﻴﻖ‬ . 2 . ‫ﺃﻭﺟﺪ‬ ‫ﺍﻟﺒﻴﺎﻥ‬ ‫ﺑﺎﺳﺘﻐﻼﻝ‬ : ‫ﺃ‬ - ‫ﺍﻟﺤﺪ‬ ‫ﺍﻟﺴﺮﻋﺔ‬ ‫ﻗﻴﻤﺔ‬ ‫ﻳﺔ‬ L v . ‫ﺏ‬ - ‫ﺍﻟﺰﻣﻦ‬ ‫ﺛﺎﺑﺖ‬  ‫ﻟﻠﺤﺮﻛﺔ‬ ‫ﺍﻟﻤﻤﻴﺰ‬ ‫ﺝ‬ - ‫ﺍﻻﺑﺘﺪﺍﺋﻲ‬ ‫ﺍﻟﺘﺴﺎﺭﻉ‬ ‫ﻗﻴﻤﺔ‬ 0 a ‫؟‬ ‫ﺗﺴﺘﻨﺘﺞ‬ ‫ﺫﺍ‬ ‫ﻣﺎ‬ ، 3 . ‫ﺃﻧﻬﺎ‬ ‫ﺑﻴﻦ‬ ‫ﻭ‬ ‫ﻟﻠﺤﺮﻛﺔ‬ ‫ﺍﻟﺘﻔﺎﺿﻠﻴﺔ‬ ‫ﺍﻟﻤﻌﺎﺩﻟﺔ‬ ‫ﺃﻭﺟﺪ‬ ‫ﺍﻟﺸﻜﻞ‬ ‫ﻋﻠﻰ‬ ‫ﺗﻜﺘﺐ‬ dv Av B dt   ‫ﺣﻴﺚ‬ ‫ﻭ‬ ‫ﻳﻄ‬ ‫ﺛﻮﺍﺑﺖ‬ ‫ﻋﺒﺎﺭﺗﻴﻬﻤﺎ‬ ‫ﺇﻳﺠﺎﺩ‬ ‫ﻠﺐ‬ . 4 . ‫ﻗﻴﻤﺔ‬ ‫ﺃﺣﺴﺐ‬ ‫ﺍﻟﻜﺮﻳﺔ‬ ‫ﻛﺘﻠﺔ‬ . II - ‫ﺍﻟﺜﺎﻧﻴﺔ‬ ‫ﺍﻟﻄﺮﻳﻘﺔ‬ : ‫ﻣﻬﺘﺰﺓ‬ ‫ﺟﻤﻠﺔ‬ ‫ﺣﺮﻛﺔ‬ ‫ﺩﺭﺍﺳﺔ‬ ) ‫ﻧﺎﺑﺾ‬ – ‫ﻛﺮﻳﺔ‬ ( ‫ﺃﻱ‬ ) ‫ﺃﻓﻘﻲ‬ ‫ﻣﺮﻥ‬ ‫ﻧﻮﺍﺱ‬ :( ‫ﻣﺮﻭﻧﺘﻪ‬ ‫ﺛﺎﺑﺖ‬ ‫ﻣﺘﻼﺻﻘﺔ‬ ‫ﻏﻴﺮ‬ ‫ﺣﻠﻘﺎﺗﻪ‬ ‫ﻣﺮﻥ‬ ‫ﺑﻨﺎﺑﺾ‬ ‫ﺍﻟﺴﺎﺑﻘﺔ‬ ‫ﺍﻟﻜﺮﻳﺔ‬ ‫ﻧﺜﺒﺖ‬ 50 / K N m  ‫ﺍﻟﺸﻜﻞ‬ ‫ﻓﻲ‬ ‫ﻣﻮﺿﺢ‬ ‫ﻫﻮ‬ ‫ﻛﻤﺎ‬ - 4 . ‫ﻋﻦ‬ ‫ﺍﻟﻜﺮﻳﺔ‬ ‫ﻧﺰﻳﺢ‬ ‫ﺑﺎﻟﻤﻘﺪﺍﺭ‬ ‫ﺍﻟﺘﻮﺍﺯﻥ‬ ‫ﻭﺿﻊ‬ (+ ) ‫ﺍﻟﻠﺤﻈﺔ‬ ‫ﻋﻨﺪ‬ ‫ﻧﺘﺮﻛﻬﺎ‬ ‫ﻭ‬ = 0 ‫ﺩﻭﻥ‬ ‫ﺗﺴﺠﻴﻞ‬ ‫ﻋﻠﻰ‬ ‫ﺑﺎﻟﺤﺼﻮﻝ‬ ‫ﻣﻨﺎﺳﺐ‬ ‫ﺗﺠﻬﻴﺰ‬ ‫ﻳﺴﻤﺢ‬ ‫ﺍﺑﺘﺪﺍﺋﻴﺔ‬ ‫ﺳﺮﻋﺔ‬ ‫ﺍﻟﺰﻣﻦ‬ ‫ﺑﺪﻻﻟﺔ‬ ‫ﺍﻟﻜﺮﻳﺔ‬ ‫ﻋﻄﺎﻟﺔ‬ ‫ﻣﺮﻛﺰ‬ ‫ﺳﺮﻋﺔ‬ t ‫ﺍﻟﺒﻴﺎﻥ‬ ‫ﻓﻲ‬ ‫ﻭﺍﻟﻤﻤﺜﻞ‬ ‫ﺍﻟﺸﻜﻞ‬ ) 05 .( 1 - ‫ﺍﻟﻔﺎﺻﻠﺔ‬ ‫ﻋﻨﺪ‬ ‫ﺍﻟﻜﺮﻳﺔ‬ ‫ﻋﻠﻰ‬ ‫ﺍﻟﻤﺆﺛﺮﺓ‬ ‫ﺍﻟﻘﻮﻯ‬ ‫ﻣﺜﻞ‬ ( > 0 ) . 2 - ‫ﻣﺘﺨﺎﻣﺪﺓ‬ ‫ﺍﻟﺠﻤﻠﺔ‬ ‫ﺣﺮﻛﺔ‬ ‫ﻫﻞ‬ ‫ﻋﻠﻞ‬ ‫؟‬ ‫ﻻ‬ ‫ﺃﻡ‬ . 3 - ‫ﺍﻟﺘﻔﺎﺿﻠﻴﺔ‬ ‫ﺍﻟﻤﻌﺎﺩﻟﺔ‬ ‫ﺃﻭﺟﺪ‬ ‫ﻟﻨﻴﻮﺗﻦ‬ ‫ﺍﻟﺜﺎﻧﻲ‬ ‫ﺍﻟﻘﺎﻧﻮﻥ‬ ‫ﺑﺘﻄﺒﻴﻖ‬ ‫ﺍﻟﻔﺎﺻﻠﺔ‬ ‫ﺑﺪﻻﻟﺔ‬ ‫ﻟﻠﺤﺮﻛﺔ‬ . 4 - ‫ﻟﻠﺤﺮﻛﺔ‬ ‫ﺍﻟﻤﻤﻴﺰﺓ‬ ‫ﺍﻟﻤﻘﺎﺩﻳﺮ‬ ‫ﺃﻭﺟﺪ‬ ‫ﺍﻟﺒﻴﺎﻥ‬ ‫ﺑﺎﺳﺘﻐﻼﻝ‬ :  ‫ﻟﻠﺤﺮﻛﺔ‬ ‫ﺍﻟﺬﺍﺗﻲ‬ ‫ﺍﻟﺪﻭﺭ‬ .  ‫ﺍﻟﺤﺮﻛﺔ‬ ‫ﻧﺒﺾ‬ .  ‫ﺍﻻﻫﺘﺰﺍﺯﺍﺕ‬ ‫ﺳﻌﺔ‬ .  ‫ﺍﻻﺑﺘﺪﺍﺋﻴﺔ‬ ‫ﺍﻟﺼﻔﺤﺔ‬ . 5 - ‫ﺍﻟﻜﺮﻳﺔ‬ ‫ﻛﺘﻠﺔ‬ ‫ﺃﺣﺴﺐ‬ ‫ﺳﺎﺑﻘﺎ‬ ‫ﺍﻟﻤﺤﺴﻮﺑﺔ‬ ‫ﺗﻠﻚ‬ ‫ﻣﻊ‬ ‫ﻗﺎﺭﻧﻬﺎ‬ ‫ﺛﻢ‬ . ‫ﻳﻌﻄﻰ‬ : = 10 ‫ﺍ‬ ‫ﻧ‬ ‫ﺘﻬ‬ ‫ﻰ‬ ‫ﺍ‬ ‫ﻟ‬ ‫ﻤ‬ ‫ﻮ‬ ‫ﺿ‬ ‫ﻮ‬ ‫ﻉ‬ ‫ﺍ‬ ‫ﻟ‬ ‫ﺜ‬ ‫ﺎ‬ ‫ﻧ‬ ‫ﻲ‬ * * * * ‫ﺃ‬ ‫ﺳ‬ ‫ﺎ‬ ‫ﺗ‬ ‫ﺬ‬ ‫ﺓ‬ ‫ﺍ‬ ‫ﻟ‬ ‫ﻤ‬ ‫ﺎ‬ ‫ﺩ‬ ‫ﺓ‬ ‫ﻳ‬ ‫ﺘ‬ ‫ﻤ‬ ‫ﻨ‬ ‫ﻮ‬ ‫ﻥ‬ ‫ﻟ‬ ‫ﻜ‬ ‫ﻢ‬ ‫ﺍ‬ ‫ﻟ‬ ‫ﺘ‬ ‫ﻮ‬ ‫ﻓ‬ ‫ﻴ‬ ‫ﻖ‬ ‫ﻭ‬ ‫ﺍ‬ ‫ﻟ‬ ‫ﻨ‬ ‫ﺠ‬ ‫ﺎ‬ ‫ﺡ‬ ‫ﻓ‬ ‫ﻲ‬ ‫ﺑ‬ ‫ﻜ‬ ‫ﺎ‬ ‫ﻟ‬ ‫ﻮ‬ ‫ﺭ‬ ‫ﻳ‬ ‫ﺎ‬ 2019 * * * * e n c y - e d u c a t i o n . c o m / e x a m s 3as.ency-education.com
Page 1 sur 9 ‫ﺍﻟﻨﻤﻮﺫﺟﻲ‬ ‫ﺍﻟﺘﺼﺤﻴﺢ‬ ‫ﺍﻟﻨﻘﻄﺔ‬ ( ) = ∆ . = 53 + (127 − 53) − ( ) . = 1050.02406 ( ) = ∆ . = 53 + (131 − 53) − ( ) . = 1081.11753 ( ) = 1050.02406 127 = 8.2673MeV/Nucl ( ) = 1050.02406 131 = 8.2528MeV/Nucl = . . = 1 × 6,02 × 10 11.54 × 24 × 3600 × 131 = 4,6 × 10 = . . = 37 × 10 × 11,54 × 24 × 3600 × 131 6,02 × 10 = 8.02 × 10 = 11.54 × 8 × 10 × 6.02 × 10 11.54 × 24 × 3600 × 24 × 10 × 131 = 4.99 ‫ﺍﻟﻤﻮﺿ‬ ‫ﻮﻉ‬ ‫ﺍﻻﻭﻝ‬ ‫ﺍﻷﻭﻝ‬ ‫ﺍﻟﺘﻤﺮﻳﻦ‬ ‫ﺣﻞ‬ : 1 . ‫ﺗﻌﺮﻳﻒ‬ : ‫ﺍﻟﻨﻮﻭﻱ‬ ‫ﺍﻟﺘﺮﺍﺑﻂ‬ ‫ﻃﺎﻗﺔ‬ l E ‫ﺳﻜﻮﻥ‬ ‫ﺣﺎﻟﺔ‬ ‫ﻓﻲ‬ ‫ﻧﻮﺍﺓ‬ ‫ﻟﺘﺤﻄﻴﻢ‬ ‫ﺍﻟﺨﺎﺭﺟﻲ‬ ‫ﺍﻟﻮﺳﻂ‬ ‫ﻳﻮﻓﺮﻫﺎ‬ ‫ﺍﻟﺘﻲ‬ ‫ﺍﻟﻄﺎﻗﺔ‬ ‫ﻫﻲ‬ ‫ﺑﺎﻟﻌﻼﻗﺔ‬ ‫ﺗﻌﻄﻰ‬ ‫ﺳﺎﻛﻨﺔ‬ ‫ﻭ‬ ‫ﻣﻨﻔﺼﻠﺔ‬ ‫ﻧﻴﻮﻛﻠﻴﻮﻧﺎﺗﻬﺎ‬ ‫ﺇﻟﻰ‬ : 2 . ‫ﺍﻟﻨﻮﻭﻱ‬ ‫ﺍﻟﺘﺮﺍﺑﻂ‬ ‫ﻃﺎﻗﺔ‬ ‫ﺣﺴﺎﺏ‬ : = ∆ . = + ( − ) − ( ) . 3 . ‫ﺍﻟﺮﺑﻂ‬ ‫ﻃﺎﻗﺔ‬ ‫ﺣﺴﺎﺏ‬ ‫ﻧﻮﻳﺔ‬ ‫ﻟﻜﻞ‬ : ‫ﺃﻥ‬ ‫ﺑﻤﺎ‬ > ‫ﻓ‬ ‫ﺎﻥ‬ ‫ﻣﻦ‬ ‫ﺍﺳﺘﻘﺮﺍﺭﺍ‬ ‫ﺃﻛﺜﺮ‬ ‫ﻫﻮ‬ ‫ﺍﻟﻤﺸﻊ‬ ‫ﺍﻟﻨﻈﻴﺮ‬ ‫ﺃﻱ‬ . 4 . ‫ﺍﻟﺘﻔﻜﻚ‬ ‫ﻣﻌﺎﺩﻟﺔ‬ ‫ﻛﺘﺎﺑﺔ‬ : → + ‫ﺍﻻﻧﺤﻔﺎﻅ‬ ‫ﻗﻮﺍﻧﻴﻦ‬ ‫ﺣﺴﺐ‬ : 131 = + 0 53 = − 1 → = 131 = 54 5 . ‫ﺍﻟﺰﻣﻦ‬ ‫ﺛﺎﺑﺖ‬ ‫ﺣﺴﺎﺏ‬ : = / = 11.54 6 . ‫ﺃ‬ . ‫ﻧﺸﺎﻁ‬ ‫ﺣﺴﺎﺏ‬ 1g ‫ﺍﻟﻴﻮﺩ‬ ‫ﻣﻦ‬ : ‫ﻟﺪﻳﻨﺎ‬ = → = . . ‫ﺏ‬ . ‫ﺍﻟﺠﺮﻋﺔ‬ ‫ﻓﻲ‬ ‫ﺍﻟﻤﺘﻮﺍﺟﺪﺓ‬ ‫ﺍﻟﻴﻮﺩ‬ ‫ﻛﺘﻠﺔ‬ ‫ﺣﺴﺎﺏ‬ : ‫ﺝ‬ . ‫ﺍﻟﻌﻼﻗﺔ‬ ‫ﺍﺛﺒﺎﺕ‬ : ‫ﻟﺪﻳﻨﺎ‬ = / → = / → ln = → = . . = . . . . 7 . ‫ﺃ‬ . ‫ﺍﻟﻴﻮﺩ‬ ‫ﻛﺘﻠﺔ‬ ‫ﺣﺴﺎﺏ‬ ‫ﺍﻟﻤﺘﺒﻘﻲ‬ : ( ) = /  ( = 4 ) = 8 × 10 . = 6 × 10  ( = 8 ) = 8 × 10 . = 4 × 10  ( = 16 ) = 8 × 10 . = 2 × 10 ‫ﺏ‬ . ‫ﻣﺮﻭﺭ‬ ‫ﺑﻌﺪ‬ ‫ﺍﻟﻤﺘﺒﻘﻴﺔ‬ ‫ﺍﻷﻧﻮﻳﺔ‬ ‫ﻋﺪﺩ‬ ‫ﺣﺴﺎﺏ‬ t1 : ‫ﻟﺪﻳﻨﺎ‬ = . = × × . × = 2.75 × 10 ‫ﺝ‬ . ‫ﻟﺘﻔﻜﻚ‬ ‫ﺍﻟﻼﺯﻡ‬ ‫ﺍﻟﺰﻣﻦ‬ ‫ﺣﺴﺎﺏ‬ 99% ‫ﺍﻻﺑﺘﺪﺍﺋﻴﺔ‬ ‫ﺍﻟﻜﺘﻠﺔ‬ ‫ﻣﻦ‬ : ‫ﻟﺪﻳﻨﺎ‬ ( ) = / → 0.01 = / → = − . ln (0.01) = 53.14 0.5 1 0.5 0.5 0.5 01 0.25 0.5 0.5 0.25 0.25 e n c y - e d u c a t i o n . c o m / e x a m s 3as.ency-education.com
Page 2 sur 9 → = → = 1 2 → − = 1 2 → = . 2 = . 2 = 1 2 = 1 2 1 − / → 2 = 1 2 1 − = 7,5 × 10 ‫ﺍﻟﺜﺎﻧﻲ‬ ‫ﺍﻟﺘﻤﺮﻳﻦ‬ ‫ﺣﻞ‬ : 1 . ‫ﺍﻟﺪﺍﺭﺓ‬ ‫ﺗﻤﺜﻴﻞ‬ : 2 . ‫ﺣ‬ ‫ﻗﻴﻤﺔ‬ ‫ﺴﺎﺏ‬ E : ‫ﺍﻟﺘﻮﺗﺮﺍﺕ‬ ‫ﺟﻤﻊ‬ ‫ﻗﺎﻧﻮﻥ‬ ‫ﺣﺴﺐ‬ = + ‫ﺍﻟﻠﺤﻈﺔ‬ ‫ﻋﻨﺪ‬ t=0 : = (0) + (0) ‫ﺍﻟﺒﻴﺎﻥ‬ ‫ﻣﻦ‬ : (0) + (0) = 10 → = 10 3 . ‫ﺃﻥ‬ ‫ﺍﺛﺒﺎﺕ‬ : + u = E ‫ﺍﻟﺘﻮﺗﺮﺍﺕ‬ ‫ﺟﻤﻊ‬ ‫ﻗﺎﻧﻮﻥ‬ ‫ﺣﺴﺐ‬ + = → + = → + u = E 4 . ‫ﻋﺒﺎﺭﺓ‬ ‫ﺍﻳﺠﺎﺩ‬ α : ‫ﻟﺪﻳﻨﺎ‬ : = (1 − ) → = αEe ‫ﻧﺠﺪ‬ ‫ﺍﻟﺘﻔﺎﺿﻠﻴﺔ‬ ‫ﺍﻟﻤﻌﺎﺩﻟﺔ‬ ‫ﻓﻲ‬ ‫ﺑﺎﻟﺘﻌﻮﻳﺾ‬ : . αEe + − e = → e − 1 = 0 5 . ‫ﻋﺒﺎﺭﺓ‬ : ‫ﻟﺪﻳﻨﺎ‬ = ‫ﻭ‬ ‫ﻟﺪﻳﻨﺎ‬ = = 1 − / → = ‫ﻭﻣﻨﻪ‬ : = = . . 6 . ‫ﺃ‬ . ‫ﺍﻟﻤﻨﺤﻨﻰ‬ ‫ﺗﺤﺪﻳﺪ‬ :  ‫ﺍﻟﻤﻨﺤﻨﻰ‬ ) 1 ( ‫ﻳﻤﺜﻞ‬ ‫ﻷﻥ‬ : (0) = 0  ‫ﺍﻟﻤﻨﺤﻨﻰ‬ ) 2 ( ‫ﻳﻤﺜﻞ‬ ‫ﻷﻥ‬ : (0) = ‫ﺏ‬ . ‫ﺍﺛﺒ‬ ‫ﺍﻥ‬ ‫ﺎﺕ‬ = . 2 ‫ﺃﻱ‬ ‫ﺍﻟﻤﻨﺤﻨﻴﻴﻦ‬ ‫ﺗﻘﺎﻃﻊ‬ ‫ﻋﻨﺪ‬ : = → . = (1 − ) → 2 . = ‫ﺝ‬ . ‫ﻗﻴﻤﺔ‬ ‫ﺍﺳﺘﻨﺘﺎﺝ‬ τ : ‫ﻟﺪﻳﻨﺎ‬ ‫ﺍﻟﺒﻴﺎﻥ‬ ‫ﻣﻦ‬ : = 1,5 ‫ﻭ‬ ‫ﻣﻨﻪ‬ : = . 2 → = = , = 2,16 ‫ﺩ‬ . ‫ﻗﻴﻤﺔ‬ L ‫ﺍﻟﻮﺷﻴﻌﺔ‬ ‫ﺫﺍﺗﻴﺔ‬ : = → = . = 2,16 × 10 × 200 = 0,4 7 . ‫ﻋﺒﺎﺭﺓ‬ : ‫ﻟﺪﻳﻨﺎ‬ : + . = ‫ﺍﻟﺪﺍﺋﻢ‬ ‫ﺍﻟﻨﻈﺎﻡ‬ ‫ﻓﻲ‬ = ‫ﻭ‬ = 0 ‫ﻭﻣﻨﻪ‬ . = E → = ‫ﻗﻴﻤﺘﻪ‬ : = = 0,05 8 . ‫ﻭﺷﻴﻌﺔ‬ ‫ﻓﻲ‬ ‫ﺍﻟﻤﺨﺰﻧﺔ‬ ‫ﺍﻟﻄﺎﻗﺔ‬ ‫ﻋﺒﺎﺭﺓ‬ : 0.5 0.5 0.5 0.5 0.5 0.5 01 0.5 0.5 01 01 E L R K ‫اﻟﺸﻜﻞ‬ 1 e n c y - e d u c a t i o n . c o m / e x a m s 3as.ency-education.com
Page 3 sur 9 − = n(I ) ⟹ = − n(I ) = 3.5 × 10 × 20 × 10 − 0.25 × 10 − = 0 ⟹ = = 0.45 × 10 0.1 = 0.045 ⁄ ‫ﺍﻟﺘﺠﺮﻳﺒﻲ‬ ‫ﺍﻟﺘﻤﺮﻳﻦ‬ ‫ﺣﻞ‬ : I . ‫ﺍﻷﻭﻝ‬ ‫ﺍﻟﻔﻮﺝ‬ 1 ( ‫ﺩﻟﺔ‬ ‫ﻣﻌﺎ‬ ‫ﺍﻟﺘﻔﺎﻋﻞ‬ : ‫ﻟﻸﻛﺴﺪﺓ‬ ‫ﺍﻟﻨﺼﻔﻴﺔ‬ ‫ﺩﻟﺔ‬ ‫ﺍﻟﻤﻌﺎ‬ : O = O + 2 + 2e ‫ﻟﻺﺭﺟﺎﻉ‬ ‫ﺍﻟﻨﺼﻔﻴﺔ‬ ‫ﺩﻟﺔ‬ ‫ﺍﻟﻤﻌﺎ‬ : I + 2e = 2 ‫ﺍﻟﺘﻔﺎﻋﻞ‬ ‫ﺩﻟﺔ‬ ‫ﻣﻌﺎ‬ O + I = O + 2 + 2 2 ( ‫ﺍﻟﺘﻔﺎﻋﻞ‬ ‫ﺗﻘﺪﻡ‬ ‫ﺟﺪﻭﻝ‬ : O + I = O + 2 + 2 ∕∕ 0 0 ∕∕ 2 − − ∕∕ 2 − − 3 ( ‫ﺍﻟﺸﺮﻭﻁ‬ : ‫ﻣﺤﻠﻮﻝ‬ ‫ﻳﻜﻮﻥ‬ ‫ﺃﻥ‬ ‫ﻳﺠﺐ‬ ‫ﺍﺫﺍ‬ ‫ِﻳﺮ‬ ‫ﻣﻌﺎ‬ ‫ﻣﺤﻠﻮﻝ‬ ‫ﺛﻴﻮﻛﺒﺮﻳﺘﺎﺕ‬ : - ‫ﻣﻌﻠﻮﻡ‬ ‫ﺗﺮﻛﻴﺰﻩ‬ . - ‫ﻣﻊ‬ ‫ﺗﻔﺎﻋﻠﻪ‬ I ‫ﺗﺎﻡ‬ ‫ﻭ‬ ‫ﺳﺮﻳﻊ‬ . 4 ( ‫ﺍﻟﻤﻌﺎﻳﺮﺓ‬ ‫ﺗﻔﺎﻋﻞ‬ ‫ﺩﻟﺔ‬ ‫ﻣﻌﺎ‬ : ‫ﻟﻸﻛﺴﺪﺓ‬ ‫ﺍﻟﻨﺼﻔﻴﺔ‬ ‫ﺩﻟﺔ‬ ‫ﺍﻟﻤﻌﺎ‬ : 2 O ⁻² = O ²⁻ + 2e ‫ﺍ‬ ‫ﻟﻺﺭﺟﺎﻉ‬ ‫ﺍﻟﻨﺼﻔﻴﺔ‬ ‫ﺩﻟﺔ‬ ‫ﻟﻤﻌﺎ‬ : I + 2e = 2 ‫ﺍﻟﺘﻔﺎﻋﻞ‬ ‫ﺩﻟﺔ‬ ‫ﻣﻌﺎ‬ 2 O ⁻² + I = O ⁻² + 2 5 ( ‫ﺍﻟﺘﻜﺎﻓﺆ‬ ‫ﻋﻨﺪ‬ : )= n( )= . × × × = 0.25× mol n( 6 ( ‫ﺣﺴﺎﺏ‬ = . × mol ‫ﺃﻥ‬ ‫ﺑﻤﺎ‬ : n (I ) ≠ 0 ‫ﻓﺈﻥ‬ : n ( O ) = 0 ‫ﻣﻨﻪ‬ ‫ﻭ‬ ‫ﻭﻣﻨﻪ‬ : 176 × 0.045 = . ⁄ = = × II . ‫ﺍﻟﺜﺎﻧﻲ‬ ‫ﺍﻟﻔﻮﺝ‬ : 1 ( ‫ﺍﻟﻤﻌﺎﻳﺮﺓ‬ ‫ﺗﻔﺎﻋﻞ‬ ‫ﺩﻟﺔ‬ ‫ﻣﻌﺎ‬ O + = O ⁻ + ₂ 2 ( = 3 ( ( = 9 ∕ P =8.2) ‫ﻟﻤﺎ‬ =4.5ml = ‫ﻓﺎﻥ‬ : P =P ‫ﻧﺠﺪ‬ ‫ﺍﻟﺒﻴﺎﻥ‬ ‫ﻋﻠﻰ‬ ‫ﺑﺎﻻﺳﻘﺎﻁ‬ : = . 4 ( ‫ﻳﻜﻮﻥ‬ ‫ﺍﻟﺘﻜﺎﻓﺆ‬ ‫ﻋﻨﺪ‬ : = ⟹ = = × × = 2.25 × 10 ⁄ ℱ = ⟹ = ℱ. =2× 2.25 × 10 = 4.5 × 10 ⁄ ‫ﻭﻣﻨﻪ‬ : 7.92 ⁄ = = × ‫ﺍﻷﻭﻝ‬ ‫ﺍﻟﻔﻮﺝ‬ ‫ﻣﻊ‬ ‫ﻋﻠﻴﻪ‬ ‫ﺍﻟﻤﺘﺤﺼﻞ‬ ‫ﺍﻟﺘﺮﻛﻴﺰ‬ ‫ﻧﻔﺲ‬ ‫ﻭﻫﻮ‬ 5 ( ‫ﺍﻷﺳﻜﻮﺭﺑﻴﻚ‬ ‫ﺣﻤﺾ‬ ‫ﺃﻥ‬ ‫ﺍﺛﺒﺎﺕ‬ AH ‫ﺿﻌﻴﻒ‬ : ‫ﺍﻟﻤﺎﺀ‬ ‫ﻓﻲ‬ ‫ﺍﻹﻧﺤﻼﻝ‬ ‫ﺩﻟﺔ‬ ‫ﻣﻌﺎ‬ , O + ₂ = O ⁻ + ₃ ⁺ ) P =3 ( ‫ﺍﻷﻭﻟﻰ‬ ‫ﺍﻟﻄﺮﻳﻘﺔ‬ = = [ ₃ ⁺] = = . × = 0.02 < 1 ‫ﺍﻟﺜﺎﻧﻴﺔ‬ ‫ﺍﻟﻄﺮﻳﻘﺔ‬ 0.5 0.25 0.5 0.5 0.5 0.5 0.5 0.5 0.5 0.5 0.5 e n c y - e d u c a t i o n . c o m / e x a m s 3as.ency-education.com
Page 4 sur 9 ‫ﺍ‬ ‫ﻟ‬ ‫ﻤ‬ ‫ﻮ‬ ‫ﺿ‬ ‫ﻮ‬ ‫ﻉ‬ ‫ﺍ‬ ‫ﻟ‬ ‫ﺜ‬ ‫ﺎ‬ ‫ﻧ‬ ‫ﻲ‬ [ O ] = − [ ₃ ⁺] = 4.5 × 10 − 10 = 4.4 × 10 ≠ 0 ⁄ 6 ( ……………….. ‫ﺇﻋﺘﺒ‬ ‫ﻳﻤﻜﻦ‬ ‫ﺗﺎﻡ‬ ‫ﺍﻟﺘﻔﺎﻋﻞ‬ ‫ﺎﺭ‬ 7 ( ( O / O⁻ ) < ( O / O⁻ ) ‫ﺍﻷﻗﻮﻯ‬ ‫ﻫﻮ‬ ‫ﺍﻷﺳﻜﻮﺭﺑﻴﻚ‬ ‫ﺣﻤﺾ‬ ‫ﻭﻣﻨﻪ‬ ‫ﻫﻮ‬ ‫ﺍﻟﻤﻨﺎﺳﺐ‬ ‫ﺍﻟﻜﺎﺷﻒ‬ : ‫ﻓﺘﺎﻟﻴﻦ‬ ‫ﺍﻟﻔﻴﻨﻮﻝ‬ 01 0.5 0.25 e n c y - e d u c a t i o n . c o m / e x a m s 3as.ency-education.com
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Studyadz 3as-physique-c3-49-64

  • 1. Page 1 sur 6 ‫ﺍﻟﺸﻌﺒﻴﺔ‬ ‫ﺍﻟﺪﻳﻤﻘﺮﺍﻃﻴﺔ‬ ‫ﺍﻟﺠﺰﺍﺋﺮﻳﺔ‬ ‫ﺍﻟﺠﻤﻬﻮﺭﻳﺔ‬ ‫ﻣﻌﻨﺼﺮ‬ ‫ﺛﺎﻧﻮﻳﺔ‬ ‫ﺍﻟﺒﻮﺍﻗﻲ‬ ‫ﺍﻡ‬ ‫ﻟﻮﻻﻳﺔ‬ ‫ﺍﻟﺘﺮﺑﻴﺔ‬ ‫ﻣﺪﻳﺮﻳﺔ‬ ‫ﻛﺮﺷﺔ‬ ‫ﻋﻴﻦ‬ ‫ﺃﻭﻧﻴﺲ‬ ‫ﺍﻟﺜﺎﻧﻮﻱ‬ ‫ﺍﻟﺘﻌﻠﻴﻢ‬ ‫ﻟﺒﻜﺎﻟﻮﺭﻳﺎ‬ ‫ﺗﺠﺮﻳﺒﻲ‬ ‫ﺍﻣﺘﺤﺎﻥ‬ . ‫ﻣﺎﻱ‬ 2019 ‫ﺍﻟﺸﻌﺒﺔ‬ : ‫ﺗﺠﺮﻳﺒﻴﺔ‬ ‫ﻋﻠﻮﻡ‬ ‫ﻣﺎﺩﺓ‬ ‫ﻓﻲ‬ ‫ﺍﺧﺘﺒﺎﺭ‬ : ‫ﺍﻟﻤﺪﺓ‬ ‫ﺍﻟﻔﻴﺰﻳﺎﺋﻴﺔ‬ ‫ﺍﻟﻌﻠﻮﻡ‬ : 03 ‫ﻭ‬ ‫ﺳﺎ‬ 30 ‫ﺩ‬ ‫ﺍﻵﺗﻴﻴﻦ‬ ‫ﺍﻟﻤﻮﺿﻮﻋﻴﻦ‬ ‫ﺃﺣﺪ‬ ‫ﻳﺨﺘﺎﺭ‬ ‫ﺃﻥ‬ ‫ﺍﻟﻤﺘﺮﺷﺢ‬ ‫ﻋﻠﻰ‬ : ‫ﺍﻷﻭﻝ‬ ‫ﺍﻟﻤﻮﺿﻮﻉ‬ ‫ﺍﻟﻤ‬ ‫ﻳﺤﺘﻮﻱ‬ ‫ﻋﻠﻰ‬ ‫ﺍﻷﻭﻝ‬ ‫ﻮﺿﻮﻉ‬ 03 ‫ﺻﻔﺤﺎﺕ‬ ) ‫ﺻﻔﺤﺔ‬ ‫ﻣﻦ‬ 1 ‫ﻣﻦ‬ 6 ‫ﺍﻟﺼﻔﺤﺔ‬ ‫ﺇﻟﻰ‬ 3 ‫ﻣﻦ‬ 6 ( ‫ﺍﻷﻭﻝ‬ ‫ﺍﻟﺠﺰﺀ‬ : ) 13 ‫ﻧﻘﻄﺔ‬ ( ‫ﺍﻷﻭﻝ‬ ‫ﺍﻟﺘﻤﺮﻳﻦ‬ : ) 06 ‫ﻧﻘﺎﻁ‬ ( ‫ﺃﺳﺎﺳﻲ‬ ‫ﺍﻟﻬﺮﻣﻮﻥ‬ ‫ﻭﻫﺬﺍ‬ ‫ﺍﻟﺪﻡ‬ ‫ﻣﻦ‬ ‫ﺍﻻﺧﻴﺮﺓ‬ ‫ﻫﺬﻩ‬ ‫ﺗﺴﺘﻨﺰﻓﻪ‬ ‫ﺣﻴﺚ‬ ‫ﺍﻟﺪﺭﻗﻴﺔ‬ ‫ﺍﻟﻐﺪﺓ‬ ‫ﻫﺮﻣﻮﻧﺎﺕ‬ ‫ﻟﺘﺮﻛﻴﺐ‬ ‫ﺍﻷﺳﺎﺳﻴﺔ‬ ‫ﺍﻟﻤﺎﺩﺓ‬ ‫ﺍﻟﻴﻮﺩ‬ ‫ﻳﻌﺘﺒﺮ‬ ‫ﻟﻠﻨﻤﻮ‬ . ‫ﺍﻟﺘﺮﻛﻴﺐ‬ ‫ﻟﻬﺬﺍ‬ ‫ﺍﻟﻼﺯﻣﺔ‬ ‫ﺍﻟﻤﺎﺩﺓ‬ ‫ﻛﻤﻴﺎﺕ‬ ‫ﻗﻠﻴﻠﺔ‬ . ‫ﻛﻞ‬ ‫ﻳﺤﺘﻮﻱ‬ ،‫ﻋﺎﺩﻱ‬ ‫ﻟﺸﺨﺺ‬ ‫ﺑﺎﻟﻨﺴﺒﺔ‬ 100mL ‫ﻋﻠﻰ‬ ‫ﺍﻟﺪﻡ‬ ‫ﻣﻦ‬ 10 ‫ﻣﻦ‬ ‫ﻫﺮﻣﻮﻧﺎﺕ‬ ‫ﺷﻜﻞ‬ ‫ﻋﻠﻰ‬ ‫ﺗﻔﺮﺯﻩ‬ ‫ﻭﺍﻟﺬﻱ‬ ‫ﻳﻮﺩﻭﺭ‬ ‫ﺷﻜﻞ‬ ‫ﻋﻠﻰ‬ ‫ﺍﻟﺪﻡ‬ ‫ﻣﻦ‬ ‫ﺍﻟﻜﻤﻴﺔ‬ ‫ﻧﻔﺲ‬ ‫ﺍﻟﺪﺭﻗﻴﺔ‬ ‫ﺍﻟﻐﺪﺓ‬ ‫ﺗﺄﺧﺬ‬ ‫ﻟﺤﻈﺔ‬ ‫ﻛﻞ‬ ‫ﻓﻲ‬ ،‫ﺍﻟﻴﻮﺩ‬ . ‫ﺛﺎﺑﺘﺔ‬ ‫ﺗﺒﻘﻰ‬ ‫ﺍﻟﺪﻡ‬ ‫ﻓﻲ‬ ‫ﺍﻟﻤﺘﻮﺍﺟﺪﺓ‬ ‫ﺍﻟﻴﻮﺩ‬ ‫ﻛﻤﻴﺎﺕ‬ . ‫ﺍﺳﺘﻘﻼﺏ‬ ‫ﺗﺘﺒﻊ‬ ‫ﺍﻟﻤﺸﻊ‬ ‫ﺍﻟﻴﻮﺩ‬ ‫ﺑﺈﺳﺘﻌﻤﺎﻝ‬ (métabolisme) ‫ﺍﻟﻴﻮ‬ ‫ﻭﻗﻴﺎﺱ‬ ‫ﺩ‬ ‫ﺍﻟﻤﺮﻛﺒﺔ‬ ‫ﺍﻟﻬﺮﻣﻮﻧﺎﺕ‬ . ‫ﻧﺠﺪ‬ ‫ﺍﻟﻴﻮﺩ‬ ‫ﻧﻈﺎﺋﺮ‬ ‫ﺑﻴﻦ‬ ‫ﻣﻦ‬ ‫ﻭ‬ ‫ﺍﻟﻨﺸﺎﻁ‬ ‫ﺇﺷﻌﺎﻋﻲ‬ ‫ﻭﺍﻵﺧﺮ‬ ‫ﻣﺴﺘﻘﺮ‬ ‫ﺃﺣﺪﻫﻤﺎ‬ β . 1 . ‫ﺍﻟﻨﻮﻭﻱ‬ ‫ﺍﻟﺘﺮﺍﺑﻂ‬ ‫ﻃﺎﻗﺔ‬ ‫ﻋﺮﻑ‬ . 2 . ‫ﺍﻟﻜﺘﺮﻭﻥ‬ ‫ﺑﺎﻟﻤﻴﻐﺎ‬ ‫ﻟﻠﻨﻮﺍﺓ‬ ‫ﺍﻟﺮﺑﻂ‬ ‫ﻃﺎﻗﺔ‬ ‫ﺃﺣﺴﺐ‬ - ‫ﺍﻟﻨﻈﻴﺮﻳﻦ‬ ‫ﻣﻦ‬ ‫ﻟﻜﻞ‬ ‫ﻓﻮﻟﻂ‬ . 3 . ‫ﺑﺎ‬ ‫ﺍﻟﺮﺑﻂ‬ ‫ﻃﺎﻗﺔ‬ ‫ﺃﺣﺴﺐ‬ ‫ﻟﻜﻞ‬ ‫ﻧﻮﻳﺔ‬ ‫ﺑﺎﻟﻨﺴﺒﺔ‬ ‫ﺍﻟﻨﻈﻴﺮ‬ ‫ﻣﻦ‬ ‫ﻟﻜﻞ‬ ‫ﺍﻟﻤﺸﻊ‬ ‫ﻭﺍﻟﻨﻈﻴﺮ‬ ‫ﺍﻟﻤﺴﺘﻘﺮ‬ ‫ﺍﻟﻨﻈﻴﺮ‬ ‫ﻭﺍﺳﺘﻨﺘﺞ‬ ، ‫ﻳﻦ‬ . 4 . ‫ﺍﻟﻤﺸﻊ‬ ‫ﻟﻠﻨﻈﻴﺮ‬ ‫ﺍﻟﺘﻔﻜﻚ‬ ‫ﻣﻌﺎﺩﻟﺔ‬ ‫ﺍﻛﺘﺐ‬ . ‫ﻋﻨﻪ‬ ‫ﻳﻨﺘﺞ‬ ‫ﺍﻟﺬﻱ‬ Xe ‫ﻣﺤﺪﺩﺍ‬ A ‫ﻭ‬ Z . ‫ﺍﻟﻌﻤﺮ‬ ‫ﻧﺼﻒ‬ ‫ﺯﻣﻦ‬ 131 ‫ﻫﻮ‬ ‫ﺍﻟﻤﺸﻊ‬ ‫ﻟﻠﻴﻮﺩ‬ / = 8 : 5 . ‫ﺍﺣﺴﺐ‬ ‫ﺯﻣﻦ‬ ‫ﺍﻟﻌﻤﺮ‬ ‫ﻧﺼﻒ‬ ‫ﻟﻨﻮﺍ‬ ‫ﺓ‬ ‫ﺛﻢ‬ ‫ﺍﻟﻤﺸﻌﺔ‬ ‫ﺍﻟﻴﻮﺩ‬ ‫ﺍﻟﺰﻣﻦ‬ ‫ﺛﺎﺑﺖ‬ ‫ﺍﺳﺘﻨﺘﺞ‬ τ . 6 . ‫ﺍﻟﻴﻮﺩ‬ ‫ﺍﺳﺘﻘﻼﺏ‬ ‫ﻟﺘﺘﺒﻊ‬ ‫ﺑﺎﻹﻳﻤﺎﺽ‬ ‫ﺍﻟﻔﺤﺺ‬ ‫ﺗﻘﻨﻴﺔ‬ ‫ﺗﻌﺘﻤﺪ‬ ‫ﺍﻟﻤﺮﻛﺒﺔ‬ ‫ﺍﻟﻬﺮﻣﻮﻧﺎﺕ‬ ‫ﻭﻗﻴﺎﺱ‬ (scintigraphie) ‫ﻟﻠﻐﺪﺓ‬ ‫ﺍﻟﻜﻈﺮﻳﺔ‬ (glandes surrénales) ‫ﻟﻠﻴﻮﺩ‬ ‫ﺑﻤﺤﻠﻮﻝ‬ ‫ﺍﻟﻤﺮﻳﺾ‬ ‫ﺣﻘﻦ‬ ‫ﻋﻠﻰ‬ ‫ﻧﺸﺎﻃﻪ‬ ‫ﺍﻹﺷﻌﺎﻋﻲ‬ 37MBq . ‫ﺃ‬ . ‫ﻧﺸﺎﻁ‬ ‫ﺍﺣﺴﺐ‬ 1g ‫ﺍﻟﻴﻮﺩ‬ ‫ﻣﻦ‬ . ‫ﺏ‬ . ‫ﺍﻟﻴﻮﺩ‬ ‫ﻣﺎﻛﺘﻠﺔ‬ ‫؟‬ ‫ﺣﻘﻨﻬﺎ‬ ‫ﺗﻢ‬ ‫ﺍﻟﺘﻲ‬ ‫ﺝ‬ . ‫ﺍﻋ‬ ‫ﺍﻟﻨﺸﺎﻁ‬ ‫ﻗﻴﺎﺱ‬ ‫ﻄﻰ‬ ‫ﺍﻹﺷﻌﺎﻋﻲ‬ ‫ﺍﻟﻠﺤﻈﺔ‬ ‫ﻋﻨﺪ‬ ‫ﺍﻟﻤﺪﺭﻭﺳﺔ‬ ‫ﻟﻠﻌﻴﻨﺔ‬ : = 24 . ‫ﺃﻥ‬ ‫ﺑﻴﻦ‬ t = τ ln . . . ‫ﺑـ‬ ‫ﻗﻴﻤﺘﻬﺎ‬ ‫ﺍﺣﺴﺐ‬ ‫ﺛﻢ‬ (jrs) . 7 . ‫ﺑﺠﺮﻋﺔ‬ ‫ﻣﺮﻳﺾ‬ ‫ﺣﻘﻦ‬ 1,00 ‫ﺍﻟﻴﻮﺩ‬ ‫ﻣﻦ‬ 131 ‫ﺍﻟﻤﺸﻊ‬ : ‫ﺃ‬ . ‫ﺍﻟﻴﻮﺩ‬ ‫ﻛﺘﻠﺔ‬ ‫ﺍﺣﺴﺐ‬ 131 ‫ﻓﻲ‬ ‫ﺍﻟﻤﺘﺒﻘﻲ‬ ‫ﻣﺮﻭﺭ‬ ‫ﺑﻌﺪ‬ ‫ﺍﻟﻤﺮﻳﺾ‬ ‫ﺟﺴﻢ‬ 4 ‫ﺑﻌﺪ‬ ‫ﺛﻢ‬ ، 8 ‫ﺑﻌﺪ‬ ‫ﺛﻢ‬، 16 . ‫ﺏ‬ . ‫ﺍﻟﻴﻮﺩ‬ ‫ﺃﻧﻮﻳﺔ‬ ‫ﻋﺪﺩ‬ ‫ﺍﺣﺴﺐ‬ 131 ‫ﻣﺮﻭﺭ‬ ‫ﺑﻌﺪ‬ ‫ﺍﻟﻤﺘﺒﻘﻴﺔ‬ 4 . ‫ﺝ‬ . ‫ﻟﺘﻔﻜﻚ‬ ‫ﺍﻟﻮﻗﺖ‬ ‫ﻣﻦ‬ ‫ﻳﻠﺰﻡ‬ ‫ﻛﻢ‬ 99% ‫ﺍﻟﻴﻮﺩ‬ ‫ﻣﻦ‬ 131 . ‫ﺍﻟﻤﻌﻄﻴﺎﺕ‬ : ‫ﺍﻟﺪﻗﻴﻘﺔ‬ ‫ﺃﻭ‬ ‫ﺍﻟﻨﻮﺍﺓ‬ 1 Xe U ‫ﺑـ‬ ‫ﻛﺘﻠﺘﻬﺎ‬ 130.9061 126.9044 1.00877 5.4858 × 10 1.00722 130.9050 ( ) = 131 . , = 6,02 × 10 , 1 = 1,6 × 10 , 1 = 931,5 e n c y - e d u c a t i o n . c o m / e x a m s 3as.ency-education.com
  • 2. ‫ﺍ‬ ‫ﺧ‬ ‫ﺘ‬ ‫ﺒ‬ ‫ﺎ‬ ‫ﺭ‬ ‫ﻓ‬ ‫ﻲ‬ ‫ﻣ‬ ‫ﺎ‬ ‫ﺩ‬ ‫ﺓ‬ ‫ﺍ‬ ‫ﻟ‬ ‫ﻌ‬ ‫ﻠ‬ ‫ﻮ‬ ‫ﻡ‬ ‫ﺍ‬ ‫ﻟ‬ ‫ﻔ‬ ‫ﻴ‬ ‫ﺰ‬ ‫ﻳ‬ ‫ﺎ‬ ‫ﺋ‬ ‫ﻴ‬ ‫ﺔ‬ / ‫ﺍ‬ ‫ﻟ‬ ‫ﺸ‬ ‫ﻌ‬ ‫ﺒ‬ ‫ﺔ‬ ، ‫ﻋ‬ ‫ﻠ‬ ‫ﻮ‬ ‫ﻡ‬ ‫ﺗ‬ ‫ﺠ‬ ‫ﺮ‬ ‫ﻳ‬ ‫ﺒ‬ ‫ﻴ‬ ‫ﺔ‬ / ‫ﺑ‬ ‫ﻜ‬ ‫ﺎ‬ ‫ﻟ‬ ‫ﻮ‬ ‫ﺭ‬ ‫ﻳ‬ ‫ﺎ‬ ‫ﺗ‬ ‫ﺠ‬ ‫ﺮ‬ ‫ﻳ‬ ‫ﺒ‬ ‫ﻲ‬ 2019 Page 2 sur 6 E L R K ‫اﻟﺸﻜﻞ‬ 1 2 ( ) 2 ( ) ‫اﻟﺸﻜﻞ‬ ) 2 ( 1 2 ‫ﺍﻟﺜﺎﻧﻲ‬ ‫ﺍﻟﺘﻤﺮﻳﻦ‬ : ) 07 ‫ﻧﻘﺎﻁ‬ ( ‫ﺍﻟﻤﻮﺿ‬ ‫ﺍﻟﺘﺮﻛﻴﺐ‬ ‫ﺑﺎﺳﺘﻌﻤﺎﻝ‬ ‫ﺍﻟﺘﺎﻟﻴﺔ‬ ‫ﺍﻟﺘﺠﺮﺑﺔ‬ ‫ﻧﻨﺠﺰ‬ ‫ﺍﻟﺸﻜﻞ‬ ‫ﻓﻲ‬ ‫ﺢ‬ ) 1 ( ‫ﻭﺍﻟﻤﻜﻮﻥ‬ ‫ﻣﻦ‬ : ‫ﻣﺜﺎﻟﻴﺔ‬ ‫ﻭﺷﻴﻌﺔ‬ (B) ‫ﺫﺍﺗﻴﺘﻬﺎ‬ L ‫ﺃﻭﻣﻲ‬ ‫ﻭﻧﺎﻗﻞ‬ (R) ‫ﻣﻘﺎﻭﻣﺘﻪ‬ = 200Ω ‫ﻭ‬ ، ‫ﺍﻟﻜﻬﺮﺑﺎﺋﻴﺔ‬ ‫ﻗﻮﺗﻪ‬ ‫ﺛﺎﺑﺖ‬ ‫ﻟﺘﻮﺗﺮ‬ ‫ﻣﻮﻟﺪ‬ E . ‫ﻭﻗﺎﻃﻌﺔ‬ K . ‫ﻋﻨﺪ‬ ‫ﺍﻟﻘﺎﻃﻌﺔ‬ ‫ﻧﻐﻠﻖ‬ ‫ﻟﻸﺯﻣﻨﺔ‬ ‫ﻣﺒﺪﺃ‬ ‫ﻧﻌﺘﺒﺮﻫﺎ‬ ‫ﻟﺤﻈﺔ‬ . 1 . ‫ﻭﻛﻴﻔﻴﺔ‬ ‫ﻓﻴﻬﺎ‬ ‫ﺍﻟﻤﺎﺭ‬ ‫ﻭﺍﻟﺘﻴﺎﺭ‬ ‫ﺍﻟﺘﻮﺗﺮﺍﺕ‬ ‫ﺟﻬﺔ‬ ‫ﻋﻠﻴﻬﺎ‬ ‫ﻣﻮﺿﺤﺎ‬ ‫ﺍﻟﺪﺍﺭﺓ‬ ‫ﺭﺳﻢ‬ ‫ﺍﻋﺪ‬ ‫ﺭﺍﺳﻢ‬ ‫ﺗﻮﺻﻴﻞ‬ ‫ﺍﻟﺘﻮﺗﺮﻳﻦ‬ ‫ﻟﻤﺸﺎﻫﺪﺓ‬ ‫ﺍﻟﻤﻬﺒﻄﺔ‬ ‫ﺍﻻﻫﺘﺰﺍﺯ‬ ‫ﻭ‬ . 2 . ‫ﻗﻴﻤﺔ‬ ‫ﺃﻥ‬ ‫ﺑﻴﻦ‬ ‫ﺍﻟﻤﻨﺤﻨﻰ‬ ‫ﻋﻠﻰ‬ ‫ﻭﺍﻋﺘﻤﺎﺩﺍ‬ ، ‫ﺍﻟﺘﻮﺗﺮﺍﺕ‬ ‫ﺟﻤﻊ‬ ‫ﻗﺎﻧﻮﻥ‬ ‫ﺑﺘﻄﻴﻖ‬ ‫ﻫﻲ‬ 10 . 3 . ‫ﻳ‬ ‫ﺍﻟﺘﻲ‬ ‫ﺍﻟﺘﻘﺎﺿﻠﻴﺔ‬ ‫ﺍﻟﻤﻌﺎﺩﻟﺔ‬ ‫ﺍﻥ‬ ‫ﺑﻴﻦ‬ ‫ﺤ‬ ‫ﺍﻟﺘﻮﺗﺮ‬ ‫ﻘﻘﻬﺎ‬ ‫ﺍﻷﻭﻣﻲ‬ ‫ﺍﻟﻨﺎﻗﻞ‬ ‫ﻃﺮﻓﻲ‬ ‫ﺑﻴﻦ‬ ‫ﺍﻟﺸﻜﻞ‬ ‫ﻋﻠﻰ‬ ‫ﺗﻜﺘﺐ‬ : + − = 0 4 . ‫ﺍ‬ ‫ﺍﻟﻤﻌﺎﺩﻟﺔ‬ ‫ﺣﻞ‬ ‫ﺍﻥ‬ ‫ﻋﻠﻤﺎ‬ ‫ﺍﻟﺸﻜﻞ‬ ‫ﻋﻠﻰ‬ ‫ﻳﻜﺘﺐ‬ ‫ﺍﻟﺴﺎﺑﻘﺔ‬ ‫ﻟﺘﻔﺎﺿﻠﻴﺔ‬ : = (1 − ) ‫ﻋﺒﺎﺭﺓ‬ ‫ﺣﺪﺩ‬ ، α ‫ﺛﻮﺍﺑﺖ‬ ‫ﺑﺪﻻﻟﺔ‬ ‫ﺍﻟﺪﺍﺭﺓ‬ . 5 . ‫ﺍﻟﻮﺷﻌﺔ‬ ‫ﻃﺮﻓﻲ‬ ‫ﺑﻴﻦ‬ ‫ﺍﻟﺘﻮﺗﺮ‬ ‫ﻋﺒﺎﺭﺓ‬ ‫ﺍﺳﺘﻨﺘﺞ‬ . 6 . ‫ﺍﻟﺸﻜﻞ‬ ‫ﻳﻤﺜﻞ‬ ) 2 ( ‫ﺍﻟﻮﺷﻴﻌﺔ‬ ‫ﻃﺮﻓﻲ‬ ‫ﺑﻴﻦ‬ ‫ﺍﻟﺘﻮﺗﺮ‬ ‫ﺗﻐﻴﺮﺍﺕ‬ ‫ﺍﻟﻤﻘﺎﺑﻞ‬ ‫ﺍﻷﻭﻣﻲ‬ ‫ﺍﻟﻨﺎﻗﻞ‬ ‫ﻭ‬ ‫ﺑﺪﻻﻟﺔ‬ ‫ﺍﻟﺰﻣﻦ‬ t . ‫ﺃ‬ . ‫ﻣﻦ‬ ‫ﻟﻜﻞ‬ ‫ﺍﻟﻤﻮﺍﻓﻖ‬ ‫ﺍﻟﻤﻨﺤﻨﻰ‬ ‫ﺣﺪﺩ‬ ‫ﻭ‬ ‫ﺍﻟﺘﻌﻠﻴﻞ‬ ‫ﻣﻊ‬ . ‫ﺏ‬ . ‫ﺑﻴﻦ‬ ‫ﺃﻥ‬ ‫ﺍﻟﺘﻮﺗﺮﻳﻦ‬ ‫ﺗﻘﺎﻃﻊ‬ ‫ﻟﺤﻈﺔ‬ ( ) ‫ﻭ‬ ( ) ‫ﺗﻜﺘﺐ‬ ‫ﺍﻟﺸﻜﻞ‬ ‫ﻋﻠﻰ‬ : = . ln (2) . ‫ﺝ‬ . ‫ﻗﻴﻤﺔ‬ ‫ﺍﺳﺘﻨﺘﺞ‬ ‫ﻟﻠﺪﺍﺭﺓ‬ ‫ﺍﻟﺰﻣﻦ‬ ‫ﺛﺎﺑﺖ‬ . ‫ﺩ‬ . ‫ﺍﻟﺬﺍﺗﻴﺔ‬ ‫ﻗﻴﻤﺔ‬ ‫ﺍﺳﺘﻨﺘﺞ‬ L ‫ﻟﻠﻮﺷﻴﻌﺔ‬ . 7 . ‫ﻋﺒﺎﺭﺓ‬ ‫ﺃﻭﺟﺪ‬ ‫ﻓﻲ‬ ‫ﺍﻟﻤﺎﺭ‬ ‫ﺍﻟﺘﻴﺎﺭ‬ ‫ﺷﺪﺓ‬ ، ‫ﺍﻟﺪﺍﺋﻢ‬ ‫ﺍﻟﻨﻈﺎﻡ‬ ‫ﻓﻲ‬ ‫ﺍﻟﺪﺍﺭﺓ‬ ‫ﻗﻴﻤﺘﻪ‬ ‫ﺍﺣﺴﺐ‬ ‫ﺛﻢ‬ . 8 . ‫ﺍﻟﺸﻜﻞ‬ ‫ﻋﻠﻰ‬ ‫ﺗﻜﺘﺐ‬ ‫ﺍﻟﺰﻣﻦ‬ ‫ﺑﺪﻻﻟﺔ‬ ‫ﺍﻟﻮﺷﻴﻌﺔ‬ ‫ﻓﻲ‬ ‫ﺍﻟﻤﺨﺰﻧﺔ‬ ‫ﺍﻟﻄﺎﻗﺔ‬ ‫ﻋﺒﺎﺭﺓ‬ ‫ﺃﻥ‬ ‫ﺑﻴﻦ‬ : = 1 − ‫ﺛﻢ‬ ، ‫ﺍﻟﻠﺤﻈﺔ‬ ‫ﻋﻨﺪ‬ ‫ﻗﻴﻤﺘﻬﺎ‬ ‫ﺃﺣﺴﺐ‬ = . ‫ﺍﻟﺜﺎﻧﻲ‬ ‫ﺍﻟﺠﺰﺀ‬ : ) 07 ‫ﻧﻘﺎﻁ‬ ( ‫ﺍﻟﺘﺠﺮﻳﺒﻲ‬ ‫ﺍﻟﺘﻤﺮﻳﻦ‬ : ‫ﻓﻲ‬ ‫ﺣﺼﺔ‬ ‫ﺍﻷﻋﻤﺎﻝ‬ ‫ﺍﻟﺘﻄﺒﻴ‬ ‫ﻘﻴﺔ‬ ‫ﺃﺭﺍﺩ‬ ‫ﻓﻮﺟﺎﻥ‬ ‫ﻣﻦ‬ ‫ﺍﻟﺘﻼﻣﻴﺬ‬ ‫ﺗﺤﺪﻳﺪ‬ ‫ﺍﻟﺘﺮﻛﻴﺰ‬ ‫ﺍﻟﻜﺘﻠﻲ‬   m C ‫ﻟﻤﺤﻠﻮﻝ‬ ‫ﺣﻤﺾ‬ ‫ﺍﻷﺳﻜﻮﺭﺑﻴﻚ‬   6 8 6 C H O ‫ﺑﻄﺮﻳﻘﺘﻴﻦ‬ . ‫ﻳﻤﻠﻚ‬ ‫ﺣﻤﺾ‬ ‫ﺍﻷﺳﻜﻮﺭﺑﻴﻚ‬ ‫ﺧﺎﺻﻴﺔ‬ ‫ﺣﻤﻀﻴﺔ‬ ‫ﻭﺧﺎﺻﻴﺔ‬ ‫ﻣﺮﺟﻌﺔ‬ . ‫ﺍﻟﺜﻨﺎﺋﻴﺎﺕ‬ ) ‫ﻣﺮ‬ / ‫ﻣﺆ‬ : (       2 2 6 6 6 6 8 6 2 4 6 2 3 C H O / C H O , I / I , S O / S O    . ‫ﺍﻟﺜﻨﺎﺋﻴﺎﺕ‬ ) ‫ﺃﺳﺎﺱ‬ / ‫ﺣﻤﺾ‬ : (     6 8 6 6 7 6 2 C H O / C H O , H O / HO   . ‫ﺍﻷﻭﻝ‬ ‫ﺍﻟﻔﻮﺝ‬ : ‫ﻗﺎﻡ‬ ‫ﺍﻟﺘﻼﻣﻴﺬ‬ ‫ﺑﺄﻛﺴﺪﺓ‬ ‫ﺣﻤﺾ‬ ‫ﺍﻷﺳﻜﻮﺭﺑﻴﻚ‬ ، ‫ﻭﺫﻟﻚ‬ ‫ﺑﺈﺿﺎﻓﺔ‬ ‫ﺔ‬ ‫ﻴ‬‫ﻛﻤ‬ ‫ﺯﺍﺋﺪﺓ‬ ‫ﻣﻦ‬ ‫ﻣﺤﻠﻮﻝ‬ ‫ﺛﻨﺎﺋﻲ‬ ‫ﺍﻟﻴﻮﺩ‬ 2 I ‫ﺇﻟﻰ‬ ‫ﺑﻴﺸﺮ‬ ‫ﻳﺤﺘﻮﻱ‬ ‫ﻋﻠﻰ‬ ‫ﺣﺠﻢ‬ 1 10 V mL  ‫ﻣﻦ‬ ‫ﺣﻤﺾ‬ ‫ﺍﻷﺳﻜﻮﺭﺑﻴﻚ‬ . ‫ﺣﺠﻢ‬ ‫ﺛﻨﺎﺋﻲ‬ ‫ﺍﻟﻴﻮﺩ‬ ‫ﺍﻟﻤﻀﺎﻑ‬ ‫ﻫﻮ‬ 2 20 V mL  ‫ﻭﺗﺮﻛﻴﺰﻩ‬ ‫ﺍﻟﻤﻮﻟﻲ‬ 2 2 3 5 10 C , mol / L    . e n c y - e d u c a t i o n . c o m / e x a m s 3as.ency-education.com
  • 3. ‫ﺍ‬ ‫ﺧ‬ ‫ﺘ‬ ‫ﺒ‬ ‫ﺎ‬ ‫ﺭ‬ ‫ﻓ‬ ‫ﻲ‬ ‫ﻣ‬ ‫ﺎ‬ ‫ﺩ‬ ‫ﺓ‬ ‫ﺍ‬ ‫ﻟ‬ ‫ﻌ‬ ‫ﻠ‬ ‫ﻮ‬ ‫ﻡ‬ ‫ﺍ‬ ‫ﻟ‬ ‫ﻔ‬ ‫ﻴ‬ ‫ﺰ‬ ‫ﻳ‬ ‫ﺎ‬ ‫ﺋ‬ ‫ﻴ‬ ‫ﺔ‬ / ‫ﺍ‬ ‫ﻟ‬ ‫ﺸ‬ ‫ﻌ‬ ‫ﺒ‬ ‫ﺔ‬ ، ‫ﻋ‬ ‫ﻠ‬ ‫ﻮ‬ ‫ﻡ‬ ‫ﺗ‬ ‫ﺠ‬ ‫ﺮ‬ ‫ﻳ‬ ‫ﺒ‬ ‫ﻴ‬ ‫ﺔ‬ / ‫ﺑ‬ ‫ﻜ‬ ‫ﺎ‬ ‫ﻟ‬ ‫ﻮ‬ ‫ﺭ‬ ‫ﻳ‬ ‫ﺎ‬ ‫ﺗ‬ ‫ﺠ‬ ‫ﺮ‬ ‫ﻳ‬ ‫ﺒ‬ ‫ﻲ‬ 2019 Page 3 sur 6 2 2 ( ) ‫ﻭﻓﻲ‬ ‫ﻧﻬﺎﻳﺔ‬ ‫ﺍﻟﺘﻔﺎﻋﻞ‬ ‫ﻗﺎﻡ‬ ‫ﺍﻟﺘﻼﻣﻴﺬ‬ ‫ﺑﻤﻌﺎﻳﺮﺓ‬ ‫ﺛﻨﺎﺋﻲ‬ ‫ﺍﻟﻴﻮﺩ‬ ‫ﻓﻲ‬ ‫ﺍﻟﺒﻴﺸﺮ‬ ‫ﺑﻮﺍﺳﻄﺔ‬ ‫ﻣﺤﻠﻮﻝ‬ ‫ﻣﺎﺋﻲ‬ ‫ﻟﺜﻴﻮﻛﺒﺮﻳﺘﺎﺕ‬ ‫ﺍﻟﺼﻮﺩﻳﻮﻡ‬   2 2 3 2Na ,S O   ‫ﺍﻟﻤﻮﻟﻲ‬ ‫ﺗﺮﻛﻴﺰﻩ‬ 2 3 2 5 10 C , mol / L    ، ‫ﻓﺎﺣﺘﺎﺟﻮﺍ‬ ‫ﺇﻟﻰ‬ ‫ﺣﺠﻢ‬ ‫ﻣﻨﻪ‬ 20 E V mL  ‫ﻻﺳﺘﻬﻼﻙ‬ ‫ﺁﻝ‬ ‫ﺛﻨﺎﺋﻲ‬ ‫ﺍﻟﻴﻮﺩ‬ ‫ﺍﻟﻤﻮﺟﻮﺩ‬ ‫ﻓﻲ‬ ‫ﺍﻟﺒﻴﺸﺮ‬ . 1 - ‫ﺍﻛﺘﺐ‬ ‫ﻣﻌﺎﺩﻟﺔ‬ ‫ﺍﻟﺘﻔﺎﻋﻞ‬ ‫ﺑﻴﻦ‬ ‫ﺣﻤﺾ‬ ‫ﺍﻷﺳﻜﻮﺭﺑﻴﻚ‬ ‫ﻭﺛﻨﺎﺋﻲ‬ ‫ﺍﻟﻴﻮﺩ‬ . 2 - ‫ﺃﻧﺸﺊ‬ ‫ﺟﺪﻭﻝ‬ ‫ﺍﻟﺘﻘﺪﻡ‬ ‫ﻟﻬﺬﺍ‬ ‫ﺍﻟﺘﻔﺎﻋﻞ‬ . 3 - ‫ﺍﺫﻛﺮ‬ ‫ﺍﻟﺸﺮﻭﻁ‬ ‫ﺍﻟﺘﻲ‬ ‫ﺗﺘﻮﻓﺮ‬ ‫ﻓﻲ‬ ‫ﻣﺤﻠﻮﻝ‬ ‫ﺛﻴﻮﻛﺒﺮﻳﺘﺎﺕ‬ ‫ﺍﻟﺼﻮﺩﻳﻮﻡ‬ ‫ﻻﺳﺘﻌﻤﺎﻟﻪ‬ ‫ﻓﻲ‬ ‫ﻫﺬﻩ‬ ‫ﺍﻟﻤﻌﺎﻳﺮﺓ‬ . 4 - ‫ﺍﻛﺘﺐ‬ ‫ﻣﻌﺎﺩﻟﺔ‬ ‫ﺗﻔﺎﻋﻞ‬ ‫ﻣﻌﺎﻳﺮﺓ‬ ‫ﺛﻨﺎﺋﻲ‬ ‫ﺍﻟﻴﻮﺩ‬ ‫ﺑﺜﻴﻮﻛﺒﺮﻳﺘﺎﺕ‬ ‫ﺍﻟﺼﻮﺩﻳﻮﻡ‬ . 5 - ‫ﺍﺣﺴﺐ‬ ‫ﻛﻤﻴﺔ‬ ‫ﻣﺎﺩﺓ‬ ‫ﺛﻨﺎﺋﻲ‬ ‫ﺍﻟﻴﻮﺩ‬ ‫ﻏﻴﺮ‬ ‫ﺍﻟﻤﺘﻔﺎﻋﻞ‬ ‫ﻣﻊ‬ ‫ﺣﻤﺾ‬ ‫ﺍﻷﺳﻜﻮﺭﺑﻴﻚ‬ . 6 - ‫ﺍﺣﺴﺐ‬ ‫ﺍﻟﺘﺮﻛﻴﺰ‬ ‫ﺍﻟﻜﺘﻠﻲ‬   m C ‫ﻟﺤﻤﺾ‬ ‫ﺍﻷﺳﻜﻮﺭﺑﻴﻚ‬ . ‫ﻗﺎﺭﻥ‬ ‫ﻧﺘﻴﺠﺘﻲ‬ ‫ﺍﻟﻔﻮﺟﻴﻦ‬ .     2 5 2 5 4 9 12 1 16 a pK C H COOH / C H COO , , C ,H O g / mol      ‫ﺍﻟﺜﺎﻧﻲ‬ ‫ﺍﻟﻔﻮﺝ‬ : ‫ﻗﺎﻡ‬ ‫ﺑﺎﻟﻤﻌﺎﻳﺮﺓ‬ ‫ﺍﻟـ‬ pH ‫ﻣﺘﺮﻳﺔ‬ ‫ﻟﺤﻤﺾ‬ ‫ﺍﻷﺳﻜﻮﺭﺑﻴﻚ‬ ، ‫ﺣﻴﺚ‬ ‫ﺃﺧﺬ‬ ‫ﺍﻟﺘﻼﻣﻴﺬ‬ ‫ﻓﻲ‬ ‫ﺑﻴﺸﺮ‬ ‫ﺣﺠﻤﺎ‬ 0 V ‫ﻣﻦ‬ ‫ﺍﻟﺤﻤﺾ‬ ‫ﻭﺃﺿﺎﻓﻮﺍ‬ ‫ﻟﻪ‬ ‫ﻧﻔﺲ‬ ‫ﺍﻟﺤﺠﻢ‬ ‫ﻣﻦ‬ ‫ﺍﻟﻤﺎﺀ‬ ،‫ﺍﻟﻤﻘﻄﺮ‬ ‫ﺛﻢ‬ ‫ﺃﺧﺬﻭﺍ‬ ‫ﻣﻦ‬ ‫ﺍﻟﻤﺤﻠﻮﻝ‬ ‫ﺍﻟﺠﺪﻳﺪ‬ ‫ﺣﺠﻤﺎ‬ 20 a V mL  ، ‫ﻭﻣﻠﺌﻮﺍ‬ ‫ﺳﺤﺎﺣﺔ‬ ‫ﺟﺔ‬ ‫ﺭ‬‫ﻣﺪ‬ ‫ﺑﻤﺤﻠﻮﻝ‬ ‫ﻣﺎﺋﻲ‬ ‫ﻟﻬﻴﺪﺭﻭﻛﺴﻴﺪ‬ ‫ﺍﻟﺒﻮﺗﺎﺳﻴﻮﻡ‬   K ,OH   ‫ﺗﺮﻛﻴﺰﻩ‬ ‫ﺍﻟﻤﻮﻟﻲ‬ 2 5 10 B C mol / L    ، ‫ﻭﺑﻌﺪ‬ ‫ﺍﻟﺤﺼﻮﻝ‬ ‫ﻋﻠﻰ‬ ‫ﺍﻟﻘﻴﺎﺳﺎﺕ‬ ‫ﻗﺎﻣﻮﺍ‬ ‫ﺑﺘﻤﺜﻴﻞ‬ ‫ﺍﻟﺒﻴﺎﻥ‬ B pH f (V )  . 1 - ‫ﺍﻛﺘﺐ‬ ‫ﻣﻌﺎﺩﻟﺔ‬ ‫ﺗﻔﺎﻋﻞ‬ ‫ﺍﻟﻤﻌﺎﻳﺮﺓ‬ . 2 - ‫ﻑ‬ ‫ﺮ‬‫ﻋ‬ ‫ﺍﻟﺘﻜﺎﻓﺆ‬ ‫ﺣﻤﺾ‬ – ‫ﺃﺳﺎﺱ‬ ، ‫ﺛﻢ‬ ‫ﺩ‬ ‫ﺪ‬‫ﺣ‬ ‫ﺇﺣﺪﺍﺛﻲ‬ ‫ﻧﻘﻄﺔ‬ ‫ﺍﻟﺘﻜﺎﻓﺆ‬ ‫ﺣﻤﺾ‬ – ‫ﺃﺳﺎﺱ‬ . 3 - ‫ﻋﻴﻦ‬ a pK ‫ﺍﻟﺜﻨﺎﺋﻴﺔ‬   6 8 6 6 7 6 C H O / C H O  . 4 - ‫ﺍﺣﺴﺐ‬ ‫ﺍﻟﺘﺮﻛﻴﺰ‬ ‫ﺍﻟﻜﺘﻠﻲ‬   m C ‫ﻟﺤﻤﺾ‬ ‫ﺍﻷﺳﻜﻮﺭﺑﻴﻚ‬ . 5 - ‫ﻦ‬ ‫ﻴ‬‫ﺑ‬ ‫ﺑﻄﺮﻳﻘﺘﻴﻦ‬ ‫ﺃﻥ‬ ‫ﺣﻤﺾ‬ ‫ﺍﻷﺳﻜﻮﺭﺑﻴﻚ‬ ‫ﺿﻌﻴﻒ‬ ‫ﺍﻟﻤﺎﺀ‬ ‫ﻓﻲ‬ . 6 - ‫ﺍﺣﺴﺐ‬ ‫ﺍﻟﺘﺮﻛﻴﺰ‬ ‫ﺍﻟﻤﻮﻟﻲ‬ ‫ﻟﺤﻤﺾ‬ ‫ﺍﻷﺳﻜﻮﺭﺑﻴﻚ‬ ‫ﻓﻲ‬ ‫ﺍﻟﺒﻴﺸﺮ‬ ‫ﻋﻨﺪ‬ ‫ﺍﻟﺘﻜﺎﻓﺆ‬ ، ‫ﺛﻢ‬ ‫ﺍﺳﺘﻨﺘﺞ‬ ‫ﺃﻧﻪ‬ ‫ﻳﻤﻜﻦ‬ ‫ﺗﻔﺎﻋﻞ‬ ‫ﺍﻋﺘﺒﺎﺭ‬ ‫ﺍﻟﻤﻌﺎﻳﺮﺓ‬ ‫ﺗﺎﻣﺎ‬ . 7 - ‫ﻗﺎﺭﻥ‬ ‫ﻗﻮﺓ‬ ‫ﺣﻤﺾ‬ ‫ﺍﻷﺳﻜﻮﺭﺑﻴﻚ‬ ‫ﻣﻊ‬ ‫ﺣﻤﺾ‬ ‫ﺍﻟﺒﺮﻭﺑﺎﻧﻮﻳﻚ‬   2 5 C H COOH . 8 - ‫ﻓﻲ‬ ‫ﺣﺎﻟﺔ‬ ‫ﺍﺳﺘﻌﻤﺎﻝ‬ ‫ﻛﺎﺷﻒ‬ ‫ﻥ‬ ‫ﻮ‬‫ﻣﻠ‬ ‫ﻟﺘﺤﺪﻳﺪ‬ ‫ﻧﻘﻄﺔ‬ ‫ﺍﻟﺘﻜﺎﻓﺆ‬ ، ‫ﻣﺎ‬ ‫ﻫﻮ‬ ‫ﺍﻟﻜﺎﺷﻒ‬ ‫ﺍﻷﻧﺴﺐ‬ ‫ﻣﻦ‬ ‫ﺑﻴﻦ‬ ‫ﺍﻟﻜﻮﺍﺷﻒ‬ ‫ﺍﻟﺘﺎﻟﻴﺔ‬ ‫ﻟﻬﺬﻩ‬ ‫ﺍﻟﻤﻌﺎﻳﺮﺓ‬ ‫؟‬ ‫ﺍﻟﻬﻠﻴﺎﻧﺘﻴﻦ‬ : ‫ﻣﺠﺎﻝ‬ ‫ﺗﻐﻴﺮ‬ ‫ﺍﻟﻠﻮﻥ‬   3 1 4 4 , ,  . ‫ﺍﻟﻔﻴﻨ‬ ‫ﻮﻝ‬ ‫ﻓﺘﺎﻟﻴﻦ‬ : ‫ﻣﺠﺎﻝ‬ ‫ﺗﻐﻴﺮ‬ ‫ﺍﻟﻠﻮﻥ‬   8 2 10 ,  . ‫ﺃﺯﺭﻕ‬ ‫ﺍﻟﺒﺮﻭﻣﻮﺗﻴﻤﻮﻝ‬ : ‫ﻣﺠﺎﻝ‬ ‫ﺗﻐﻴﺮ‬ ‫ﺍﻟﻠﻮﻥ‬   6 7 6 ,  . ‫ﺍ‬ ‫ﻧ‬ ‫ﺘ‬ ‫ﻬ‬ ‫ﻰ‬ ‫ﺍ‬ ‫ﻟ‬ ‫ﻤ‬ ‫ﻮ‬ ‫ﺿ‬ ‫ﻮ‬ ‫ﻉ‬ ‫ﺍ‬ ‫ﻷ‬ ‫ﻭﻝ‬ e n c y - e d u c a t i o n . c o m / e x a m s 3as.ency-education.com
  • 4. ‫ﺍ‬ ‫ﺧ‬ ‫ﺘ‬ ‫ﺒ‬ ‫ﺎ‬ ‫ﺭ‬ ‫ﻓ‬ ‫ﻲ‬ ‫ﻣ‬ ‫ﺎ‬ ‫ﺩ‬ ‫ﺓ‬ ‫ﺍ‬ ‫ﻟ‬ ‫ﻌ‬ ‫ﻠ‬ ‫ﻮ‬ ‫ﻡ‬ ‫ﺍ‬ ‫ﻟ‬ ‫ﻔ‬ ‫ﻴ‬ ‫ﺰ‬ ‫ﻳ‬ ‫ﺎ‬ ‫ﺋ‬ ‫ﻴ‬ ‫ﺔ‬ / ‫ﺍ‬ ‫ﻟ‬ ‫ﺸ‬ ‫ﻌ‬ ‫ﺒ‬ ‫ﺔ‬ ، ‫ﻋ‬ ‫ﻠ‬ ‫ﻮ‬ ‫ﻡ‬ ‫ﺗ‬ ‫ﺠ‬ ‫ﺮ‬ ‫ﻳ‬ ‫ﺒ‬ ‫ﻴ‬ ‫ﺔ‬ / ‫ﺑ‬ ‫ﻜ‬ ‫ﺎ‬ ‫ﻟ‬ ‫ﻮ‬ ‫ﺭ‬ ‫ﻳ‬ ‫ﺎ‬ ‫ﺗ‬ ‫ﺠ‬ ‫ﺮ‬ ‫ﻳ‬ ‫ﺒ‬ ‫ﻲ‬ 2019 Page 4 sur 6 ‫ﺍﻟﺜﺎﻧﻲ‬ ‫ﺍﻟﻤﻮﺿﻮﻉ‬ ‫ﺍﻟ‬ ‫ﺍﻟﻤﻮﺿﻮﻉ‬ ‫ﻳﺤﺘﻮﻱ‬ ‫ﺜﺎﻧﻲ‬ ‫ﻋﻠﻰ‬ 3 ‫ﺻﻔﺤﺎﺕ‬ ) ‫ﺻﻔﺤﺔ‬ ‫ﻣﻦ‬ 4 ‫ﻣﻦ‬ 6 ‫ﺍﻟﺼﻔﺤﺔ‬ ‫ﺇﻟﻰ‬ 6 ‫ﻣﻦ‬ 6 ( ‫ﺍﻷﻭﻝ‬ ‫ﺍﻟﺠﺰﺀ‬ ) 13 ‫ﻥ‬ ( ‫ﺍﻷﻭﻝ‬ ‫ﺍﻟﺘﻤﺮﻳﻦ‬ : ) 06 ‫ﻧﻘﺎﻁ‬ ( ‫ﻧﻌ‬ ‫ﺘﺒﺮ‬ ‫ﺍﻟﺪﺍﺭﺓ‬ ‫ﺍﻟﻜﻬﺮﺑﺎﺋﻴﺔ‬ ‫ﺍﻟﻤﻤﺜﻠﺔ‬ ‫ﻓﻲ‬ ‫ﺍﻟﺸﻜﻞ‬ (1) ، ‫ﺍﻟﻤﻜﻮﻧﺔ‬ ‫ﻣﻦ‬ :  ‫ﻣﻮﻟﺪ‬ ‫ﺗﻮﺗﺮ‬ ‫ﻗﻮﺗﻪ‬ ‫ﺍﻟﻤﺤﺮﻛﺔ‬ ‫ﺍﻟﻜﻬﺮﺑﺎﺋﻴﺔ‬ .  ‫ﻧﺎﻗﻼﻥ‬ ‫ﺃﻭﻣﻴﺎﻥ‬ ‫ﻣﻘﺎﻭﻣﺘﻴﻬﻤﺎ‬ 1 =75 Ω ‫ﻭ‬ 2 ‫ﻣﺠﻬﻮﻟﺔ‬ .  ‫ﻣﻜﺜﻔﺔ‬ ‫ﺳﻌﺘﻬﺎ‬ ، ‫ﻏﻴﺮ‬ ‫ﻣﺸﺤﻮﻧﺔ‬ .  ‫ﺑﺎﺩﻟﺔ‬ 1 . ‫ﻋﻨﺪ‬ ‫ﺍﻟﻠﺤﻈﺔ‬ =0 ، ‫ﻧﻀﻊ‬ ‫ﺍﻟﺒﺎﺩﻟﺔ‬ ‫ﺍﻟﻮﺿﻊ‬ ‫ﻋﻠﻰ‬ 1 . ‫ﺃﻋﺪ‬ ‫ﺍﻟﻜﻬﺮﺑﺎ‬ ‫ﺍﻟﺪﺍﺭﺓ‬ ‫ﻧﻘﻞ‬ ‫ﺋﻴﺔ‬ ‫ﺟﻬﺔ‬ ‫ﻭ‬ ،‫ﺍﻟﺘﻴﺎﺭ‬ ‫ﺟﻬﺔ‬ ‫ﻋﻠﻴﻬﺎ‬ ‫ﻭﻣﺜﻞ‬ ‫ﺍﻟﺘﻮﺗﺮﺍﺕ‬ . ‫ﺃ‬ - ‫ﺃﻭﺟﺪ‬ ‫ﺍﻟﻤﻌﺎﺩﻟﺔ‬ ‫ﺍﻟﺘﻔﺎﺿﻠﻴﺔ‬ ‫ﺍﻟﺘﻲ‬ ‫ﺷﺪﺓ‬ ‫ﺗﺤﻘﻘﻬﺎ‬ ‫ﺍﻟﻜﻬﺮﺑﺎﺋﻲ‬ ‫ﺍﻟﺘﻴﺎﺭ‬ . ‫ﻋﻦ‬ ‫ﺍﻟﻤﻌﺒﺮﺓ‬ ‫ﺗﻠﻚ‬ ‫ﻣﻨﻬﺎ‬ ‫ﻭﺍﺳﺘﻨﺘﺞ‬ ‫ﻃﺮﻓﻲ‬ ‫ﺑﻴﻦ‬ ‫ﺍﻻﻭﻣﻲ‬ ‫ﺍﻟﻨﺎﻗﻞ‬ .R2 ‫ﺏ‬ - ‫ﺍﻟﺘﻔﺎﺿﻠﻴﺔ‬ ‫ﺍﻟﻤﻌﺎﺩﻟﺔ‬ ‫ﺣﻞ‬ ‫ﺍﻟﺸﻜﻞ‬ ‫ﻣﻦ‬ ‫ﻛﺘﺎﺑﺘﻬﺎ‬ ‫ﻳﻤﻜﻦ‬ = . ‫ﻋﺒ‬ ‫ﻋﻦ‬ ‫ﺮ‬ ‫ﻭ‬ ‫ﻋﻨﺎﺻﺮ‬ ‫ﻣﻤﻴﺰﺍﺕ‬ ‫ﺑﺪﻻﻟﺔ‬ ‫ﺍﻟﺪﺍﺭﺓ‬ . ‫ﺝ‬ - ‫ﺍﻟﻜﻬﺮﺑﺎﺋ‬ ‫ﺍﻟﺘﻮﺗﺮ‬ ‫ﻋﺒﺎﺭﺓ‬ ‫ﺍﺳﺘﻨﺘﺞ‬ ‫ﺍﻟﻤﻜﺜﻔﺔ‬ ‫ﻃﺮﻓﻲ‬ ‫ﺑﻴﻦ‬ ‫ﻲ‬ . ( ) 2 . ‫ﺫﺍﻛﺮﺓ‬ ‫ﺫﻭ‬ ‫ﺍﻟﻤﻬﺒﻄﻲ‬ ‫ﺍﻻﻫﺘﺰﺍﺯ‬ ‫ﺭﺍﺳﻢ‬ ‫ﻳﺴﻤﺢ‬ ‫ﺍﻟﺴﺎﺑﻘﻴﻦ‬ ‫ﺍﻟﺘﻮﺗﺮﻳﻦ‬ ‫ﺑﻤﻌﺎﻳﻨﺔ‬ ‫ﻭ‬ ). ‫ﺍﻟﺸﻜﻞ‬ - 2 - ( ‫ﺃ‬ - ‫ﺍﻟﺠﻬﺎﺯ‬ ‫ﺭﺑﻂ‬ ‫ﻛﻴﻔﻴﺔ‬ ‫ﻛﻴﻔﻲ‬ ‫ﺑﺮﺳﻢ‬ ‫ﻭﺿﺢ‬ ‫ﻟﻤﻌﺎﻳﻨﺔ‬ ‫ﺍﻟﻤﺪﺧﻞ‬ ‫ﻋﻞ‬ y1 ‫ﻭ‬ ‫ﻋﻠﻰ‬ y1 ‫ﺏ‬ - ‫ﺍﻟﻤﻨﺎﺳﺐ‬ ‫ﺍﻟﺘﻮﺗﺮ‬ ‫ﻣﺪﺧﻞ‬ ‫ﻟﻜﻞ‬ ‫ﺃﻧﺴﺐ‬ . ‫ﺝ‬ - ‫ﻣﻦ‬ ‫ﻛﻞ‬ ‫ﻗﻴﻢ‬ ‫ﺣﺪﺩ‬ ‫ﺍﻟﺸﻜﻞ‬ ‫ﻋﻠﻰ‬ ‫ﺍﻋﺘﻤﺎﺩﺍ‬ : E ، R2 ‫ﻭ‬ C . 3 . ‫ﺍﻟﺒﺎﺩﻟﺔ‬ ‫ﻧﻨﻘﻞ‬ ‫ﻣﺸﺤﻮﻧﺔ‬ ‫ﺍﻟﻤﻜﺜﻔﺔ‬ ‫ﺗﺼﺒﺢ‬ ‫ﻋﻨﺪﻣﺎ‬ ‫ﺍﻟﻮﺿﻊ‬ ‫ﺇﻟﻰ‬ ) 2 ( ‫ﻣﺒﺪﺍ‬ ‫ﻧﻌﺘﺒﺮﻫﺎ‬ ‫ﻟﺤﻈﺔ‬ ‫ﻓﻲ‬ ‫ﺍﻟﻌﺒﺎﺭﺓ‬ ‫ﺗﺼﺒﺢ‬ ، ‫ﻟﻠﺰﻣﻦ‬ ‫ﺟﺪﻳﺪ‬ ‫ﺍﻟﻠﺤﻈﻴﺔ‬ : = − . ‫ﺃ‬ - ‫ﺗ‬ ‫ﻛﻴﻒ‬ ‫ﺍﻟﺘﻮﺗﺮ‬ ‫ﺇﺷﺎﺭﺓ‬ ‫ﻔﺴﺮ‬ . ‫ﺏ‬ - ‫ﻛﻞ‬ ‫ﺗﻮﺟﻴﻪ‬ ‫ﺍﻟﺸﻜﻞ‬ ‫ﻋﻠﻰ‬ ‫ﻭﺿﺢ‬ ‫ﺍﻟﺤﺎﻟﺔ‬ ‫ﻫﺬﻩ‬ ‫ﻓﻲ‬ ‫ﺍﻟﻜﻬﺮﺑﺎﺋﻲ‬ ‫ﻭﺍﻟﺘﻮﺗﺮ‬ ‫ﺍﻟﺘﻴﺎﺭ‬ ‫ﺷﺪﺓ‬ ‫ﻣﻦ‬ . ‫ﺝ‬ - ‫ﺍﻟﻠﺤﻈﺔ‬ ‫ﺣﺪﺩ‬ t1 ‫ﺍ‬ ‫ﺍﻟﻄﺎﻗﺔ‬ ‫ﻓﻴﻬﺎ‬ ‫ﺗﺼﺒﺢ‬ ‫ﻟﺘﻲ‬ ‫ﺍﻟﻨﺎﻗﻞ‬ ‫ﻓﻲ‬ ‫ﺟﻮﻝ‬ ‫ﺑﻔﻌﻞ‬ ‫ﺍﻟﻤﺤﻮﻟﺔ‬ ‫ﺍﻻﻭﻣﻲ‬ R2 ‫ﻫﻲ‬ : We =0.32J ‫ﺍﻟﺜﺎﻧﻲ‬ ‫ﺍﻟﺘﻤﺮﻳﻦ‬ ) 7 ‫ﻧﻘﺎﻁ‬ :( ‫ﺍﻟﻤﻴﺜﺎﻧﻮﻳﻚ‬ ‫ﺣﻤﺾ‬ ‫ﻳﻌﺘﺒﺮ‬ ) ‫ﺣﻤﺾ‬ ‫ﺍﻟﻨﻤﻞ‬ ( ‫ﻟﻠﻨﻤﻞ‬ ‫ﺍﻟﺪﻓﺎﻉ‬ ‫ﻭﺳﺎﺋﻞ‬ ‫ﻣﻦ‬ . ‫ﻣﺤﻠﻮﻟﻬﺎﻟﻤـﺎﺋﻲ‬ ‫ﺧﻮﺍﺹ‬ ‫ﺑﻌﺾ‬ ‫ﺩﺭﺍﺳﺔ‬ ‫ﻧﺮﻳﺪ‬ . I - ‫ﺣﺠﻤﺎ‬ ‫ﻧﻀﻊ‬ = 2 ‫ﺍﻟﻤﻮﻟﻲ‬ ‫ﺍﻟﺘﺮﻛﻴﺰ‬ ‫ﺫﻱ‬ ‫ﺍﻟﻨﻤﻞ‬ ‫ﺣﻤﺾ‬ ‫ﻣﻦ‬ ‫ﺳﻌﺔ‬ ‫ﺫﺍﺕ‬ ‫ﻋﻴﺎﺭﻳﺔ‬ ‫ﺣﻮﺟﻠـﺔ‬ ‫ﻓـــﻲ‬ = 100 ‫ﺍﻟﺤﺠﻢ‬ ‫ﺛﻢ‬ ‫ﺍﻟﻌﻴﺎﺭ‬ ‫ﺧﻂ‬ ‫ﺇﻟﻰ‬ ‫ﺍﻟﻤﻘﻄﺮ‬ ‫ﺑﺎﻟﻤﺎﺀ‬ . ‫ﻣﺤﻠﻮﻝ‬ ‫ﻋﻠﻰ‬ ‫ﻓﻨﺤﺼﻞ‬ ‫ﺟﻴﺪﺍ‬ ‫ﺍﻟﻤﺤﻠﻮﻝ‬ ‫ﻧﺮﺝ‬ ( ) ‫ﺍﻟﻤﻮ‬ ‫ﺗﺮﻛﻴﺰ‬ ‫ﺫﻱ‬ ‫ﻟﻲ‬ ‫ﻗﻴﺎﺱ‬ ‫ﻋﻨﺪ‬ ‫ﻧﺎﻗﻠﻴﺘﻬﺎﻟﻨﻮﻋﻴ‬ ‫ﻧﺠﺪ‬ ‫ﺔ‬ = 0,25 / . ‫ﻳﻌﻄﻰ‬ : 3 2 5,46 10 . / HCOO S m mol      ، 3 3 2 35,00 10 . / H O S m mol      e n c y - e d u c a t i o n . c o m / e x a m s 3as.ency-education.com
  • 5. ‫ﺍ‬ ‫ﺧ‬ ‫ﺘ‬ ‫ﺒ‬ ‫ﺎ‬ ‫ﺭ‬ ‫ﻓ‬ ‫ﻲ‬ ‫ﻣ‬ ‫ﺎ‬ ‫ﺩ‬ ‫ﺓ‬ ‫ﺍ‬ ‫ﻟ‬ ‫ﻌ‬ ‫ﻠ‬ ‫ﻮ‬ ‫ﻡ‬ ‫ﺍ‬ ‫ﻟ‬ ‫ﻔ‬ ‫ﻴ‬ ‫ﺰ‬ ‫ﻳ‬ ‫ﺎ‬ ‫ﺋ‬ ‫ﻴ‬ ‫ﺔ‬ / ‫ﺍ‬ ‫ﻟ‬ ‫ﺸ‬ ‫ﻌ‬ ‫ﺒ‬ ‫ﺔ‬ ، ‫ﻋ‬ ‫ﻠ‬ ‫ﻮ‬ ‫ﻡ‬ ‫ﺗ‬ ‫ﺠ‬ ‫ﺮ‬ ‫ﻳ‬ ‫ﺒ‬ ‫ﻴ‬ ‫ﺔ‬ / ‫ﺑ‬ ‫ﻜ‬ ‫ﺎ‬ ‫ﻟ‬ ‫ﻮ‬ ‫ﺭ‬ ‫ﻳ‬ ‫ﺎ‬ ‫ﺗ‬ ‫ﺠ‬ ‫ﺮ‬ ‫ﻳ‬ ‫ﺒ‬ ‫ﻲ‬ 2019 Page 5 sur 6 1 - ‫ﺍﻟﻤﺎﺀ‬ ‫ﻓﻲ‬ ‫ﺍﻟﻤﻴﺜﺎﻧﻮﻳﻚ‬ ‫ﺣﻤﺾ‬ ‫ﺍﻧﺤﻼﻝ‬ ‫ﻣﻌﺎﺩﻟﺔ‬ ‫ﺃﻛﺘﺐ‬ . 2 - ‫ﺃﻭﺟﺪ‬ ‫ﺟﺪ‬ ‫ﺑﻴﻦ‬ ‫ﺍﻟﻌﻼﻗﺔ‬ ‫ﻭ‬ . 1 - ‫ﻗﻴﻤﺔ‬ ‫ﺃﺣﺴﺐ‬ ‫ﺍﻟﻤﺤﻠﻮﻝ‬ ( ) . 2 - ‫ﻋﺒﺎﺭﺓ‬ ‫ﺃﻛﺘﺐ‬ ‫ﺍﻟﻨﻬﺎﺋﻲ‬ ‫ﺍﻟﺘﻘﺪﻡ‬ ‫ﻧﺴﺒﺔ‬ f  ‫ﺍﻟﻤﺤﻠﻮﻝ‬ ‫ﻓﻲ‬ ‫ﺍﻟﻤﺎﺀ‬ ‫ﻣﻊ‬ ‫ﺍﻟﻨﻤﻞ‬ ‫ﻟﺤﻤﺾ‬ ‫ﺍﻟﺤﺎﺻﻞ‬ ‫ﻟﻠﺘﺤﻮﻝ‬ ( ) ‫ﺑﺪﻻﻟﺔ‬ . -II ‫ﺍﻟﻤﻴﺜﺎﻧﻮﻳﻚ‬ ‫ﺣﻤﺾ‬ ‫ﺑﻴﻦ‬ ‫ﻳﺤﺪﺙ‬ ‫ﺍﻟﺬﻱ‬ ‫ﺍﻟﻜﻴﻤﻴﺎﺋﻲ‬ ‫ﺍﻟﺘﻔﺎﻋﻞ‬ ‫ﺩﺭﺍﺳﺔ‬ ‫ﻧﺮﻳﺪ‬ ‫ﺍﻟﻤﺠﻤﻠﺔ‬ ‫ﺍﻟﺠﺰﻳﺌﻴﺔ‬ ‫ﺻﻴﻐﺘﻪ‬ ‫ﻛﺤﻮﻝ‬ ‫ﻭ‬ . ‫ﺃ‬ ‫ﺛﻤﺎﻧﻴﺔ‬ ‫ﻓﻲ‬ ‫ﻧﻀﻊ‬ ‫ﻣﻦ‬ ‫ﻣﺮﻗـــﻤﺔ‬ ‫ﺍﺧـﺘﺒﺎﺭ‬ ‫ﻧﺎﺑـﻴﺐ‬ 01 ‫ﺇﻟﻰ‬ 08 ‫ﻣﻦ‬ ‫ﺍﻟﻤﺘﻜﻮﻥ‬ ‫ﺍﻟﻤﺰﻳﺞ‬ ‫ﻧﻔﺲ‬ 0,2 ‫ﺍﻟﺤـﻤﻀﻮ‬ ‫ﻣﻦ‬ 0,2 ‫ﺣﺮﺍﺭﺗﻪ‬ ‫ﺩﺭﺟﺔ‬ ‫ﻣﺎﺋﻲ‬ ‫ﺣﻤﺎﻡ‬ ‫ﻓﻲ‬ ‫ﺍﻷﻧﺎﺑﻴﺐ‬ ‫ﻫﺬﻩ‬ ‫ﻧﺪﺧﻞ‬ ‫ﺛﻢ‬ ‫ﺍﻟﻜﺤﻮﻝ‬ ‫ﻣﻦ‬ (180° ) ‫ﺳﺎﻋﺔ‬ ‫ﻛﻞ‬ ‫ﺑﻌﺪ‬ ‫ﻭ‬ ‫ﻣﻦ‬ ‫ﺑﺎﻟﺘﺮﺗﻴﺐ‬ ‫ﺍﻷﻧﺎﺑﻴﺐ‬ ‫ﻫﺬﻩ‬ ‫ﺃﺣﺪ‬ ‫ﻧﺨﺮﺝ‬ 01 ‫ﺇﻟﻰ‬ 08 ‫ﻣ‬ ‫ﺑﻮﺍﺳﻄﺔ‬ ‫ﻓﻴﻪ‬ ‫ﺍﻟﻤﺘﺒﻘﻲ‬ ‫ﺍﻟﺤﻤﺾ‬ ‫ﻧﻌﺎﻳﺮ‬ ‫ﺛﻢ‬ ، ‫ﺍﻟﺼﻮﺩﻳﻮﻡ‬ ‫ﻫﻴﺪﺭﻭﻛﺴﻴﺪ‬ ‫ﺤﻠﻮﻝ‬ . ( ( ) + ( )) ‫ﺍﻟﻨﺘﺎﺋﺞ‬ ‫ﻋﻠﻴﻬﺎ‬ ‫ﺍﻟﻤﺘﺤﺼﻞ‬ ‫ﺍﻟﺘﺎﻟﻲ‬ ‫ﺍﻟﺠﺪﻭﻝ‬ ‫ﻓﻲ‬ ‫ﻣﺪﻭﻧﺔ‬ : 08 07 06 05 04 03 02 01 ‫ﺍﻷﻧﺒﻮﺏ‬ ‫ﺭﻗﻢ‬ 7 6 5 4 3 2 1 0 (ℎ ) 0,067 0,067 0,067 0,068 0,074 0,084 0,114 0,200 (‫)ﺣﻤﺾ‬ (‫)ﺃﺳﺘﺮ‬ 1 - ‫ﺃﻋﻼﻩ‬ ‫ﺍﻟﺠﺪﻭﻝ‬ ‫ﺃﻛﻤﻞ‬ . 2 - ‫ﺍﻟﺒﻴﺎﻧﻲ‬ ‫ﺍﻟﻤﻨﺤﻨﻰ‬ ‫ﺃﺭﺳﻢ‬ (‫)ﺃﺳﺘﺮ‬ = ( ) . ‫ﻭﻓﻖ‬ ‫ﺍﻟﺴﻠﻢ‬ : ( 1 → 1ℎ  ‫ﻭ‬ 1 → 0,01 ) 3 - ‫ﺍﻟﺤﻤﺾ‬ ‫ﺑﻴﻦ‬ ‫ﺍﻟﺘﻔﺎﻋﻞ‬ ‫ﺗﻘﺪﻡ‬ ‫ﺟﺪﻭﻝ‬ ‫ﺃﻧﺸﺊ‬ ‫ﺍﻟﻜﺤﻮﻝ‬ ‫ﻭ‬ . 4 - ‫ﺍﻟﺒﻴﺎﻥ‬ ‫ﻣﻦ‬ ‫ﺍﺳﺘﻨﺘﺞ‬ : ‫ﺃ‬ - ‫ﺍﻟﻠﺤﻈﺔ‬ ‫ﻋﻨﺪ‬ ‫ﺍﻟﺘﻔﺎﻋﻞ‬ ‫ﺳﺮﻋﺔ‬ = 2ℎ . ‫ﺏ‬ - ‫؟‬ ‫ﻫﺬﺍﺍﻟﺘﺤﻮﻝ‬ ‫ﻟﻨﻬﺎﻳﺔ‬ ‫ﺍﻟﻤﻮﺍﻓﻘﺔ‬ ‫ﺣﺪﺩﺍﻟﻠﺤﻈﺔ‬ ‫ﺟـ‬ - ‫ﺍﻷﺳﺘﺮﺓ‬ ‫ﻣﺮﺩﻭﺩ‬ . - ‫ﺍﻟﻤﻤﻜﻨﺔ‬ ‫ﺍﻟﻤﻔﺼﻠﺔ‬ ‫ﻧﺼﻒ‬ ‫ﺻﻴﻐﻪ‬ ‫ﻭ‬ ‫ﺍﻟﻤﺴﺘﻌﻤﻞ‬ ‫ﺍﻟﻜﺤﻮﻝ‬ ‫ﺻﻨﻒ‬ ‫ﺍﺳﺘﻨﺘﺞ‬ . 5 - ‫ﺍﻟﻤﺘﻔﺮﻋﺔ‬ ‫ﺍﻟﺼﻴﻐﺔ‬ ‫ﺫﻱ‬ ‫ﺍﻟﻜﺤﻮﻝ‬ ‫ﻭ‬ ‫ﺍﻟﺤﻤﺾ‬ ‫ﺑﻴﻦ‬ ‫ﺍﻟﺤﺎﺻﻞ‬ ‫ﻟﻠﺘﺤﻮﻝ‬ ‫ﺍﻟﻤﻨﻤﺬﺝ‬ ‫ﺍﻟﺘﻔﺎﻋﻞ‬ ‫ﻣﻌﺎﺩﻟﺔ‬ ‫ﺃﻛﺘﺐ‬ . ‫ﺍﻷﺳﺘﺮ‬ ‫ﺗﺴﻤﻴﺔ‬ ‫ﻣﻊ‬ ‫ﺍﻟﻨﺎﺗﺞ‬ 6 - ‫ﺍﻷ‬ ‫ﻧﺨﺮﺝ‬ ‫ﺭﻗﻢ‬ ‫ﻧﺒﻮﺏ‬ 07 ‫ﺍﻟﻠﺤﻈﺔ‬ ‫ﻋﻨﺪ‬ = 6 ℎ ‫ﻣﺒﺎﺷﺮﺓ‬ ‫ﻟﻪ‬ ‫ﻧﻀﻴﻒ‬ ‫ﺛﻢ‬ 0,2 ‫ﺍﻷﺳﺘﺮ‬ ‫ﻣﻦ‬ .  ‫ﻋﻠﻞ‬ ‫؟‬ ‫ﺍﻟﻜﻴﻤﻴﺎﺋﻴﺔ‬ ‫ﺍﻟﺠﻤﻠﺔ‬ ‫ﺗﻄﻮﺭ‬ ‫ﺗﺘﻮﻗﻊ‬ ‫ﺟﻬﺔ‬ ‫ﺃﻱ‬ ‫ﻓﻲ‬ . ‫ﺍﻟﺜــﺎﻧﻲ‬ ‫ﺍﻟﺠﺰﺀ‬ ) 7 ‫ﻥ‬ ( ‫ﺍﻟﺘﺠﺮﻳﺒﻲ‬ ‫ﺍﻟﺘﻤﺮﻳﻦ‬ : ) 7 ‫ﻥ‬ ( ‫ﻛﺮﻳﺔ‬ ( ) ‫ﻛﺘﻠﺘﻬﺎ‬ ‫ﻧﻘﺘﺮﺡ‬ ‫ﻗﻴﻤﺘﻬﺎ‬ ‫ﻟﺘﺤﺪﻳﺪ‬ ‫ﻣﺠﻬﻮﻟﺔ‬ . I - ‫ﺍﻷﻭﻟﻰ‬ ‫ﺍﻟﻄﺮﻳﻘﺔ‬ : ‫ﺍﻟﺴﻘ‬ ‫ﺣﺮﻛﺔ‬ ‫ﺩﺭﺍﺳﺔ‬ ‫ﺍﻟﻬﻮﺍﺀ‬ ‫ﻓﻲ‬ ‫ﻟﻠﻜﺮﻳﺔ‬ ‫ﺍﻟﺸﺎﻗﻮﻟﻲ‬ ‫ﻮﻁ‬ : ‫ﺍﻟﻨﻘﻄﺔ‬ ‫ﻣﻦ‬ ‫ﺍﺑﺘﺪﺍﺀ‬ ‫ﺍﻟﻬﻮﺍﺀ‬ ‫ﺍﺑﺘﺪﺍﺋﻴﺔﻓﻲ‬ ‫ﺳﺮﻋﺔ‬ ‫ﺩﻭﻥ‬ ‫ﺍﻟﻜﺮﻳﺔ‬ ‫ﺗﺴﻘﻂ‬ ‫ﻗﻮﺓ‬ ‫ﺣﺮﻛﺘﻬﺎ‬ ‫ﺗﻌﻴﻖ‬ ، ‫ﺍﻟﺪﺭﺍﺳﺔ‬ ‫ﻣﻌﻠﻢ‬ ‫ﺍﺣﺪﺍﺛﻴﺎﺕ‬ ‫ﻣﺒﺪﺃ‬ ‫ﻣﻦ‬ ‫ﻋﺒﺎﺭﺗﻬﺎ‬ ‫ﺍﺣﺘﻜﺎﻙ‬ ‫ﺍﻟﺸﻜﻞ‬ : f Kv  ). ‫ﺃﺭﺧﻤﻴﺪﺱ‬ ‫ﺩﺍﻓﻌﺔ‬ ‫ﻧﻬﻤﻞ‬ ( ‫ﺍﻟﺸﻜﻞ‬ ‫ﺍﻟﺒﻴﺎﻥ‬ ‫ﻳﻤﺜﻞ‬ - 3 ‫ﻋﻄﺎﻟ‬ ‫ﻣﺮﻛﺰ‬ ‫ﺳﺮﻋﺔ‬ ‫ﺗﻐﻴﺮﺍﺕ‬ ‫ﺍﻟﺰﻣﻦ‬ ‫ﺑﺪﻻﻟﺔ‬ ‫ﺍﻟﻜﺮﻳﺔ‬ ‫ﺔ‬ . e n c y - e d u c a t i o n . c o m / e x a m s 3as.ency-education.com
  • 6. ‫ﺍ‬ ‫ﺧ‬ ‫ﺘ‬ ‫ﺒ‬ ‫ﺎ‬ ‫ﺭ‬ ‫ﻓ‬ ‫ﻲ‬ ‫ﻣ‬ ‫ﺎ‬ ‫ﺩ‬ ‫ﺓ‬ ‫ﺍ‬ ‫ﻟ‬ ‫ﻌ‬ ‫ﻠ‬ ‫ﻮ‬ ‫ﻡ‬ ‫ﺍ‬ ‫ﻟ‬ ‫ﻔ‬ ‫ﻴ‬ ‫ﺰ‬ ‫ﻳ‬ ‫ﺎ‬ ‫ﺋ‬ ‫ﻴ‬ ‫ﺔ‬ / ‫ﺍ‬ ‫ﻟ‬ ‫ﺸ‬ ‫ﻌ‬ ‫ﺒ‬ ‫ﺔ‬ ، ‫ﻋ‬ ‫ﻠ‬ ‫ﻮ‬ ‫ﻡ‬ ‫ﺗ‬ ‫ﺠ‬ ‫ﺮ‬ ‫ﻳ‬ ‫ﺒ‬ ‫ﻴ‬ ‫ﺔ‬ / ‫ﺑ‬ ‫ﻜ‬ ‫ﺎ‬ ‫ﻟ‬ ‫ﻮ‬ ‫ﺭ‬ ‫ﻳ‬ ‫ﺎ‬ ‫ﺗ‬ ‫ﺠ‬ ‫ﺮ‬ ‫ﻳ‬ ‫ﺒ‬ ‫ﻲ‬ 2019 Page 6 sur 6 ‫اﻟﺷﻛل‬ – 04 0 0,1 1,88 ( ) ( / ) ‫اﻟﺷﻛ‬ ‫ل‬ - 5 - ‫ﻳﻌﻄﻰ‬ : 2 3.57 10 / K Kg s    ، 2 10 / g m s  . 1 . ‫؟‬ ‫ﻫﺬﻫﺎﻟﺤﺮﻛﺔ‬ ‫ﻟﺪﺭﺍﺳﺔ‬ ‫ﺍﻟﻤﻨﺎﺳﺐ‬ ‫ﺍﻟﻤﺮﺟﻊ‬ ‫ﻫﻮ‬ ‫ﻣﺎ‬ - ‫ﺗﺴﻤﺢ‬ ‫ﺍﻟﺘﻲ‬ ‫ﻭ‬ ‫ﺍﻟﻤﺮﺟﻊ‬ ‫ﺑﻬﺬﺍ‬ ‫ﺍﻟﻤﺘﻌﻠﻘﺔ‬ ‫ﺍﻟﻔﺮﺿﻴﺔ‬ ‫ﻫﻲ‬ ‫ﻣﺎ‬ ‫ﻟﻨﻴﻮﺗﻦ؟‬ ‫ﺍﻟﺜﺎﻧﻲ‬ ‫ﺍﻟﻘﺎﻧﻮﻥ‬ ‫ﺑﺘﻄﺒﻴﻖ‬ . 2 . ‫ﺃﻭﺟﺪ‬ ‫ﺍﻟﺒﻴﺎﻥ‬ ‫ﺑﺎﺳﺘﻐﻼﻝ‬ : ‫ﺃ‬ - ‫ﺍﻟﺤﺪ‬ ‫ﺍﻟﺴﺮﻋﺔ‬ ‫ﻗﻴﻤﺔ‬ ‫ﻳﺔ‬ L v . ‫ﺏ‬ - ‫ﺍﻟﺰﻣﻦ‬ ‫ﺛﺎﺑﺖ‬  ‫ﻟﻠﺤﺮﻛﺔ‬ ‫ﺍﻟﻤﻤﻴﺰ‬ ‫ﺝ‬ - ‫ﺍﻻﺑﺘﺪﺍﺋﻲ‬ ‫ﺍﻟﺘﺴﺎﺭﻉ‬ ‫ﻗﻴﻤﺔ‬ 0 a ‫؟‬ ‫ﺗﺴﺘﻨﺘﺞ‬ ‫ﺫﺍ‬ ‫ﻣﺎ‬ ، 3 . ‫ﺃﻧﻬﺎ‬ ‫ﺑﻴﻦ‬ ‫ﻭ‬ ‫ﻟﻠﺤﺮﻛﺔ‬ ‫ﺍﻟﺘﻔﺎﺿﻠﻴﺔ‬ ‫ﺍﻟﻤﻌﺎﺩﻟﺔ‬ ‫ﺃﻭﺟﺪ‬ ‫ﺍﻟﺸﻜﻞ‬ ‫ﻋﻠﻰ‬ ‫ﺗﻜﺘﺐ‬ dv Av B dt   ‫ﺣﻴﺚ‬ ‫ﻭ‬ ‫ﻳﻄ‬ ‫ﺛﻮﺍﺑﺖ‬ ‫ﻋﺒﺎﺭﺗﻴﻬﻤﺎ‬ ‫ﺇﻳﺠﺎﺩ‬ ‫ﻠﺐ‬ . 4 . ‫ﻗﻴﻤﺔ‬ ‫ﺃﺣﺴﺐ‬ ‫ﺍﻟﻜﺮﻳﺔ‬ ‫ﻛﺘﻠﺔ‬ . II - ‫ﺍﻟﺜﺎﻧﻴﺔ‬ ‫ﺍﻟﻄﺮﻳﻘﺔ‬ : ‫ﻣﻬﺘﺰﺓ‬ ‫ﺟﻤﻠﺔ‬ ‫ﺣﺮﻛﺔ‬ ‫ﺩﺭﺍﺳﺔ‬ ) ‫ﻧﺎﺑﺾ‬ – ‫ﻛﺮﻳﺔ‬ ( ‫ﺃﻱ‬ ) ‫ﺃﻓﻘﻲ‬ ‫ﻣﺮﻥ‬ ‫ﻧﻮﺍﺱ‬ :( ‫ﻣﺮﻭﻧﺘﻪ‬ ‫ﺛﺎﺑﺖ‬ ‫ﻣﺘﻼﺻﻘﺔ‬ ‫ﻏﻴﺮ‬ ‫ﺣﻠﻘﺎﺗﻪ‬ ‫ﻣﺮﻥ‬ ‫ﺑﻨﺎﺑﺾ‬ ‫ﺍﻟﺴﺎﺑﻘﺔ‬ ‫ﺍﻟﻜﺮﻳﺔ‬ ‫ﻧﺜﺒﺖ‬ 50 / K N m  ‫ﺍﻟﺸﻜﻞ‬ ‫ﻓﻲ‬ ‫ﻣﻮﺿﺢ‬ ‫ﻫﻮ‬ ‫ﻛﻤﺎ‬ - 4 . ‫ﻋﻦ‬ ‫ﺍﻟﻜﺮﻳﺔ‬ ‫ﻧﺰﻳﺢ‬ ‫ﺑﺎﻟﻤﻘﺪﺍﺭ‬ ‫ﺍﻟﺘﻮﺍﺯﻥ‬ ‫ﻭﺿﻊ‬ (+ ) ‫ﺍﻟﻠﺤﻈﺔ‬ ‫ﻋﻨﺪ‬ ‫ﻧﺘﺮﻛﻬﺎ‬ ‫ﻭ‬ = 0 ‫ﺩﻭﻥ‬ ‫ﺗﺴﺠﻴﻞ‬ ‫ﻋﻠﻰ‬ ‫ﺑﺎﻟﺤﺼﻮﻝ‬ ‫ﻣﻨﺎﺳﺐ‬ ‫ﺗﺠﻬﻴﺰ‬ ‫ﻳﺴﻤﺢ‬ ‫ﺍﺑﺘﺪﺍﺋﻴﺔ‬ ‫ﺳﺮﻋﺔ‬ ‫ﺍﻟﺰﻣﻦ‬ ‫ﺑﺪﻻﻟﺔ‬ ‫ﺍﻟﻜﺮﻳﺔ‬ ‫ﻋﻄﺎﻟﺔ‬ ‫ﻣﺮﻛﺰ‬ ‫ﺳﺮﻋﺔ‬ t ‫ﺍﻟﺒﻴﺎﻥ‬ ‫ﻓﻲ‬ ‫ﻭﺍﻟﻤﻤﺜﻞ‬ ‫ﺍﻟﺸﻜﻞ‬ ) 05 .( 1 - ‫ﺍﻟﻔﺎﺻﻠﺔ‬ ‫ﻋﻨﺪ‬ ‫ﺍﻟﻜﺮﻳﺔ‬ ‫ﻋﻠﻰ‬ ‫ﺍﻟﻤﺆﺛﺮﺓ‬ ‫ﺍﻟﻘﻮﻯ‬ ‫ﻣﺜﻞ‬ ( > 0 ) . 2 - ‫ﻣﺘﺨﺎﻣﺪﺓ‬ ‫ﺍﻟﺠﻤﻠﺔ‬ ‫ﺣﺮﻛﺔ‬ ‫ﻫﻞ‬ ‫ﻋﻠﻞ‬ ‫؟‬ ‫ﻻ‬ ‫ﺃﻡ‬ . 3 - ‫ﺍﻟﺘﻔﺎﺿﻠﻴﺔ‬ ‫ﺍﻟﻤﻌﺎﺩﻟﺔ‬ ‫ﺃﻭﺟﺪ‬ ‫ﻟﻨﻴﻮﺗﻦ‬ ‫ﺍﻟﺜﺎﻧﻲ‬ ‫ﺍﻟﻘﺎﻧﻮﻥ‬ ‫ﺑﺘﻄﺒﻴﻖ‬ ‫ﺍﻟﻔﺎﺻﻠﺔ‬ ‫ﺑﺪﻻﻟﺔ‬ ‫ﻟﻠﺤﺮﻛﺔ‬ . 4 - ‫ﻟﻠﺤﺮﻛﺔ‬ ‫ﺍﻟﻤﻤﻴﺰﺓ‬ ‫ﺍﻟﻤﻘﺎﺩﻳﺮ‬ ‫ﺃﻭﺟﺪ‬ ‫ﺍﻟﺒﻴﺎﻥ‬ ‫ﺑﺎﺳﺘﻐﻼﻝ‬ :  ‫ﻟﻠﺤﺮﻛﺔ‬ ‫ﺍﻟﺬﺍﺗﻲ‬ ‫ﺍﻟﺪﻭﺭ‬ .  ‫ﺍﻟﺤﺮﻛﺔ‬ ‫ﻧﺒﺾ‬ .  ‫ﺍﻻﻫﺘﺰﺍﺯﺍﺕ‬ ‫ﺳﻌﺔ‬ .  ‫ﺍﻻﺑﺘﺪﺍﺋﻴﺔ‬ ‫ﺍﻟﺼﻔﺤﺔ‬ . 5 - ‫ﺍﻟﻜﺮﻳﺔ‬ ‫ﻛﺘﻠﺔ‬ ‫ﺃﺣﺴﺐ‬ ‫ﺳﺎﺑﻘﺎ‬ ‫ﺍﻟﻤﺤﺴﻮﺑﺔ‬ ‫ﺗﻠﻚ‬ ‫ﻣﻊ‬ ‫ﻗﺎﺭﻧﻬﺎ‬ ‫ﺛﻢ‬ . ‫ﻳﻌﻄﻰ‬ : = 10 ‫ﺍ‬ ‫ﻧ‬ ‫ﺘﻬ‬ ‫ﻰ‬ ‫ﺍ‬ ‫ﻟ‬ ‫ﻤ‬ ‫ﻮ‬ ‫ﺿ‬ ‫ﻮ‬ ‫ﻉ‬ ‫ﺍ‬ ‫ﻟ‬ ‫ﺜ‬ ‫ﺎ‬ ‫ﻧ‬ ‫ﻲ‬ * * * * ‫ﺃ‬ ‫ﺳ‬ ‫ﺎ‬ ‫ﺗ‬ ‫ﺬ‬ ‫ﺓ‬ ‫ﺍ‬ ‫ﻟ‬ ‫ﻤ‬ ‫ﺎ‬ ‫ﺩ‬ ‫ﺓ‬ ‫ﻳ‬ ‫ﺘ‬ ‫ﻤ‬ ‫ﻨ‬ ‫ﻮ‬ ‫ﻥ‬ ‫ﻟ‬ ‫ﻜ‬ ‫ﻢ‬ ‫ﺍ‬ ‫ﻟ‬ ‫ﺘ‬ ‫ﻮ‬ ‫ﻓ‬ ‫ﻴ‬ ‫ﻖ‬ ‫ﻭ‬ ‫ﺍ‬ ‫ﻟ‬ ‫ﻨ‬ ‫ﺠ‬ ‫ﺎ‬ ‫ﺡ‬ ‫ﻓ‬ ‫ﻲ‬ ‫ﺑ‬ ‫ﻜ‬ ‫ﺎ‬ ‫ﻟ‬ ‫ﻮ‬ ‫ﺭ‬ ‫ﻳ‬ ‫ﺎ‬ 2019 * * * * e n c y - e d u c a t i o n . c o m / e x a m s 3as.ency-education.com
  • 7. Page 1 sur 9 ‫ﺍﻟﻨﻤﻮﺫﺟﻲ‬ ‫ﺍﻟﺘﺼﺤﻴﺢ‬ ‫ﺍﻟﻨﻘﻄﺔ‬ ( ) = ∆ . = 53 + (127 − 53) − ( ) . = 1050.02406 ( ) = ∆ . = 53 + (131 − 53) − ( ) . = 1081.11753 ( ) = 1050.02406 127 = 8.2673MeV/Nucl ( ) = 1050.02406 131 = 8.2528MeV/Nucl = . . = 1 × 6,02 × 10 11.54 × 24 × 3600 × 131 = 4,6 × 10 = . . = 37 × 10 × 11,54 × 24 × 3600 × 131 6,02 × 10 = 8.02 × 10 = 11.54 × 8 × 10 × 6.02 × 10 11.54 × 24 × 3600 × 24 × 10 × 131 = 4.99 ‫ﺍﻟﻤﻮﺿ‬ ‫ﻮﻉ‬ ‫ﺍﻻﻭﻝ‬ ‫ﺍﻷﻭﻝ‬ ‫ﺍﻟﺘﻤﺮﻳﻦ‬ ‫ﺣﻞ‬ : 1 . ‫ﺗﻌﺮﻳﻒ‬ : ‫ﺍﻟﻨﻮﻭﻱ‬ ‫ﺍﻟﺘﺮﺍﺑﻂ‬ ‫ﻃﺎﻗﺔ‬ l E ‫ﺳﻜﻮﻥ‬ ‫ﺣﺎﻟﺔ‬ ‫ﻓﻲ‬ ‫ﻧﻮﺍﺓ‬ ‫ﻟﺘﺤﻄﻴﻢ‬ ‫ﺍﻟﺨﺎﺭﺟﻲ‬ ‫ﺍﻟﻮﺳﻂ‬ ‫ﻳﻮﻓﺮﻫﺎ‬ ‫ﺍﻟﺘﻲ‬ ‫ﺍﻟﻄﺎﻗﺔ‬ ‫ﻫﻲ‬ ‫ﺑﺎﻟﻌﻼﻗﺔ‬ ‫ﺗﻌﻄﻰ‬ ‫ﺳﺎﻛﻨﺔ‬ ‫ﻭ‬ ‫ﻣﻨﻔﺼﻠﺔ‬ ‫ﻧﻴﻮﻛﻠﻴﻮﻧﺎﺗﻬﺎ‬ ‫ﺇﻟﻰ‬ : 2 . ‫ﺍﻟﻨﻮﻭﻱ‬ ‫ﺍﻟﺘﺮﺍﺑﻂ‬ ‫ﻃﺎﻗﺔ‬ ‫ﺣﺴﺎﺏ‬ : = ∆ . = + ( − ) − ( ) . 3 . ‫ﺍﻟﺮﺑﻂ‬ ‫ﻃﺎﻗﺔ‬ ‫ﺣﺴﺎﺏ‬ ‫ﻧﻮﻳﺔ‬ ‫ﻟﻜﻞ‬ : ‫ﺃﻥ‬ ‫ﺑﻤﺎ‬ > ‫ﻓ‬ ‫ﺎﻥ‬ ‫ﻣﻦ‬ ‫ﺍﺳﺘﻘﺮﺍﺭﺍ‬ ‫ﺃﻛﺜﺮ‬ ‫ﻫﻮ‬ ‫ﺍﻟﻤﺸﻊ‬ ‫ﺍﻟﻨﻈﻴﺮ‬ ‫ﺃﻱ‬ . 4 . ‫ﺍﻟﺘﻔﻜﻚ‬ ‫ﻣﻌﺎﺩﻟﺔ‬ ‫ﻛﺘﺎﺑﺔ‬ : → + ‫ﺍﻻﻧﺤﻔﺎﻅ‬ ‫ﻗﻮﺍﻧﻴﻦ‬ ‫ﺣﺴﺐ‬ : 131 = + 0 53 = − 1 → = 131 = 54 5 . ‫ﺍﻟﺰﻣﻦ‬ ‫ﺛﺎﺑﺖ‬ ‫ﺣﺴﺎﺏ‬ : = / = 11.54 6 . ‫ﺃ‬ . ‫ﻧﺸﺎﻁ‬ ‫ﺣﺴﺎﺏ‬ 1g ‫ﺍﻟﻴﻮﺩ‬ ‫ﻣﻦ‬ : ‫ﻟﺪﻳﻨﺎ‬ = → = . . ‫ﺏ‬ . ‫ﺍﻟﺠﺮﻋﺔ‬ ‫ﻓﻲ‬ ‫ﺍﻟﻤﺘﻮﺍﺟﺪﺓ‬ ‫ﺍﻟﻴﻮﺩ‬ ‫ﻛﺘﻠﺔ‬ ‫ﺣﺴﺎﺏ‬ : ‫ﺝ‬ . ‫ﺍﻟﻌﻼﻗﺔ‬ ‫ﺍﺛﺒﺎﺕ‬ : ‫ﻟﺪﻳﻨﺎ‬ = / → = / → ln = → = . . = . . . . 7 . ‫ﺃ‬ . ‫ﺍﻟﻴﻮﺩ‬ ‫ﻛﺘﻠﺔ‬ ‫ﺣﺴﺎﺏ‬ ‫ﺍﻟﻤﺘﺒﻘﻲ‬ : ( ) = /  ( = 4 ) = 8 × 10 . = 6 × 10  ( = 8 ) = 8 × 10 . = 4 × 10  ( = 16 ) = 8 × 10 . = 2 × 10 ‫ﺏ‬ . ‫ﻣﺮﻭﺭ‬ ‫ﺑﻌﺪ‬ ‫ﺍﻟﻤﺘﺒﻘﻴﺔ‬ ‫ﺍﻷﻧﻮﻳﺔ‬ ‫ﻋﺪﺩ‬ ‫ﺣﺴﺎﺏ‬ t1 : ‫ﻟﺪﻳﻨﺎ‬ = . = × × . × = 2.75 × 10 ‫ﺝ‬ . ‫ﻟﺘﻔﻜﻚ‬ ‫ﺍﻟﻼﺯﻡ‬ ‫ﺍﻟﺰﻣﻦ‬ ‫ﺣﺴﺎﺏ‬ 99% ‫ﺍﻻﺑﺘﺪﺍﺋﻴﺔ‬ ‫ﺍﻟﻜﺘﻠﺔ‬ ‫ﻣﻦ‬ : ‫ﻟﺪﻳﻨﺎ‬ ( ) = / → 0.01 = / → = − . ln (0.01) = 53.14 0.5 1 0.5 0.5 0.5 01 0.25 0.5 0.5 0.25 0.25 e n c y - e d u c a t i o n . c o m / e x a m s 3as.ency-education.com
  • 8. Page 2 sur 9 → = → = 1 2 → − = 1 2 → = . 2 = . 2 = 1 2 = 1 2 1 − / → 2 = 1 2 1 − = 7,5 × 10 ‫ﺍﻟﺜﺎﻧﻲ‬ ‫ﺍﻟﺘﻤﺮﻳﻦ‬ ‫ﺣﻞ‬ : 1 . ‫ﺍﻟﺪﺍﺭﺓ‬ ‫ﺗﻤﺜﻴﻞ‬ : 2 . ‫ﺣ‬ ‫ﻗﻴﻤﺔ‬ ‫ﺴﺎﺏ‬ E : ‫ﺍﻟﺘﻮﺗﺮﺍﺕ‬ ‫ﺟﻤﻊ‬ ‫ﻗﺎﻧﻮﻥ‬ ‫ﺣﺴﺐ‬ = + ‫ﺍﻟﻠﺤﻈﺔ‬ ‫ﻋﻨﺪ‬ t=0 : = (0) + (0) ‫ﺍﻟﺒﻴﺎﻥ‬ ‫ﻣﻦ‬ : (0) + (0) = 10 → = 10 3 . ‫ﺃﻥ‬ ‫ﺍﺛﺒﺎﺕ‬ : + u = E ‫ﺍﻟﺘﻮﺗﺮﺍﺕ‬ ‫ﺟﻤﻊ‬ ‫ﻗﺎﻧﻮﻥ‬ ‫ﺣﺴﺐ‬ + = → + = → + u = E 4 . ‫ﻋﺒﺎﺭﺓ‬ ‫ﺍﻳﺠﺎﺩ‬ α : ‫ﻟﺪﻳﻨﺎ‬ : = (1 − ) → = αEe ‫ﻧﺠﺪ‬ ‫ﺍﻟﺘﻔﺎﺿﻠﻴﺔ‬ ‫ﺍﻟﻤﻌﺎﺩﻟﺔ‬ ‫ﻓﻲ‬ ‫ﺑﺎﻟﺘﻌﻮﻳﺾ‬ : . αEe + − e = → e − 1 = 0 5 . ‫ﻋﺒﺎﺭﺓ‬ : ‫ﻟﺪﻳﻨﺎ‬ = ‫ﻭ‬ ‫ﻟﺪﻳﻨﺎ‬ = = 1 − / → = ‫ﻭﻣﻨﻪ‬ : = = . . 6 . ‫ﺃ‬ . ‫ﺍﻟﻤﻨﺤﻨﻰ‬ ‫ﺗﺤﺪﻳﺪ‬ :  ‫ﺍﻟﻤﻨﺤﻨﻰ‬ ) 1 ( ‫ﻳﻤﺜﻞ‬ ‫ﻷﻥ‬ : (0) = 0  ‫ﺍﻟﻤﻨﺤﻨﻰ‬ ) 2 ( ‫ﻳﻤﺜﻞ‬ ‫ﻷﻥ‬ : (0) = ‫ﺏ‬ . ‫ﺍﺛﺒ‬ ‫ﺍﻥ‬ ‫ﺎﺕ‬ = . 2 ‫ﺃﻱ‬ ‫ﺍﻟﻤﻨﺤﻨﻴﻴﻦ‬ ‫ﺗﻘﺎﻃﻊ‬ ‫ﻋﻨﺪ‬ : = → . = (1 − ) → 2 . = ‫ﺝ‬ . ‫ﻗﻴﻤﺔ‬ ‫ﺍﺳﺘﻨﺘﺎﺝ‬ τ : ‫ﻟﺪﻳﻨﺎ‬ ‫ﺍﻟﺒﻴﺎﻥ‬ ‫ﻣﻦ‬ : = 1,5 ‫ﻭ‬ ‫ﻣﻨﻪ‬ : = . 2 → = = , = 2,16 ‫ﺩ‬ . ‫ﻗﻴﻤﺔ‬ L ‫ﺍﻟﻮﺷﻴﻌﺔ‬ ‫ﺫﺍﺗﻴﺔ‬ : = → = . = 2,16 × 10 × 200 = 0,4 7 . ‫ﻋﺒﺎﺭﺓ‬ : ‫ﻟﺪﻳﻨﺎ‬ : + . = ‫ﺍﻟﺪﺍﺋﻢ‬ ‫ﺍﻟﻨﻈﺎﻡ‬ ‫ﻓﻲ‬ = ‫ﻭ‬ = 0 ‫ﻭﻣﻨﻪ‬ . = E → = ‫ﻗﻴﻤﺘﻪ‬ : = = 0,05 8 . ‫ﻭﺷﻴﻌﺔ‬ ‫ﻓﻲ‬ ‫ﺍﻟﻤﺨﺰﻧﺔ‬ ‫ﺍﻟﻄﺎﻗﺔ‬ ‫ﻋﺒﺎﺭﺓ‬ : 0.5 0.5 0.5 0.5 0.5 0.5 01 0.5 0.5 01 01 E L R K ‫اﻟﺸﻜﻞ‬ 1 e n c y - e d u c a t i o n . c o m / e x a m s 3as.ency-education.com
  • 9. Page 3 sur 9 − = n(I ) ⟹ = − n(I ) = 3.5 × 10 × 20 × 10 − 0.25 × 10 − = 0 ⟹ = = 0.45 × 10 0.1 = 0.045 ⁄ ‫ﺍﻟﺘﺠﺮﻳﺒﻲ‬ ‫ﺍﻟﺘﻤﺮﻳﻦ‬ ‫ﺣﻞ‬ : I . ‫ﺍﻷﻭﻝ‬ ‫ﺍﻟﻔﻮﺝ‬ 1 ( ‫ﺩﻟﺔ‬ ‫ﻣﻌﺎ‬ ‫ﺍﻟﺘﻔﺎﻋﻞ‬ : ‫ﻟﻸﻛﺴﺪﺓ‬ ‫ﺍﻟﻨﺼﻔﻴﺔ‬ ‫ﺩﻟﺔ‬ ‫ﺍﻟﻤﻌﺎ‬ : O = O + 2 + 2e ‫ﻟﻺﺭﺟﺎﻉ‬ ‫ﺍﻟﻨﺼﻔﻴﺔ‬ ‫ﺩﻟﺔ‬ ‫ﺍﻟﻤﻌﺎ‬ : I + 2e = 2 ‫ﺍﻟﺘﻔﺎﻋﻞ‬ ‫ﺩﻟﺔ‬ ‫ﻣﻌﺎ‬ O + I = O + 2 + 2 2 ( ‫ﺍﻟﺘﻔﺎﻋﻞ‬ ‫ﺗﻘﺪﻡ‬ ‫ﺟﺪﻭﻝ‬ : O + I = O + 2 + 2 ∕∕ 0 0 ∕∕ 2 − − ∕∕ 2 − − 3 ( ‫ﺍﻟﺸﺮﻭﻁ‬ : ‫ﻣﺤﻠﻮﻝ‬ ‫ﻳﻜﻮﻥ‬ ‫ﺃﻥ‬ ‫ﻳﺠﺐ‬ ‫ﺍﺫﺍ‬ ‫ِﻳﺮ‬ ‫ﻣﻌﺎ‬ ‫ﻣﺤﻠﻮﻝ‬ ‫ﺛﻴﻮﻛﺒﺮﻳﺘﺎﺕ‬ : - ‫ﻣﻌﻠﻮﻡ‬ ‫ﺗﺮﻛﻴﺰﻩ‬ . - ‫ﻣﻊ‬ ‫ﺗﻔﺎﻋﻠﻪ‬ I ‫ﺗﺎﻡ‬ ‫ﻭ‬ ‫ﺳﺮﻳﻊ‬ . 4 ( ‫ﺍﻟﻤﻌﺎﻳﺮﺓ‬ ‫ﺗﻔﺎﻋﻞ‬ ‫ﺩﻟﺔ‬ ‫ﻣﻌﺎ‬ : ‫ﻟﻸﻛﺴﺪﺓ‬ ‫ﺍﻟﻨﺼﻔﻴﺔ‬ ‫ﺩﻟﺔ‬ ‫ﺍﻟﻤﻌﺎ‬ : 2 O ⁻² = O ²⁻ + 2e ‫ﺍ‬ ‫ﻟﻺﺭﺟﺎﻉ‬ ‫ﺍﻟﻨﺼﻔﻴﺔ‬ ‫ﺩﻟﺔ‬ ‫ﻟﻤﻌﺎ‬ : I + 2e = 2 ‫ﺍﻟﺘﻔﺎﻋﻞ‬ ‫ﺩﻟﺔ‬ ‫ﻣﻌﺎ‬ 2 O ⁻² + I = O ⁻² + 2 5 ( ‫ﺍﻟﺘﻜﺎﻓﺆ‬ ‫ﻋﻨﺪ‬ : )= n( )= . × × × = 0.25× mol n( 6 ( ‫ﺣﺴﺎﺏ‬ = . × mol ‫ﺃﻥ‬ ‫ﺑﻤﺎ‬ : n (I ) ≠ 0 ‫ﻓﺈﻥ‬ : n ( O ) = 0 ‫ﻣﻨﻪ‬ ‫ﻭ‬ ‫ﻭﻣﻨﻪ‬ : 176 × 0.045 = . ⁄ = = × II . ‫ﺍﻟﺜﺎﻧﻲ‬ ‫ﺍﻟﻔﻮﺝ‬ : 1 ( ‫ﺍﻟﻤﻌﺎﻳﺮﺓ‬ ‫ﺗﻔﺎﻋﻞ‬ ‫ﺩﻟﺔ‬ ‫ﻣﻌﺎ‬ O + = O ⁻ + ₂ 2 ( = 3 ( ( = 9 ∕ P =8.2) ‫ﻟﻤﺎ‬ =4.5ml = ‫ﻓﺎﻥ‬ : P =P ‫ﻧﺠﺪ‬ ‫ﺍﻟﺒﻴﺎﻥ‬ ‫ﻋﻠﻰ‬ ‫ﺑﺎﻻﺳﻘﺎﻁ‬ : = . 4 ( ‫ﻳﻜﻮﻥ‬ ‫ﺍﻟﺘﻜﺎﻓﺆ‬ ‫ﻋﻨﺪ‬ : = ⟹ = = × × = 2.25 × 10 ⁄ ℱ = ⟹ = ℱ. =2× 2.25 × 10 = 4.5 × 10 ⁄ ‫ﻭﻣﻨﻪ‬ : 7.92 ⁄ = = × ‫ﺍﻷﻭﻝ‬ ‫ﺍﻟﻔﻮﺝ‬ ‫ﻣﻊ‬ ‫ﻋﻠﻴﻪ‬ ‫ﺍﻟﻤﺘﺤﺼﻞ‬ ‫ﺍﻟﺘﺮﻛﻴﺰ‬ ‫ﻧﻔﺲ‬ ‫ﻭﻫﻮ‬ 5 ( ‫ﺍﻷﺳﻜﻮﺭﺑﻴﻚ‬ ‫ﺣﻤﺾ‬ ‫ﺃﻥ‬ ‫ﺍﺛﺒﺎﺕ‬ AH ‫ﺿﻌﻴﻒ‬ : ‫ﺍﻟﻤﺎﺀ‬ ‫ﻓﻲ‬ ‫ﺍﻹﻧﺤﻼﻝ‬ ‫ﺩﻟﺔ‬ ‫ﻣﻌﺎ‬ , O + ₂ = O ⁻ + ₃ ⁺ ) P =3 ( ‫ﺍﻷﻭﻟﻰ‬ ‫ﺍﻟﻄﺮﻳﻘﺔ‬ = = [ ₃ ⁺] = = . × = 0.02 < 1 ‫ﺍﻟﺜﺎﻧﻴﺔ‬ ‫ﺍﻟﻄﺮﻳﻘﺔ‬ 0.5 0.25 0.5 0.5 0.5 0.5 0.5 0.5 0.5 0.5 0.5 e n c y - e d u c a t i o n . c o m / e x a m s 3as.ency-education.com
  • 10. Page 4 sur 9 ‫ﺍ‬ ‫ﻟ‬ ‫ﻤ‬ ‫ﻮ‬ ‫ﺿ‬ ‫ﻮ‬ ‫ﻉ‬ ‫ﺍ‬ ‫ﻟ‬ ‫ﺜ‬ ‫ﺎ‬ ‫ﻧ‬ ‫ﻲ‬ [ O ] = − [ ₃ ⁺] = 4.5 × 10 − 10 = 4.4 × 10 ≠ 0 ⁄ 6 ( ……………….. ‫ﺇﻋﺘﺒ‬ ‫ﻳﻤﻜﻦ‬ ‫ﺗﺎﻡ‬ ‫ﺍﻟﺘﻔﺎﻋﻞ‬ ‫ﺎﺭ‬ 7 ( ( O / O⁻ ) < ( O / O⁻ ) ‫ﺍﻷﻗﻮﻯ‬ ‫ﻫﻮ‬ ‫ﺍﻷﺳﻜﻮﺭﺑﻴﻚ‬ ‫ﺣﻤﺾ‬ ‫ﻭﻣﻨﻪ‬ ‫ﻫﻮ‬ ‫ﺍﻟﻤﻨﺎﺳﺐ‬ ‫ﺍﻟﻜﺎﺷﻒ‬ : ‫ﻓﺘﺎﻟﻴﻦ‬ ‫ﺍﻟﻔﻴﻨﻮﻝ‬ 01 0.5 0.25 e n c y - e d u c a t i o n . c o m / e x a m s 3as.ency-education.com
  • 11. Page 5 sur 9 e n c y - e d u c a t i o n . c o m / e x a m s 3as.ency-education.com
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