141 66 1 MB
Arabic Pages 153
ﺣﺪﻭﺩ ﺍﻟﻔﻴﺰﻳﺎء ﺍﻟﻜﻼﺳﻴﻜﻴﺔ THE LIMITS OF CLASSICAL PHYSICS ﻤﻴﻼﺩ ﺍﻟﻨﻅﺭﻴﺔ ﺍﻟﻜﻤﻴﺔ The Dawn of the Quantum Theory
ﻓﻲ ﻨﻬﺎﻴﺔ ﺍﻟﻘﺭﻥ ﺍﻟﺘﺎﺴﻊ ﻋﺸﺭ ،ﺍﻋﺘﻘﺩ ﻜﺜﻴﺭ ﻤﻥ ﺍﻟﻌﻠﻤﺎﺀ ﺃﻥ ﻜل ﺍﻻﻜﺘـﺸﺎﻓﺎﺕ
ﺍﻟﻌﻠﻤﻴﺔ ﻗﺩ ﺘﻡ ﺇﻨﺠﺎﺯﻫﺎ ﻭﻓﻬﻤﻬﺎ ﻭﺃﻨﻪ ﻟﻡ ﻴﺒﻘﻰ ﺇﻻ ﺒﻌﺽ ﺍﻟﻤـﺴﺎﺌل ﺍﻟﺒـﺴﻴﻁﺔ ﺍﻟﺘـﻲ
ﺘﺤﺘﺎﺝ ﻟﻤﺯﻴﺩ ﻤﻥ ﺍﻹﻴﻀﺎﺡ .ﺇﻥ ﻫﺫﻩ ﺍﻟﻘﻨﺎﻋﺔ ﻭﻜﺎﻨﺕ ﻨﺎﺸﺌﺔ ﻤﻥ ﺍﻟﺘﻘﺩﻡ ﺍﻟﻌﻠﻤﻲ ﻓـﻲ
ﻤﺠﺎﻻﺕ ﺸﺘﻰ ﻭﺍﻟﺫﻱ ﺘﻤﺜل – ﻋﻠﻰ ﺴﺒﻴل ﺍﻟﻤﺜﺎل – ﻓﻲ ﻤﻴﻜﺎﻨﻴﻜﺎ ﻨﻴـﻭﺘﻥ Newton
ﻭﺍﻟﺘﻲ ﻁﹸﻭﺭﺕ ﺒﻭﺍﺴﻁﺔ ﺍﻟﻌﺎﻟﻤﺎﻥ ﻻﺠـﺭﺍﻨﺞ ﻭﻫـﺎﻤﻠﺘﻭﻥ J. LaGrange and W.
.Hamiltonﺤﻴﺙ ﺘﻡ ﺍﺴﺘﺨﺩﺍﻡ ﻫﺫﻩ ﺍﻟﻨﻅﺭﻴﺔ ﻟﻭﺼﻑ ﺤﺭﻜﺔ ﺍﻟﻜﻭﺍﻜﺏ ﻭﻜﺫﻟﻙ ﻓﻬـﻡ
ﻜﺜﻴﺭ ﻤﻥ ﺍﻟﻅﻭﺍﻫﺭ ﺍﻟﻤﻌﻘﺩﺓ ﻤﺜل ﻨﻅﺭﻴﺔ ﺍﻟﻤﺭﻭﻨـﺔ elasticity theoryﻭﺩﻴﻨﺎﻤﻴﻜـﺎ ﺍﻟﻤﻭﺍﺌﻊ hydrodynamicﺇﻨﺠﺎﺯﺍﺕ ﺍﻟﻌﺎﻟﻡ ﺠﻭل ﻭﺒﻴﺎﻥ ﺘﻜﺎﻓﺅ ﺍﻟـﺸﻐل ﻭﺍﻟﺤـﺭﺍﺭﺓ،
ﺃﺒﺤﺎﺙ ﻜﺎﺭﻨﻭﺕ Carnotﻭﺍﻟﺘﻲ ﺃﺩﺕ ﻟﻔﻬﻡ ﺍﻹﻨﺘﺭﻭﺒﻲ ﻭﺍﻟﻘﺎﻨﻭﻥ ﺍﻟﺜـﺎﻨﻲ ﻟﻠـﺩﻴﻨﺎﻤﻴﻜﺎ ﺍﻟﺤﺭﺍﺭﻴﺔ ،ﻭﻤﺎ ﻴﺘﺒﻊ ﻫﺫﻩ ﺍﻷﺒﺤﺎﺙ ﻤﻥ ﺘﻁﻭﻴﺭ ﻋﻠﻰ ﻴﺩ ﺍﻟﻌـﺎﻟﻡ ﺠـﺒﺱ J. Gibbs
ﻹﺭﺴﺎﺀ ﺃﺴﺱ ﻋﻠﻡ ﺍﻟﺩﻴﻨﺎﻤﻴﻜﺎ ﺍﻟﺤﺭﺍﺭﻴﺔ.
ﻭﺸﻬﺩﺕ ﻤﺠـﺎﻻﺕ ﺃﺨـﺭﻯ ﻤـﻥ ﺍﻟﻔﻴﺯﻴـﺎﺀ )ﻤﺜـل ﺍﻟـﻀﻭﺀ ﻭﺍﻟﻨﻅﺭﻴـﺔ
ﺍﻟﻜﻬﺭﻭﻤﻐﻨﺎﻁﻴﺴﻴﺔ (optics and electromagnetic theoryﺇﻨﺠﺎﺯﺍﺕ ﻤﻠﺤﻭﻅـﺔ. ﻓﻤﺜﻼﹰ ،ﺍﻻﺴﺘﻨﺘﺎﺠﺎﺕ ﺍﻟﻬﺎﻤﺔ ﺍﻟﺘﻲ ﺘﻭﺼل ﺇﻟﻴﻬﺎ ﺍﻟﻌﺎﻟﻡ ﻤﺎﻜﺴﻭﻴل J. Maxwellﻤﺘﻤﺜﻠﺔ
ﺒﻤﻌﺎﺩﻻﺘﻪ ﺍﻟﺸﻬﻴﺭﺓ "ﻭﺍﻟﺒﺴﻴﻁﺔ" ﻭﺍﻟﺘـﻲ ﻭﺤـﺩﺕ ﻤﺠـﺎﻻﺕ ﺍﻟـﻀﻭﺀ ﻭﺍﻟﻜﻬﺭﺒﻴـﺔ
ﻭﺍﻟﻤﻐﻨﺎﻁﻴﺴﻴﺔ ﻭﻤﺎ ﻴﺘﺒﻊ ﻫﺫﻩ ﺍﻷﺒﺤﺎﺙ ﻤﻥ ﺍﻟﺘﺠﺎﺭﺏ ﺍﻟﻤﻌﻤﻠﻴﺔ ﺒﻭﺍﺴﻁﺔ ﺍﻟﻌﺎﻟﻡ ﻫﻴﺭﺘﺯ H. Hertzﻓﻲ ﻋﺎﻡ 1887ﻭﺍﻟﺘﻲ ﺃﺩﺕ ﺇﻟﻰ ﺇﺜﺒﺎﺕ ﺍﻟﻁﺒﻴﻌﺔ ﺍﻟﻤﻭﺠﺒﺔ ﻟﻠﻀﻭﺀ.
-٢PDF created with pdfFactory Pro trial version www.pdffactory.com
ﻜل ﻫﺫﻩ ﺍﻹﻨﺠﺎﺯﺍﺕ ﻓﻲ ﺍﻟﻤﺠﺎﻻﺕ ﺍﻟﻤﺨﺘﻠﻔﺔ ﻟﻠﻔﻴﺯﻴﺎﺀ ﻜﻭﻨﺕ ﻤﺎ ﻴﻌـﺭﻑ ﺍﻵﻥ ﺒﺎﻟﻔﻴﺯﻴﺎﺀ ﺍﻟﺘﻘﻠﻴﺩﻴﺔ Classical physicsﻭﻤﻊ ﺒﺩﺍﻴﺔ ﺍﻟﻘﺭﻥ ﺍﻟﻌﺸﺭﻴﻥ ،ﻭﺠﺩﺕ ﺒﻌﺽ ﺍﻟﻨﺘﺎﺌﺞ ﺍﻟﺘﺠﺭﻴﺒﻴﺔ ﺍﻟﺠﺩﻴﺩﺓ ﻭﺍﻟﺘﻲ ﺍﺴﺘﻠﺯﻡ ﺘﻔﺴﻴﺭﻫﺎ ﻤﻔﺎﻫﻴﻡ ﻓﻴﺯﻴﺎﺌﻴﺔ ﺠﺩﻴﺩﺓ ﺘﺘﻨـﺎﻗﺽ
ﻤﻊ ﻤﺒﺎﺩﺉ ﺍﻟﻔﻴﺯﻴﺎﺀ ﺍﻟﺠﺩﻴﺩﺓ ﻭﻟﺩ ﻤﺎ ﻴﺴﻤﻰ ﺍﻵﻥ ﺒﺎﻟﻨﻅﺭﻴﺔ ﺍﻟﻜﻤﻴﺔ quantum theory
ﻭﺴﻨﺤﺎﻭل ﻓﻲ ﻫﺫﺍ ﺍﻟﻔﺼل ﺃﻥ ﻨﺼﻑ ﺨﻠﻔﻴﺔ ﻫﺫﻩ ﺍﻷﺯﻤﺎﺕ ﻟﻨـﺼل ﻤـﻥ ﺨﻼﻟﻬـﺎ ﻟﻤﻌﺭﻓﺔ ﺍﻟﻨﻅﺭﻴﺔ ﺍﻟﻜﻤﻴﺔ.
ﻭﻴﻤﻜﻨﻨﺎ ﺘﻠﺨﻴﺹ ﺍﻟﻤﻔﺎﻫﻴﻡ ﺍﻟﻔﻴﺯﻴﺎﺌﻴﺔ ﺍﻟﺠﺩﻴـﺩﺓ ﻓـﻲ :ﺍﻟﺨـﻭﺍﺹ ﺍﻟﺠـﺴﻴﻤﻴﺔ ﻟﻺﺸﻌﺎﻉ ،the particle properties of radiationﺍﻟﺨﻭﺍﺹ ﺍﻟﻤﻭﺠﻴﺔ ﻟﻠﻤﺎﺩﺓ the
،wave properties of matterﻭﺘﻜﻤﻴﻡ ﺍﻟﻜﻤﻴﺎﺕ ﺍﻟﻔﻴﺯﻴﺎﺌﻴﺔ the quantization of
.physical quantitiesﻭﺴﻨﻘﻭﻡ ﻓﻲ ﻫﺫﺍ ﺍﻟﻔﺼل ﺒﻤﻨﺎﻗﺸﺔ ﻫﺫﻩ ﺍﻟﻤﻔﺎﻫﻴﻡ. .1-1ﺇﺸﻌﺎﻉ ﺍﻟﺠﺴﻡ "ﺍﻷﺴﻭﺩ"
1-1 Blackbody Radiation
ﺍﻟﻔﻴﺯﻴﺎﺀ ﺍﻟﺘﻘﻠﻴﺩﻴﺔ ﻟﻡ ﺘﺘﻤﻜﻥ ﻤﻥ ﺸﺭﺡ ﺇﺸﻌﺎﻉ ﺍﻟﺠﺴﻡ ﺍﻷﺴﻭﺩ- Blackbody Radiation could not be explained by classical physics
ﻤﻥ ﺃﻫﻡ ﺍﻟﻨﺘﺎﺌﺞ ﺍﻟﺘﺠﺭﻴﺒﻴﺔ ﺍﻟﺘﻲ ﺃﺤﺩﺜﺕ ﺜﻭﺭﺓ ﻓﻲ ﺍﻟﻤﻔﺎﻫﻴﻡ ﺍﻟﻔﻴﺯﻴﺎﺌﻴﺔ ﺍﻟﺘﻘﻠﻴﺩﻴﺔ ﺘﻠﻙ ﺍﻟﻤﺘﻌﻠﻘﺔ ﺒﺎﻹﺸﻌﺎﻉ ﺍﻟﺼﺎﺩﺭ ﻤﻥ ﺍﻷﺠﺴﺎﻡ ﻋﻨﺩ ﺘﺴﺨﻴﻨﻬﺎ .ﻓﻤـﻥ ﺍﻟﻤﻌﻠـﻭﻡ ﻋﻨـﺩ
ﺘﺴﺨﻴﻥ ﺠﺴﻡ ﻤﺎ ،ﻨﺠﺩ ﺃﻥ ﻟﻭﻨﻪ ﻴﺘﻐﻴﺭ ﻤﻊ ﺯﻴﺎﺩﺓ ﺩﺭﺠﺔ ﺍﻟﺤﺭﺍﺭﺓ ﺤﻴﺙ ﻴﺒﺩﺃ ﺒﺎﻷﺤﻤﺭ
ﺜﻡ ﺍﻷﺒﻴﺽ ﺜﻡ ﺍﻷﺯﺭﻕ .ﻭﺒﺩﻻﻟﺔ ﺍﻟﺘﺭﺩﺩ ،ﻨﻘﻭل ﺃﻥ ﺍﻹﺸﻌﺎﻉ ﺍﻟﻤﻨﺒﻌﺙ ﻤﻥ ﻫﺫﺍ ﺍﻟﺠﺴﻡ
ﻴﺒﺩﺃ ﺒﺘﺭﺩﺩﺍﺕ ﻤﻨﺨﻔﻀﺔ ،ﻭﻋﻨﺩ ﺍﺭﺘﻔﺎﻉ ﺩﺭﺠﺔ ﺍﻟﺤﺭﺍﺭﺓ ،ﺘﺯﺩﺍﺩ ﺍﻟﺘﺭﺩﺩﺍﺕ ،ﺤﻴﺙ ﺃﻥ ﺍﻟﻠﻭﻥ ﺍﻷﺤﻤﺭ ﺫﻭ ﺘﺭﺩﺩ ﻗﻠﻴل ﻓﻲ ﻤﻨﻁﻘﺔ ﻁﻴﻑ ﺍﻹﺸﻌﺎﻉ ﻭﺫﻟـﻙ ﻤﻘﺎﺭﻨـﺔ ﺒـﺎﻟﻠﻭﻥ
ﺍﻷﺯﺭﻕ .ﺇﻥ ﻁﻴﻑ ﺍﻟﺘﺭﺩﺩ ﻟﻺﺸﻌﺎﻉ ﺍﻟﻤﻨﺒﻌﺙ ﻤﻥ ﺠﺴﻡ ﻤﺎ ﻴﻌﺘﻤﺩ ﻋﻠﻰ ﻁﺒﻴﻌﺔ ﺍﻟﺠﺴﻡ ﻨﻔﺴﻪ ،ﻭﻟﻜﻥ ﺍﻟﺠﺴﻡ ﺍﻟﻤﺜﺎﻟﻲ ،deal bodyﻭﺍﻟﺫﻱ ﻴﻤﺘﺹ ﺃﻭ ﻴﺒﻌﺙ ﻜل ﺍﻟﺘـﺭﺩﺩﺍﺕ
-٣PDF created with pdfFactory Pro trial version www.pdffactory.com
ﻴﺴﻤﻰ ﺒﺎﻟﺠﺴﻡ ﺍﻷﺴﻭﺩ ﻭﻴﻌﺘﺒﺭ ﺤﺎﻟﺔ ﻤﺜﺎﻟﻴﺔ ﻷﻱ ﻤﺎﺩﺓ ﺘﹸـﺼﺩﺭ ﺇﺸـﻌﺎﻉ .ﺍﻹﺸـﻌﺎﻉ ﺍﻟﻤﻨﺒﻌﺙ ﻤﻥ "ﺠﺴﻡ ﺃﺴﻭﺩ" ﻴﺴﻤﻰ ﺇﺸﻌﺎﻉ ﺍﻟﺠﺴﻡ ﺍﻷﺴﻭﺩ. an ideal body, which absorbs and emits all frequencies, is called a blackbody and serves as an idealization for any radiating material, the radiation emitted by a blackbody is called blackbody radiation.
ﺸﻜل 1-1ﻴﻭﻀﺢ ﺘﻐﻴﺭ ﺸﺩﺓ ﺍﻹﺸﻌﺎﻉ ﺍﻟﺼﺎﺩﺭ ﻤﻥ ﺠﺴﻡ ﺃﺴﻭﺩ ﻤﻊ ﺍﻟﺘـﺭﺩﺩ
ﻭﺫﻟﻙ ﻋﻨﺩ ﺩﺭﺠﺎﺕ ﺤﺭﺍﺭﺓ ﻤﺨﺘﻠﻔﺔ .ﻭﻗﺩ ﺤﺎﻭل ﺍﻟﻌﺩﻴﺩ ﻤﻥ ﺍﻟﻔﻴﺯﻴـﺎﺌﻴﻴﻥ ﺍﺴـﺘﻨﺘﺎﺝ ﻤﻌﺎﺩﻟﺔ ﺭﻴﺎﻀﻴﺔ ﺘﺸﺭﺡ ﺍﻟﻨﺘﺎﺌﺞ ﺍﻟﺘﺠﺭﻴﺒﻴﺔ )ﻓﻲ ﺸﻜل (1-1ﻭﻟﻜﻥ ﺒﺩﻭﻥ ﺘﻭﺍﻓﻕ ﻜﺎﻤل
ﻭﺃﻭﻟﻰ ﺍﻟﻤﺤﺎﻭﻻﺕ ﻟﻭﺼﻑ ﻫﺫﻩ ﺍﻟﻨﺘﺎﺌﺞ ﻗﺎﻡ ﺒﻬﺎ ﻜل ﻤﻥ ﺍﻟﻌﺎﻟﻤﻴﻥ ﺭﺍﻟـﻲ ﻭﺠﻴﻨـﺯ Rayleigh and Jeansﻭﺍﻟﺘﻲ ﺍﺸﺘﻘﺕ ﺒﻨﺎﺀﺍﹰ ﻋﻠﻰ ﻗﻭﺍﻨﻴﻥ ﺍﻟﻘﺭﻥ ﺍﻟﺘﺎﺴﻊ ﻋﺸﺭ ﻭﻴﻤﻜﻥ ﻜﺘﺎﺒﺔ ﻫﺫﻩ ﺍﻟﻌﻼﻗﺔ ﺒﺎﻟﺼﻴﻐﺔ: )(1-1
ν2
ﺤﻴﺙ
)u (ν , T
8D K B T 3
C
= )u (ννT
ﻜﺜﺎﻓﺔ ﺍﻟﻁﺎﻗﺔ energy densityﻭﻭﺤﺩﺘﻬﺎ ﺠـﻭل ﻟﻜـل ﻤﺘـﺭ
ﻤﻜﻌﺏ ) .(J/m3ﻓﻲ ﺍﻟﻤﻌﺎﺩﻟﺔ )T ،(1-1ﺩﺭﺠﺔ ﺍﻟﺤﺭﺍﺭﺓ ﺒﺎﻟﻜﻴﻠـﭭﻥ c ،ﺴﺭﻋﺔ ﺍﻟﻀﻭﺀ،
KBﺜﺎﺒﺕ ﺒﻭﻟﺘﺯﻤﺎﻥ .ﺍﻟﺨﻁ ﺍﻟﻤﺘﻘﻁﻊ ﻓﻲ ﺸﻜل 1-1ﻴﺒﻴﻥ ﺍﻟﻌﻼﻗﺔ ﺤـﺴﺏ ﻤﻌﺎﺩﻟـﺔ
ﺭﺍﻟﻲ -ﺠﻴﻨﺯ .ﻻﺤﻅ ﺍﻟﺘﻭﺍﻓﻕ ﺒﻴﻥ ﻫﺫﻩ ﺍﻟﻌﻼﻗﺔ ﻭﺍﻟﻨﺘﺎﺌﺞ ﺍﻟﺘﺠﺭﻴﺒﻴﺔ ﻋﻨﺩ ﺍﻟﺘـﺭﺩﺩﺍﺕ ﺍﻟﻤﻨﺨﻔﻀﺔ .ﻋﻨﺩ ﺍﻟﺘﺭﺩﺩﺍﺕ ﺍﻟﻌﺎﻟﻴﺔ ،ﻭﺤﺴﺏ ﻤﻌﺎﺩﻟﺔ ﺭﺍﻟـﻲ-ﺠﻴﻨـﺯ ،ﻓـﺈﻥ ﻜﺜﺎﻓـﺔ ﺍﻹﺸﻌﺎﻉ ﺘﺯﺩﺍﺩ ﻁﺒﻘﺎﹰ ﻟـ
ν2
ﻭﺘﺼل ﺇﻟﻰ ﻤﺎﻻ ﻨﻬﺎﻴﺔ ﻭﺫﻟﻙ ﻋﻨﺩﻤﺎ ﺘﺼل ﺍﻟﺘـﺭﺩﺩﺍﺕ
ﺇﻟﻰ ﻤﺎﻻ ﻨﻬﺎﻴﺔ ﻭﻫﺫﺍ ﻴﺤﺩﺙ ﻓﻲ ﻤﻨﻁﻘﺔ ﺍﻷﺸﻌﺔ ﻓﻭﻕ ﺍﻟﺒﻨﻔﺴﺠﻴﺔ ultravioletﻭﻫـﺫﺍ
ﻤﺎ ﻴﻌﺭﻑ ﺒﺎﻻﻨﻬﻴﺎﺭ ﻓﻭﻕ ﺒﻨﻔﺴﺠﻲ .ultraviolet catastropheﻤﻥ ﺍﻟﻤﻌﺎﺩﻟﺔ )(1-1
ﻭﻟﻜل ﺍﻟﺘﺭﺩﺩﺍﺕ ﻤﻥ ﺼﻔﺭ ﺇﻟﻰ ﻤﺎﻻ ﻨﻬﺎﻴﺔ.
8 Dk B T 2 ∞ → ν dν C3
∞
∞
°
°
∫ = ∫ u (ν1T) dν
-٤PDF created with pdfFactory Pro trial version www.pdffactory.com
ﻭﻫﺫﻩ ﺍﻟﻨﺘﻴﺠﺔ ﺘﺘﻨﺎﻗﺽ ﻤﻊ ﺍﻟﻨﺘﺎﺌﺞ ﺍﻟﺘﺠﺭﻴﺒﻴﺔ ﺤﻴﺙ ﺃﻥ ﺸﺩﺓ ﺍﻹﺸﻌﺎﻉ ﺘﺯﺩﺍﺩ ﻤﻊ ﺜﻡ ﺘﻘل ﺇﻟﻰ
ν max
ﺯﻴﺎﺩﺓ ﺍﻟﺘﺭﺩﺩ ﻟﺘﺼل ﺇﻟﻰ ﺃﻗﺼﻰ ﻗﻴﻤﺔ ﻋﻨﺩ ﺘﺭﺩﺩ ﻤﻌﻴﻥ ﻨﺭﻤﺯ ﻟﻪ ﺒـ
ﻭﻤﻥ ﺍﻟﺠـﺩﻴﺭ ﺒﺎﻟـﺫﻜﺭ.ﺍﻟﺼﻔﺭﺓ ﻫﺫﺍ ﻴﻌﻨﻲ ﺃﻥ ﻗﻴﻤﺔ ﺍﻟﺘﻜﺎﻤل ﻻ ﺘﺴﺎﻭﻱ ﻤﺎﻻ ﻨﻬﺎﻴﺔ .ﺘﺘﻐﻴﺭ ﻤﻊ ﺘﻐﻴﺭ ﺩﺭﺠﺔ ﺍﻟﺤﺭﺍﺭﺓ ﻜﻤﺎ ﻫﻭ ﻤﻭﻀﺢ ﺒﺎﻟﺸﻜل
ν max
ﻜﺫﻟﻙ ﻤﻼﺤﻅﺔ ﺃﻥ
FIGURE 1.1 Spectral distribution of the intensity of blackbody radiation as a function of frequency for several temperatures. The intensity is given in arbitrary units. The dashed line is the prediction of classical physics. As the temperature increases. The maximum shifts to higher frequencies and the total radiated energy (the area under each curve) increases significantly. Note that the horizontal axis is labeled by 14 -1 ν /10 s . This notation means that the dimensionless numbers on that axis re frequencies divided by 1014 s-1. We shall use this notation to label columns in tables and axes in figures because of its unambiguous nature and algebraic convenience.
-٥PDF created with pdfFactory Pro trial version www.pdffactory.com
ﻜﻴﻑ ﺘﻡ ﺤل ﻫﺫﺍ ﺍﻹﺸﻜﺎل ﺒﻴﻥ ﺍﻟﻨﻅﺭﻴﺔ ﻭﺍﻟﺘﺠﺎﺭﺏ ﺍﻟﻌﻤﻠﻴﺔ؟ 2-1ﺘﻭﺯﻴﻊ ﺒﻼﻨﻙ ﻭﺘﻜﻤﻴﻡ ﺍﻟﻁﺎﻗﺔ 1-2 The Planck Distribution and the Quantum of Energy
ﺇﻥ ﺃﻭل ﻤﻥ ﻗﺩﻡ ﺘﻔﺴﻴﺭ ﺼﺤﻴﺢ ﻹﺸﻌﺎﻉ ﺍﻟﺠﺴﻡ ﺍﻷﺴﻭﺩ ﻫﻭ ﺍﻟﻌﺎﻟﻡ ﺍﻷﻟﻤـﺎﻨﻲ
ﻤﺎﻜﺱ ﺒﻼﻨﻙ Max Planckﻓﻲ ﻋﺎﻡ .1900ﻭﻓﻲ ﻨﻅﺭﻴﺘﻪ ،ﺍﻓﺘـﺭﺽ ﺒﻼﻨـﻙ ﺃﻥ ﺍﻹﺸﻌﺎﻉ ﺍﻟﻤﻨﺒﻌﺙ ﻤﻥ ﺍﻟﺠﺴﻡ ﺍﻷﺴﻭﺩ ﻤﻥ ﺍﻫﺘﺯﺍﺯ ﺍﻹﻟﻜﺘﺭﻭﻨـﺎﺕ ﺍﻟﻤﻜﻭﻨـﺔ ﻟﻤـﺎﺩﺓ ﺍﻟﺠﺴﻡ .ﻭﻟﻜﻭﻥ ﻫﺫﻩ ﺍﻻﻫﺘﺯﺍﺯﺍﺕ ﺫﺍﺕ ﺘﺭﺩﺩﺍﺕ ﻋﺎﻟﻴﺔ ﻓﺈﻨﻨﺎ ﻨﺠﺩ ﻓﻲ ﻁﻴﻑ ﺍﻹﺸﻌﺎﻉ
ﺍﻟﻤﻨﺒﻌﺙ ﺘﺭﺩﺩﺍﺕ ﻓﻲ ﻤﻨﻁﻘﺔ ﺍﻟﻀﻭﺀ ﺍﻟﻤﺭﺌﻲ ﻭﺍﻷﺸﻌﺔ ﺘﺤـﺕ ﺍﻟﺤﻤـﺭﺍﺀ ﻭﻓـﻭﻕ
ﺍﻟﺒﻨﻔﺴﺠﻴﺔ ﺒﻴﻨﻤﺎ ﻻ ﻨﺠﺩ ﺃﻱ ﻤﻥ ﺘﺭﺩﺩﺍﺕ ﺍﻟﺭﺍﺩﻴﻭ ﻓﻲ ﻫﺫﺍ ﺍﻟﻁﻴﻑ .ﻁﺒﻘـﺎﹰ ﻟﻨﻅﺭﻴـﺔ
ﺭﺍﻟﻲ-ﺠﻴﻨﺯ ،ﻓﺈﻨﻪ ﻤﻔﻬﻭﻡ ﻅﻤﻨﺎﹰ ﺃﻥ ﻁﺎﻗﺔ ﺍﻹﻟﻜﺘﺭﻭﻨﺎﺕ ﺍﻟﻤﻬﺘﺯﺓ ،ﻭﺍﻟﺘﻲ ﻫﻲ ﺴـﺒﺏ ﺍﻨﺒﻌﺎﺙ ﺍﻹﺸﻌﺎﻉ ﻤﻥ ﺍﻟﻤﺎﺩﺓ -ﻤﺴﻤﻭﺡ ﻟﻬﺎ ﺃﻥ ﺘﺄﺨﺫ ﺃﻱ ﻗﻴﻤﺔ ﻤﻥ ﺍﻟﻁﺎﻗـﺔ .ﻭﻫـﺫﻩ
ﺍﻟﻔﺭﻀﻴﺔ ﻫﻲ ﺇﺤﺩﻯ ﺍﻷﺴﺎﺴﻴﺎﺕ ﺍﻟﻔﺭﻀﻴﺔ ﻓﻲ ﺍﻟﻔﻴﺯﻴﺎﺀ ﺍﻟﺘﻘﻠﻴﺩﻴـﺔ .ﻓـﻲ ﺍﻟﻔﻴﺯﻴـﺎﺀ ﺍﻟﺘﻘﻠﻴﺩﻴﺔ ،ﺍﻟﻜﻤﻴﺎﺕ ﺍﻟﻔﻴﺯﻴﺎﺌﻴﺔ ﺍﻟﻤﺘﻐﻴﺭﺓ ﻭﺍﻟﺘﻲ ﺘﹸﻤﺜـل ﻤـﺸﺎﻫﺩﺍﺕ )ﻤﺜـل ﺍﻟﻤﻭﻗـﻊ
،Positionﻜﻤﻴﺔ ﺍﻟﺤﺭﻜﺔ Momentumﻭﺍﻟﻁﺎﻗﺔ (energyﻴﻤﻜﻥ ﺘﻤﻠﻙ ﻗﻴﻡ ﻤﺘﺼﻠﺔ. In classical physics, the variables that represent observables (such as position, momentum and energy) can take on a continuum of values.
ﻭﻟﻘﺩ ﺃﺩﺭﻙ ﺍﻟﻌﺎﻟﻡ ﺒﻼﻨﻙ – ﺒﻌﻤﻕ ﺘﻔﻜﻴﺭﻩ – ﺒﻀﺭﻭﺭﺓ ﺇﺤﺩﺍﺙ ﺘﻐﻴﻴﺭ ﺠﺫﺭﻱ ﻭﺠﻭﻫﺭﻱ ﻓﻲ ﻫﺫﺍ ﺍﻟﻤﻔﻬﻭﻡ ﺍﻟﻔﻴﺯﻴﺎﺌﻲ ﻓﻜﺎﻨﺕ ﻓﺭﻀﻴﺘﻪ ﺍﻻﻨﻘﻼﺒﻴـﺔ ﻓـﻲ ﺍﻟﻔﻴﺯﻴـﺎﺀ
ﺍﻟﺤﺩﻴﺜﺔ :ﻁﺎﻗﺔ ﺍﻹﻟﻜﺘﺭﻭﻨﺎﺕ ﺍﻟﻤﻬﺘﺯﺓ ﻤﻜﻤﻤﺔ ﻭﻗﻴﻤﻬﺎ ﻏﻴﺭ ﻤﺘﺼﻠﺔ ﻭﺘﺘﻨﺎﺴﺏ ﺒـﺭﻗﻡ
ﻜﻤﻲ ﺼﺤﻴﺢ ﻤﻊ ﺍﻟﺘﺭﺩﺩ ﻭﺫﻟﻙ ﻤﻥ ﺨﻼل ﺍﻟﻤﻌﺎﺩﻟﺔ
E = nhν
-٦PDF created with pdfFactory Pro trial version www.pdffactory.com
Energies of the oscillators were discrete and had to be proportional to an integral multiple of the frequently in an equation of the form E = nhv
ﺤﻴﺙ Eﻫﻲ ﻁﺎﻗﺔ ﺍﻟﻤﺘﺫﺒﺫﺏ n ،ﻫﻭ ﺭﻗﻡ ﺼﺤﻴﺢ h ،ﺜﺎﺒﺕ ﺍﻟﺘﻨﺎﺴﺏ ﻭﻴﻌﺭﻑ ﺒﺜﺎﺒﺕ ﺒﻼﻨﻙ ،ﻭ vﻫﻭ ﺘﺭﺩﺩ ﺍﻟﻤﺘﺫﺒﺫﺏ. ﻭﺒﻨﺎﺀﺍﹰ ﻋﻠﻰ ﻤﺒﺩﺃ ﺘﻜﻤﻴﻡ ﺍﻟﻁﺎﻗﺔ ﻭﻤﻔﺎﻫﻴﻡ ﺩﻴﻨﺎﻤﻴﻜﺎ ﺤﺭﺍﺭﻴﺔ ﺇﺤﺼﺎﺌﻴﺔ ،ﺘﻤﻜـﻥ
ﺒﻼﻨﻙ ﻤﻥ ﺍﺴﺘﻨﺘﺎﺝ ﺍﻟﻌﻼﻗﺔ ﺍﻟﺭﻴﺎﻀﻴﺔ ﺍﻟﺘﺎﻟﻴﺔ:
8 Dh ν3 u (ν1 T) = 3 hv/k T c e B −1
)(1-2
ﻭﻫﺫﻩ ﺍﻟﻌﻼﻗﺔ ﺘﺘﻔﻕ ﺘﻤﺎﻤﺎﹰ ﻤﻊ ﺍﻟﻨﺘﺎﺌﺞ ﺍﻟﺘﺠﺭﻴﺒﻴﺔ ﻋﻨﺩ ﻜل ﺍﻟﺘﺭﺩﺩﺍﺕ ﻭﺩﺭﺠﺎﺕ ﺍﻟﺤﺭﺍﺭﺓ ﺍﻟﻤﻌﺎﺩﻟﺔ ) (1-2ﺘﹸﻌﺭﻑ ﺒﺘﻭﺯﻴﻊ ﺒﻼﻨﻙ ﻹﺸﻌﺎﻉ ﺍﻟﺠﺴﻡ ﺍﻷﺴﻭﺩ Planck distribution lance for black body radiation.
) (iﻋﻨﺩ ﺍﻟﺘﺭﺩﺩﺍﺕ ﺍﻟﻤﻨﺨﻔﻀﺔ ،ﺘﺅﻭل ﻤﻌﺎﺩﻟﺔ ﺒﻼﻨﻙ ﺇﻟﻰ ﻤﻌﺎﺩﻟﺔ ﺭﺍﻟﻲ-ﺠﻴﻨﺯ ﺒﺎﻟﻨﻅﺭ ﺇﻟﻰ ﺍﻟﻨﺴﺒﺔ
hv k BT
ﻓﻌﻨﺩﻤﺎ ﺘﻜﻭﻥ vﺼﻐﻴﺭﺓ ﻓﺈﻥ
ﻤﻥ ﻤﻜﻨﻭﻥ ﺘﺎﻴﻠﻭﺭ ﻟﻠﺩﺍﻟﺔ ﺍﻷﺴﻴﺔ
v2 + ... 2i
hv 〈〈 1 k BT e v =1+ v +
ﺇﺫﺍ ﻜﺎﻨﺕ vﺼﻐﻴﺭﺓ ،ﻴﻤﻜﻨﻨﺎ ﺇﻫﻤﺎل ﺍﻟﺤﺩﻭﺩ ﺫﺍﺕ ﺍﻷﺴـﺱ ﺍﻟﻌﻠﻴـﺎ ﻭﻋﻠﻴـﻪ e v ≈1+ v
ﻭﺫﻟﻙ ﺇﺫﺍ ﻜﺎﻨﺕ
v 〈〈 1
ﻓﻲ ﻫﺫﻩ ﺍﻟﺤﺎﻟﺔ ﻴﻤﻜﻨﻨﺎ ﻜﺘﺎﺒﺔ ﻤﻌﺎﺩﻟﺔ ﺒﻼﻨﻙ
8Dh v3 8Dh v 3 u (v1 T) = 3 = c 1 + hν − 1 c 3 hv vT k BT 8D K T v3 c3 B
)(1-3
= )u (v1 T
ﻭﻫﺫﻩ ﻫﻲ ﻤﻌﺎﺩﻟﺔ ﺭﺍﻟﻲ -ﺠﻴﻨﺯ. ) (iiﺍﻟﻁﺎﻗﺔ ﺍﻟﻜﻠﻴﺔ
-٧PDF created with pdfFactory Pro trial version www.pdffactory.com
∞
∞
v3 8Dh u (t) = ∫ u (v1T) dv = 3 ∫ dv hv/k T c ° e B −1 °
ﻭﻹﺠﺭﺍﺀ ﻫﺫﺍ ﺍﻟﺘﻜﺎﻤل ﻨﻌﻭﺽ ﻋـﻥ ﻭﻴﻤﻜﻨﻨﺎ ﻜﺘﺎﺒﺔ dvﺒﺩﻻﻟﺔ dvﻭ ﻋﻥ vﺒـ
v
k BT h
dv
k BT h
hν k BT
= ، vﻫـﺫﺍ ﻴﻌﻨـﻲ ﺃﻥ
h dv k BT
= dv
= . dvﺤﺩﻭﺩ ﺍﻟﺘﻜﺎﻤل ﻟﻡ ﺘﺘﻐﻴﺭ ،ﺒـﺎﻟﺘﻌﻭﻴﺽ
ﻭﻜﺫﻟﻙ dvﺒﺩﻻﻟﺔ dvﻨﺘﺤﺼل ﻋﻠﻰ ∞
8Dh kT v 3 kT u (v, T) = 3 ∫ ( ) 3 v ( ) dv c ° h e −1 h ∞ ∞ 8Dh kT 4 v 3 8Dh k B T 4 v 3 ( ) dv = ( ) ∫° e v −1 dv D c 3 ∫° h e v − 1 c3
ﺍﻟﺘﻜﺎﻤل ∫ v dvﻗﻴﻤﺘﻪ ﺘﺴﺎﻭﻱ e −1 ∞
3
v
°
D4 15
8Dh k B T 4 D 4 8D 5 k B 4 4 ( ) T ( ) = h 15 15 c 3 h c3 )(1 − 4
ﺤﻴﺙ:
=
= )∴u (T
= aT 4 ............. 8D 5 k B 4 ) ( 15c 3 h
=a
ﺍﻟﻤﻌﺎﺩﻟﺔ ) (1-4ﺘﻤﺜل ﻤﺎ ﻴﻌـﺭﻑ ﺒﻘـﺎﻨﻭﻥ ﺴـﺘﻴﻔﺎﻥ -ﺒﻭﻟﺘﺯﻤـﺎﻥ Stefan-
Boltzmanﻟﻠﻁﺎﻗﺔ ﺍﻟﻜﻠﻴﺔ ﻟﻜل ﻭﺤﺩﺓ ﺍﻟﺤﺠﻭﻡ.
) (iiiﻤﻌﺎﺩﻟﺔ ﺒﻼﻨﻙ ﺘﻤﻜﻨﻨﺎ ﻤﻥ ﺇ – ﺝ ﻗﺎﻨﻭﻥ ﻭﺍﻴﻥ Wienﻭﺍﻟﺫﻱ ﻴﻨﺹ ﻋﻠﻰ ﺃﻨﻪ ﺇﺫﺍ ﻜﺎﻨﺕ
λ max
ﻀﺭﺏ )..(1-5
ﺘﻤﺜل ﺍﻟﻁﻭل ﺍﻟﻤﻭﺠﻲ ﺍﻟﻤﻘﺎﺒل ﻷﻗﺼﻰ ﻗﻴﻤﺔ ﻟﻜﺜﺎﻓﺔ ﺍﻟﻁﺎﻗﺔ uﻓﺈﻥ ﺤﺎﺼل
λ max
ﻓﻲ ﺩﺭﺠﺔ ﺍﻟﺤﺭﺍﺭﺓ Tﻴﻌﻁﻴﻨﺎ ﻗﻴﻤﺔ ﺜﺎﺒﺘﺔ ،ﺃﻱ: λ max T = 2.90 × 10 −3 m.K
ﻭﻟﻠﺤﺼﻭل ﻋﻠﻰ ﻤﻌﺎﺩﻟﺔ ) (1-5ﻤﻥ ﻤﻌﺎﺩﻟﺔ ﺒﻼﻨﻙ. ﻟﻜﻲ ﻨﻭﺠﺩ ﺍﻟﻘﻴﻤﺔ ﺍﻟﻘﺼﻭﻯ ﻟـ vﻭﺘﺴﺎﻭﻱ ﻫﺫﺍ ﺍﻟﺘﻔﺎﻀل ﺒﺎﻟﺼﻔﺭ ،ﺃﻱ
) u (ν
ﻴﻠﺯﻡ ﺃﻥ ﺘﻔﺎﻀل ﺍﻟﺩﺍﻟﺔ
) u (ν
ﺒﺎﻟﻨﺴﺒﺔ ﻟـ
) dv (ν =0 dv
-٨PDF created with pdfFactory Pro trial version www.pdffactory.com
or d 8Dh v3 =0 3 hv/k BT dv c e − 1
ﻭﻤﺭﺓ ﺃﺨﺭﻯ ﻨﻌﻭﺽ ﻋﻥ
hv k BT
= dv , v
k BT h
vν
ﻭﺒﻘﺴﻤﺔ ﻁﺭﻓﻲ ﺍﻟﻤﻌﺎﺩﻟﺔ ﻋﻠﻰ ﻜل ﺍﻟﺜﻭﺍﺒﺕ ﻨﺘﺤﺼل ﻋﻠﻰ. 3 (e v − 1) − ve v = o 3 e v − 3 = ve v
ﺒﺎﻟﻘﺴﻤﺔ ﻋﻠﻰ ev 3 − 3 e −v = v or 3 - v = 3e - v
ﻭﻴﻤﻜﻥ ﺤل ﻫﺫﻩ ﺍﻟﻤﻌﺎﺩﻟﺔ ﻤﻥ ﺨﻼل ﻨﻘﻁﺔ ﺍﻟﺘﻘﺎﻁﻊ ﺒﻴﻥ ﺍﻟﺩﺍﻟﺘﻴﻥ ) (3-vﻭ ) (3e-vﻜﻤﺎ ﻫﻭ ﻤﻭﻀﺢ ﺒﺎﻟﺭﺴﻡ ﻓﻲ ﺸﻜل )(١-٤
ﺸﻜل :1-2ﺤل ﺍﻟﻤﻌﺎﺩﻟﺔ 3 – x = 3 e-xﺒﺎﻟﺭﺴﻡ
-٩PDF created with pdfFactory Pro trial version www.pdffactory.com
ﻤﻥ ﺍﻟﺸﻜل ،ﻨﺠﺩ ﺃﻥ v = 2.82 or hv mar = 2.82 k BT or k BT
= v max
)(2.82 h h T c ∴ )= b (2.82 λ max h hc )(6.63 × 10 −34 J.s) (3.0 × 10 8 m/s = = 2.9 × 10 −3 m.K − 23 −1 2.82K B ) (2.82) (1.38× 10 J.v
= or λ max T
ﻭﻤﻥ ﺍﻟﺠﺩﻴﺭ ﺒﺎﻟﺫﻜﺭ ،ﺃﻥ ﻨﻅﺭﻴﺔ ﺇﺸﻌﺎﻉ ﺍﻟﺠﺴﻡ ﺍﻷﺴﻭﺩ ﺘﺴﺘﺨﺩﻡ ﻓﻲ ﻋﻠﻡ ﺍﻟﻔﻠﻙ ﻭﺫﻟﻙ ﻟﺘﺤﺩﻴﺩ ﺩﺭﺠﺔ ﺤﺭﺍﺭﺓ ﺍﻟﻨﺠﻭﻡ .ﺒﺎﻟﻨﻅﺭ ﻟﺸﻜل ﻭﺍﻟﺫﻱ ﻴﻭﻀﺢ ﻁﻴﻑ ﺇﺸـﻌﺎﻉ ﺍﻟﺸﻤﺱ ﻤﻘﺎﺱ ﻤﻥ ﻁﺒﻘﺎﺕ ﺍﻟﺠﻭ ﺍﻟﻌﻠﻴﺎ ﻟﻸﺭﺽ ،ﻨﺠﺩ ﺃﻥ ﻗﻴﻤﺔ
λ max = 500nm
،(1nm = 10-9 mﻭﺒﺎﺴﺘﺨﺩﺍﻡ ﻗﺎﻨﻭﻥ ﻭﺍﻴﻥ ﻴﺘﺒﻴﻥ ﺃﻥ ﺩﺭﺠﺔ ﺤﺭﺍﺭﺓ ﺍﻟﺸﻤﺱ. 2.9 ×10 −3 m.k 2.9 × 10 −3 m.k = = 5800k λ max 500 × 10 −3 m
)ﺤﻴﺙ
=T
ﺍﻟﻁﻴﻑ ﺍﻟﻜﻬﺭﻭﻤﻐﻨﺎﻁﻴﺴﻲ ﻷﺸﻌﺔ ﺍﻟﺸﻤﺱ.
- ١٠PDF created with pdfFactory Pro trial version www.pdffactory.com
FIGURE 1.3 The electromagnetic spectrum of the sun as measured in the upper atmosphere of the earth. A comparison of this figure with Figure 1.2 shows that the sun's surface radiates as a blackbody at a temperature of about 6000 k.
ﻭﺒﻌﺩﻤﺎ ﻋﺭﻓﻨﺎ ﺍﻟﻁﺒﻴﻌﺔ ﺍﻟﻜﻤﻴﺔ ﻟﻠﻁﺎﻗﺔ )ﻤﻤﺜﻠﺔ (quantaﻭﻫﻲ nvﻓﺈﻥ ﻫـﺫﺍ
ﻴﻌﻨﻲ ﺃﻥ ﺸﻌﺎﻉ ﻁﻭﻟﻪ ﺍﻟﻤﻭﺠﻲ
°
λ = 6000 A
)ﺤﻴﺙ
m
−10
(1 A =10ﻁﺎﻗﺘﻪ ﺘﺴﺎﻭﻱ. °
)hc (6.63 × 10 34 J.s) (3 × 10 8 m/s = = E = hv λ (6000 ×10 −10 m = 3.3 × 10 −19 J = 2.06 ev
ﻓﺈﺫﺍ ﻜﺎﻥ ﻫﺫﺍ ﺍﻟﺸﻌﺎﻉ ﺼﺎﺩﺭ ﻤﻥ ﻤﺼﺩﺭ ﻗﺩﺭﺘـﻪ 100ﻭﺍﻁ )) (Wﻭﺤـﺩﺓ
ﺍﻟﻭﺍﻁ Wattﻫﻲ ﺠﻭل ﻟﻜل ﺜﺎﻨﻴﺔ ،(W: Joul/secﻫﺫﺍ ﻤﻌﻨﺎﻩ ﺃﻥ ﺇﺫﺍ ﻜﺎﻨﺕ Nﺘﺭﻤﺯ
ﻟﻌﺩﺩ ﺍﻟﻜﻤﺎﺕ ﻓﻲ ﺍﻟﺜﺎﻨﻴﺔ ﺍﻟﻭﺍﺤﺩﺓ ،ﻓﺈﺫﺍ ﻜﺎﻨﺕ ﻜل ﻜﻡ ﻤﻥ ﺍﻟﻁﺎﻗﺔ ﺘـﺴﺎﻭﻱ hvﻓـﺈﻥ ﺍﻟﻘﺩﺭﺓ ).Power = N (hv ∴ P = 100 = W = N hv P 100W = =∴N = 3×10 20 guanta/sec −19 E 3.3× 10 J
ﻭﻫﺫﺍ ﺒﻼ ﺸﻙ ﻋﺩﺩ ﻜﺒﻴﺭ ﺠﺩﺍﹰ ﻤﻥ ﺍﻟﻜﻤﺎﺕ ﻤﻤﺎ ﻴﺅﺩﻱ ﺇﻟﻰ ﻋـﺩﻡ ﺇﺤـﺴﺎﺴﻨﺎ ﺒﺎﻟﻁﺒﻴﻌﺔ ﺍﻟﺠﺴﻴﻤﻴﺔ ﻟﻠﻀﻭﺀ )ﻋﻠﻰ ﺍﻓﺘﺭﺍﺽ ﺃﻥ ﻜل ﻜﻡ ﻤﻥ ﺍﻟﻁﺎﻗﺔ ﻫﻭ ﺠﺴﻴﻡ(. 3-1ﺍﻟﺘﺄﺜﻴﺭ ﺍﻟﻜﻬﺭﻭﻀﻭﺌﻲ
1-3 The Photoelectric Effect
ﻓﻲ ﻋﺎﻡ 1887, 1886ﻭﺒﻴﻨﻤﺎ ﻜﺎﻥ ﻴﺠﺭﻱ ﺘﺠﺎﺭﺒﻪ ﺍﻟﺘـﻲ ﺃﻜـﺩﺕ ﻨﻅﺭﻴـﺔ
ﻤﺎﻜﺴﻭﻴل ﺍﻟﺨﺎﺼﺔ ﺒﺎﻟﻁﺒﻴﻌﺔ ﺍﻟﻤﻭﺠﻴﺔ ﻟﻠﻀﻭﺀ ،ﺍﻜﺘﺸﻑ ﺍﻟﻔﻴﺯﻴﺎﺌﻲ ﺍﻷﻟﻤﺎﻨﻲ ﻫﻴﻨـﺭﻙ
ﻫﻴﺭﺘﺯ Heinrich Hertzﺃﻥ ﺍﻷﺸﻌﺔ ﺍﻟﻀﻭﺌﻴﺔ ﺍﻟﻔﻭﻕ ﺒﻨﻔـﺴﺠﻴﺔ ultraviolet light
ﺘﺴﺒﺏ ﺍﻨﻁﻼﻕ )ﺍﻨﺒﻌﺎﺙ( emissionﺍﻹﻟﻜﺘﺭﻭﻨﺎﺕ ﻤﻥ ﺴﻁﺢ ﻤﻌﺩﻥ ﺒﺎﻹﺸﻌﺎﻉ ﻴﺴﻤﻰ
ﺒﺎﻟﺘﺄﺜﻴﺭ ﺍﻟﻜﻬﺭﻭﻀﻭﺌﻲ.
ﻭﻁﺒﻘﺎﹰ ﻟﻘﻭﺍﻨﻴﻥ ﺍﻟﻔﻴﺯﻴﺎﺀ ﺍﻟﺘﻘﻠﻴﺩﻴﺔ ،ﻓﺈﻥ ﺍﻹﺸﻌﺎﻉ ﺍﻟﻜﻬﺭﻭﻤﻐﻨﺎﻁﻴﺴﻲ ﻋﺒﺎﺭﺓ ﻋﻥ ﻤﺠﺎل ﻜﻬﺭﺒﻲ ﻴﺘﺫﺒﺫﺏ ﻋﻤﻭﺩﻴﺎﹰ ﻋﻠﻰ ﺍﺘﺠﺎﻩ ﺍﻨﺒﻌﺎﺙ ﺍﻹﺸﻌﺎﻉ )ﻫﻨﺎ ﺃﻫﻤﻠﻨـﺎ ﺍﻟﻤﺠـﺎل - ١١PDF created with pdfFactory Pro trial version www.pdffactory.com
ﺍﻟﻤﻐﻨﺎﻁﻴﺴﻲ( .ﻭﺍﻟﺫﻱ ﻨﺭﻴﺩ ﺃﻥ ﻨﺭﻜﺯ ﻋﻠﻴﻪ ﻫﻨﺎ ﺃﻥ ﺸﺩﺓ ﺍﻷﺸﻌﺔ intensity of the
radiationﺘﺘﻨﺎﺴﺏ ﻤﻊ ﻤﺭﺒﻊ ﺴﻌﺔ ﺍﻟﻤﺠﺎل ﺍﻟﻜﻬﺭﺒﻲ. The intensity of the radiation is proportioned to the square of the amplitude of the .oscillating electric field.
ﻭﻴﻤﻜﻥ ﻟﻼﻟﻜﺘﺭﻭﻨﺎﺕ ﺍﻟﺘﻲ ﻋﻠﻰ ﺴﻁﺢ ﺍﻟﻤﻌﺩﻥ ﺃﻥ ﺘﺘﺫﺒـﺫﺏ ﻤـﻊ ﺍﻟﻤﺠـﺎل ﺍﻟﻜﻬﺭﺒﻲ ﺍﻟﺴﺎﻗﻁ ﻋﻠﻴﻬﺎ ،ﻭﻜﻠﻤﺎ ﺍﺯﺩﺍﺩﺕ ﺸـﺩﺘﻪ )ﺴـﻌﺘﻪ( ﺘـﺯﺩﺍﺩ ﺴـﻌﺔ ﺘﺫﺒـﺫﺏ
ﺍﻹﻟﻜﺘﺭﻭﻨﺎﺕ ﻜﺜﻴﺭﺍﹰ ﻤﻤﺎ ﻴﺅﺩﻱ ﻓﻲ ﺍﻟﻨﻬﺎﻴﺔ ﺇﻟﻰ ﻜﺴﺭ ﺍﺭﺘﺒﺎﻁﻬﺎ ﺒﺎﻟﺴﻁﺢ ﻭﺍﻨﻁﻼﻗﻬـﺎ
ﺒﻁﺎﻗﺔ ﺤﺭﻜﻴﺔ kinetic energyﻭﺍﻟﺘﻲ ﺴﺘﻌﺘﻤﺩ ﻋﻠﻰ ﺴﻌﺔ )ﺸﺩﺓ( ﺍﻟﻤﺠﺎل ﺍﻟﻜﻬﺭﺒﻲ ﻟﻺﺸﻌﺎﻉ ﺍﻟﺴﺎﻗﻁ .ﺇﻥ ﻫﺫﺍ ﺍﻟﺘﻔﺴﻴﺭ ﺍﻟﻔﻴﺯﻴﺎﺌﻲ )ﺍﻟﺘﻘﻠﻴـﺩﻱ( ﻴﺘﻌـﺎﺭﺽ ﺘﻤﺎﻤـﺎﹰ ﻤـﻊ ﺍﻟﻤﺸﺎﻫﺩﺍﺕ ﺍﻟﻔﻴﺯﻴﺎﺌﻴﺔ ﻟﻬﺫﻩ ﺍﻟﻅﺎﻫﺭﺓ ﻭﺍﻟﺘﻲ ﺘﻤﺜﻠﺕ ﻓﻲ:
-1ﻭﺠﺩ ﺃﻥ ﺍﻟﻁﺎﻗﺔ ﺍﻟﺤﺭﻜﻴﺔ ﻟﻺﻟﻜﺘﺭﻭﻨﺎﺕ ﺍﻟﻤﻨﺒﻌﺜﺔ ﻤﻥ ﺍﻟﺴﻁﺢ ﻻ ﺘﻌﺘﻤﺩ ﻋﻠﻰ ﺸﺩﺓ ﺍﻻﻨﺒﻌﺎﺙ .independent of the intensity of incidental -2ﻭﺠﺩ ﺘﺠﺭﻴﺒﻴﺎﹰ ﺃﻥ ﺍﻹﻟﻜﺘﺭﻭﻨﺎﺕ ﻻ ﺘﻨﺒﻌﺙ ﻤﻥ ﺍﻟﺴﻁﺢ ﺇﻻ ﺇﺫﺍ ﻜﺎﻥ ﺘﺭﺩﺩ ﺍﻹﺸﻌﺎﻉ
ﺍﻟﺴﺎﻗﻁ ﺃﻜﺒﺭ ﻤﻥ ﺘﺭﺩﺩ ﻤﻌﻴﻥ ﺍﻟﺘﺭﺩﺩ
ν°
ν°
ﺒﻐﺽ ﺍﻟﻨﻅﺭ ﻋﻥ ﺸﺩﺓ ﺍﻹﺸـﻌﺎﻉ ﺴﻨـﺴﻤﻲ ﻫـﺫﺍ
ﺒﻌﺘﺒﺔ ﺍﻟﺘﺭﺩﺩ threshold frequencyﻭﻗﻴﻤﺔ
ν°
ﺘﻌﺘﻤﺩ ﻋﻠﻰ ﻨﻭﻉ ﺍﻟﻤﻌﺩﻥ.
ﻭﻜﺫﻟﻙ ﻭﺠﺩ ﺃﻥ ﺍﻟﻁﺎﻗﺔ ﺍﻟﺤﺭﻜﻴﺔ ﻟﻼﻟﻜﺘﺭﻭﻨﺎﺕ ﺍﻟﻤﻨﺒﻌﺜﺔ ﺘﺘﻨﺎﺴﺏ ﺨﻁﻴﺎﹰ ﻤﻊ ﺍﻟﺘﺭﺩﺩ ﻭﺫﻟﻙ ﺇﺫﺍ ﻜﺎﻨﺕ
ν
ﺃﻜﺒﺭ
ﻤﻥ ν °
ν
ﻜﻤﺎ ﻫﻭ ﻤﻭﻀﺢ ﻓﻲ ﺸﻜل .1-4
ﺸﻜل :1-4ﺘﻐﻴﺭ ﻗﻴﻡ ﺍﻟﻁﺎﻗﺔ ﺍﻟﺤﺭﻜﻴﺔ ﻟﻺﻟﻜﺘﺭﻭﻨﺎﺕ ﻤﻊ ﺍﻟﺘﺭﺩﺩ. - ١٢PDF created with pdfFactory Pro trial version www.pdffactory.com
FIGURE 1.4 The kinetic energy of electrons ejected from the surface of
sodium metal versus the frequency of the incident ultraviolet radiation. The )threshold frequency here is 4.40 × 1014 Hz (1 Hz = 1 s-1
ﻭﻟﺘﻔﺴﻴﺭ ﻫﺫﻩ ﺍﻟﻨﺘﺎﺌﺞ ،ﻁﻭﺭ ﺍﻟﻌﺎﻟﻡ ﺍﻴﻨﺸﺘﺎﻴﻥ ﻓﺭﻀﻴﺔ ﺒﻼﻨﻙ ﻤﻥ ﺘﻜﻤﻴﻡ ﺍﻟﻁﺎﻗﺔ.
ﻭﻫﻨﺎ ﻨﺫﻜﺭ ﺃﻥ ﺒﻼﻨﻙ ﻁﺒﻕ ﻤﻔﻬﻭﻡ ﺘﻜﻤﻴﻡ ﺍﻟﻁﺎﻗﺔ ﻓﻘﻁ ﻋﻠﻰ ﺍﻟﻤﺘﺫﺒﺫﺏ ﺍﻟﺫﻱ ﻴﻤـﺘﺹ ﺃﻭ ﻴﺒﻌﺙ ﺍﻟﻁﺎﻗﺔ )ﻓﺈﺫﺍ ﻤﺼﺩﺭ ﺘﺭﺩﺩﻩ
ν
ﻓﺈﻨﻪ ﻴﻤﻠﻙ ﻁﺎﻗﺔ
E = nhν
( ﺃﻤﺎ ﺇﺫﺍ ﻤﺎ ﺍﻨﺒﻌﺙ
ﺍﻹﺸﻌﺎﻉ ﻤﻥ ﻫﺫﺍ ﺍﻟﻤﺘﺫﺒﺫﺏ ﻓﺈﻨﻪ ﻴﺴﻠﻙ )ﻜﻤﺎ ﺍﻗﺘﺭﺡ ﺒﻼﻨﻙ( ﻜﻤﻭﺠﺔ ﻋﺎﺩﻴﺔ ﺘﺨـﻀﻊ
ﻟﻠﻤﻔﺎﻫﻴﻡ ﺍﻟﻔﻴﺯﻴﺎﺌﻴﺔ ﺍﻟﺘﻘﻠﻴﺩﻴﺔ .ﺃﻤﺎ ﺍﻟﻌﺎﻟﻡ ﺍﻴﻨﺸﺘﺎﻴﻥ ﻓﻘﺩ ﺫﻫﺏ ﺇﻟﻰ ﻤﺎ ﻫﻭ ﺃﺒﻌﺩ ﻤﻥ ﻫﺫﺍ،
ﻓﻘﺎل ﺇﻥ ﺍﻟﺘﻜﻤﻴﻡ ﻻ ﻴﻨﻁﺒﻕ ﻓﻘﻁ ﻋﻠﻰ ﺍﻟﻤﺘﺫﺒﺫﺏ ﺒل ﺇﻥ ﺍﻹﺸﻌﺎﻉ ﺍﻟﻤﻨﺒﻌﺙ ﻤﻨﻪ ﻴﻨﺒﻌﺙ
ﻜﻤﺎﺕ ﻤﻥ ﺍﻟﻁﺎﻗﺔ ) ﺃﻭ ﺤﺯﻡ ﻤﻥ ﺍﻟﻁﺎﻗﺔ( ﻤﻨﻔﺼﻠﺔ ﻋﻥ ﺒﻌﻀﻬﺎ ﻭﻁﺎﻗـﺔ ﻜـل ﻜـﻡ E = hν
ﻭﺃﻁﻠﻕ ﺍﺴﻡ ﺍﻟﻔﻭﺘﻭﻥ photonﻋﻠﻴﻪ.
Einstein proposed that the radiation itself existed as smell packers of energy, E = hν , now known as photons.
ﻭﺒﻨﺎﺀﺍﹰ ﻋﻠﻰ ﻤﺒﺩﺃ ﺤﻔﻅ ﺍﻟﻁﺎﻗﺔ ،Conservation of energyﻭﻀﺢ ﺍﻴﻨﺸﺘﺎﻴﻥ ﺃﻥ ﺍﻟﻁﺎﻗﺔ ﺍﻟﺤﺭﻜﻴﺔ
1 mv 2 2
= K. E
)ﺍﻟﺘﻲ ﻴﻨﺒﻌـﺙ ﺒﻬـﺎ ﺍﻹﻟﻜﺘـﺭﻭﻥ ﺫﻭ ﺍﻟﻜﺘﻠـﺔ m
ﻭﺍﻟﺴﺭﻋﺔ (vﻤﻥ ﺴﻁﺢ ﺍﻟﻤﻌﺩﻥ ﻋﺒﺎﺭﺓ ﻋﻥ ﺍﻟﻔﺭﻕ ﺒﻴﻥ ﺍﻟﻁﺎﻗﺔ ﺍﻟـﺴﺎﻗﻁﺔ ﻟﻠﻔﻭﺘـﻭﻥ hν
ﻭﺃﻗل ﻁﺎﻗﺔ ﻻﺯﻤﺔ ﻟﻨﺯﻉ ﺍﻹﻟﻜﺘﺭﻭﻥ )ﻭﺘﺤﺭﻴﺭﻩ( ﻤﻥ ﻁﺎﻗﺔ ﺭﺒﻁﺔ ﺒﺎﻟﻤﻌـﺩﻥ W
)ﻭﻴﺭﻤﺯ ﻟﻬﺎ ﺃﺤﻴﺎﻨﺎﹰ ﺒـ ( ο/ﻭﺍﻟﺘﻲ ﺘﺴﻤﻰ ﺩﺍﻟﺔ ﺍﻟﺸﻐل .Work function ﻭﻴﻤﻜﻨﻨﺎ ﺼﻴﺎﻏﺔ ﻤﺎ ﺴﺒﻕ ﺫﻜﺭﻩ ﺒﺎﻟﻤﻌﺎﺩﻟﺔ ﺍﻟﺘﺎﻟﻴﺔ: 1 mv 2 = hv − φ 2
)(1-5
ﻻﺤﻅ ﺃﻥ ﻗﻴﻤﺔ
1 2 mc 2
= K. E
ﻤﻭﺠﺒﺔ ﻭﻋﻠﻴﻪ ﻓـﺎﻟﻔﺭﻕ
ﺴﺎﻟﺏ .ﺃﻱ ﺃﻨﻪ . hν ≥ φﺇﻥ ﺍﻗل ﻗﻴﻤﺔ ﻟـ
hν
hν − φ
ﻫـﻲ ﺍﻟﻁﺎﻗـﺔ ﺍﻟﻼﺯﻤـﺔ ﻟﺘﺤﺭﻴـﺭ
ﺍﻹﻟﻜﺘﺭﻭﻥ ﻤﻥ ﺭﺒﻁ ﺍﻟﻨﻭﺍﺓ ﻭﻫﺫﺍ ﻴﺤﺩﺙ ﻋﻥ ﻋﺘﺒﺔ ﺍﻟﺘﺭﺩﺩ )(1-6
ﻻ ﻴﻤﻜـﻥ ﺃﻥ ﻴﻜـﻭﻥ
ν°
ﺃﻱ ﺃﻥ:
hν ° = φ
- ١٣PDF created with pdfFactory Pro trial version www.pdffactory.com
ﻭﻴﻤﻜﻨﻨﺎ ﻜﺘﺎﺒﺔ ﻤﻌﺎﺩﻟﺔ ) (1-5ﻋﻠﻰ ﺍﻟﺼﻭﺭﺓ. 1 mc 2 = hv − hv ° 2
)(1-7
ﻭﻫﺫﻩ ﻤﻌﺎﺩﻟﺔ ﺨﻁ ﻤﺴﺘﻘﻴﻡ )ﻋﻠﻰ ﺍﻟﺼﻭﺭﺓ (y = mx-cﻭﻫﻭ ﺘﻤﺎﻤﺎﹰ ﻤﺎ ﻴﺸﺎﻫﺩ
ﻤﻥ ﺍﻟﻨﺘﺎﺌﺞ ﺍﻟﺘﺠﺭﻴﺒﻴﺔ ﺍﻟﻤﻭﻀﺤﺔ ﻓﻲ ﺸﻜل .1-5
Figure 1-5. Photoelectric effect data showing a plot of retarding potential necessary to stop electron flow from a metal (lithium), or equivalently, electron kinetic energy, as a function of frequency of the incident light, the slope of the line is h/e.
ﻭﻗﺒل ﺃﻥ ﻨﻨﺘﻘل ﻟﻤﻭﻀﻭﻉ ﺁﺨﺭ ،ﻤﻥ ﺍﻟﻤﻬﻡ ﺃﻥ ﻨﻭﻀﺢ ﺍﻟﻭﺤﺩﺍﺕ ﺍﻟﻤـﺴﺘﺨﺩﻤﺔ ﻓـﻲ ﺍﻟﻤﻌﺎﺩﻟﺔ .ﺒﺎﻟﻨﺴﺒﺔ ﻟـ
φ
ﻋﺎﺩﺓ ﺘﻌﻁﻲ ﺒﻭﺤﺩﺓ ﺍﻹﻟﻜﺘﺭﻭﻥ ﻓﻭﻟﺕ ،eVﻭﻫـﺫﻩ ﻭﺤـﺩﺓ
ﻁﺎﻗﺔ ،ﻭﻟﺤﺴﺎﺏ )ﻤﺜﻼﹰ( ﺴﺭﻋﺔ ﺍﻹﻟﻜﺘﺭﻭﻥ ﺍﻟﻤﻨﺒﻌﺙ ﺒﻭﺤﺩﺓ m/sﻓﻤـﻥ ﺍﻟﻤﻬـﻡ ﺃﻥ ﻴﻌﺭﻑ ﺍﻟﻁﺎﻟﺏ ﻜﻴﻑ ﻴﻘﻭﻡ ﺒﻌﻤﻠﻴﺔ ﺍﻟﺘﺤﻭﻴل ﺒﻁﺭﻴﻘﺔ ﺼﺤﻴﺤﺔ. - ١٤ -
PDF created with pdfFactory Pro trial version www.pdffactory.com
ﺃﻭﻻﹰ :ﻤﻥ ﺘﻌﺭﻴﻑ ﺍﻹﻟﻜﺘﺭﻭﻥ ﻓﻭﻟﺕ (1 coulomb) (1 volt) = 1 Joule
ﻭﺍﻟﺸﺤﻨﺔ ﺍﻷﻭﻟﻴﺔ ﻟﻠﺒﺭﻭﺘﻭﻥ ﺘﺴﺎﻭﻱ ﻭﺍﻟﺸﺤﻨﺔ ﺍﻷﻭﻟﻴﺔ ﻟﻠﺒﺭﻭﺘﻭﻥ ﺘﺴﺎﻭﻱ
1.6 ×10 −19 C
ﻋﻨﺩﺌﺫ
) Jﺘﺭﻤﺯ ﻟﻠﺠﻭل (
)1 eV = (1.6 × 10 −19 C) (1 V = 1.6 ×10 −19 J
ﻤﺜﺎل :2ﺩﺍﻟﺔ ﺍﻟﺸﻐل ﻟﻠﺼﻭﺩﻴﻭﻡ ﺘﺴﺎﻭﻱ ،1.82 enﺍﺤﺴﺏ ﻋﺘﺒﺔ ﺍﻟﺘـﺭﺩﺩ
ν°
ﻟﻠﺼﻭﺩﻴﻭﻡ؟ ﺍﻟﺤل :ﺃﻭﻻﹰ :ﻴﻠﺯﻡ ﺘﺤﻭﻴل
φ
ﻤﻥ eVﺇﻟﻰ ﺠﻭل: φ = 1.82 eV
J ) eV
= (1.82 eV) (1.6 × 10 −19 = 2.92 ×10 −19 J
ﻭﻤﻥ ﻤﻌﺎﺩﻟﺔ )(1-6
]= φ
°
[hνﻴﻤﻜﻨﻨﺎ
ﺤﺴﺎﺏ ν °
)φ (2.92 ×10 −19 J = )h (6.63×10 −3 J.S 1 = 4.40 ×1014 H 2 5
ﺤﻴﺙ Hzﺘﺭﻤﺯ ﻟﻠﻬﺭﺘﺯ ﻭﻫﻭ
= ν°
= 4.40 ×1014
1 ec 5
ﻤﺜﺎل:3 ﺃﺸﻌﺔ ﻓﻭﻕ ﺒﻨﻔﺴﺠﻴﺔ ﻁﻭﻟﻬﺎ ﺍﻟﻤﻭﺠﻲ 3500 Åﺘﺴﻘﻁ ﻋﻠﻰ ﺴﻁﺢ ﺒﻭﺘﺎﺴﻴﻭﻡ. ﺃﻗﺼﻰ ﻁﺎﻗﺔ ﻟﻺﻟﻜﺘﺭﻭﻨﺎﺕ ﺍﻟﻀﻭﺌﻴﺔ ﺘﺴﺎﻭﻱ .1.6 eV
- ١٥PDF created with pdfFactory Pro trial version www.pdffactory.com
ﺍﺤﺴﺏ ﺩﺍﻟﺔ ﺍﻟﺸﻐل ﻟﻠﺒﻭﺘﺎﺴﻴﻭﻡ؟ ﺍﻟﺤل: ﻤﻥ ﻤﻌﺎﺩﻟﺔ )(1-5 K.E = hν − ο/ ∴φ = hν − k.E
ﻨﺤﺴﺏ ﺃﻭﻻﹰ ﻗﻴﻤﺔ
hν )hc (6.63×10 −34 J.s) (3 × 10 8 m/s = = hν λ )(3500 × 10 −10 m
ﻭﻟﺘﺤﻭﻴل ﻫﺫﻩ ﺍﻟﻘﻴﻤﺔ ﻤﻥ ﺍﻟﺠﻭل ﻟﻺﻟﻜﺘﺭﻭﻥ ﻓﻭﻟﺕ ∴φ = 3. eV − 1.6 eV = 1.95 eV
ﻤﺜﺎل :ﻋﻨﺩ ﺘﻌﺭﺽ ﺴﻁﺢ ﻤﺎﺩﺓ ﺍﻟﻠﻴﺜﻴﻭﻡ ﻟﻺﺸﻌﺎﻉ ،ﻓﺈﻥ ﺍﻟﻁﺎﻗﺔ ﺍﻟﺤﺭﻜﻴﺔ ﻟﻺﻟﻜﺘﺭﻭﻨﺎﺕ ﺍﻟﻤﻨﺒﻌﺜﺔ
2.935 ×10 −19 J
ﻭﺫﻟﻙ ﺇﺫﺍ ﻜﺎﻨﺕ . λ = 300.0nm =
ﺇﻤﺎ ﺇﺫﺍ ﻜﺎﻨﺕ
λ
ﻓﺈﻥ ﺍﻟﻁﺎﻗﺔ ﺍﻟﺤﺭﻜﻴﺔ ﺘﺴﺎﻭﻱ
1.28 × 10 −19 J
ﺍﺤﺴﺏ ) (aﺜﺎﺒﺕ ﺒﻼﻨﻙ؟ ) (bﻋﺘﺒﺔ ﺍﻟﺘﺭﺩﺩ؟ ) (cﺍﻟﺸﻐل؟ ﺍﻟﺤل :ﻤﻥ ﻤﻌﺎﺩﻟﺔ ) (1-5ﻓﻲ ﺤﺎﻟﺔ ﺍﻟﻁﻭل ﺍﻟﻤﻭﺠﻲ ﺍﻷﻭل ﻭﺍﻟﻁﻭل ﺍﻟﻤﻭﺠﻲ
ﺍﻟﺜﺎﻨﻲ ﻨﺘﺤﺼل ﻋﻠﻰ ﺒﺎﻟﺘﻌﻭﻴﺽ:
1 1 ) − λ1 λ 2
( (K.E)1 − (K.E) 2 = h (ν1 − ν 2 ) = hc
1 1 2.935×10 −19 J − 1.280 × 10 −19 J = h (3×10 8 m/s) − −9 −9 3 ×10 m 400 ×10 m
- ١٦PDF created with pdfFactory Pro trial version www.pdffactory.com
1.655 ×1019 J ∴h = 6.625 ×10 −34 J.S 14 −1 2.498 ×10 s
) (bﻟﺤﺴﺎﺏ ﻋﺘﺒﺔ ﺍﻟﺘﺭﺩﺩ ،ﻨﺄﺨﺫ ﺍﻟﻁﻭل ﺍﻟﻤﻭﺠﻲ )ﻤﺜﻼﹰ(
λ = 300 nm
∴ = hv − hv °
hc − hv ° 300 ×10 −9 m
= ∴ 2.935 ×10 −19 J
ﻭﺒﺎﻟﺘﻌﻭﻴﺽ ﻋﻥ ﻗﻴﻡ c,hﻨﺠﺩ ﺃﻥ ﻗﻴﻤﺔ
v°
v ° = 5.564 × 1014 Hz
) (٢ﺩﺍﻟﺔ ﺍﻟﺸﻐل ﺘﺤﺴﺏ ﻤﺒﺎﺸﺭﺓ ﻤﻥ ﺍﻟﻌﻼﻗﺔ
φ = hv°
= 3.687 ×10 −19 J 3.687 ×1019 J/ = = 2.30/eV J/ 1.6 ± ×10 −19 eV
4-1ﺘﺄﺜﻴﺭ ﻜﻭﻤﺒﺘﻭﻥ: 1-4 The Compton Effect ﺍﻟﺘﺠﺭﺒﺔ ﺍﻟﺘﻲ ﺘﺅﻜﺩ ﺒﻭﻀﻭﺡ ﺍﻟﻁﺒﻴﻌﺔ ﺍﻟﺠـﺴﻴﻤﻴﺔ ﻟﻺﺸـﻌﺎﻉ the particle
nature of radiationﺘﺴﻤﻰ ﺒﺘﺄﺜﻴﺭ )ﻅﺎﻫﺭﺓ( ﻜﻭﻤﺒﺘﻭﻥ ﻨﺴﺒﺔ ﻟﻠﻌـﺎﻟﻡ ﻜﻭﻤﺒﺘـﻭﻥ
Arthur H. Comptonﻟﻘﺩ ﺍﻜﺘﺸﻑ ﻜﻭﻤﺒﺘﻭﻥ ﺃﻨﻪ ﺇﺫﺍ ﺍﺨﺘﺭﻕ ﺇﺸـﻌﺎﻉ ﺫﻭ ﻁـﻭل
ﻤﻭﺠﻲ )ﻓﻲ ﻤﻨﻁﻘﺔ ﺍﻷﺸﻌﺔ ﺍﻟﺴﻴﻨﻴﺔ (X-rayﺸﺭﻴﺤﺔ ﻤﻌﺩﻨﻴﺔ ﻓـﺴﻴﺘﺒﻌﺜﺭ scattered
ﺒﻁﺭﻴﻘﺔ ﻻ ﻴﻤﻜﻥ ﺘﻔﺴﻴﺭﻫﺎ ﺤﺴﺏ ﺍﻟﻨﻅﺭﻴﺔ ﺍﻟﺘﻘﻠﻴﺩﻴﺔ ﻟﻺﺸﻌﺎﻉ. ﺍﻟﺫﻱ ﺘﺨﺒﺭﻨﺎ ﺒﻪ ﻗﻭﺍﻨﻴﻥ ﺍﻟﻔﻴﺯﻴﺎﺀ ﺍﻟﺘﻘﻠﻴﺩﻴﺔ ﺃﻥ ﺸﺩﺓ ﺍﻹﺸﻌﺎﻉ Iﺍﻟﻤﻨﺒﻌﺙ ﻤـﻥ
ﻤﺎﺩﺓ ﻨﺘﻴﺠﺔ ﺘﺄﺜﺭﻫﺎ ﺒﺈﺸﻌﺎﻉ ﺴﻘﻁ ﻋﻠﻴﻬﺎ )ﻤﻤﺎ ﻴﺅﺩﻱ ﺇﻟﻰ ﺍﻫﺘﺯﺍﺯ ﺇﻟﻜﺘﺭﻭﻨﺎﺘﻬﺎ ﻭﺍﻟﺘـﻲ ﺒﺩﻭﺭﻫﺎ ﺴﺘﺒﻌﺙ ﺇﺸﻌﺎﻉ( ﻋﻨﺩﻤﺎ ﺘﻘﺎﺱ ﻋﻨﺩ ﺯﺍﻭﻴﺔ ﺍﻟﺴﺎﻗﻁﺔ( ﻓﺈﻥ Iﺘﺘﻐﻴﺭ ﻤﻊ
θ
ﺤﺴﺏ ﺍﻟﻌﻼﻗﺔ
θ
)ﺒﺎﻟﻨـﺴﺒﺔ ﻻﺘﺠـﺎﻩ ﺍﻷﺸـﻌﺔ
- ١٧PDF created with pdfFactory Pro trial version www.pdffactory.com
)Ι ≈ (1 + cos 2 θ
)(1-8
ﻭﻫﺫﺍ ﻴﻌﻨﻲ ﺃﻥ Iﻻ ﺘﻌﺘﻤﺩ ﻋﻠﻰ ﺍﻟﻁﻭل ﺍﻟﻤﻭﺠﻲ ﻟﻸﺸﻌﺔ ﺍﻟـﺴﺎﻗﻁﺔ ،ﻭﻫـﺫﺍ ﻴﺘﻌﺎﺭﺽ ﺒﻭﻀﻭﺡ ﻤﻊ ﺍﻟﻨﺘﺎﺌﺞ ﺍﻟﺘﺠﺭﻴﺒﻴﺔ )ﺸﻜل (1-6ﻭﺍﻟﺘﻲ ﻴﺘﺒﻴﻥ ﺒﻭﻀﻭﺡ ﺘﻐﻴﺭ I
ﺒﺘﻐﻴﺭ . λ
Figure 1-6. The spectrum of radiation scattered by carbon, showing the unmodified line at 0.7078 Å on the left and the shifted line at 0.7314 Å on the right. The former is the wave-length of the primary radiation.
ﻨﺘﺎﺌﺞ ﺘﺠﺭﺒﺔ ﻜﻭﻤﺒﺘﻭﻥ: ﻟﻘﺩ ﻭﺠﺩ ﻜﻭﻤﺒﺘﻭﻥ ﺃﻥ ﺍﻹﺸﻌﺎﻉ ﺍﻟﻤﺘﺒﻌﺜﺭ ﻟﻪ ﻤﺭﻜﺒﺘﻴﻥ؛ ﻤﺭﻜﺒﺔ ﻁﻭﻟﻬﺎ ﺍﻟﻤﻭﺠﻲ
ﻤﺴﺎﻭٍ ﻟﻁﻭل ﻤﻭﺠﺔ ﺍﻹﺸﻌﺎﻉ ﺍﻟﺴﺎﻗﻁ ﻭﻤﺭﻜﺒﺔ ﺃﺨﺭﻯ ﺘﺨﺘﻠﻑ ﻓﻲ ﻁﻭﻟﻬﺎ ﺍﻟﻤـﻭﺠﻲ ﻋﻥ ﺍﻟﻁﻭل ﺍﻟﻤﻭﺠﻲ ﻟﻺﺸﻌﺎﻉ ﺍﻟﺴﺎﻗﻁ ﻭﺘﻌﺘﻤﺩ ﻋﻠﻰ ﺯﺍﻭﻴﺔ ﺍﻟﺒﻌﺜـﺭﺓ ﻭﻗـﺩ ﺘﻤﻜـﻥ ﻜﻭﻤﺒﺘﻭﻥ ﻤﻥ ﺸﺭﺡ ﻭﺠﻭﺩ ﻤﺭﻜﺒﺔ ﺍﻹﺸـﻌﺎﻉ ﺍﻟﻤﺘﺒﻌﺜـﺭﺓ ﺫﺍﺕ ﺍﻟﻁـﻭل ﺍﻟﻤـﻭﺠﻲ
ﺍﻟﻤﺨﺘﻠﻑ ﻋﻥ ﺍﻟﻁﻭل ﺍﻟﻤﻭﺠﻲ ﺍﻟﺴﺎﻗﻁ ﻭﺫﻟﻙ ﺒﺎﻋﺘﺒﺎﺭ ﺍﻟﺸﻌﺎﻉ ﺍﻟﺴﺎﻗﻁ ﻋﺒﺎﺭﺓ ﻋـﻥ
ﺸﻌﺎﻉ ﻤﻥ ﺍﻟﻔﻭﺘﻭﻨﺎﺕ ﺒﻁﺎﻗﺔ
hv
ﺤﻴﺙ ﻴﻌﺎﻨﻲ ﻜل ﻓﻭﺘﻭﻥ ﻤﻥ ﺘﺒﻌﺜﺭ )ﺘﺸﺘﺕ( ﻤـﺭﻥ
elastic scatteringﻤﻊ ﻜل ﺇﻟﻜﺘﺭﻭﻥ.
ﻭﻜﻤﺎ ﻫﻭ ﻤﻌﻠﻭﻡ ،ﻓﻲ ﺤﺎﻟﺔ ﺍﻟﺘﺸﺘﺕ ﺍﻟﻤﺭﻥ ﻓﺈﻥ ﻜﻤﻴﺔ ﺍﻟﺤﺭﻜـﺔ momentum ﻭﺍﻟﻁﺎﻗﺔ energyﻜﻤﻴﺎﺕ ﺘﺨﻀﻊ ﻟﻘﺎﻨﻭﻥ ﺍﻟﺤﻔﻅ )ﺍﻟﺒﻘﺎﺀ( momentum and energy
.must be conserved
- ١٨PDF created with pdfFactory Pro trial version www.pdffactory.com
ﻭﻟﺘﻔﺴﻴﺭ ﻫﺫﻩ ﺍﻟﻅﺎﻫﺭﺓ ﺭﻴﺎﻀﻴﺎﹰ ،ﺍﻓﺘﺭﺽ ﻜﻭﻤﺒﺘﻭﻥ ﺃﻥ ﺍﻟﻔﻭﺘﻭﻥ ﻟـﻪ ﻜﻤﻴـﺔ ﺤﺭﻜﺔ pﺘﻌﻁﻲ ﺒﺎﻟﻌﻼﻗﺔ hv c
)(1-9
=p
ﺤﻴﺙ ﺘﻡ ﺍﻋﺘﺒﺎﺭ ﺍﻟﺤﺭﻜﺔ ﺍﻟﺩﻴﻨﺎﻤﻴﻜﻴﺔ ﻟﻠﻔﻭﺘﻭﻥ ﻜﺠﺴﻴﻡ ﻴﺨﻀﻊ ﻟﻘﻭﺍﻨﻴﻥ ﺍﻟﻨﻅﺭﻴﺔ ﺍﻟﻨﺴﺒﻴﺔ ﻭﺍﻟﺘﻲ ﺘﻭﻀﺢ ﺍﻟﻌﻼﻗﺔ ﺒﻴﻥ ﺍﻟﻁﺎﻗﺔ ﻭﻜﻤﻴﺔ ﺍﻟﺤﺭﻜﺔ.
)] (1-10 2
1
[
E = (m o c 2 ) 2 + (pc) 2
ﺤﻴﺙ moﻫﻲ ﺍﻟﻜﺘﻠﺔ ﺍﻟﺴﻜﻭﻨﻴﺔ rest massﻟﻠﺠﺴﻴﻡ ،ﻭﺴﺭﻋﺔ ﺍﻟﺠﺴﻴﻡ vﻋﻨﺩ
ﻜﻤﻴﺔ ﺍﻟﺤﺭﻜﺔ pﺘﻌﻁﻰ ﺒﺎﻟﻌﻼﻗﺔ 2
]
1
[
aE d = (m ο c 2 ) 2 + (pc) 2 ap dp
d )(pc ap
1 2
−
(pc)c 2
)(1-11
]
[
1 = = (m ο c 2 ) 2 + (pc) 2 2
]
1
=v
1 1 2 (m c 2 ) 2 + (pc) 2 °
[
pc 2 pc 2 = E (m 2 c 4 + p 2 C 2 ) 12 °
ﻭ
ﻓﻲ ﺤﺎﻟﺔ ﺍﻟﻔﻭﺘﻭﻥ ،ﻨﻌﻭﺽ ﻋﻥ
m° = o
)(1-12
pc 2 ⇒E=pc E
v= c
=
= ∴v
ﻓﻲ ﻤﻌﺎﺩﻟﺔ )(1-11
= ∴C/
ﻤﻥ ﻤﻌﺎﺩﻟﺔ ) (1-12ﻨﺘﺤﺼل ﻋﻠﻰ ﻤﻌﺎﺩﻟﺔ ) (1-9ﺤﻴﺙ E hv = C C
E = hv
=p
ﺩﻋﻨﺎ ﻨﻔﺘﺭﺽ ﺍﻵﻥ ﻭﺠﻭﺩ ﻓﻭﺘﻭﻥ ﺒﻜﻤﻴﺔ ﺤﺭﻜﺔ ﺍﺒﺘﺩﺍﺌﻴـﺔ
p°
ﺴـﺎﻗﻁ ﻋﻠـﻰ
ﺇﻟﻜﺘﺭﻭﻥ ﺴﺎﻜﻥ .ﻭﺒﻌﺩ ﺍﻟﺘﺼﺎﺩﻡ ،ﻨﻔﺘﺭﺽ ﺃﻥ ﻜﻤﻴـﺔ ﺍﻟﺤﺭﻜـﺔ ﻟﻠﻔﻭﺘـﻭﻥ
p
ﺃﻤـﺎ
- ١٩PDF created with pdfFactory Pro trial version www.pdffactory.com
ﺍﻹﻟﻜﺘﺭﻭﻥ ﻓﻴﺤﺩﺙ ﻟﻪ ﺍﺭﺘﺩﺍﺩ recoilﺒﻜﻤﻴﺔ ﺤﺭﻜﺔ . pﻭﺒﺘﻁﺒﻴﻕ ﻗﺎﻨﻭﻥ ﺒﻘﺎﺀ ﻜﻤﻴـﺔ e
ﺍﻟﺤﺭﻜﺔ )ﺍﻨﻅﺭ ﺸﻜل .(1-7 )(1-13
pe = p ° + p
ﺒﺘﺭﺒﻴﻊ ﻁﺭﻓﻲ ﺍﻟﻤﻌﺎﺩﻟﺔ.
2 2 r r 2 p e = p ° − p − 2p ° p
)(1-14
ﻨﻁﺒﻕ ﺍﻵﻥ ﻗﺎﻨﻭﻥ ﺒﻘﺎﺀ ﺍﻟﻁﺎﻗﺔ
ﺃﻭﻻﹰ ،ﻁﺎﻗﺔ ﺍﻹﻟﻜﺘﺭﻭﻥ Eﻗﺒل ﺍﻟﺘﺼﺎﺩﻡ ] 2
ﺒﺎﻟﺘﻌﻭﻴﺽ ﻋﻥ
m° = 0
ﺘﺴﺎﻭﻱ ﺼﻔﺭ(.
1
[
E = (m ° c 2 ) 2 + (pc) 2
ﻷﻥ ﺍﻹﻟﻜﺘﺭﻭﻥ ﺴـﺎﻜﻥ ﻗﺒـل ﺍﻟﺘـﺼﺎﺩﻡ )ﻷﻥ ﺃﻥ P ∴E = m ° c = mc 2
)(1-15
ﻗﺎﻨﻭﻥ ﺒﻘﺎﺀ ﺍﻟﻁﺎﻗﺔ :ﻁﺎﻗﺔ ﺍﻹﻟﻜﺘﺭﻭﻥ ﻭﺍﻟﻔﻭﺘﻭﻥ ﺍﻟﺴﺎﻗﻁ ﻗﺒل ﺍﻟﺘﺼﺎﺩﻡ = ﻁﺎﻗﺔ
ﺍﻹﻟﻜﺘﺭﻭﻥ ﻭﺍﻟﻔﻭﺘﻭﻥ ﺒﻌﺩ ﺍﻟﺘﺼﺎﺩﻡ 2
1
) hv ° + mc 2 = hv + (m 2 c 4 + p e c 2 2
- ٢٠PDF created with pdfFactory Pro trial version www.pdffactory.com
ﺒﻨﻘل hvﻟﻠﻁﺭﻑ ﺍﻷﻴﺴﺭ ﻭﺘﺭﺒﻴﻊ ﻁﺭﻓﻲ ﺍﻟﻤﻌﺎﺩﻟﺔ ) (hv ° + mc 2 − hv) 2 = (m 2c 4 + pe c 2 2
or m 2c 4 + pe c 2 = (hv ° − hv + mc 2 )2 2
= (hν ° − hv) 2 + 2m c 2 (hv ° − hv) + m 2c 4
)(1-16
ﻤﻥ ﻤﻌﺎﺩﻟﺔ ) ،(1-14ﺒﺎﻟﺘﻌﻭﻴﺽ ﻋﻥ
hv hv , p° = ° c c
hv ° 2 hv 2 hv hv ) + ( ) − 2 ( ° ) ( ) cosθ c c c c
ﺤﻴﺙ
θ
=p
( = ∴p e 2
ﻫﻲ ﺍﻟﺯﺍﻭﻴﺔ ﺍﻟﻤﺤﺼﻭﺭﺓ ﺒﻴﻥ ﺍﺘﺠﺎﻩ ﺍﻟـﺸﻌﺎﻉ ﺍﻟﻤـﺸﺘﺕ ﻭﺍﻟـﺸﻌﺎﻉ
ﺍﻟﺴﺎﻗﻁ .ﺒﻀﺭﺏ ﻁﺭﻓﻲ ﺍﻟﻤﻌﺎﺩﻟﺔ ﻓﻲ C2
pe c 2 = (hv° )2 + (hv)2 − 2 (hv ° ) (hv) cosθ 2
)(1-17
ﻭﻟﻠﺤﺼﻭل ﻋﻠﻰ ﻤﺭﺒﻊ ﻜﺎﻤل ،ﻨﻀﻴﻑ ﻭﻨﻁﺭﺡ ] [2hv° hvﻟﻤﻌﺎﺩﻟﺔ )(1-17 pe c 2 (hv ° ) 2 + (hv) 2 − 2hv °hv + 2(hv ° ) (hv) − 2(hv° ) (hv) cosθ 1444 424444 3 14444 4244444 3 2
)2hv ° hv (1− cos θ
)∴pe c 2 = (hv° − hv) 2 + 2hv °hv (1 − cos θ 2
)(1-18
ﺒﺎﻟﺘﻌﻭﻴﺽ ﻋﻥ ﻗﻴﻤﺔ
ﻋﻠﻰ
(hv ° − hv) 2
(hv ° − hv) 2
ﻤﻥ ﻤﻌﺎﺩﻟﺔ ) (1-18ﻓﻲ ﻤﻌﺎﺩﻟﺔ ) (1-18ﻨﺘﺤـﺼل
m 2 c 4 + pe c 2 = p e c 2 − 2hv °hv (1 − cos θ) + 2mc 2 (hv ° − hv) + m 2c 4 2
2
) ..(1-19ﺒﺤﺫﻑ ﺍﻟﺤﺩﻭﺩ ﺍﻟﻤﺘﺸﺎﺒﻬﺔ ﻤﻥ ﻁﺭﻓﻲ ﺍﻟﻤﻌﺎﺩﻟﺔ:
- ٢١PDF created with pdfFactory Pro trial version www.pdffactory.com
2hv °hv (1 − cos θ) = 2mc 2 (hv ° − hv) − )= 2mc 2 h (v° − v 1 1 c c ) − ) 2mc 2 hc ( − λ° λ λ° λ
( = 2mc 2 h
λ − λ° c c h (1 − cos θ) = 2mc 2 h λ°λ λ° λ
2h or
λ − λ° c/ 2 (1 − cos θ) 2/ mc 3/ h/ 2/ h λ/ °λ/ λ °λ 2/
) h (1 − cos θ) = mc (λ − λ °
)...(1-20 ﻻﺤﻅ ﺃﻥ
h )(1 − cos θ cm h mc
= or λ − λ °
ﻓﻲ ﻤﻌﺎﺩﻟﺔ ) (1-20ﻟﻪ ﺒﻌﺩ ﺍﻟﻁﻭل ﻭﻫﺫﺍ ﺍﻟﺤﺩ ﻴـﺴﻤﻰ ﻁـﻭل
ﻤﻭﺠﺔ ﻜﻭﻤﺒﺘﻭﻥ Compton wavelengthﻟﻼﻟﻜﺘﺭﻭﻥ ﻭﻤﻘﺩﺍﺭﻩ h ≅ 2.4 × 10 −10 cm mc
)(1-21
ﻭﻗﺩ ﺘﺒﻴﻥ ﻤﻥ ﺍﻟﻘﻴﺎﺴﺎﺕ ﺍﻟﻤﻌﻤﻠﻴﺔ ﺃﻥ ﻁﻭل ﻤﻭﺠﺔ ﺍﻟﻔﻭﺘـﻭﻥ ﻭﺍﻟﻤﺘـﺸﺘﺕ ﺘﺘﻁﺎﺒﻕ ﻤﻊ ﺍﻟﻘﻴﻤﺔ ﺍﻟﻨﻅﺭﻴﺔ .ﺃﻤﺎ ﺍﻟﻤﺭﻜﺒﺔ ﺍﻟﺜﺎﻨﻴﺔ ﻟـ
ﻴﺒﻴﻥ ﻤﺭﻜﺒﺘﻴﻥ ﻟـ
λ
ﺇﺤﺩﺍﻫﻤﺎ ﺘﺨﺘﻠﻑ ﻋﻥ
λ°
λ
)ﺍﻨﻅﺭ ﺸﻜل
λ
ﻭﺍﻟـﺫﻱ
ﻭﺍﻷﺨﺭﻯ ﻤﺴﺎﻭﻴﺔ ﻟـ ( λﻭﺍﻟﺘﻲ ﻻ °
ﺘﺘﻐﻴﺭ ﺒﺎﻟﻨﺴﺒﺔ ﻓﺈﻥ ﻤﻨﺸﺄﻫﺎ ﻫﻭ ﺘﺼﺎﺩﻡ ﺍﻟﻔﻭﺘﻭﻥ ﺍﻟﺴﺎﻗﻁ ﻤﻊ ﺍﻟﺫﺭﺓ ﻜﻜل ،ﻓﻠﻭ ﻋﻭﻀﻨﺎ
ﻋﻥ mﺒﻜﺘﻠﺔ ﺍﻟﺫﺭﺓ )ﺒﺩﻻﹰ ﻤﻥ ﻜﺘﻠﺔ ﺍﻹﻟﻜﺘﺭﻭﻥ( ﻭﺤﻴﺙ ﺃﻥ ﻫﺫﻩ ﺍﻟﻘﻴﻤﺔ ﻓـﻲ ﺍﻟﻤﻘـﺎﻡ )ﻭﻫﻲ ﻜﺒﻴﺭﺓ ﺠﺩﺍﹰ ﺒﺎﻟﻨﺴﺒﺔ ﻟﻜﺘﻠﺔ ﺍﻹﻟﻜﺘﺭﻭﻥ( ﻓﺈﻥ ﺍﻟﺤﺩ ﺠﺩﺍﹰ ﻗﺭﻴﺒﺔ ﻤﻥ ﺍﻟﺼﻔﺭ ،ﺃﻱ ﺃﻥ
λ − λ° ≅ ο
ﻭﻫﺫﺍ ﻴﻌﻨﻲ
h mc
ﺴﺘﻜﻭﻥ ﻗﻴﻤﺘﻪ ﺼـﻐﻴﺭﺓ
λ ≈ λ°
ﻭﺃﺨﻴـﺭﺍﹰ ﻤـﺎ ﺍﻟـﺫﻱ
ﻨﺴﺘﻨﺘﺠﻪ ﻤﻥ ﺘﺄﺜﻴﺭ ﻜﻭﻤﺒﺘﻭﻥ؟ ﺇﻥ ﺍﻟﻘﻴﺎﺴﺎﺕ ﺍﻟﺘﻲ ﺃﺠﺭﻴﺕ ﻋﻠﻰ ﺍﻹﻟﻜﺘﺭﻭﻥ ﺍﻟﻤﺭﺘﺩ ﻭﺍﻟﻔﻭﺘﻭﻥ ﺍﻟﻤﺘﺒﻌﺜﺭ ﻤﻨـﻪ
ﺘﺅﻜﺩ -ﺒﻤﺎ ﻻ ﻴﺩﻉ ﻤﺠﺎﻻﹰ ﻟﻠﺸﻙ – ﺒﺄﻥ ﻫﺫﺍ ﺍﻟﺘﺼﺎﺩﻡ ﻤﻤﺎﺜل ﻟﻠﺘﺼﺎﺩﻡ ﺍﻟﺫﻱ ﻴﺤـﺩﺙ
- ٢٢PDF created with pdfFactory Pro trial version www.pdffactory.com
ﺒﻴﻥ ﻜﹸﺭﺘﻲ ﺒﻠﻴﺎﺭﺩﻭ ،ﺃﻱ ﺃﻥ ﺍﻟﻔﻭﺘﻭﻥ )ﺃﻭ ﺍﻟﺸﻌﺎﻉ ﺍﻟﺴﺎﻗﻁ( ﻴﺠﺏ ﺃﻥ ﺘﺘﻌﺎﻤـل ﻤﻌـﻪ ﻋﻠﻰ ﺃﺴﺎﺱ ﺃﻨﻪ ﺠﺴﻴﻡ ،ﻭﻫﺫﺍ ﻴﺅﻜﺩ ﺍﻟﻁﺒﻴﻌﺔ ﺍﻟﺠﺴﻴﻤﻴﺔ ﻟﻺﺸﻌﺎﻉ. 1-5ﺍﻟﺨﺼﺎﺌﺹ ﺍﻟﻤﻭﺠﻴﺔ ﻟﻠﻤﺎﺩﺓ ﻭﺤﻴﻭﺩ ﺍﻹﻟﻜﺘﺭﻭﻥ 1-5 Wave Prosperities and Electron Diffraction
ﻭﺍﺠﻪ ﺍﻟﻌﻠﻤﺎﺀ ﻜﺜﻴﺭﺍﹰ ﻤﻥ ﺍﻟﺼﻌﻭﺒﺎﺕ ﻓﻲ ﻭﺼﻑ ﻁﺒﻴﻌـﺔ ﺍﻟـﻀﻭﺀ .ﺒﻌـﺽ ﺍﻟﺘﺠﺎﺭﺏ ﺘﺒﻴﻥ ﺍﻟﻁﺒﻴﻌﺔ ﺍﻟﻤﻭﺠﻴﺔ ﻟﻠﻀﻭﺀ )ﻤﺜـل؟ dispersionﺍﻟـﻀﻭﺀ ﺍﻷﺒـﻴﺽ
ﻟﻤﺭﻜﺒﺎﺕ ﻁﻴﻔﻪ ﻋﻨﺩ ﻤﺭﻭﺭﻩ ﺩﺍﺨل ﻤﻨﺸﻭﺭ (prismﻭﺍﻟﺒﻌﺽ ﺍﻵﺨﺭ ﻤﻥ ﺍﻟﺘﺠـﺎﺭﺏ
ﻴﺜﺒﺕ ﺍﻟﻁﺒﻴﻌﺔ ﺍﻟﺠﺴﻴﻤﻴﺔ ﻟﻠﻀﻭﺀ )ﻤﺜل ﺍﻟﺘﺄﺜﻴﺭ ﺍﻟﻜﻬﺭﻭﻀﻭﺌﻲ(..
ﻭﻫﺫﺍ ﻤﺎ ﻴﻌﺭﻑ ﺒﺎﻟﻁﺒﻴﻌﺔ ﺍﻟﻤﺯﺩﻭﺠﺔ ﻟﻠـﻀﻭﺀ ware- particle duality of
.lights
ﻓﻲ ﻋﺎﻡ ،1924ﻗﺩﻡ ﻋﺎﻟﻡ ﻓﺭﻨﺴﻲ ﻴﺩﻋﻰ ﺩﻴﺒﺭﻭﻟـﻲ – Louis de Broglie ﻭﺒﻨﺎﺀﺍﹰ ﻋﻠﻰ ﻤﺎ ﺴﺒﻕ ﺫﻜﺭﻩ ﺒﺎﻟﻨﺴﺒﺔ ﻟﻠﻀﻭﺀ -ﻨﻤﻭﺫﺠﺎﹰ ﻴﺒﻴﻥ ﻓﻴﻪ ﺍﻟﻁﺒﻴﻌـﺔ ﺍﻟﻤﻭﺠﻴـﺔ ﻟﻠﻤﺎﺩﺓ؛ ﻓﺈﺫﺍ ﻜﺎﻥ ﺍﻟﻀﻭﺀ -ﺫﻭ ﺍﻟﻁﺒﻴﻌﺔ ﺍﻟﻤﻭﺠﻴﺔ – ﻴﺴﻠﻙ ﺃﺤﻴﺎﻨﺎﹰ ﻜﻤـﺎ ﻟـﻭ ﻜـﺎﻥ
ﺠﺴﻴﻤﺎﺕ ،ﻓﻠﻤﺎﺫﺍ ﻻ ﻴﻜﻭﻥ ﻟﻠﻤﺎﺩﺓ ﻁﺒﻴﻌﺔ ﻤﻭﺠﻴﺔ؟!! ﻭﻗﺩ ﺼﺎﻍ ﺩﻴﺒﺭﻭﻟﻲ ﻓﻜﺭﺘﻪ ﻫﺫﻩ ﺒﺼﻴﻐﺔ ﺭﻴﺎﻀﻴﺔ -ﺍﻗﺘﺒﺴﻬﺎ ﻤﻥ ﻋﻼﻗﺔ ﺍﻴﻨﺸﺘﺎﻴﻥ ﺍﻟﺘﻲ ﺘﺭﺘﺒﻁ ﺒـﻴﻥ ﻁـﻭل ﻤﻭﺠـﺔ ﺍﻟﻔﻭﺘﻭﻥ
ﺤﺭﻜﺘﻪ ﻤﻭﺠﻲ
λ
ﻭﻜﻤﻴﺔ ﺤﺭﻜﺘﻪ P
p = mv
λ
)...(1-22
)(١
،ﻓﺈﺫﺍ ﻜﺎﻥ ﺠﺴﻴﻡ ﻜﺘﻠﺘﻪ mﻭﺴﺭﻋﺘﻪ vﻓﺈﻥ ﻜﻤﻴﺔ
ﻋﻨﺩﺌﺫ ﻓﺈﻥ ﺍﻟﺠﺴﻴﻡ – ﻭﺤﺴﺏ ﻨﻤﻭﺫﺝ ﺩﻴﺒﺭﻭﻟﻲ – ﺴﻴﻜﻭﻥ ﻟﻪ ﻁـﻭل
ﻴﻌﻁﻲ ﺒﺎﻟﻌﻼﻗﺔ. h p
λ
) (١ﺍﻟﻌﻼﻗﺔ ﺍﻟﺘﻲ ﻗﺩﻤﻬﺎ ﺍﻴﻨﺸﺘﺎﻴﻥ ﻟﻠﻔﻭﺘﻭﻥ ﺍﻟﺫﻱ ﺘﺭﺩﺩﻩ vﻭﻁﻭل ﻤﻭﺠﺘﻪ λﻭﻜﻤﻴﺔ ﺤﺭﻜﺘﻪ p c hc hc h h = = = = v hv E E p ) ( c
=λ
- ٢٣PDF created with pdfFactory Pro trial version www.pdffactory.com
ﺤﻴﺙ hﻫﻭ ﺜﺎﺒﺕ ﺒﻼﻨﻙ. ﻤﺜﺎل :5 ﺍﺤﺴﺏ ﺍﻟﻁﻭل ﺍﻟﻤﻭﺠﻲ ﻟﻜﺭﺓ ﻜﺘﻠﺘﻬﺎ 0.14 kgﻭﺴﺭﻋﺘﻬﺎ 40 m/sﻭﻗﺎﺭﻥ ﻫﺫﺍ ﺍﻟﻁﻭل ﺍﻟﻤﻭﺠﻲ ﻤﻊ ﻁﻭل ﻤﻭﺠﺔ ﺇﻟﻜﺘﺭﻭﻥ ﺴﺭﻋﺘﻪ 1.00%ﻤﻥ ﺴﺭﻋﺔ ﺍﻟﻀﻭﺀ؟ ﺍﻟﺤل: ﻨﻭﺠﺩ ﺃﻭﻻﹰ ﻜﻤﻴﺔ ﺍﻟﺤﺭﻜﺔ ﻟﻠﻜﺭﺓ )p = mco = (0.14kg) (40m/s = 5.6 kg.m.s -1
ﺍﻟﻁﻭل ﺍﻟﻤﻭﺠﻲ
λ
ﺤﺴﺏ ﻤﻌﺎﺩﻟﺔ ﺩﻴﺒﺭﻭﻟﻲ h 6.63×10 −34 J.s = =λ = 1.2 × 10 −34 m −1 p 5.6 kg ms
!!!
ﻻﺤﻅ ﺃﻥ ﻫﺫﺍ ﺍﻟﻁﻭل ﺍﻟﻤﻭﺠﻲ ﻤﺘﻨﺎﻫﻲ ﻓﻲ ﺍﻟﺼﻐﺭ ﻨﻭﺠﺩ ﺍﻵﻥ ﻜﻤﻴﺔ ﺤﺭﻜﺔ ﺍﻹﻟﻜﺘﺭﻭﻥ ) p = m e v = (9.1× 10 −31 kg) (2.998× 10 6 ms −1
)ﻻﺤﻅ ﺃﻥ ﺍﻟﺴﺭﻋﺔ vﻟﻺﻟﻜﺘﺭﻭﻥ ﻫﻲ 1%ﻤﻥ ﺴﺭﻋﺔ ﺍﻟـﻀﻭﺀ ﻜﻤـﺎ ﻫـﻭ ﻤﻌﻁﻰ ﻓﻲ ﺍﻟﺴﺅﺍل(. ∴ p = 2.73× 10 −24 kg.m.s −1
ﻁﻭل ﻤﻭﺠﺔ ﺩﻴﺒﺭﻭﻟﻲ ﻟﻺﻟﻜﺘﺭﻭﻥ. h 6.63 ×10 −34 J.s = =λ = 2.43× 10 −10 m p 2.73× 10 − 24 kg.m.s −1 = 243 pm
)ﺤﻴﺙ pmﻴﺴﺎﻭﻱ
m
−12
..(10ﻭﻫﺫﻩ ﺍﻟﻘﻴﻤﺔ 243ﻤﻘﺎﺭﻨﺔ ﻟﻸﺒﻌﺎﺩ ﺍﻟﺫﺭﻴﺔ..
- ٢٤PDF created with pdfFactory Pro trial version www.pdffactory.com
ﻤﻥ ﺨﻼل ﻫﺫﻩ ﺍﻟﻘﻴﻡ ،ﻴﺘﻀﺢ ﻟﻨﺎ ﺃﻥ ﻁﻭل ﻤﻭﺠﺔ ﺩﻴﺒﺭﻭﻟﻲ ﻟﻺﻟﻜﺘﺭﻭﻥ ﻤﻘﺎﺭﺒﺔ ﻟﻁﻭل ﻤﻭﺠﺔ ﺍﻷﺸﻌﺔ ﺍﻟﺴﻴﻨﻴﺔ .ﻭﻫﺫﺍ ﻴﻌﻨﻲ ﺃﻥ ﺍﻹﻟﻜﺘﺭﻭﻥ ﺴﻴﺴﻠﻙ ﻜﻤﺎ ﻟﻭ ﻜﺎﻥ ﺃﺸﻌﺔ
ﺴﻴﻨﻴﺔ!!.
ﺃﻤﺎ ﺒﺎﻟﻨﺴﺒﺔ ﻟﻠﻜﺭﺓ ﻓﺈﻥ ﻁﻭل ﻤﻭﺠﺔ ﺩﻴﺒﺭﻭﻟﻲ ﻟﻬﺎ ﻗﺼﻴﺭ ﺠﺩﺍﹰ ﻤﻘﺎﺭﻨﺔ ﺒﺎﻷﺒﻌﺎﺩ ﺍﻟﺫﺭﻴﺔ... ﻤﻭﺠﺎﺕ ﺩﻴﺒﺭﻭﻟﻲ ﻴﻤﻜﻥ ﻤﺸﺎﻫﺩﺘﻬﺎ ﺘﺠﺭﻴﺒﻴﺎﹰ de Broglie Waves are observed Experimentally
ﻋﻨﺩ ﺍﺼﻁﺩﺍﻡ ﺃﺸﻌﺔ Xﺒﻤﺎﺩﺓ ﺒﻠﻭﺭﻴﺔ ﻓﺈﻨﻪ ﻴﺘﺸﺘﺕ ﺒﻁﺭﻴﻘـﺔ ﻤﻌﻴﻨـﺔ ﺘﻌﻜـﺱ ﻁﺒﻴﻌﺔ ﺍﻟﺘﺭﺘﻴﺏ ﺍﻟﺩﻭﺭﻱ ﺍﻟﺒﻠﻭﺭﻱ ﻟﻠﻤﺎﺩﺓ ﻫﺫﻩ ﺍﻟﻅﺎﻫﺭﺓ ﺘﹲﻌﺭﻑ ﺒﺤﻴـﻭﺩ ﺃﺸـﻌﺔ .X
X-ray diffractionﻭﺴﺒﺏ ﺤﺩﻭﺙ ﻫﺫﻩ ﺍﻟﻅﺎﻫﺭﺓ ﻫﻭ ﺃﻥ ﺍﻷﺒﻌﺎﺩ ﺍﻟﺒﻠﻭﺭﻴﺔ )ﺍﻟﻤﺴﺎﻓﺔ
ﺒﻴﻥ ﺍﻟﻤﺴﺘﻭﻴﺎﺕ ﺍﻟﺒﻠﻭﺭﻴﺔ( ﻤﺘﻘﺎﺭﺒﺔ ﻤﻊ ﺍﻟﻁﻭل ﺍﻟﻤﻭﺠﻲ ﻷﺸﻌﺔ .X
ﻭﻤﻥ ﺍﻟﻨﺎﺤﻴﺔ ﺍﻟﺘﺠﺭﻴﺒﻴﺔ ،ﻓﻘﺩ ﻭﺠﺩ ﺃﻥ ﺘﺒﻌﺜﺭ ﺸﻌﺎﻉ ﻤﻥ ﺍﻹﻟﻜﺘﺭﻭﻨﺎﺕ ﺒﻤـﺎﺩﺓ
ﺒﻠﻭﺭﻴﺔ ﻴﻨﺸﺄ ﻋﻨﻪ ﺤﻴﻭﺩ ﻤﺸﺎﺒﻪ ﻟﺤﻴﻭﺩ ﺃﺸﻌﺔ Xﻤﻤﺎ ﺒﻴﻥ -ﺒﻤﺎ ﻻ ﻴﺩﻉ ﻤﺠﺎﻻﹰ ﻟﻠﺸﻙ- ﺍﻟﻁﺒﻴﻌﺔ ﺍﻟﻤﻭﺠﻴﺔ ﻟﻠﺠﺴﻴﻤﺎﺕ.
ﺸﻜل 1-8ﻴﻭﻀﺢ ﺘﺸﺘﺕ ﻤﻭﺠـﺎﺕ ﻨﺘﻴﺠـﺔ ﺍﺼـﻁﺩﺍﻤﻬﺎ ﺒﺘﺭﻜﻴـﺏ ﺩﻭﺭﻱ
) periodic Structureﻤﺜﻼﹰ ﺸﻜل ﺒﻠﻭﺭﻱ( .ﻭﻤﻥ ﺍﻟﺸﻜل ﻴﺘﻀﺢ ﺃﻨﻪ ﺴﻴﻜﻭﻥ ﻓـﺭﻕ ﻓﻲ ﺍﻟﻁﻭﺭ phase differenceﺒﻴﻥ ﺍﻟﻤﻭﺠﺎﺕ ﺍﻟﻤﻨﻌﻜﺴﺔ ﻤﻥ ﻤـﺴﺘﻭﻴﺎﺕ ﺒﻠﻭﺭﻴـﺔ ﻤﺘﺠﺎﻭﺭﺓ ﻭﻤﻘﺩﺍﺭ ﻓﺭﻕ ﺍﻟﻁﻭﺭ . ( 2D ) 2a sin θ λ
ﻭﻋﻨﺩﻤﺎ ﻴﺴﺎﻭﻱ ﻓﺭﻕ ﺍﻟﻁﻭﺭ
2 Dn
)ﺤﻴﺙ nﻫﻭ ﺭﻗﻡ ﺼـﺤﻴﺢ( ﻓـﺴﻴﺘﺤﻘﻕ
ﺸﺭﻁ ﺍﻟﺘﺩﺍﺨل ﺍﻟﺒﻨﱠﺎﺀ ،ﺃﻱ ﺃﻥ. 2D 2a sinθ = 2D n λ
- ٢٥PDF created with pdfFactory Pro trial version www.pdffactory.com
Figure 1-8 the scattering of X-rays by crystal planes when the angle 0 between the scattered X-rays and the crystal plane equals the angle between the incident X-rays and the plane. λ=
2a sin θ n
... (1-23)
ﺇﻥ ﺸﻜل ﺍﻟﺘﺩﺍﺨل ﺍﻟﻤﺸﺎﻫﺩ ﺘﺠﺭﻴﺒﻴﺎﹰ ﻤﻥ ﺘﺸﺘﺕ ﺍﻹﻟﻜﺘﺭﻭﻨﺎﺕ ﻨﺘﻴﺠﺔ ﺍﺴـﺘﺩﺍﻤﻬﺎ (Davisson and ﺒﺒﻠﻭﺭﺓ )ﻜﻤﺎ ﺘﺤﻘﻕ ﻋﻠﻰ ﻴﺩﻱ ﺍﻟﻌـﺎﻟﻤﻴﻥ ﺩﺍﻓﻴـﺴﻭﻥ ﻭ ﺠﻴﺭﻤـﺭ (1- ﺘﻌﻁﻲ ﺒﺎﻟﻤﻌﺎﺩﻟﺔ
λ
( ﻭﺫﻟﻙ ﺒﺸﺭﻁ ﺃﻥ1-23) ﻴﻤﻜﻥ ﺘﻔﺴﻴﺭﻩ ﺒﺎﻟﻤﻌﺎﺩﻟﺔGermer
ﺇﻥ ﻫﺫﻩ ﺍﻟﻨﺘﻴﺠﺔ ﺍﻟﺘﺠﺭﻴﺒﻴﺔ ﺴﺎﻫﻤﺕ ﺒﺨﻁـﻭﺓ.1-9 ﻜﻤﺎ ﻫﻭ ﻤﻭﻀﺢ ﻓﻲ ﺸﻜل22) .Wave mechanics ﻜﺒﻴﺭﺓ ﻓﻲ ﺘﻁﻭﻴﺭ ﻋﻠﻡ ﻤﻴﻜﺎﻨﻴﻜﺎ ﺍﻟﻤﻭﺠﺎﺕ
FIGURE 1.9 (a) the X-ray diffraction pattern of aluminum foil. (b) the electron diffraction pattern of aluminum foil. The similarity of these two patterns shows that electrons can behave like X-rays and display wavelike properties.
- ٢٦ PDF created with pdfFactory Pro trial version www.pdffactory.com
1-6ﺫﺭﺓ ﺒﻭﺭ 1-6 The Bohr Atom "The Bohr Theory of the Hydrogen Atom Can be Used "Drive the Rydberg For mole
ﻓﻲ ﻋﺎﻡ ،1911ﻗﺩﻡ ﺍﻟﻌﺎﻟﻡ ﺍﻟﺩﻨﻤﺎﺭﻜﻲ ﻨﻴﻠﺱ ﺒـﻭﺭ Niels Bohrﻨﻅﺭﻴﺘـﻪ
ﺍﻟﺸﻬﻴﺭﺓ ﻟﺫﺭﺓ ﺍﻟﻬﻴﺩﺭﻭﺠﻴﻥ ﻭﺍﻟﺘﻲ ﺘﺸﺭﺡ ﻭﺘﻔﺴﺭ ﺒﺒﺴﺎﻁﺔ ﺍﻟﻁﻴـﻑ ﺍﻟﻤﻨﺒﻌـﺙ ﻤـﻥ ﺍﻟﺫﺭﺓ. ﻁﺒﻘﺎﹰ ﻟﻠﻨﻤﻭﺫﺝ ﺍﻟﻨﻭﻭﻱ ﻟﻠﺫﺭﺓ ،ﻭﺍﻟﺫﻱ ﻴﻘﻭﻡ ﻋﻠﻰ ﺍﻟﻨﺘﺎﺌﺞ ﺍﻟﺘﺠﺭﻴﺒﻴﺔ ﻟﺘﻁـﺎﻴﺭ
ﺠﺴﻴﻤﺎﺕ ، αﻴﻤﻜﻥ ﺍﻋﺘﺒﺎﺭ ﺃﻥ ﻜﺘﻠﺔ ﺍﻟﺫﺭﺓ ﻤﺘﺭﻜﺯﺓ ﻓﻲ ﺍﻟﻨﻭﺍﺓ ﻭﺍﻟﺘﻲ ﺘﻌﺘﺒـﺭ ﺜﺎﺒﺘـﺔ ﻭﻴﺩﻭﺭ ﺤﻭﻟﻬﺎ ﺇﻟﻜﺘﺭﻭﻥ .ﺍﻟﻘﻭﺓ Fﺍﻟﺘﻲ ﻴﺭﺘﺒﻁ ﺒﻬﺎ ﺍﻹﻟﻜﺘﺭﻭﻥ ﻓﻲ ﻤﺩﺍﺭ ﺩﺍﺌﺭﻱ ﻫـﻲ ﻗﻭﺓ ﻜﻭﻟﻭﻡ ﺤﺴﺏ ﻗﺎﻨﻭﻥ ﻜﻭﻟﻭﻡ.
)... (1-24
)1 (Ze) (e 4Dε ° r 2
=F
ﺤﻴﺙ ) (Zeﻫﻲ ﺸﺤﻨﺔ ﺍﻟﻨﻭﺍﺓ )ﻟﺫﺭﺓ ﺍﻟﻬﻴﺩﺭﻭﺠﻴﻥ ( Z =1ﻭ ) (eﻫﻲ ﺸﺤﻨﺔ
ﺍﻹﻟﻜﺘﺭﻭﻥ.
ε ° = 8.85× 10 −12 c 2 N −1 m −2
,
r
ﻨﺼﻑ ﻗﻁﺭ ﺍﻟﺫﺭﺓ .ﺘﺘﻭﺍﺯﻥ ﻗﻭﺓ ﻜﻭﻟﻭﻡ ﻤﻊ ﺍﻟﻘﻭﺓ
ﺍﻟﻁﺎﺭﻗﺔ ﺍﻟﻤﺭﻜﺯﻴﺔ. )...(1-25
mv 2 r
=F
ﺤﻴﺙ vﻫﻲ ﺍﻟﺴﺭﻋﺔ ﺍﻟﺨﻁﻴﺔ ﻟﻺﻟﻜﺘﺭﻭﻥ. ﺒﻤﺴﺎﻭﺍﺓ ﺍﻟﻘﻭﺘﺎﻥ ﻨﺠﺩ: )...(1-26
1 e 2 mv 2 = 4Dε ° r 2 r
- ٢٧PDF created with pdfFactory Pro trial version www.pdffactory.com
ﻁﺒﻘﺎﹰ ﻟﻘﻭﺍﻨﻴﻥ ﺍﻟﻔﻴﺯﻴﺎﺀ ﺍﻟﻜﻼﺴﻴﻜﻴﺔ ،ﻓﺈﻥ ﺍﻟﺠﺴﻡ ﺍﻟﻤﺸﺤﻭﻥ ﺍﻟﻤﺘﺴﺎﺭﻉ .ﻴﺼﺩﺭ ﺇﺸﻌﺎﻉ ﻤﻤﺎ ﻴﺅﺩﻱ ﺇﻟﻰ ﻓﻘﺩﺍﻥ ﻟﻁﺎﻗﺘﻪ ،ﻭﻟﻬﺫﺍ ﺍﻟﺴﺒﺏ ﻓﺈﻥ ﺍﻹﻟﻜﺘﺭﻭﻥ ﺴـﻴﻔﻘﺩ ﻁﺎﻗﺘـﻪ
ﺨﻼل ﺩﻭﺭﺍﻨﻪ ﺤﻭل ﺍﻟﻨﻭﺍﺓ ﻭﺴﻴﺩﻭﺭ ﻓﻲ ﺸﻜل ﺤﻠﺯﻭﻨﻲ ﻭﻴﺘﻼﺸﻰ ﺩﺍﺨـل ﺍﻟﻨـﻭﺍﺓ،
ﻭﻋﻠﻴﻪ ﻓﻼ ﻴﻤﻜﻥ ﺃﻥ ﻴﻭﺠﺩ ﻤﺩﺍﺭ ﻤﺴﺘﻘﺭ stable orbitﻟﻼﻟﻜﺘﺭﻭﻥ .ﻭﻟﻠﺨﺭﻭﺝ ﻤـﻥ ﻫﺫﺍ ﺍﻹﺸﻜﺎل ﺍﻗﺘﺭﺡ ﺒﻭﺭ ﻓﺭﻀﻴﺎﺘﻪ ﺍﻟﺘﻲ ﺘﺨﺎﻟﻑ ﻗﻭﺍﻨﻴﻥ ﺍﻟﻔﻴﺯﻴﺎﺀ ﺍﻟﺘﻘﻠﻴﺩﻴﺔ. ﺍﻟﻔﺭﻀﻴﺔ ﺍﻷﻭﻟﻰ: ﺍﻟﻤﺩﺍﺭﺍﺕ ﺍﻹﻟﻜﺘﺭﻭﻨﻴﺔ ﺍﻟﻤﺴﺘﻘﺭﺓ )ﺍﻟﺜﺎﺒﺘﺔ( Stationary electron orbits
ﻭﻫﺫﻩ ﺍﻟﻔﺭﻀﻴﺔ ﻫﻲ ﺘﹶﺤﺩٍ ﺒﺎﻟﻤﻔﺎﻫﻴﻡ ﺍﻟﺘﻘﻠﻴﺩﻴﺔ ﻟﻠﻔﻴﺯﻴﺎﺀ .ﻭﻟﻘﺩ ﺤﺩﺩ ﺒـﻭﺭ ﻫـﺫﻩ
ﺍﻟﻤﺩﺍﺭﺍﺕ ﺒﺎﺴﺘﺤﺩﺍﺙ ﺸﺭﻁ ﺍﻟﺘﻜﻤﻴﻡ quantization conditionﻭﺍﻓﺘﺭﺽ ﺃﻥ ﻜﻤﻴﺔ ﺍﻟﺤﺭﻜﺔ ﺍﻟﺯﺍﻭﻴﺔ angular momentum Lﻤﻜﻤﻠﺔ ﺤﺴﺏ ﺍﻟﻌﻼﻗﺔ. L = mvr = nh , n = 1, 2, ........
)...(1-27 ﻤﻥ ﻫﺫﻩ ﺍﻟﻌﻼﻗﺔ ﻨﺠﺩ ﺃﻥ
hh mr
=v
ﺒﺎﻟﺘﻌﻭﻴﺽ ﻓﻲ )(1-26 1 e 2 m nh 2 m n 2 h 2 ) ( = 4Dε ° r 2 r mr m2 r3
ﻭﻓﻴﻬﺎ ﻨﺠﺩ ﺃﻥ. )...(1-28
4D ε ° h 2 2 n2 h2 1 2 .n = e =⇒ r 4Dε ° mr me 2
ﻭﻤﻥ ﻫﺫﻩ ﺍﻟﻌﻼﻗﺔ ﻨﺠﺩ ﺃﻥ ﺃﻨﺼﺎﻑ ﺃﻗﻁﺎﺭ ﺍﻟﻤﺩﺍﺭﺍﺕ ﻟﻬﺎ ﻤﻘﺎﺩﻴﺭ ﻤﺤـﺩﺩﺓ ﺃﻭ ﻤﻜﻤﻤﺔ .ﺇﻥ ﺃﻗل ﻨﺼﻑ ﻗﻁﺭ ﻟﻺﻟﻜﺘﺭﻭﻥ ﻴﻭﺠﺩ ﺒﺎﻟﺘﻌﻭﻴﺽ ﻋﻥ .n = 1 - ٢٨PDF created with pdfFactory Pro trial version www.pdffactory.com
( u D ) ( 8.85×10−12 c 2 N −1 m −2 ) (1.055× 10−34 J.s) 2 =r (9.11× 10 − 31 kg) (1.602 × 10 −19 c) 2 °
= 5.29 ×10 −11 m = 52.9 pm ≈ 0.53 A
ﻭﻫﺫﻩ ﺍﻟﻘﻴﻤﺔ ﻴﺭﻤﺯ ﻟﻬﺎ ﻋﺎﺩﺓ ﺒـ
a°
ﺍﻟﻁﺎﻗﺔ ﺍﻟﻜﻠﻴﺔ ﻟﻺﻟﻜﺘﺭﻭﻥ Eﻫﻲ ﻋﺒﺎﺭﺓ ﻋﻥ ﻤﺠﻤﻭﻉ ﻁـﺎﻗﺘﻲ ﺍﻟﺤﺭﻜـﺔ k.E
ﻭﺍﻟﺘﻲ ﺘﺴﺎﻭﻱ )...(1-29
1 mv 2 2
e2 1 4D ε ° r
ﻭﻁﺎﻗﺔ ﺍﻟﻭﻀﻊ ) V(rﻭﺼﻴﻐﺘﻬﺎ.
V (r) = −
ﺍﻹﺸﺎﺭﺓ ﺍﻟﺴﺎﻟﺒﺔ ﻓﻲ ) (1-29ﺘﻌﻨﻲ ﺃﻥ ﺍﻟﺒﺭﻭﺘﻭﻥ ﻭﺍﻹﻟﻜﺘﺭﻭﻥ ﻴﺠـﺫﺏ ﻜـل
ﻤﻨﻬﻤﺎ ﺍﻵﺨﺭ.
ﻻﺤﻅ ﺃﻥ ﻁﺎﻗﺔ ﺍﻟﺘﺠﺎﺫﺏ ﺒﻴﻥ ﺍﻟﺒﺭﻭﺘﻭﻥ ﻭﺍﻹﻟﻜﺘﺭﻭﻥ ﺘﻘل ﻜﻠﻤﺎ ﺯﺍﺩﺕ ﺍﻟﻤﺴﺎﻓﺔ
ﺒﻴﻨﻬﻤﺎ ﻭﻋﻨﺩ )...(1-30
∞= r
ﻓﺈﻥ
V (∞ ) = 0
1 e2 1 mv 2 + (− ) 2 4Dε ° r
ﻤﻥ ﺍﻟﻌﻼﻗﺔ )(1-26
= )E = KE + V (r
1 e 2 mv 2 = 2 r 4Dε ° r
ﻴﻤﻜﻥ ﻜﺘﺎﺒﺔ mv2ﺒﺎﻟﺼﻴﻐﺔ
1 e2 4Dε ° r
= mv 2
ﺒﺎﻟﺘﻌﻭﻴﺽ ﻓﻲ )(1-30 1 1 e2 1 e2 ( = ∴E )− 2 4D ε ° r 4Dε ° r )......(1 − 3
1 1 e2 1 e2 =− 2 4Dε ° r 8Dε ° r
=−
ﺒﺎﻟﺘﻌﻭﻴﺽ ﻋﻥ rﻤﻥ ) (1-28ﻓﻲ ) (1-31ﻨﺠﺩ: h 2D me 4 1 2 2 2 2 32 D ε ° h n
=− n2
1 e2 8Dε ° 4Dε ° h 2 2
=h
E=−
me
- ٢٩PDF created with pdfFactory Pro trial version www.pdffactory.com
ﺍﻻﺨﺘﺼﺎﺭ
h 2h
=h 1 n2 n =1,2,.............
me 4
E=−
2
h 4D 2 me 4 1 ∴E n = − 2 2 2 8ε ° h n 2
(32)D 2 ε °
)...(1-32 ﺍﻹﺸﺎﺭﺓ ﺍﻟﺴﺎﻟﺒﺔ ﻓﻲ ﻫﺫﻩ ﺍﻟﻤﻌﺎﺩﻟﺔ ﺘﺩل ﻋﻠﻰ ﺃﻥ ﺤﺎﻻﺕ ﺍﻟﻁﺎﻗﺔ ﺤﺎﻻﺕ ﻤﻘﻴﺩﺓ
.bound states
ﻻﺤﻅ ﺃﻥ ﻓﻲ ﺤﺎﻟﺔ n = 1ﻓﺈﻥ ﻤﻌﺎﺩﻟﺔ ) (1- 32ﺘﻌﻁﻴﻨﺎ ﺍﻟﺤﺎﻟﺔ ﺫﺍﺕ ﺃﻗل ﻁﺎﻗﺔ state of lowest energyﻭﺘﺴﻤﻰ ﻫﺫﻩ ﺍﻟﻁﺎﻗﺔ ﺒﺎﻟﻁﺎﻗﺔ ﺍﻷﺭﻀﻴﺔ ﺃﻭ ﻁﺎﻗﺔ ﺍﻟﺤﺎﻟـﺔ
ﺍﻷﺭﻀﻴﺔ .ground-state energyﺃﻤﺎ ﺤﺎﻻﺕ ﺍﻟﻁﺎﻗﺔ ﺍﻷﻋﻠﻰ ﺘـﺴﻤﻰ ﺍﻟﺤـﺎﻻﺕ ﺍﻟﻤﺴﺘﺜﺎﺭﺓ excited statesﻭﻋﺎﺩﺓ ﻤﺎ ﺘﻜﻭﻥ ﻏﻴﺭ ﻤﺴﺘﻘﺭﺓ ﺒﺎﻟﻨﺴﺒﺔ ﻟﻠﺤﺎﻟﺔ ﺍﻷﺭﻀﻴﺔ،
ﻭﻋﻨﺩﻤﺎ ﺘﻜﻭﻥ ﺍﻟﺫﺭﺓ )ﺃﻭ ﺍﻟﺠﺯﻱﺀ( ﻓﻲ ﺍﻟﺤﺎﻟﺔ ﺍﻟﻤﺴﺘﺜﺎﺭﺓ ﻓﺈﻨﻬﺎ ﺘـﺴﺘﺭﺨﻲ ﻭﺘﺭﺠـﻊ
ﻟﻠﺤﺎﻟﺔ ﺍﻷﺭﻀﻴﺔ ﻭﺘﻌﻁـﻲ ﺃﻭ ﺘـﺘﺨﻠﺹ ﻤـﻥ ﻁﺎﻗﺘﻬـﺎ ﻓـﻲ ﺼـﻭﺭﺓ ﻤﻭﺠـﺎﺕ
ﻜﻬﺭﻭﻤﻐﻨﺎﻁﻴﺴﻴﺔ ﻜﻤﺎ ﻫﻭ ﻤﻭﻀﺢ ﻓﻲ ﺍﻟﺸﻜل ).(1-10
ﺍﻟﻔﺭﻀﻴﺔ ﺍﻟﺜﺎﻨﻴﺔ :ﻴﻤﻜﻥ ﻟﻼﻟﻜﺘﺭﻭﻨﺎﺕ ﺃﻥ ﺘﻨﺘﻘل )ﺃﻭ ﺘﻘﻔـﺯ( ﻤـﻥ ﻤـﺩﺍﺭﺍﺘﻬﺎ
ﺒﻁﺭﻴﻘﺔ ﻏﻴﺭ ﻤﺘﺼﻠﺔ discontinuous transitionsﻭﺃﻥ ﺍﻟﺘﻐﻴﺭ ﻓـﻲ ﺍﻟﻁﺎﻗـﺔ ΔE
ﻴﺅﺩﻱ ﻻﻨﺒﻌﺎﺙ ﺇﺸﻌﺎﻉ ﻟﻪ ﺘﺭﺩﺩ . hvﻭﻟﻬﺫﺍ ﺍﻟﺴﺒﺏ ﻟﻭ ﺍﻨﺘﻘل ﺇﻟﻜﺘﺭﻭﻥ ﻤـﻥ ﺍﻟﻤـﺩﺍﺭ ﺍﻟﺫﻱ ﻟﻪ n2 =1ﺇﻟﻰ ﺍﻟﻤﺩﺍﺭ ﺫﻭ n1 = 2ﻓﺈﻥ ﺍﻟﻔﺭﻕ ﻓﻲ ﺍﻟﻁﺎﻗﺔ. me 4 1 me 4 1 − − ( ) 2 2 2 2 8ε ° h 2 n 2 8 ε ° h 2 n1 )...............(1 − 33
ΔE = E 2 − E 1 = −
me 4 1 1 ΔE = 2 2 ( 2 − 2 ) = hv 8ε ° h n 1 n 2
- ٣٠PDF created with pdfFactory Pro trial version www.pdffactory.com
FIGURE 1.10 The energy level diagram for the hydrogen atom, showing how transitions from higher states into some particular state lead to the observed spectral series for hydrogen.
( ﻴﻤﻜﻨﻨﺎ ﺍﻟﺤﺼﻭل ﻋﻠﻰ ﺍﻟﺼﻴﻐﺔ ﺍﻟﺭﻴﺎﻀﻴﺔ ﻟﻠﻌﻼﻗﺔ ﺍﻟﻌﺩﺩﻴﺔ1-33) ﻤﻥ ﻤﻌﺎﺩﻟﺔ
.ﺍﻟﻤﻌﺭﻭﻓﺔ ﺒﻤﻌﺎﺩﻟﺔ ﺭﻴﺩﺒﺭﺝ ﻭﺍﻟﺘﻲ ﺘﺼﻑ ﺠﻤﻴﻊ ﺨﻁﻭﻁ ﺍﻟﻁﻴﻑ ﻟﺫﺭﺓ ﺍﻟﻬﻴﺩﺭﻭﺠﻴﻥ .( ﻨﺠﺩ1-33) ﻓﻲ
hc λ
ﺒـhv ﺒﺎﻟﺘﻌﻭﻴﺽ ﻋﻥ
hc me 4 1 1 = 2 2 ( 2 − 2) λ 8ε ° h n 1 n 2 1 λ 1 me 4 1 1 = 2 3 ( 2 − 2) λ 8ε ° h c n 1 n 2 1 1 1 = R H ( 2 − 2 ) ..............(1 − 34) λ n1 n2
- ٣١ PDF created with pdfFactory Pro trial version www.pdffactory.com
ﻭﻓﻴﻬﺎ ﻨﺠﺩ
ﺤﻴﺙ RHﻫﻭ ﺜﺎﺒﺕ ﺭﻴﺩﺒﺭﺝ Rydberg constant me 4 (9.11× 10 −31 kg) (1.60× 10−19 c)4 = 2 )8ε ° h 3c (8) (8.85× 10 −12 c 2 N −1m − 2 ) (6.63 × 10 − 34 J.s)3 (2.0 × 108 m/s
= RH
R H =1.089 × 107 m −1 )(1
ﺒﺎﻟﻨﻅﺭ ﻟﺨﻁﻭﻁ ﺍﻟﻁﻴﻑ ﻓﻲ ﺸﻜل 1-10ﻴﻤﻜﻥ ﺃﻥ ﻨﻼﺤﻅ ﺍﻟﺘﻭﺍﻓﻕ ﺒﻴﻥ ﻨﻤﻭﺫﺝ ﺒﻭﺭ ﻭﻫﺫﻩ ﺍﻟﺨﻁﻭﻁ ﺍﻟﻤﺘﻔﻘﺔ ﻤﻊ ﺍﻟﻨﺘﺎﺌﺞ ﺍﻟﺘﺠﺭﻴﺒﻴﺔ .ﻓﻤﺜﻼﹰ ﺨﻁﻭﻁ ﺴﻠـﺴﻠﺔ ﻟﻴﻤـﺎﻥ Lymanﺘﻨﺸﺄ ﻤﻥ ﺭﺠﻭﻉ )ﺍﺴﺘﺭﺨﺎﺀ( ﺍﻹﻟﻜﺘﺭﻭﻨﺎﺕ ﺍﻟﻤﺴﺘﺜﺎﺭﺓ ﻤﻥ ﺍﻟﻤﺴﺘﻭﻴﺎﺕ ﺍﻟﻌﻠﻴﺎ
ﻟﻠﻤﺩﺍﺭ ﺍﻷﻭل ) ،(n = 1ﻭﻜﺫﻟﻙ ﺨﻁﻭﻁ ﺴﻠﺴﻠﺔ ﺒﺎﻟﻤﺭ Balmer seriesﺘﺤﺩﺙ ﻤـﻥ ﺍﺴﺘﺭﺨﺎﺀ ﺍﻹﻟﻜﺘﺭﻭﻨﺎﺕ ﺍﻟﻤﺴﺘﺜﺎﺭﺓ ﻤﻥ ﻜل ﺍﻟﻤﺴﺘﻭﻴﺎﺕ ﺍﻟﻌﻠﻴـﺎ ﻟﻠﻤـﺴﺘﻭﻯ ﺍﻟﺜـﺎﻨﻲ
).(n=2
٢
ﻤﺜﺎل 6
ﺍﺤﺴﺏ ﻁﺎﻗﺔ ﺍﻟﺘﺄﻴﻥ ﻟﺫﺭﺓ ﺍﻟﻬﻴﺩﺭﻭﺠﻴﻥ؟ ?Calculate the ionization energy of the hydrogen atom
ﺍﻟﺤل: ﻁﺎﻗﺔ ﺍﻟﺘﺄﻴﻥ ﻫﻲ ﺍﻟﻁﺎﻗﺔ ﺍﻟﻼﺯﻤﺔ ﻹﺯﺍﻟﺔ ﺍﻹﻟﻜﺘﺭﻭﻥ ﻤـﻥ ﻤـﺴﺘﻭﻯ ﺍﻟﺤﺎﻟـﺔ
ﺍﻷﺭﻀﻴﺔ n1 = 1ﺇﻟﻰ ﺍﻟﺤﺎﻟﺔ ﻏﻴﺭ ﺍﻟﻤﻘﻴﺩﺓ ﺃﻱ
∞= n2
me 4 1 1 )( 2 − 2 2 2 8ε ° h n 1 n 2
= ΔE
(9.1 × 10 - 34 kg) (1.9 × 10 −19 c) 4 1 1 = )( 2 − 2 − 12 2 −1 −2 − 34 2 ∞ (8) (8.85 × 10 c N m ) (6.63 × 10 J.s) 1 = 2.18 × 10 -18 J = 13.6 eV
ﻓﻲ ﻋﺎﻡ ،1925ﻅﻬﺭﺕ ﺍﻟﻨﻅﺭﻴﺔ ﺍﻟﺤﺩﻴﺜﺔ ﻟﻤﻴﻜﺎﻨﻴﻜﺎ ﺍﻟﻜﻡ ﻭﺫﻟﻙ ﺒﻨﺎﺀﺍﹰ ﻋﻠـﻰ ﺍﻷﺒﺤﺎﺙ ﺍﻟﺘﻲ ﺭﺴﺦ ﺩﻋﺎﺌﻤﻬﺎ ﻜل ﻤﻥ ﺍﻟﻌﺎﻟﻡ ﻫﻴﺯﻨﺒﺭﺝ W. Heisenbergﻭﺍﻟﻌـﺎﻟﻡ (2 ) The exact value of RH = 1.09736 × 107 m-1 = 109.736 cm-1
- ٣٢PDF created with pdfFactory Pro trial version www.pdffactory.com
ﻤﺎﻜﺱ ﺒﻭﺭﻥ ،M. Bornﻭﺍﻟﻌﺎﻟﻡ ﺸـﺭﻭﺩﻨﺠﺭ E. Shrödingerﻭﺍﻟﻌـﺎﻟﻡ ﺩﻴـﺭﺍﻙ .P. Dirac ﻭﺘﺸﺭﺡ ﺍﻟﻨﻅﺭﻴﺔ ﺍﻟﻜﻤﻴﺔ ﺍﻟﺤﺩﻴﺜﺔ ﺍﻟﻤﻔﺎﻫﻴﻡ ﺍﻟﻤﺤﻴﺭﺓ ﺒﺸﺭﻁ ﺃﻥ ﻨﺘﺨﻠـﻰ ﻋـﻥ
ﺒﻌﺽ ﺍﻟﻤﻔﺎﻫﻴﻡ ﺍﻟﺘﻲ ﺘﺭﺴﺨﺕ ﻓﻲ ﺃﺫﻫﺎﻨﻨﺎ ﻤﻥ ﻋﻠﻡ ﺍﻟﻔﻴﺯﻴﺎﺀ ﺍﻟﺘﻘﻠﻴﺩﻴﺔ!. (Q1ﺇﺫﺍ ﻋﻠﻤﺕ ﺃﻥ ﻜﺜﺎﻓﺔ ﺍﻹﺸﻌﺎﻉ ﺘﻌﻁﻰ ﺒﺎﻟﻤﻌﺎﺩﻟﺔ
8Dh v3 c 3 e hv/KT − 1
) (aﺍﺤﺴﺏ ﻜﺜﺎﻓﺔ ﺍﻟﻁﺎﻗﺔ ﻓﻲ ﻤﺩﻯ ﻁﻭل ﻤﻭﺠﻲ ، Δλﺃﻱ ) (bﺍﺴﺘﺨﺩﻡ ﺍﻟﻨﺘﻴﺠﺔ ﻓﻲ ﺠﺯﺀ ) (aﻹﻴﺠﺎﺩ ﻗﻴﻤﺔ
max
λ =λ
ﺍﻹﺸﻌﺎﻉ ﺃﻗﺼﻰ ﻤﺎ ﻴﻤﻜﻥ؟
) (cﻭﻀﺢ ﺃﻥ
λ max
ﻴﻤﻜﻨﻨﺎ ﻜﺘﺎﺒﺘﻬﺎ ﻋﻠﻰ ﺍﻟﺼﻴﻐﺔ
b T
max
, T)dλ
= )u (v, T
u (λ؟
ﻭﺍﻟﺘﻲ ﻋﻨﺩﻫﺎ ﺘﻜﻭﻥ ﻜﺜﺎﻓﺔ
، λﻭﺍﺤﺴﺏ ﻗﻴﻤـﺔ bﻓـﻲ
ﺤﺎﻟﺔ ﺴﻁﺢ ﺍﻟﺸﻤﺱ ﻋﻠﻤﺎﹰ ﺒﺄﻥ ﺩﺭﺠﺔ ﺤﺭﺍﺭﺓ ﺴﻁﺢ ﺍﻟﺸﻤﺱ ﺘﺴﺎﻭﻱ 5620k؟ )ﺘﻨﺒﻴﻪ :ﺤل ﺍﻟﻤﻌﺎﺩﻟﺔ s-x = se-xﺒﺎﻟﺭﺴﻡ ﺍﻟﺒﻴﺎﻨﻲ(. ) (dﻴﻨﺒﻌﺙ ﻤﻥ ﺃﺸﺩ ﺍﻟﻨﺠﻭﻡ ﺤﺭﺍﺭﺓ )ﻨﺠﻡ ﺍﻟﺸﹼﻌﺭﻱ (Siriusﻁﻴﻑ ﺇﺸﻌﺎﻉ ﺍﻟﺠـﺴﻡ ﺍﻷﺴﻭﺩ ﻭﺍﻟﺫﻱ ﻟﻪ
= 260nm
max
. λﺍﺤﺴﺏ ﺩﺭﺠﺔ ﺤﺭﺍﺭﺓ ﺴﻁﺢ ﻫﺫﺍ ﺍﻟﻨﺠﻡ؟
) (eﺇﺫﺍ ﻋﻠﻤﺕ ﺃﻥ ﻤﺘﻭﺴﻁ ﺩﺭﺠﺔ ﺤﺭﺍﺭﺓ ﺴﻁﺢ ﺍﻷﺭﺽ .288 kﺍﺴﺤﺏ ﺍﻟﻁـﻭل ﺍﻟﻤﻭﺠﻲ ﻷﻗﺼﻰ ﻜﺜﺎﻓﺔ ﺇﺸﻌﺎﻉ ﻟﻠﺠﺴﻡ ﺍﻷﺴﻭﺩ ﻟﻸﺭﺽ .ﺤﺩﺩ ﻷﻱ ﺠﺯﺉ ﻤـﻥ
ﺍﻟﻁﻴﻑ ﻴﻘﺎﺒل ﻫﺫﺍ ﺍﻟﻁﻭل ﺍﻟﻤﻭﺠﻲ؟ (Q2ﺃﻗﺼﻰ ﻁﺎﻗﺔ ﺤﺭﻜﻴﺔ ﻹﻟﻜﺘﺭﻭﻨﺎﺕ ﻤﻨﺒﻌﺜﺔ ﻤﻥ ﺴﻁﺢ ﺃﻟﻭﻤﻨﻴﻭﻡ ﺘﺴﺎﻭﻱ 2.3 eV
ﻭﺫﻟﻙ ﻋﻨﺩ ﺘﻌﺭﺽ ﻫﺫﺍ ﺍﻟﺴﻁﺢ ﻷﺸﻌﺔ ﺫﺍﺕ ﻁﻭل ﻤﻭﺠﻲ .2000Åﺃﻤﺎ ﻋﻨـﺩ ﺘﻌﺭﺽ ﺍﻟﺴﻁﺢ ﻷﺸﻌﺔ ﻁﻭﻟﻬﺎ ﺍﻟﻤﻭﺠﻲ 2580 Åﻓـﺈﻥ ﺍﻟﻁﺎﻗـﺔ ﺍﻟﺤﺭﻜﻴـﺔ
ﻟﻺﻟﻜﺘﺭﻭﻨﺎﺕ ﺍﻟﻤﻨﺒﻌﺜﺔ ﺘﺴﺎﻭﻱ .0.9 eVﺍﺤﺴﺏ ﻗﻴﻤﺔ ﺜﺎﺒـﺕ ﺒﻼﻨـﻙ ﻭﺩﺍﻟـﺔ ﺍﻟﺸﻐل ﻟﻸﻟﻤﻭﻨﻴﻭﻡ؟
- ٣٣PDF created with pdfFactory Pro trial version www.pdffactory.com
(Q3ﺍﺤﺴﺏ ) (aﺍﻟﻁﻭل ﺍﻟﻤﻭﺠﻲ ﻭﺍﻟﻁﺎﻗﺔ ﺍﻟﺤﺭﻜﻴﺔ ﻹﻟﻜﺘﺭﻭﻥ ﻤﻌﺠل ﺘﺤﺕ ﺘـﺄﺜﻴﺭ ﻓﺭﻕ ﺠﻬﺩ 100 Vﻭ ) (bﻁﺎﻗﺔ ﺍﻟﺤﺭﻜﺔ ﺍﻹﻟﻜﺘﺭﻭﻥ ﻟﻪ ﻁﻭل ﻤﻭﺠﻲ ﺩﻴﺒﺭﻭﻟﻲ
) 200 pmﺤﻴﺙ (1 pm = 10-12 m؟
(Q4ﺃﺸﻌﺔ Xﺍﺴﺘﻁﺎﺭﺕ ﺒﺈﻟﻜﺘﺭﻭﻥ ﺴﺎﻜﻥ .ﺍﺤﺴﺏ ﻁﺎﻗﺔ ﺃﺸـﻌﺔ Xﺍﻟـﺴﺎﻗﻁﺔ ﺇﺫﺍ ﻋﻠﻤﺕ ﺃﻥ ﻁﻭل ﻤﻭﺠﺔ ﺍﻷﺸﻌﺔ ﺍﻟﻤﺴﺘﻁﺎﺭﺓ ﻋﻨﺩ ﺯﺍﻭﻴﺔ
60°
ﺘﺴﺎﻭﻱ 0.035 A؟ °
(Q5ﺍﻟﻤﺴﺎﻓﺔ ﺒﻴﻥ ﻤﺴﺘﻭﻴﻴﻥ ﻤﺘﺠﺎﻭﺭﻴﻥ ﻤﻥ ﺍﻟﻤﺴﺘﻭﻴﺎﺕ ﺍﻟﺒﻠﻭﺭﻴـﺔ ﻴـﺭﺍﺩ ﻗﻴﺎﺴـﻬﺎ ﺒﺎﺴﺘﺨﺩﺍﻡ ﺃﺸﻌﺔ Xﻁﻭﻟﻬﺎ ﺍﻟﻤﻭﺠﻲ 0.5 Åﻭﺍﻟﺘﻲ ﺘﻡ ﻗﻴﺎﺴﻬﺎ ﻋﻨﺩ ﺯﺍﻭﻴـﺔ . 5 °
ﺍﺤﺴﺏ ﻗﻴﻤﺔ ﺍﻟﻤﺴﺎﻓﺔ ﺒﻴﻥ ﻫﺫﻴﻥ ﺍﻟﻤﺴﺘﻭﻴﻴﻥ؟ ﻋﻨﺩ ﺃﻱ ﺯﺍﻭﻴﺔ ﻴﻤﻜﻨﻨـﺎ ﻗﻴـﺎﺱ
ﺍﻟﻘﻴﻤﺔ ﺍﻟﺜﺎﻨﻴﺔ؟
ﺇﺠﺎﺒﺎﺕ ﺍﻷﺴﺌﻠﺔ :ﺍﻹﺠﺎﺒﺔ ﺍﻷﺨﻴﺭﺓ ﻤﻥ ﻜل ﺴﺅﺍل: 8Dhc hc/λc/ (e − T) −1 λs
= )u (λ , T
)Q1) (a
ﻨﻅﺭﻱ λ max = b / Tﺍﻟﻤﻁﻠﻭﺏ ﺇﺜﺒﺎﺕ ﺍﻟﻘﺎﻨﻭﻥ )(b °
λ mex = 5160 A
)(c
T = 1.12 ×10 4 k 1
)(d
λ = 1.01× 10 −5 m
)(e
W = 3.92 eV
)(Q2
K.E = 6.02 × 10-18 J
)(Q3
E = 5.4× 105 eV
)(Q4
θ =10 °
)(Q5
- ٣٤PDF created with pdfFactory Pro trial version www.pdffactory.com
ﺍﻟﺤﺯﻡ ﺍﻟﻤﻭﺠﻴﺔ ﻭﻋﻼﻗﺎﺕ ﺍﻟﻼﺘﺤﺩﻴﺩ WAVE PACKETS AND THE UNCERTAINTY RELATIONS
ﺇﻥ ﻋﻠﻡ ﻤﻴﻜﺎﻨﻴﻜﺎ ﺍﻟﻜﻡ ﻴﺯﻭﺩﻨﺎ ﺒﻔﻬﻡ ﺼﺤﻴﺢ ﻟﻜل ﺍﻟﻅﻭﺍﻫﺭ ﺍﻟﻔﻴﺯﻴﺎﺌﻴﺔ ﺍﻟﺘﻲ ﺘـﻡ
ﻤﻨﺎﻗﺸﺘﻬﺎ ﻓﻲ ﺍﻟﻔﺼل ﺍﻷﻭل .ﻭﻫﺫﺍ ﺍﻟﻌﻠﻡ ﻀﺭﻭﺭﻱ ﻟﻔﻬﻡ ﺴﻠﻭﻙ ﺍﻟﺫﺭﺍﺕ ،ﺍﻟﺠﺯﻴﺌﺎﺕ، ﺃﻨﻭﻴﺔ ﺍﻟﺫﺭﺍﺕ ﻭﺘﺠﻤﻌﺎﺕ ﻤﻥ ﻫﺫﻩ ﻭﺘﻠﻙ .ﻭﻋﺎﺩﺓ ﻤﺎ ﺘﺘﻡ ﺍﻟﺩﺭﺍﺴﺔ ﺒﺎﺴﺘﺨﺩﺍﻡ ﻤﻌﺎﺩﻟـﺔ
ﺸﺭﻭﺩﻨﺠﺭ Shrodinger equationﻭﺍﻟﺘﺄﻭﻴﻼﺕ ﺍﻟﺼﺤﻴﺤﺔ ﻟﺤﻠﻭﻟﻬﺎ .ﻭﻫﺫﻩ ﺍﻟﻤﻌﺎﺩﻟﺔ ﻻ ﻴﻭﺠﺩ ﻟﻬﺎ ﺇﺜﺒﺎﺕ! ﻭﻟﻜﻥ ﺍﺴﺘﻁﺎﻉ ﺸﺭﻭﺩﻨﺠﺭ ﺍﻟﺘﻭﺼل ﺇﻟﻴﻬﺎ ﻋﻥ ﻁﺭﻴـﻕ ﺇﺘﺒـﺎﻉ
ﺘﻭﺠﻴﻬﺎﺕ )ﺃﻭ ﻓﺭﺍﺴﺎﺕ (insightﺍﻟﻌﺎﻟﻡ ﺍﻟﻔﺭﻨﺴﻲ ﺩﻴﺒﺭﻭﻟﻲ .ﻭﻤﻥ ﺍﻟﺠﺩﻴﺭ ﺒﺎﻟﺫﻜﺭ ﺃﻥ ﻫﺫﻩ ﺍﻟﻤﻌﺎﺩﻟﺔ ﺘﻭﺠﺩ ﺨﺎﺭﺝ ﻨﻁﺎﻕ ﺍﻟﻔﻴﺯﻴﺎﺀ ﺍﻟﺘﻘﻠﻴﺩﻴﺔ ،ﺍﻷﻤﺭ ﺍﻟﺫﻱ ﻴﺠﻌل ﺍﺴـﺘﻨﺒﺎﻁﻬﺎ
ﺼﻌﺏ .ﻭﺨﻼل ﺍﻟﻤﻨﺎﻗﺸﺔ ﻓﻲ ﻫﺫﺍ ﺍﻟﻔﺼل ﺴﻨﺤﺎﻭل ﺍﻟﺘﻭﻓﻴﻕ ﺒﻴﻥ ﺍﻟﺨﻭﺍﺹ ﺍﻟﺠﺴﻴﻤﻴﺔ ﻭﺍﻟﻤﻭﺠﻴﺔ ﻟﻼﻟﻜﺘﺭﻭﻨﺎﺕ.
ﺇﻨﻪ ﻤﻥ ﺍﻟﺼﻌﺏ ﺘﺨﻴل ﺍﻹﻟﻜﺘﺭﻭﻨﺎﺕ ﻋﻠﻰ ﺃﺴﺎﺱ ﺃﻥ ﻟﻬﺎ ﺴـﻠﻭﻙ ﺍﻟﻤﻭﺠـﺎﺕ
ﻭﻟﻜﻥ ﺘﺠﺎﺭﺏ ﺍﻟﺤﻴﻭﺩ ﺍﻟﺘﻲ ﺃﺠﺭﺍﻫﺎ ﻜـل ﻤـﻥ Freshet and Youngﺃﺩﺕ ﺇﻟـﻰ ﺍﻹﺠﻤﺎﻉ )ﺃﻭ ﺍﺘﺤﺎﺩ ﺍﻵﺭﺍﺀ( ﺒﻘﺒﻭل ﺍﻟﻨﻅﺭﻴﺔ ﺍﻟﻤﻭﺠﻴﺔ ﻟﻠﻀﻭﺀ .ﻭﻤﻥ ﺠﺎﻨـﺏ ﺁﺨـﺭ
ﺒﺎﻹﻤﻜﺎﻥ ﺃﻥ ﻨﺘﺼﻭﺭ ﺘﺠﻤﻌﺎﺕ )ﺃﻭ ﻫﻴﺌﺎﺕ( ﻟﻠﻤﻭﺠﺎﺕ ﺍﻟﻤﺤﺼﻭﺭﺓ ﺃﻭ ﺍﻟﻤﺘﺤﻴﺯﺓ )ﺃﻭ
ﺍﻟﻤﺘﻤﺭﻜﺯﺓ ﺠﺩﺍﹰ( Very localizedﻓﻌﻠﻰ ﺴﺒﻴل ﺍﻟﻤﺜﺎل ﺘﻌﺘﺒﺭ ﻗﺭﻗﻌﺔ ﺍﻟﺭﻋﺩ a clap
of thunderﻤﺜﺎل ﻟﺘﺩﺍﺨل ﺍﻟﻤﻭﺠﺎﺕ ﻭﺘﺭﺍﻜﺒﻬﺎ ﻤﻤﺎ ﻴﺅﺩﻱ ﺇﻟﻰ ﺘﺄﺜﻴﺭ ﻤﺘﻤﺭﻜﺯ ﻤـﻊ
ﺍﻟﺯﻤﻥ .localized in timeﻤﺜل ﻫﺫﻩ ﺍﻟﺤﺯﻡ ﺍﻟﻤﻭﺠﻴـﺔ ﺍﻟﻤﺘﻤﺭﻜـﺯﺓ localized
" "wane packetsﻴﻤﻜﻥ ﺍﻟﺤﺼﻭل ﻋﻠﻴﻬـﺎ ﻋـﻥ ﻁﺭﻴـﻕ ﺘﺭﺍﻜـﺏ ﺍﻟﻤﻭﺠـﺎﺕ superposing wavesﺒﺘﺭﺩﺩﺍﺕ ﻤﺨﺘﻠﻔﺔ ﺒﺤﻴﺙ ﻴﺘﻡ ﺘﺩﺍﺨل ﻜل ﻤﻭﺠﺔ ﻤﻊ ﺍﻷﺨـﺭﻯ ﺨﺎﺭﺝ ﻤﻨﻁﻘﺔ ﻤﻜﺎﻨﻴﺔ. They interfere with each other almost completely outside of a given spatial region.
- ٣٥PDF created with pdfFactory Pro trial version www.pdffactory.com
ﻭﺍﻟﻁﺭﻴﻘﺔ ﺍﻟﺘﻲ ﻴﻤﻜﻥ ﺇﺘﺒﺎﻋﻬﺎ )ﺃﻭ ﺍﻷﺩﻭﺍﺕ ﺍﻟﺭﻴﺎﻀﻴﺔ ﺍﻟﻤﺴﺘﺨﺩﻤﺔ( ﻫﻲ ﻋﻤل ﺘﻜﺎﻤﻼﺕ ﻓﻭﺭﻴﺭ .Fourier integrals ﻭﻗﺒل ﺍﻟﺒﺩﺀ ﻓﻲ ﺍﻟﻤﻌﺎﻟﺠﺔ ﺍﻟﺭﻴﺎﻀﻴﺔ ﻟﻠﺤﺯﻡ ﺍﻟﻤﻭﺠﻴﺔ ،ﻟﻌﻠﻪ ﻤﻥ ﺍﻟﻤﻔﻴﺩ ﻤﺭﺍﺠﻌﺔ
ﺒﻌﺽ ﺃﺴﺎﺴﻴﺎﺕ ﺍﻟﺩﻭﺍل ﺍﻟﻤﺭﻜﺒﺔ ﻭﺘﻜﺎﻤﻼﺕ ﻓﻭﺭﻴﺭ ﻭﻜﻴﻔﻴﺔ ﺇﻴﺠﺎﺩ ﻋﺭﺽ ﺍﻟـﺩﻭﺍل ﻭﻜﺫﻟﻙ ﻤﺭﺍﺠﻌﺔ ﺒﻌﺽ ﺍﻟﺘﻜﺎﻤﻼﺕ. ﻴﻤﻜﻥ ﻜﺘﺎﺒﺔ ﺃﻱ ﺭﻗﻡ ﻤﺭﻜﺏ zﺒﺎﻟﺼﻴﻐﺔ. z= x+i y
)......(A : 2 − 1
ﺤﻴﺙ xﺘﻤﺜل ﺍﻟﺠﺯﺀ ﺍﻟﺤﻘﻴﻘﻲ ﻟـ zﺃﻱ
z
) Re ( z
ﺍﻟﻭﺤﺩﺓ ﺍﻟﺘﺨﻴﻠﻴﺔ Iﻭﻫﻲ:
)(A : 2 − 2
ﻭ yﺘﻤﺜل ﺍﻟﺠﺯﺀ ﺍﻟﺘﺨﻴﻠﻲ Im
i = −1 ⇒ i2 = −1
ﻟﻘﺴﻤﺔ ﺩﻭﺍل ﺘﺨﻴﻠﻴﺔ )ﺃﻭ ﺃﺭﻗﺎﻡ ﺘﺨﻴﻠﻴﺔ( ﻴﺤﺴﻥ ﺒﻨﺎ ﺍﺴﺘﺤﺩﺍﺙ ﺍﻟﻤﺭﺍﻓﻕ ﺍﻟﻤﺭﻜﺏ
complex conjugateﻭﻴﺭﻤﺯ ﻟﻪ * zﺤﻴﺙ:
z * = v − iy
)(A : 2 − 3
ﻻﺤﻅ ﺃﻨﻨﺎ ﺍﺴﺘﺒﺩﻟﻨﺎ iﺏ –iﺍﻟﺭﻗﻡ ﺍﻟﺘﺨﻴﻠﻲ ﻤﻀﺭﻭﺏ ﻓﻲ ﻤﺭﺍﻓﻘﻪ ﺃﻱ * zzﻴﻌﻁﻴﻨﺎ )........... (A : 2 − 4
ﻻﺤﻅ ﺃﻥ
*z z
zz * = (x + iy) (x − iy) = x 2 − i 2 y 2 = x 2 + y 2
ﺘﺴﺎﻭﻱ ﻤﺭﺒﻊ ﺍﻟﻘﻴﻤﺔ ﺍﻟﻤﻁﻠﻘﺔ
2
z
ﻭﺍﻵﻥ ﻹﻴﺠﺎﺩ ﺍﻟﻘﻴﻤﺔ ﺍﻟﺤﻘﻴﻘﻴﺔ ﻭﺍﻟﺘﺨﻴﻠﻴﺔ ﻟـ
1 1 = z x + iy
1 1 x − iy x − iy x y = = 2 2 = 2 2 −i 2 2 z x + iy x − iy x + y x + y x +y 1 x 1 y ⇒ Re ( ) = 2 2 and In ( ) = 2 2 z x +y z x +y
ﻤﻥ ﺍﻟﻤﻔﻴﺩ ﻜﺘﺎﺒﺔ
z
ﺒﺩﻻﻟﺔ
θ, r
ﺒﺎﺴﺘﺨﺩﺍﻡ ﻤﻌﺎﺩﻟﺔ Euler
- ٣٦PDF created with pdfFactory Pro trial version www.pdffactory.com
eiθ = cos θ + i sin θ
(A : 1 − 5)
x = r cos θ and y = r sin θ ⇒ z = x + iy = r cos θ + i r sin θ = r (cos θ + i sin θ ) = reiθ
ﻫﻭ
e iθ . e − iθ = e ° = 1⇐ e − iθ
e iθ
ﻻﺤﻅ ﺃﻥ ﺍﻟﻤﺭﺍﻓﻕ ﻟـ ﻭﻜﺫﻟﻙ
z* = r e − iθ
ﻭﻤﻨﻬﺎ ﻨﺠﺩ
zz * = (re−iθ ) (re−iθ ) = r 2
ﻭﻤﻥ ﺍﻟﻤﻔﻴﺩ ﺘﺫﻜﺭ ﺃﻥ e iθ + e − iθ 2
(A : 2 − 6)
eiθ − e − iθ sinθ = 2i
............ (A : 2 − 7)
cos θ = and
ϕ (θ ) = 0 ≤ θ ≤ 2D
1 imθ e 2D
ﻫﻲ
ϕ ( θ) θ
ﻭ ﺤﺩﻭﺩ
ﺇﺫﺍ ﻜﺎﻨﺕ ﻟﺩﻴﻨﺎ ﺩﺍﻟﺔ
m = 0 ± 1 ± 2 .......
:ﻓﺈﻥ 1 imθ 1 eimθ dθ (θ ) dθ e ϕ = = ∫° ∫° 2D 2D im ° let m ≠ o 1 1 i2 Dm ° = e −e 2D im 2D
2D
[
2D
]
1 1 cos 2Dm + i sin 2Dm − 1 424 3 1 424 3 2D im 1 1 0 1 1 = [1 − 1]= 0 2D im =
let m = 0 2D
2D
°
°
2D
2D 1 i(0) θ 1 1 e = dθ [ θ ] = ° 2D 2D ∫° 2D 1 2D . (2D − 0) = = 2D 2D 2D
∫ dθ ϕ (θ ) = ∫ dθ =
- ٣٧ PDF created with pdfFactory Pro trial version www.pdffactory.com
ﺤﻴﺙ
ﻴﻤﻜﻥ ﺇﺠﺭﺍﺀ ﺍﻟﺘﻜﺎﻤل ﺍﻷﻭل ﺒﺼﻭﺭﺓ ﺃﺴﻬل ﻜﺎﻟﺘﺎﻟﻲ: 2D
2D
2D
i 1 1 imθ sin mθ dθ = 0 ∫ cos m θ dθ + ∫ = e dθ 2D 2D 2D ° °
∫ °
ﻜل ﺤﺩ ﻤﻥ ﺍﻟﺘﻜﺎﻤل ﻴﺴﺎﻭﻱ ﺼﻔﺭ ﻷﻨﻨﺎ ﻨﺠﺭﻱ ﺤﺩﻭﺩ ﺍﻟﺘﻜﺎﻤـل ﻋﺒـﺭ ﺩﻭﺭﺓ
ﻜﺎﻤﻠﺔ ﻭﻨﺭﺠﻊ ﻟﻨﻔﺱ ﺍﻟﺒﺩﺍﻴﺔ ﻭﻟﻬﺫﺍ ﻨﺴﺘﺨﻠﺹ ﺃﻨﻪ ﺇﺫﺍ ﻜﺎﻨﺕ m = 0ﻓـﺈﻥ ﺍﻟﺘﻜﺎﻤـل
ﻴﺴﺎﻭﻱ
2D
ﺃﻤﺎ ﻟﻠﻘﻴﻡ ﺍﻷﺨﺭﻯ ﻟـ mﻓﺈﻥ ﺍﻟﺘﻜﺎﻤل ﻴﺴﺎﻭﻱ ﺼﻔﺭ ﻭﻴﻤﻜﻥ ﺼـﻴﺎﻏﺔ
ﺍﻟﻨﺘﻴﺠﺔ ﺒﻬﺫﻩ ﺍﻟﺼﻭﺭﺓ. 2D
∫ dθϕ (θ ) = 0
)for all values of m ≠ 0 ..................(A : 2 - 8
°
for m = 0
)(A : 2 - 9
= 2D
ﻭﺒﻨﻔﺱ ﺍﻟﻁﺭﻴﻘﺔ ﻴﻤﻜﻨﻨﺎ ﺍﺴﺘﻨﺘﺎﺝ ﺍﻟﻌﻼﻗﺔ 2D
)..................(A : 2 - 10
m≠n
(θ )ϕ n (θ ) = 0
)(A : 2 -11
m= n
=1
* m
∫ dθϕ °
1 − imθ 1 inθ e . e 2D 2D 2D
∫ dθ °
2D
2D 1 1 1 i (n − m)θ ° [ ] ⇒m=n dθ e = dθ e θ° = 2D ∫° 2D ∫° 2D
1 (2D − 0) = 1 2D
2D
Now, Let
= 2D
1 dθ ei (n − m) θ 2D ∫°
⇒ Let m ≠ n
2D
1 ] dθ [cos ( n − m ) θ + i sin ( n − m ) θ = 2D ∫°
ﻗﻴﻤﺔ ﻫﺫﺍ ﺍﻟﺘﻜﺎﻤل ﺘﺴﺎﻭﻱ ﺼﻔﺭ ﻭﺫﻟﻙ ﻷﻥ ﺍﻟﺘﻜﺎﻤل ﻴﺠﺭﻱ ﻋﻠﻰ ﺩﻭﺭﺓ ﻜﺎﻤﻠﺔ
ﻟﻠﺩﺍﻟﺔ )ﺃﻱ ﺃﻨﻪ ﻴﺒﺩﺃ ﺜﻡ ﻴﻨﺘﻬﻲ ﻋﻨﺩ ﻨﻔﺱ ﺍﻟﻘﻴﻤﺔ(.
ﻭﻤﻥ ﺍﻟﺩﻭﺍل ﺍﻟﻤﺭﻜﺒﺔ ﺍﻟﻤﻬﻤﺔ sin h x :ﻭ cos h xﻭﺘﻌﺭﻑ ﺒـ )..................(A : 2 - 12 )(A : 2 - 13
ex − e− x ⇒ sin h i x = i sin x 2 ex + e− x cos h x ⇒ cos h i x = cos x 2
= sin h x
- ٣٨PDF created with pdfFactory Pro trial version www.pdffactory.com
ﻨﺒﺫﺓ ﻤﺨﺘﺼﺭﺓ ﻋﻥ ﺘﺤﻠﻴﻼﺕ ﻓﻭﺭﻴﺭ ﻭﺘﻜﺎﻤﻼﺘﻪ Short Introduction about Fourier Series and its integral.
ﺇﻥ ﺃﻱ ﺩﺍﻟﺔ ﻟﻬﺎ ﺼﻭﺭﺓ ﻤﻭﺠﻴﺔ ﺃﻭ ﺘﻜﺭﺭ ﻨﻔﺴﻬﺎ ﺒﺩﻭﺭﺓ ﻤﻌﻴﻨﺔ ،ﻤﺜل ﺍﻟـﺩﻭﺍل
ﺍﻟﻤﻭﺠﻴﺔ ﺍﻟﻤﻭﻀﺤﺔ ﻓﻲ ﺸﻜل ،A:2.1ﻴﻤﻜﻥ ﻭﺼﻔﻬﺎ ﺭﻴﺎﻀـﻴﺎﹰ ﺒﺎﺴـﺘﺨﺩﺍﻡ ﻤﺒـﺩﺃ ﺍﻟﺘﺭﺍﻜﺏ ،Principle of super positionﺃﻱ ﻴﻤﻜﻥ ﺍﻟﺘﻌﺒﻴﺭ ﻋﻨﻬﺎ ﻋﻥ ﻁﺭﻴﻕ ﺠﻤﻊ
)ﺃﻭ ﺘﺭﺍﻜﺏ( ﻤﻭﺠﺎﺕ ﺘﻜﺘﺏ ﺭﻴﺎﻀﻴﺎﹰ ﺒﺩﻻﻟﺔ ﺍﻟـ :ﺠﺎ sinﻭﺍﻟـ ﺠﺘـﺎ .cosﻫـﺫﻩ
ﺍﻟﻁﺭﻴﻘﺔ ﺍﻟﺭﻴﺎﻀﻴﺔ ﺘﹸﻌﺭﻑ ﺒﺘﺤﻠﻴل ﻓﻭﺭﻴﺭ Fourier analysisﻨﺴﺒﺔ ﻟﻠﻌﺎﻟﻡ ﻓـﻭﺭﻴﺭ.
ﻭﺘﻌﺘﻤﺩ ﻫﺫﻩ ﺍﻟﻁﺭﻴﻘﺔ ﻋﻠﻰ ﺃﺴﺎﺱ ﺃﻨﻬﺎ ﻴﻤﻜﻥ ﺼﻴﺎﻏﺔ ﺃﻱ ﺩﺍﻟﺔ )) f (xﺤﻴﺙ xﻴﻤﻜﻥ ﺃﻥ ﺘﻤﺜل ﺍﻟﻤﻭﻗﻊ ﺃﻭ ﺍﻟﺯﻤﻥ( ﻴﻤﻜﻥ ﺃﻥ ﺘﻜﺘﺏ ﻓﻲ ﺼﻭﺭﺓ ﻤﺘﺴﻠﺴﻠﺔ ﻤﻥ ﺩﻭﺍل ﺩﻭﺭﻴﺔ
)ﺘﻌﺭﻑ ﺒﺴﻠﺴﻠﺔ ﻓﻭﺭﻴﺭ( .Fourier series
1 f (x) = a ° + a 1 cos x + a 2 cos 2x + .....+ a n cos n x + b1 sin x + b 2 sin 2x + .....+ b n sin n x + ........ 2 ∞ ∞ 1 )= a ° + ∑ a u cos n x + ∑ b n sin v x ..............(A : 2 − 14 2 n =1 n =1
ﺘﺘﻜﻭﻥ ﻫﺫﻩ ﺍﻟﻤﺘﺴﻠﺴﻠﺔ ﻤﻥ ﺤﺩ ﺜﺎﺒﺕ ﻭﺠﻴﻭﺏ ﺍﻟﺘﻤﺎﻡ
1 a 2 °
ﺒﺎﻹﻀﺎﻓﺔ ﺇﻟﻰ ﺤﺩﻭﺩ ﻤﻥ ﺍﻟﺠﻴـﺏ
(sine and cosine termsﻟﻬـﺎ ﺴـﻌﺎﺕ ﻤﺨﺘﻠﻔـﺔ a1, a2, ….
.b1, b2, ……… Different amplitudesﺘﺭﺩﺩﺍﺕ ﺍﻟﺠﺎ ﻭﺍﻟﺠﺘﺎ ﻓﻲ ﻜل ﺤﺩ ﻤـﻥ ﺍﻟﺤﺩﻭﺩ ﻋﺒﺎﺭﺓ ﻋﻥ harmonicsﻭ ﻤﻀﺭﻭﺒﺎﺕ multiplesﺍﻟﺘﺭﺩﺩ ﺍﻷﺴﺎﺴﻲ .x
.Fundamental frequency xﻟﻠﺘﻌﺒﻴﺭ ﻋﻥ ﻤﻌﻴﻨﺔ ﻤﻥ ﺨﻼل ﺘﺤﻠﻴل ﻓﻭﺭﻴﺭ ﻓـﺈﻥ
ﺍﻟﻤﻌﺎﻤﻼﺕ ……… a1, a2,ﻭ b1, b2, ……..ﻴﻠﺯﻡ ﺘﺤﺩﻴﺩﻫﺎ ﻭﺫﻟﻙ ﺒﺘﻜﺎﻤل ﺍﻟﺩﺍﻟﺔ
) f(xﺨﻼل ﺩﻭﺭﺓ ﻜﺎﻤﻠﺔ ﺃﻱ ﻤﻥ ﺼﻔﺭ ﺇﻟﻰ
2D
ﻜﻤﺎ ﻴﻠﻲ:
2D
1 )b n = ∫ f (x) sin n x dx .....................(A : 2 − 15 D °
2D
,
1 a n = ∫ f (x) cos n x dx D °
ﺍﻟﺸﻜل ﻴﻭﻀﺢ ﺃﻥ ﺘﻭﻟﻴﺩ ﺩﺍﻟﺔ ﻤﻭﺠﻴﺔ ﻤﺭﺒﻌﺔ: - ٣٩PDF created with pdfFactory Pro trial version www.pdffactory.com
ﻴﺘﻡ ﻋﻥ ﻁﺭﻴﻕ ﺘﺩﺍﺨل )ﺃﻭ ﺘﺭﺍﻜﺏ( ﺜﻼﺙ ﻤﻭﺠﺎﺕ ﺠﻴﺒﻴﺔ ﻫﻲ ﺃﻭل ﺍﻟﺤﺩﻭﺩ ﻓﻲ ﻤﺘﺴﻠﺴﻠﺔ ﻓﻭﺭﻴﺭ ﺒﺎﻟﺼﻭﺭﺓ.
1 a 2 °
1 1 sin 2x , sin 3x , sinx 2 3
ﻴﻤﻜﻥ ﻜﺘﺎﺒﺘﻪ )ﻤﻥ ﻤﻌﺎﺩﻟـﺔ (A: 2-15
2D
1 f (x) dx 2D ∫°
ﻭﻫﺫﻩ ﺍﻟﻤﻌﺎﺩﻟﺔ ﺘﻤﺜل ﺒﺒﺴﺎﻁﺔ ﻤﺘﻭﺴﻁ ﺍﻟﺩﺍﻟﺔ ) f(xﻋﺒﺭ ﺍﻟﻔﺘﺭﺓ . 2D
ﺘﺤﻠﻴل ﻓﻭﺭﻴﺭ ﻭﺍﻟﺤﺯﻡ ﺍﻟﻤﻭﺠﻴﺔ: ﻴﻤﻜﻥ ﺍﻻﺴﺘﻔﺎﺩﺓ ﻤﻥ ﺘﺤﻠﻴل ﻓﻭﺭﻴﺭ ﻟﺘﻤﺜﻴل ﺍﻟﺤﺯﻡ ﺍﻟﻤﻭﺠﻴـﺔ Wave packets
ﻭﻫﻲ ﻋﺒﺎﺭﺓ ﻋﻥ ﻨﺒﻀﺎﺕ ﻤﻭﺠﻴﺔ Wave pulsesﻋﻨﺩ ﺩﺭﺍﺴﺔ ﺍﻟﺤﺭﻜـﺔ ﺍﻟﻤﻭﺠﻴـﺔ ﻟﻨﺘﺤﺼل ﻋﻠﻰ ﻨﺒﻀﺔ ﻤﻭﺠﻴﺔ ﻴﻠﺯﻡ ﺘﺭﺍﻜﺏ ﻤﻭﺠﺎﺕ ﺠﻴﺒﻴﺔ ﻋﻠﻰ ﻤﺩﻯ ﻤﺘﺼل ﻤـﻥ
ﺍﻷﺭﻗﺎﻡ ﺍﻟﻤﻭﺠﻴﺔ kﻓﻲ ﻤﺩﻯ ﻗﺩﺭﺘﻪ . Δkﻴﻤﻜﻨﻨﺎ ﻓﻬﻡ ﻜﻴﻔﻴﺔ ﺘﻜﻭﻥ ﺤﺯﻤﺔ ﻤﻭﺠﻴـﺔ،
- ٤٠PDF created with pdfFactory Pro trial version www.pdffactory.com
ﻋﻠﻰ ﺴﺒﻴل ﺍﻟﻤﺜﺎل ،ﺨﺫ ﺘﺭﺍﻜﺏ ﻋﺩﺩ ﺴﺒﻌﺔ ﻤﻥ ﺍﻟﻤﻭﺠﺎﺕ ﺍﻟﺠﻴﺒﻴﺔ ﺍﻟﺘﻲ ﻟﻜـل ﻤﻨﻬـﺎ ﺍﻟﺼﻴﻐﺔ ﺍﻟﺭﻴﺎﻀﻴﺔ. ).......... .......... .(A : 2 − 16
)y = A k cos ( k x − ω t
ﻭﻋﻨــﺩ ،t = 0ﺍﻷﺭﻗــﺎﻡ ﺍﻟﻤﻭﺠﻴــﺔ ) k (Wave umbersﺘــﺴﺎﻭﻱ 45, 42, 39, 36, 33, 30 , 27ﻭﻜل ﻤﻥ ﻫﺫﻩ ﺍﻷﺭﻗﺎﻡ ﻴﻘﺎﺒﻠﻬﺎ ﺴـﻌﺔ Akﺘـﺴﺎﻭﻱ
0.25 , 0.33 , 0.5 , 1.0, 0.5 , 0.33 , 0. 25ﺸﻜل A: 2-3ﻴﻭﻀﺢ ﺘﻐﻴﺭ ﺍﻟﺴﻌﺔ
Akﻤﻊ kﻭﻫﻲ ﻤﺎ ﻴﺴﻤﻰ ﺒﻁﻴﻑ ﺍﻟﺭﻗﻡ ﺍﻟﻤﻭﺠﻲ .Wave umber spectrum
Figure A:2-3 the frequency spectrum used to generate the we packet shown in Figure 12.84.
ﺍﻟﺸﻜل ﺍﻟﻤﻭﺠﻲ ﻟﻜل ﻤﻭﺠﺔ ﻤﻥ ﺍﻟﻤﻭﺠﺎﺕ ﺍﻟﺴﺒﻌﺔ ﻭﺤﺎﺼـل ﺘﺭﺍﻜﺒﻬـﺎ )ﺃﻭ
ﻤﺤﺼﻠﺔ ﺘﺭﺍﻜﺒﻬﺎ( ﻤﻭﻀﺤﺔ ﻓﻲ ﺸﻜل ).(A: 2-4
- ٤١PDF created with pdfFactory Pro trial version www.pdffactory.com
ﺸﻜل ) (A: 2-4ﻜﻴﻔﻴﺔ ﺘﺅﻜﺩ ﺤﺯﻤﺔ ﻤﻭﺠﻴﺔ )ﺃﺴﻔل ﺍﻟﺸﻜل( ﻤﻥ ﺘﺩﺍﺨل ﺴـﺒﻌﺔ
ﻤﻭﺠﺎﺕ ﻜﺎﻟﻤﻭﻀﺤﺔ ﻓﻲ ﺃﻋﻠﻰ ﺍﻟﺤﺯﻤﺔ ﺍﻟﻤﻭﺠﻴﺔ.
ﺃﻤﺎ ﻓﻲ ﺤﺎﻟﺔ ﺘﺭﺍﻜﺏ ﻋﺩﺩ ﻜﺒﻴﺭ ﺠﺩﺍﹰ ﻤﻥ ﺍﻟﻤﻭﺠﺎﺕ ﺒﺤﻴﺙ ﻴﻤﺘﺩ ﺘﻐﻴﺭ ﺍﻷﺭﻗﺎﻡ ﺍﻟﻤﻭﺠﻴﺔ ﻋﺒﺭ ﻋﺭﺽ ﻗﺩﺭﻩ
ﻤﻭﻀﺢ ﻓﻲ ﺍﻟﺸﻜل .A: 2-5
ΔK
ﺴﻨﺘﺤﺼل ﻋﻠﻰ ﺤﺯﻤﺔ ﻤﻭﺠﻴﺔ ﻭﺍﺤﺩﺓ ﻜﻤـﺎ ﻫـﻭ
ﻻﺤﻅ ﺃﻥ ﻋﺭﺽ ﺍﻟﺤﺯﻤﺔ ﺍﻟﻤﻭﺠﻴﺔ ...... ΔK
- ٤٢PDF created with pdfFactory Pro trial version www.pdffactory.com
ﻭﻴﻤﻜﻥ ﺃﻥ ﻨﻼﺤﻅ ﻨﻘﻁﺔ ﻫﺎﻤﺔ ﺠﺩﺍﹰ ﻭﻫﻲ ﺃﻨﻪ ﻜﻠﻤﺎ ﺍﺯﺩﺍﺩﺕ ﺍﻟﺤﺯﻤﺔ
ΔK
ΔK
ﻓﺈﻥ ﻋﺭﺽ
ﻴﻘل .ﻭﻫﺫﻩ ﺍﻟﻨﺘﻴﺠﺔ ﻫﺎﻤﺔ ﻷﻨﻬﺎ ﺍﻷﺴﺎﺱ ﺍﻟﺫﻱ ﻴﺒﻨﻰ ﻋﻠﻴﻪ ﻤﺒﺩﺃ ﺍﻟﻼﺘﺤﺩﻴﺩ
ﻓﻲ ﻋﻠﻡ ﻤﻴﻜﺎﻨﻴﻜﺎ ﺍﻟﻜﻡ.
Figure A: 2-5 (a) the wave number spectrum and (b) the wave packet. Produced when ∆k the wave number spread shown in Figure A:2-4 becomes broader.
ﺇﻥ ﺘﺭﺍﻜﺏ ﻋﺩﺩ ﻻﻨﻬﺎﺌﻲ infinite numberﻤﻥ ﺍﻟﻤﻭﺠﺎﺕ ﻴﻤﻜﻥ ﺼـﻴﺎﻏﺘﻪ
ﺭﻴﺎﻀﻴﺎﹰ ﺒﺘﻜﺎﻤل ﻓﻭﺭﻴﺭ Fourier integralﺒﺩﻻﹰ ﻤﻥ ﻤﺘﺴﻠﺴﻠﺔ ﻓﻭﺭﻴﺭ .ﻭﻫﻜﺫﺍ ،ﻓﻌﻨﺩ ﺯﻤﻥ ﺜﺎﺒﺕ ،ﻓﺈﻥ ﻤﺠﻤﻭﻋﺔ ﻤﻥ ﺍﻟﻤﻭﺠﺎﺕ ﺘﻌﻁﻲ ﺒـ: )..........(A : 2 − 17
∞
y (x) ∫ A (k) cos k x dk °
ﺤﻴﺙ ) A(kﻫﻲ ﺍﻟﺴﻌﺔ ﻭﺍﻟﺘﻲ ﺘﺘﻐﻴﺭ ﻤﻊ .kﻭﺴﻨﻨﺎﻗﺵ ﻓﻲ ﻫﺫﺍ ﺍﻟﻔﺼل ﺒﻌـﺩ
ﻫﺫﻩ ﺍﻟﻤﻘﺩﻤﺔ ،ﻜﻴﻑ ﻴﻤﻜﻨﻨﺎ ،ﺒﺎﺴﺘﺨﺩﺍﻡ ﺘﻜﺎﻤﻼﺕ ﻓـﻭﺭﻴﺭ ،ﺃﻥ ﻨﻭﻀـﺢ ﺃﻥ
Δx , Δk
ﻤﺭﺘﺒﻁﺘﺎﻥ ﺒﺎﻟﺩﺍﻟﺔ.
)....................(A : 2 − 18
(Δ x) (Δ k) = constant
- ٤٣PDF created with pdfFactory Pro trial version www.pdffactory.com
ﻭﻟﻌﻠﻪ ﻤﻥ ﺍﻟﻤﻔﻴﺩ ﺼﻴﺎﻏﺔ ﺍﻟﻤﻌﺎﺩﻟﺔ ) (A: 2-17ﻟﻠﺤﺯﻤﺔ ﺍﻟﻤﻭﺠﻴﺔ ﺒﺩﻻﻟﺔ ﺍﻟﺩﺍﻟﺔ ﺍﻟﻤﺭﻜﺒﺔ ﻋﻠﻰ ﺍﻟﺼﻭﺭﺓ ﺍﻟﻌﺎﻤﺔ. ∞
).............(A : 2 − 19
ikx
∫ dk A (k) e
∞−
1 2D
=)f (x
ﻭﺇﺫﺍ ﺃﺩﺨﻠﻨﺎ ﺍﻟﺯﻤﻥ ﻜﻤﺘﻐﻴﺭ ﻟﺩﺭﺍﺴﺔ ﺘﺄﺜﻴﺭ ﺍﻟﺯﻤﻥ ﻋﻠـﻰ ﺍﻨﺘـﺸﺎﺭ ﺍﻟﺤﺯﻤـﺔ
ﺍﻟﻤﻭﺠﻴﺔ ﻴﻤﻜﻥ ﺼﻴﺎﻏﺔ ﺍﻟﻤﻌﺎﺩﻟﺔ ) (A:2-19ﺒﺎﻟﺼﻭﺭﺓ.
∞
).............(A : 2 − 20
ikx - ω (k) t
∫ dk A (k) e
= ) f ( x, t
∞−
ﻋﻨﺩ ﺘﺩﺍﺨل ﻤﻭﺠﺎﺕ ﺴﺭﻋﺔ ﻜل ﻤﻨﻬﺎ vﻓﺈﻥ ﺍﻟﺤﺯﻤﺔ ﺍﻟﻤﻭﺠﻴﺔ ﺍﻟﻤﺘﻜﻭﻨﺔ ﺘﻜﻭﻥ
ﺴﺭﻋﺘﻬﺎ ﺘﻤﺜل ﺴﺭﻋﺔ ﻤﺠﻤﻭﻋﺔ ﻤﻥ ﺍﻟﻤﻭﺠﺎﺕ ﺍﻟﻤﻜﻭﻨﺔ ﻟﻠﺤﺯﻤﺔ ﻭﺘـﺴﻤﻰ ﺒـﺴﺭﻋﺔ ﺍﻟﻤﺠﻤﻭﻋﺔ group velocityﻭﻴﺭﻤﺯ ﻟﻬﺎ ﺒـ vgﻭﺘﻌﻁﻰ ﺒﺎﻟﻌﻼﻗﺔ. )...................... (A : 2 − 21
dω dk
= vg
ﻜﻴﻔﻴﺔ ﺇﻴﺠﺎﺩ ﻋﺭﺽ ﺩﺍﻟﺔ: ﻤﻥ ﺍﻟﻨﻘﺎﻁ ﺍﻟﺘﻲ ﺘﻬﻤﻨﺎ ﻓﻲ ﻫﺫﺍ ﺍﻟﻔﺼل ﺇﻴﺠﺎﺩ ﻋﺭﺽ ﺩﺍﻟـﺔ ﻭﺫﻟـﻙ ﻷﻫﻤﻴـﺔ
ﺍﻟﻤﻭﻀﻭﻉ ﺒﻤﺒﺩﺃ ﺍﻟﻼﺘﺤﺩﻴﺩ ﻓﻲ ﻤﻴﻜﺎﻨﻴﻜﺎ ﺍﻟﻜﻡ.
ﺨﺫ ﻋﻠﻰ ﺴﺒﻴل ﺍﻟﻤﺜﺎل ﺍﻟﺩﺍﻟﺔ ﺍﻟﻤﻭﻀﺤﺔ ﻓﻲ ﺍﻟﺸﻜل A: 2-6
- ٤٤PDF created with pdfFactory Pro trial version www.pdffactory.com
ﻜﻠﺘﺎ ﺍﻟﺩﺍﻟﺘﺎﻥ ﻟﻬﻤﺎ ﻨﻔﺱ ﺍﻟﻌﺭﺽ ،ﻭﻟﻜﻥ ﺇﺤﺩﺍﻫﻤﺎ ﻤﺘﻤﺎﺜﻠﺔ ﺤﻭل x = oﻭﺍﻷﺨـﺭﻯ ﻤﺘﻤﺎﺜﻠﺔ ﺤﻭل .xoﺍﻷﻭل ﻓﻲ ﺍﻟﺸﻜل ) (aﻴﻤﻜﻥ ﻜﺘﺎﺒﺘﻬﺎ ﺭﻴﺎﻀﻴﺎﹰ ﺒﺎﻟﺼﻴﻐﺔ
2
y = A e− α x
ﺤﻴﺙ αﺜﺎﺒﺕ ﻭﻤﺭﺘﺒﻁ ﺒﻌﺭﺽ ﺍﻟﺩﺍﻟﺔ ﻭ Aﻴﻤﺜل ﺴﻌﺔ ﺍﻟﺩﺍﻟـﺔ ﺃﻱ ﻗﻴﻤـﺔ y
ﻋﻨﺩﻤﺎ .x = o
ﺇﺤﺩﻯ ﺍﻟﻁﺭﻕ ﻹﻴﺠﺎﺩ ﻋﺭﺽ ﺍﻟﺩﺍﻟﺔ ﺃﻥ ﻨﺄﺨﺫ ﺍﻟﻌﺭﺽ ﻋﻨﺩﻤﺎ ﺘﻘل ﺍﻟﺩﺍﻟﺔ )ﺃﻱ
ﻋﻨﺩﻤﺎ ﺘﻘل ﻗﻴﻤﺔ .yﻤﻥ ﺃﻗﺼﻰ ﻗﻴﻤﺔ )ﻋﻨﺩﻤﺎ (y = Aﺇﻟﻰ ﺃﻥ ﺘـﺼل ﺇﻟـﻰ ، y = A 2
ﻋﻨﺩﻫﺎ ﻨﺄﺨﺫ ﻋﺭﺽ ﺍﻟﺩﺍﻟﺔ . ∆x ﻜﻤﺎ ﻫﻭ ﻤﻭﻀﺢ ﺒﺎﻟﺸﻜل. ln y = ln A − α x 2 ⇒ α x 2 = ln A − ln y ﻧﻌﻮض ﻋﻦ
1 A 2
=y
A = ln 2 = 0.693 A/2 0.693 0.693 = ⇒ Δx ° ⇒ Δx = 2 α α
1 α x 2 = ln A − ln A 2 = ln 1 ⇒ x 2 = 0.693 α
- ٤٥PDF created with pdfFactory Pro trial version www.pdffactory.com
ﻴﻤﻜﻥ ﺇﻴﺠﺎﺩ ﺘﻌﺭﻴﻑ ﺁﺨﺭ ﻟﻌﺭﺽ ﺍﻟﺩﺍﻟﺔ ﻭﻫﻭ ﺃﺴﻬل ﻤﻥ ﺍﻟﻁﺭﻴﻘﺔ ﺍﻟـﺴﺎﺒﻘﺔ. ﻨﺄﺨﺫ ﻋﺭﺽ ﺍﻟﺩﺍﻟﺔ ﻋﻨﺩﻤﺎ ﺘﻘل ﻗﻴﻤﺔ yﻤﻥ ﺍﻟﻘﻴﻤﺔ ﺍﻟﻘﺼﻭﻯ ﺇﻟﻰ ﺃﻥ ﺘﺼل ﺇﻟـﻰ ﺤﻴﺙ e = 2.718ﻭﻋﻠﻴﻪ
1 = 0.368 e
1 e
ﺃﻱ ﺒﺩل ﺃﻥ ﻨﺄﺨﺫ ﺍﻟﻌﺭﺽ ﻋﻨﺩﻤﺎ ﺘﻘـل yﺇﻟـﻰ
ﺍﻟﻨﺼﻑ )ﻜﻤﺎ ﻓﻲ ﺍﻟﻁﺭﻴﻘﺔ ﺍﻟﺴﺎﺒﻘﺔ( ﻨﺄﺨﺫ ﺍﻟﻌﺭﺽ ﻋﻨﺩﻤﺎ ﺘﻘل yﺇﻟﻰ 0.368ﻤـﻥ
ﻗﻴﻤﺘﻬﺎ ﺍﻟﻘﺼﻭﻯ ..ﻭﺴﻬﻭﻟﺔ ﻫﺫﻩ ﺍﻟﻁﺭﻴﻘﺔ ﻤﻥ ﺍﻟﻨﺎﺤﻴﺔ ﺍﻟﺭﻴﺎﻀﻴﺔ.
2
y = A e− α x
ﺨﺫ ﻤﺜﻼﹰ ﺍﻟﺩﺍﻟﺔ ﺍﻟﺴﺎﺒﻘﺔ
ﻗﻴﻤﺔ yﺍﻟﻘﺼﻭﻯ ﻫﻲ y = Aﻟﻜﻲ ﺘﻘل yﺇﻟﻰ ﻫﻲ ﻗﻴﻤﺔ x2ﺍﻟﺘﻲ ﺘﺠﻌل ﺍﻷﺱ
ﻭﺍﺤﺩ ﻓﺈﻥ
) (α x 2
1 e
ﻤﻥ ﻗﻴﻤﺘﻬﺎ ﺍﻟﻘﺼﻭﻯ ،ﺃﻱ ﻤﺎ
ﻴﺴﺎﻭﻱ ﺍﻟﻭﺍﺤﺩ .ﻋﻨﺩﻤﺎ ﻴﺴﺎﻭﻱ ﺍﻷﺱ ﻗﻴﻤـﺔ
1 y = ( ) A ⇐ y = A e −1 e 1 2 = ⇒ Δx = 2Δ x ° α α
= ⇒ α x 2 =1 ⇒ Δ x °
ﻭﻫﺫﻩ ﻫﻲ ﺍﻟﻁﺭﻴﻘﺔ ﺍﻟﺘﻲ ﺴﻨﺴﺘﺨﺩﻤﻬﺎ ﻹﻴﺠﺎﺩ ﻋﺭﺽ ﺍﻟﺩﻭﺍل ﺍﻟﺘﻲ ﻟﻬﺎ ﺍﻟﺼﻴﻐﺔ
ﺍﻷﺴﻴﺔ .ﻭﻫﻨﺎﻙ ﻁﺭﻕ ﺃﺨﺭﻯ ﻹﻴﺠﺎﺩ ﻋﺭﺽ ﺍﻟﺩﻭﺍل ﺍﻟﺠﻴﺒﻴﺔ.
ﻭﺒﻌﺩ ﻫﺫﺍ ﺍﻻﺴﺘﻁﺭﺍﺩ ﻨﻌﻭﺩ ﻟﻤﻨﺎﻗﺸﺔ ﻤﻭﻀﻭﻉ ﻫﺫﺍ ﺍﻟﻔـﺼل ﻭﻫـﻭ ﺍﻟﺤـﺯﻡ
ﺍﻟﻤﻭﺠﻴﺔ ﻭﻋﻼﻗﺎﺕ ﺍﻟﻼﺘﺤﺩﻴﺩ..
ﺨﺫ ﻋﻠﻰ ﺴﺒﻴل ﺍﻟﻤﺜﺎل ﺍﻟﺩﺍﻟﺔ ) f (xﺍﻟﻤﻌﺭﻓﺔ ﺒﺎﻟﻌﻼﻗﺔ. - ٤٦PDF created with pdfFactory Pro trial version www.pdffactory.com
∞
f (x) = ∫ dk g (k) e ikv
)....(2 − 1
∞−
ﻫﺫﻩ ﺍﻟﻌﻼﻗﺔ ﺘﹸﻌﺒﺭ ﻋﻥ ﺘﺭﺍﻜﺏ ﺨﻁﻲ ﻤﻥ ﻤﻭﺠﺎﺕ ﺫﺍﺕ ﻁﻭل ﻤﻭﺠﻲ ، λ = 2D k
ﻭﻟﻜل ﻗﻴﻤﺔ ﻟـ kﻓﻜل ﻤﻭﺠﺔ ﺘﻜﺭﺭ ﻨﻔﺴﻬﺎ ﻋﻨﺩﻤﺎ ﺘﺘﻐﻴﺭ xﺇﻟﻰ . v + 2D k
ﻭﻟﺘﻭﻀﻴﺢ ﻁﺒﻴﻌﺔ ﻫﺫﻩ ﺍﻟﺤﺯﻤﺔ ﺍﻟﻤﻭﺠﻴﺔ ،ﺩﻋﻨﺎ ﻨﺨﺘﺎﺭ ﺼـﻴﻐﺔ ﻟﻠﺩﺍﻟـﺔ )g (k
ﺤﻴﺙ.
2
) g (k) = e −α (k − k °
)...(2-2 Let k 1 = k − k ° ⇒ k = k1 + k ° ⇒ dk = dk 1 i (u1 + k ° )v
12
⇒ g (k) → g (k 1 ) = e −αk and e ikv = e
ﺒﻌﺩ ﻫﺫﻩ ﺍﻟﺘﻌﻭﻴﻀﺎﺕ ﻴﻤﻜﻥ ﻜﺘﺎﺒﺔ ) (2-1ﺒﺎﻟﺼﻴﻐﺔ i (k1 + k ° )v
∞
⇒ f (x) = ∫ dk 1 e −αk . e 12
∞−
1
∞
° {= ∫ dk e − αk e ik v . e ik v
1
12
ﺛﺎﺑﺖ
1
∞−
12
e − αk e ik v
∞
1
∫ dk
= e ikov
∞−
ﻟﺤﺴﺎﺏ ﻗﻴﻤﺔ ﺍﻟﺘﻜﺎﻤل ﻴﺠﺏ ﺘﺤﻭﻴﺭ ﺼﻴﻐﺘﻪ ﺍﻟﺭﻴﺎﻀﻴﺔ ﻟﻠـﺼﻭﺭﺓ ﻭﺫﻟﻙ ﻷﻥ ﻗﻴﻤﺔ ﺍﻟﺘﻜﺎﻤل
D α
).................(2 − 3
1
dk 1
−αk − 2
∞
∫e
∞−
∞
− αk 1 ∫ e dk = 2 12
°
ﻭﻟﻠﻭﺼﻭل ﻟﻬﺫﻩ ﺍﻟﺼﻭﺭﺓ ﻴﻠﺯﻡ ﺃﻥ ﻨﺠﻌل ﺍﻷﺱ ﻤﺭﺒﻊ ﻜﺎﻤل ﺤﺴﺏ ﺍﻟﺨﻁﻭﺍﺕ ﺍﻟﺘﺎﻟﻴﺔ. 12 + ik iv
∞
1 − αk ik v 1 − αk ∫ dk e . e = ∫ dk e 1
∞−
ﺨﺫ
) ( −α
∞
12
∞−
ﻋﺎﻤل ﻤﺸﺘﺭﻙ ﻓﻲ ﺍﻷﺱ. - ٤٧PDF created with pdfFactory Pro trial version www.pdffactory.com
i )− α(k12 − k1v α
∞
= ∫ dk e
− αk12 + ik1v
1
∞
1
e
∫ dk
∞−
∞−
ﻟﻜﻲ ﻨﺘﺤﺼل ﻋﻠﻰ ﻤﺭﺒﻊ ﻜﺎﻤل ﻟﻸﺱ
ﻨﻁﺭﺡ ﻭﻨـﻀﻴﻑ ﺍﻟﺤـﺩ
i )(k 12 − u 1 v α
v2 4α 2 v2 v2 i i u 12 − k 1 v = k 12 − 2 − k 1 v + 2 α 4α 4α { 1 23 α v2 v2 i 1 k )v − + 4α 2 4α 2 α
= (k 12 −
2
v2 iv = k 1 − ( ) + 2 2α 4α
ﻓﻴﺼﺒﺢ ﺍﻟﺘﻜﺎﻤل ﻋﻠﻰ ﺍﻟﺼﻭﺭﺓ 2 v 2 1 iv −α k − ( ) + 4α 2 1422α 4 3 q
∞
1
e
∫ dk
∞−
∞ v2 4α
1 − αq ∫−∞dk e 14243 2
−
=e
v2 4α 2
−α
− αq 2
.e
= ∫ dk e 1
∞−
ﺍﻟﺩﺍﻟﺔ ﺯﻭﺠﻴﺔ ﻭﻋﻠﻴﻪ ﻴﻤﻜﻥ ﻜﺘﺎﺒﺔ ﺍﻟﺘﻜﺎﻤل ﻭﻗﻴﻤﺔ ﺍﻟﺘﻜﺎﻤل ﻴﺴﺎﻭﻱ
∞
2
∞
2∫ du 1 e −αq °
∫ 1 2 α
)..............(2 − 4
v2 4α
−
.e
D ∴f (x) = e{ . α phase factor ikov
ﻻﺤﻅ ﺃﻥ ﺍﻟﺩﺍﻟﺔ ) f (xﻓﻲ ) (2-4ﺩﺍﻟﺔ ﺘﺨﻴﻠﻴﺔ .ﻭﻟﻜﻲ ﺘﺒﻌﺩ ﻫﺫﻩ ﺍﻟﺩﺍﻟﺔ ﻋﻠـﻰ ﺠﺴﻡ ﻴﻠﺯﻡ ﺃﻥ ﺘﻜﻭﻥ ﺤﻘﻴﻘﻴﺔ .ﻭﻟﻬﺫﺍ ﺍﻟﺴﺒﺏ ﻓﺎﻟﺫﻱ ﻴﻬﻤﻨﺎ ﻟﻴﺱ ) f (xﺒل ﻤﺭﺒﻊ ﺍﻟﻘﻴﻤﺔ ﺍﻟﻤﻁﻠﻘﺔ ﻷﻨﻬﺎ ﺤﻘﻴﻘﻴﺔ.
- ٤٨PDF created with pdfFactory Pro trial version www.pdffactory.com
)f (x) = f * (x) f (x 2
x2
x2
D − 4α D − 4α e ) (e +ikox ) e α α
− kox
= (e
v2
D − f ( x ) = e 2α α
)........... (2 − 5
2
ﻫﺫﻩ ﺍﻟﺩﺍﻟﺔ ﺘﻤﺜل ﺠﺴﻴﻡ. ﺍﻟﺩﺍﻟﺔ ) (2-5ﻟﻬﺎ ﻗﻴﻤﺔ ﻗﺼﻭﻯ )ﺃﻭ ﻗﻴﻤﺔ( ﻋﻨﺩ v = 0ﻭﺘﻌﺘﻤﺩ ﻋﻠـﻰ ﻗﻴﻤـﺔ ﺍﻟﺜﺎﺒﺕ ، αﺇﺫﺍ ﻜﺎﻨﺕ
α
ﻜﺒﻴﺭﺓ ﺘﻜﻭﻥ ﺍﻟﺩﺍﻟﺔ ﻋﺭﻴﻀﺔ ﻭﺍﻟﻌﻜﺱ ﺼﺤﻴﺢ ﺇﺫﺍ ﻜﺎﻨﺕ
α
ﺼﻐﻴﺭﺓ ﻓﺈﻥ ﺍﻟﺩﺍﻟﺔ ﺘﻌﺒﺭ ﻋﻥ ﺤﺯﻤﺔ ﻤﻭﺠﻴﺔ ﻀـﻴﻘﺔ harrowﻭﻜﻤـﺎ ﻗﻠﻨـﺎ ﻓـﺈﻥ 2
)f (x
ﺘﻌﺒﺭ ﻋﻥ ﺠﺴﻴﻡ.
ﺒﺈﻤﻜﺎﻨﻨﺎ ﺃﻥ ﻨﻭﺠﺩ ﻋﺭﺽ ﺍﻟﺩﺍﻟﺔ
)ﻤﻠﺤﻕ (Aﻨﺄﺨﺫ ﻋﺭﺽ ﺍﻟﺩﺍﻟﺔ
∆v
ﻟﻜﻲ ﺘﻨﺫل ﺍﻟﺩﺍﻟﺔ ﻤﻥ ﺍﻟﻘﻴﻤﺔ ﺇﻟﻰ
1
e
2
)f (x
،ﻜﻤﺎ ﺸﺭﺤﻨﺎ ﻓﻲ ﺍﻟﺠـﺯﺀ ﺍﻟـﺴﺎﺒﻕ
ﻋﻨﺩﻤﺎ ﺘﻘل ﻗﻴﻤﺘﻬﺎ ﺇﻟﻰ
e
1
ﻋﻥ ﺃﻗﺼﻰ ﻗﻴﻤﺔ ﻟﻬﺎ
ﻫﺫﺍ ﻴﺤﺩﺙ ﻋﻨـﺩﻤﺎ ﻴـﺴﺎﻭﻱ ﺍﻷﺱ ﻭﺍﺤـﺩ ﺃﻱ
v2 ﻋﻨﺩﻤﺎ . =1 2α
⇒ v 2 = 2α ⇒ Δv ° = ± 2α )................(2 − 6
⇒ Δv = 2 2α
ﻨﻭﺠﺩ ﺍﻵﻥ ﻋﺭﺽ ﺍﻟﺩﺍﻟﺔ ) g (kﻭﺍﻟﺘﻲ ﺘﻌﻁﻲ ﺒــ
ﻴﻬﻤﻨﺎ ﻫﻭ ﻋﺭﺽ ﺍﻟﺩﺍﻟﺔ ) g2 (kﻭﻟـﻴﺱ ﺍﻟﺩﺍﻟـﺔ
)2 g (k
2
) g (k) = e − α (k − k °
) g 2 (k) = e − 2 α (k − k °
ﻭﺍﻟـﺫﻱ
ﻫـﺫﻩ ﺍﻟﺩﺍﻟـﺔ
- ٤٩PDF created with pdfFactory Pro trial version www.pdffactory.com
ﻤﺘﻤﺭﻜﺯﺓ ﺤﻭل
k°
ﻭﻟﻴﺱ ﺤﻭل k = 0ﺒل ﺤﻭل
ﺍﻟﻁﺭﻴﻘﺔ ،ﻨﻭﺠﺩ ﺍﻟﻌﺭﺽ ﻋﻨﺩﻤﺎ ﺘﻘل ﺍﻟﺩﺍﻟﺔ ﺇﻟﻰ )...................(2 − 7
)
1 2α
e
1
k =k°
ﻤﻥ ﻗﺴﻤﺘﻬﺎ ﺍﻟﻘﺼﻭﻯ.
( )Δk = 2ΔΔ° = (2
ﻻﺤﻅ ﺍﻟﺘﻨﺎﺴﺏ ﺍﻟﻌﻜﺴﻲ ﻓﻲ ﻋﺭﺽ ﺍﻟﺩﺍﻟﺔ
ﻓﻲ ) (2-6ﻭﺍﻟﻌﺭﺽ ﻓﻲ
∆v
ﻓﻲ ﺍﻟﻌﻼﻗﺔ ) (2-7ﻫﺫﺍ ﻴﻌﻨﻲ ﺃﻨﻪ ﻜﻠﻤﺎ ﺘﻤﻴﺯﺕ ﺍﻟﺩﺍﻟﺔ
)f (x
ﺍﻨﺒﺴﻁﺕ ﻭﺍﺴﺘﻌﺭﻀﺕ ﻓﻲ ﻓﻀﺎﺀ kﻭﺍﻟﻌﻜﺱ ﺼﺤﻴﺢ. ﻭﻤﻥ ﺍﻟﻤﻬﻡ ﺃﻥ ﻨﻨﻅﺭ ﺇﻟﻰ ﺤﺎﺼل ﻀﺭﺏ ﺍﻟﻌﺭﻀﻴﻥ )...............(2 − 8
2 2α = 4
ﻋﺭﻀﻬﺎ ﻴﻭﺠـﺩ ﺒـﻨﻔﺱ
.
1 2α
∆k
ﻓـﻲ ﻓـﻀﺎﺀ xﻜﻠﻤـﺎ ∆ u , ∆v
Δ u . Δv = 2
ﺍﻟﺫﻱ ﻴﻬﻤﻨﺎ ﻓﻲ ﺍﻟﻌﻼﻗﺔ ) (2-8ﻟﻴﺱ ﺍﻟﻘﻴﻤﺔ ﺍﻟﻌﺩﺩﻴﺔ ،ﺍﻟﺫﻱ ﻴﻬﻡ ﻫﻭ ﺃﻥ ﺤﺎﺼل
ﺍﻟﻀﺭﺏ
∆ u , ∆v
ﻻ ﻴﻌﺘﻤﺩ ﻋﻠﻰ ﻗﻴﻤﺔ ﺍﻟﺜﺎﺒﺕ .αﻫﺫﻩ ﺨﺎﺼﻴﺔ ﻋﺎﻤﺔ ﻟﻜـل ﺍﻟـﺩﻭﺍل
ﺘﺭﺘﺒﻁ ﺒﺒﻌﻀﻬﺎ ﺒﺘﺤﻭﻴﻼﺕ ﻓﻭﺭﻴﺭ ﻜﻤﺎ ﻫﻭ ﻤﻭﻀﺢ ﻓﻲ ﺸﻜل .2-1 This is a general property of functions that are Fourier trans forms of each other.
- ٥٠PDF created with pdfFactory Pro trial version www.pdffactory.com
ﻭﻨﺨﻠﺹ ﻤﻥ ﻫﺫﺍ ﺍﻟﺠﺯﺀ ﺒﺎﻻﺴﺘﻨﺘﺎﺝ ﺃﻨﻪ ﻤﻥ ﺍﻟﻤﺴﺘﺤﻴل ﺃﻥ ﻨﺠﻌل ﻜـﻼﹰ ﻤـﻥ ﺍﻷﻤﺭ ﺍﻟـﺫﻱ، ﻭﻫﺫﻩ ﺨﺎﺼﻴﺔ ﻋﺎﻤﺔ ﻟﻠﺤﺯﻡ ﺍﻟﻤﻭﺠﻴﺔ.ﺫﺍﺕ ﻗﻴﻡ ﺼﻐﻴﺭﺓ
∆k
ﻭ
∆v
.ﻴﺘﺭﺘﺏ ﻋﻠﻴﻪ ﻨﺘﺎﺌﺞ ﻫﺎﻤﺔ ﺠﺩﺍﹰ ﻓﻲ ﻤﻴﻜﺎﻨﻴﻜﺎ ﺍﻟﻜﻡ
It is impossible to make both Δ v and Δk small. This is a general feature of wouae packets, but we shall soon see rat it has some very deep implications for quantum mechanics.
Figure 2-4 Relation between wave packet and its Fourier transform for a square-shaped wave packet.
- ٥١ PDF created with pdfFactory Pro trial version www.pdffactory.com
ﺍﻨﺘﺸﺎﺭ ﺍﻟﺤﺯﻡ ﺍﻟﻤﻭﺠﻴﺔ THE PROPAGATION OF WAVE PACKETS
ﺒﻌﺩﻤﺎ ﺩﺭﺴﻨﺎ ﺘﺭﺍﻜﺏ ﻤﻭﺠﺎﺕ ﻤﻥ ﻨﻭﻉ
e ikx
ﻟﻨﺘﺤﺼل ﻋﻠﻰ ﺩﺍﻟﺔ ) ،f (xﻨـﻭﺩ
ﺍﻵﻥ ﻤﻨﺎﻗﺸﺔ ﻜﻴﻔﻴﺔ ﺇﻨﺘﺸﺎﺭ ﻫﺫﻩ ﺍﻟﻤﻭﺠﺎﺕ؟ ﺒﻼ ﺸﻙ ﺇﻥ ﺍﻨﺘﺸﺎﺭ ﺤﺯﻤﺔ ﻤﻭﺠﻴﺔ ﻴﻌﺘﻤﺩ
ﻋﻠﻰ ﺇﻨﺘﺸﺎﺭ ﻜل ﻤﻥ ﺍﻟﻤﻭﺠﺎﺕ ﺍﻟﻤﻜﻭﻨﺔ ﻟﻬﺎ .ﻓﻠﻭ ﺃﺨﺫﻨﺎ ﻤﻭﺠﺔ ﺘﺘﻐﻴﺭ ﻤﻜﺎﻨﻴﺎﹰ ﻋﻠـﻰ
ﻤﺤﻭﺭ ) Xﺒﻴﻨﻤﺎ ﻻ ﺘﻌﺘﻤﺩ ﺍﻟﻤﻭﺠﺔ ﻤﻥ ﺤﻴﺙ ﺍﻹﺤﺩﺍﺜﻴﺎﺕ t , yﻋﻠﻰ ﺍﻟﺯﻤﻥ( .ﻫـﺫﻩ ﺍﻟﻤﻭﺠﺎﺕ ﺘﺴﻤﻰ ﻤﻭﺠﺎﺕ ﻤﺴﺘﻭﻴﺔ ،plane wanesﻴﻤﻜﻥ ﻜﺘﺎﺒﺔ ﻫﺫﻩ ﺍﻟﻤﻭﺠﺎﺕ ﻜﺩﺍﻟﺔ
ﻓﻲ ﺍﻟﺯﻤﻥ tﻭﺍﻟﻤﻭﺠﺔ Xﺒﺎﻟﺼﻴﻐﺔ. )...... (2-9 ﺤﻴﺙ
e iuv − iwt
w = 2Dv 2D =k λ
ﺒﺎﻟﺘﻌﻭﻴﺽ ﻓﻲ )(2-9 )................(2 − 10
v ) 2 Di ( − vt D
=e
2D v − i 2 Dvt λ
ei
ﻟﻭ ﺃﺨﺫﻨﺎ ﻜﺤﺎﻟﺔ ﺨﺎﺼﺔ؛ ﺇﻨﺘﺸﺎﺭ ﻤﻭﺠﺎﺕ ﺍﻟﻀﻭﺀ ﻓﻲ ﺍﻟﻔـﻀﺎﺀ ﺤﻴـﺙ ﻟﻠﻀﻭﺀ ،ﺒﺎﻟﺘﻌﻭﻴﺽ ﻓﻲ ).(2-10 ).................(2 − 12
) = e ik ( v − ct
2D ) i ( v − ct λ
=e
v c )− t λ λ
( 2 Di
c λ
=v
e
ﺨﺫ ﺍﻵﻥ ﺘﺭﺍﻜﺏ ﺍﻟﻤﻭﺠﺎﺕ ﺴﻌﺘﻬﺎ ) g (kﻤﻥ ﻫﺫﺍ ﺍﻟﻨﻭﻉ ﻤﻥ ﺍﻟﻤﻭﺠﺎﺕ ،ﻓﺒﻌﺩ
ﻤﺭﻭﺭ ﺯﻤﻥ ﻗﺩﺭﻩ t
∞
)f (x 1 t) = ∫ du g (u) e iu (v −ct) = eik (v − ct ∞−
- ٥٢PDF created with pdfFactory Pro trial version www.pdffactory.com
ﺇﻥ ﻫﺫﻩ ﺍﻟﺩﺍﻟﺔ ) f (x – ctﻟﻬﺎ ﻨﻔﺱ ﺍﻟﺸﻜل ﺍﻟﺫﻱ ﺒﺩﺃﺕ ﺒﻪ ﻗﺒل ﺍﻻﻨﺘﺸﺎﺭ ﻤـﺎ ﻋﺩﺍ ﺃﻥ ﻤﻭﻗﻌﻬﺎ ﺒﺩﻻﹰ ﻤﻥ ﺃﻥ ﻴﻜﻭﻥ Xﺃﺼـﺒﺢ ﺍﻵﻥ ﻤﺘﺭﻜـﺯ )ﺃﻭ ﻤﺘﺤﻴـﺯ( ﻋﻨـﺩ v = ct ⇐ v − ct = o
ﻫﺫﺍ ﻴﻌﻨﻲ ﺃﻥ ﺤﺯﻡ ﺍﻟﻀﻭﺀ ﺍﻟﻤﻭﺠﻴﺔ ﺘﻨﺘﺸﺭ ﻓﻲ ﺍﻟﻔﻀﺎﺀ ﺒﻼ ﺘﺸﻭﻴﻪ. Thus the wave packet go light waves propagates, without any distortion, with velocity, the velsal'y of light.
ﻭﻟﻜﻥ ﺍﻟﺫﻱ ﻴﻌﻨﻴﻨﺎ ﻫﻨﺎ ﺍﻨﺘﺸﺎﺭ ﻤﻭﺠﺎﺕ ،ﻴﻔﺘﺭﺽ ﺃﻨﻬﺎ ﺘﹸﻌﺒﺭ ﻋﻥ ﺠﺴﻴﻤﺎﺕ ،ﻓﻲ
ﻫﺫﻩ ﺍﻟﺤﺎﻟﺔ ﻟﻥ ﻨﺴﺘﻁﻴﻊ ﺍﺴﺘﺨﺩﺍﻡ ﺍﻟﻌﻼﻗﺔ. )... (2-13
w = kc W (k) = kc
ﻻﺤﻅ ﻫﺫﻩ ﺍﻟﻌﻼﻗﺔ ﺼﺤﻴﺤﺔ ﻓﻘﻁ ﻓﻲ ﺤﺎﻟﺔ ﺍﻨﺘﺸﺎﺭ ﺤﺯﻤﺔ ﻀﻭﺀ ﻤﻭﺠﻴﺔ ﻓـﻲ
ﺍﻟﻔﻀﺎﺀ .ﻭﻟﻜﻲ ﺘﻨﻁﺒﻕ ﺩﺭﺍﺴﺘﻨﺎ ﻋﻠﻰ ﺍﻨﺘﺸﺎﺭ ﺤﺯﻤﺔ ﻤﻭﺠﻴﺔ )ﺠﺴﻴﻡ( ﻴﻠﺯﻡ ﺍﺴـﺘﺨﺩﺍﻡ
ﺍﻟﻌﻼﻗﺔ ﺍﻟﻌﻼﻤﺔ ﻟـ ) w (kﺤﻴﺙ
w = (k) ≠ kc or kv
ﻓﻲ ﻫﺫﻩ ﺍﻟﺤﺎﻟﺔ ﻨﻌﻴﺩ ﻜﺘﺎﺒﺔ ﻤﻌﺎﺩﻟﺔ ) (2-11ﺒﺎﻟﺼﻴﻐﺔ. )ik (v − ct )f (x 1 t) = ∫ dk g (k) e ik (v − ct
ﻟﺘﺼﺒﺢ ﻓﻲ ﺤﺎﻟﺔ ﺠﺴﻡ ﺤﺭ ﺍﻟﺤﺭﻜﺔ = dk g(k) e ikv − iw (k)t
)..........(2 − 14
ﻻﺤﻅ ﺃﻨﻨﺎ ﻟﻡ ﻨﻌﺭﻑ ﺒﻌﺩ ﺍﻟﺼﻴﻐﺔ ﺍﻟﺤﻘﻴﻘﻴﺔ ﻟﻁﺒﻴﻌﺔ ﺍﻟﺩﺍﻟﺔ ) ،w (kﻭﻟﻜﻥ ﺨـﺫ ﺍﻵﻥ ﺤﺯﻤﺔ ﻤﻭﺠﻴﺔ ﻤﺘﻤﺭﻜﺯﺓ ﻭﻤﺘﺤﺒﺯﺓ ﺠﺩﺍﹰ ﺤﻭل ﻫﺫﺍ ﻴﻌﻨﻲ ﺃﻥ αﻓﻲ ﺍﻟﻌﻼﻗﺔ
2
) g (k) = e −α(k − k °
k°
ﻓﻲ ﻓﻀﺎﺀ .u
ﺴﺘﻜﻭﻥ ﻜﺒﻴﺭﺓ .ﻫﺫﺍ ﻗﺩ ﻻ ﻴﻌﻨﻲ ﺃﻥ
) f (xﺴﺘﻜﻭﻥ ﻤﺘﺤﺒﺯﺓ ﻓﻲ ﻓﻀﺎﺀ vﻭﻟﻜﻥ ﻫﺫﺍ ﺍﻟﺘﻘﺭﻴﺏ ﺴﻴﺴﻬل ﺤﺴﺎﺒﺎﺘﻨﺎ. - ٥٣ -
PDF created with pdfFactory Pro trial version www.pdffactory.com
ﺒﻤﺎ ﺃﻥ ﺍﻟﺘﻜﺎﻤل ﻓﻲ ) (2- 14ﻤﺘﻤﺭﻜﺯ ﺤﻭل
ﻤﻊ ﺍﻓﺘـﺭﺍﺽ ﺃﻥ
k°
)w (k
ﻻ
ﺘﺘﻐﻴﺭﻩ ﺒﺸﺩﺓ ﻤﻊ ﺘﻐﻴﺭ .k )............(2 − 15
ﺍﻟﺤﺩ ﺍﻷﻭل
) w (k °
)dw(k 1 )d 2 w (k ( ) k + (k − k ° ) 2 ) ° dk 2 dk 2 k°
( ) w (k) ≅ w (k ° ) + (k − k °
ﺜﺎﺒﺕ ﻷﻨﻪ ﻻ ﻴﻌﺘﻤﺩ ﻋﻠﻰ .kﺍﻟﺤﺩ ﺍﻟﺜﺎﻨﻲ
dw ) dk k = k °
(
ﻴﻤﺜل ﺴﺭﻋﺔ
ﺍﻟﻤﺠﻤﻭﻋﺔ -group velocityﻜﻤﺎ ﺸﺭﺤﻨﺎ ﻓﻲ ﺍﻟﺘﻤﻬﻴﺩ -ﻫﻲ ﺍﻟﺴﺭﻋﺔ ﺍﻟﺘﻲ ﺘـﺴﻴﺭ
ﺒﻬﺎ ﺍﻟﺤﺯﻤﺔ ﺍﻟﻤﻭﺠﻴﺔ.
dw ) = vg dk k = k°
⇒
(
)(2-16
and 1 d2w ) =β ( 2 dk 2 k = k °
ﺒﺎﻓﺘﺭﺍﺽ ﺃﻥ: k1 = k − k ° ⇒ dk 1 = dk
ﻓﺈﻥ ﺍﻟﺤﺯﻤﺔ ﺍﻟﻤﻭﺠﻴﺔ ﺘﻌﺘﻤﺩ ﻋﻠﻰ ﺍﻟﺯﻤﻥ ﺒﺎﻟﺼﻴﻐﺔ. i (k1 + k ° ) v − i w (k ° ) + k1vg + k12β t
∞
f (x 1 t) = ∫ dk 1 e −αk e 12
∞−
ﺨﺫ ﺍﻟﺤﺩﻭﺩ ﺍﻟﺜﺎﺒﺘﺔ ﺨﺎﺭﺝ ﺍﻟﺘﻜﺎﻤل. 1 1
12βt
12
1
e −αk . e ik v. e −ik vgt. e −ik 12βt
e −ik
)1 (v − vgt
e ik
∞
∫ dk
∞−
ik ° v − i w(k ° )t
∞
∫ dk e
1 − αk12
=e
ikov − iw(k ° )t
=e
∞−
)........(2-17 ﻭﺒﻨﻔﺱ ﺍﻟﻁﺭﻴﻘﺔ ﺍﻟﺘﻲ ﺃﺠﺭﻴﻨﺎﻫﺎ ﻓﻲ ﺍﻟﺤﺎﻟﺔ ﺍﻷﻭﻟﻰ ،ﻴﻤﻜﻥ ﻋﻤل ﺍﻟﺘﻜﺎﻤل ﻋﻥ
ﻁﺭﻴﻕ ﺇﻜﻤﺎل ﺍﻟﻤﺭﺒﻊ ﻓﻲ ﺍﻷﺱ ،ﺨﺫ ﺍﻷﺱ ﻓﻘﻁ. )− (α + iβ t ) k12 + ik1 (v − μ g t
=e
12 + ik1 (v − μgt) − ik12 βt
e − αk
- ٥٤PDF created with pdfFactory Pro trial version www.pdffactory.com
.ﻋﺎﻤل ﻤﺸﺘﺭﻙ =e
− (α + iβ t)
ﺨﺫ
(v − ν g t) − (α + iβ t) k12 − ik1 (α = iβ t )
.ﻟﻠﺤﺼﻭل ﻋﻠﻰ ﻤﺭﺒﻊ ﻜﺎﻤل ﻨﻁﺭﺡ ﻭﻨﻀﻴﻑ ﺍﻟﺤﺩ (v − ν g t)2 4(α + iβ t)
e
iu1 (u − v g t) (x − v g t) 2 ( x − v g t) 2 + − (α + iβ t) u12 − − α + iββ 4 (α + iββ) 4 (α + iβ t)
ﺨﺫ ﻫﺫﻩ ﺍﻹﺸﺎﺭﺓ ﻤﺸﺘﺭﻙ ﻤﻊ ﺍﻟﺤﺩ ﺍﻷﻭل ik 1 (u − v g t) (u − v g t) 2 − (α + iβ t) k12 − − α + iβ t 4 (α + iββ)
e ⇒ e
i (u − v g t) − (α + iβ t) k1 − 2 ( α + iβ t
2 +
( x − v t) 2 g − 4 (α + iβ t)
2 1 ( x − v g t) 4 (α + iβ t)
............(2 − 18)
:( ﻨﺘﺤﺼل ﻋﻠﻰ2-17) ( ﻓﻲ2-18) ﺒﺘﻌﻭﻴﺽ f ( x1t ) = e
∞
∫ dk
i ( k° x − w ( k° ) t )
1
e
(v − µ g t ) − (α + i β t ) k1 − 2 (α = i β t )
−∞
= e i ( k° x − w ( k° ) t e
−
( x− y g t ) 2 4 (α + iβ t )
2 −
ﺛﺎﺑﺖ
( x − vg t ) 2
e 4 ( α + i βt ) 1424 3 اﻟﺤﺪ
ھﺬا
اﺑﺖ
i ( x −u gt) 2 − ( α + i β t ) k1 − 2 (α + iβ t ) 1
∞
∫ dk e
−∞
1444424444 3
− (x − u g t) 2
f (x 1 t) = e
i (u ° x − w (k ° ) t
e
D (α + iβ t)
4 (α + iβ t)
2
f (x 1 t) = f (n 1 t) f * (x 1 t) = e ° . e
−
(v − u g t) 2 4 (α + iβ t)
−
( v − u g t) 2 4 (α − iβ t)
D D (α + i β t ) ( α − i β t)
− (v − u g t) 2 ( 2α ) D 2 α 2 + β 2t 2 = 1. e (α 2 + β 2 t 2 ) =e
−
(v − u g t) 2
D
2 (α + β 2 t 2 /αα (α 2 + β 2 t 2 )
- ٥٥ PDF created with pdfFactory Pro trial version www.pdffactory.com
ﺒﻘﺴﻤﺔ ﺍﻷﺱ ﻓﻲ ﺍﻟﺒﺴﻁ ﻭﺍﻟﻤﻘﺎﻡ ﻋﻠﻰ . α ﺒﻤﻘﺎﺭﻨﺔ ﺍﻟﺼﻴﻐﺔ ﻓﻲ ) (2-19ﺒـ )(2-5
v2 D − 2α 2 f (x) = e α
ﻨﺠﺩ ﺃﻥ ﻫﻨﺎﻙ ﺘﻤﺎﺜل ﻤﻊ ﻤﻼﺤﻅﺔ ﺃﻥ vﻓﻲ ) (2-5ﺘﻐﻴﺭﺕ ﻟـ ﺘﺤﻭﻟﺕ ﻟـ
β2 t 2 α
ﻭﻟﻬﺫﺍ ﻓﺈﻥ
)(v − u g t
ﻭα
α+ 2
) f (x 1 t
ﺘﻤﺜل ﺤﺯﻤﺔ ﻤﻭﺠﻴﺔ ﺘﺴﻴﺭ ﺒﺴﺭﻋﺔ
ug
ﻭﻟﻜـﻥ ﻋـﺭﺽ
ﻫﺫﻩ ﺍﻟﺤﺯﻤﺔ ﻏﻴﺭ ﻤﺤﺩﺩ ﻭﻟﻜﻥ ﻴﺘﻐﻴﺭ ﻤﻊ ﺍﻟﺯﻤﻥ ،ﻻﺤﻅ ﺃﻥ ﻋﺭﺽ ﺍﻟﺩﺍﻟﺔ ).(2-19 Δv = 2. Δ v ° = 2. 2 (α + β 2 t 2 /αα or )...............(2 − 20
β 2 t 2 12 ) α2
Δv = 8α (1 +
ﻤﻥ ﺍﻟﻤﻌﺎﺩﻟﺔ ) (2-20ﻴﺘﻀﺢ ﺃﻥ ﻋﺭﺽ ﺍﻟﺩﺍﻟﺔ ﻴﺯﺩﺍﺩ ﻤﻊ ﺍﻟﺯﻤﻥ ﺃﻱ ﺃﻥ ﺍﻟﺩﺍﻟﺔ ﺘﻨﺒﺴﻁ ﻤﻊ ﻤﺭﻭﺭ ﺍﻟﺯﻤﻥ ﻭﺫﻟﻙ ﻤﻘﺎﺭﻨﺔ ﺒﻌﺭﻀﻬﺎ ﻋﻨﺩ t = 0 Δv = ( t = 0 ) = 8α
ﺃﻤﺎ ﻋﻨﺩ ﺃﻱ ﺯﻤﻥ ﺁﺨﺭ ﻓﺈﻥ ﻋﺭﺽ ﺍﻟﺩﺍﻟﺔ ﻴﻌﻁﻲ ﺒﺎﻟﺼﻴﻐﺔ ) (2- 20ﺇﻥ ﻤﻌﺩل
ﺍﻻﻨﺒﺴﺎﻁ ﻓﻲ ﺍﻟﺩﺍﻟﺔ ﺴﻴﻜﻭﻥ ﺼﻐﻴﺭﺍﹰ ﺇﺫﺍ ﻜﺎﻨﺕ αﻜﺒﻴﺭﺓ ﺃﻱ ﺇﺫﺍ ﻜﺎﻨﺕ ﺍﻟﺩﺍﻟﺔ ﻤﻨﺒﺴﻁﺔ ﻜﺜﻴﺭﺍﹰ ﻋﻨﺩ ﺍﻟﺒﺩﺍﻴﺔ.
- ٥٦PDF created with pdfFactory Pro trial version www.pdffactory.com
ﻤﻥ ﺍﻟﺤﺯﻡ ﺍﻟﻤﻭﺠﻴﺔ ﺇﻟﻰ ﻤﻌﺎﺩﻟﺔ ﺸﺭﻭﺩﻨﺠﺭ: FROM WAVE PACKETS TO THE SCHRöDINGER EQUATION
ﺇﺫﺍ ﻜﺎﻨﺕ ﺍﻟﺩﺍﻟﺔ }
.........
iku −iw ( k ) t
{f ( x t ) = ∫ dk g (u) e 1
ﺘﻤﺜل ﺠﺴﻴﻡ ﻜﻤﻴﺔ ﺍﻟﺤﺭﻜﺔ ﻟﻪ
ﺘﺴﺎﻭﻱ pﻭﻁﺎﻗﺘﻪ ﺍﻟﺤﺭﻜﻴﺔ ) ،(p2 /2mﻓﺈﻥ ﻫﺫﺍ ﻴﺴﺘﻠﺯﻡ ﺃﻥ ﺴﺭﻋﺔ ﺍﻟﻤﺠﻤﻭﻋـﺔ ug
ﺘﻌﻁﻰ ﺒﺎﻟﻤﻌﺎﺩﻟﺔ. dw p )= particle velocity ⇒ = .................(2 − 21 dk m
= ug
ﻭﻤﻥ ﺍﻓﺘﺭﺍﻀﺎﺕ ﻤﻴﻜﺎﻨﻴﻜﺎ ﺍﻟﻜﻡ ،ﻓﺈﻥ ﺃﻱ ﺇﺸﻌﺎﻉ ﻟﻪ ﺘﺭﺩﺩ wﻓﺈﻨﻪ ﻴﻘﺎﺒﻠﻪ ﻁﺎﻗﺔ
Eﺘﻌﻁﻲ ﺒﺎﻟﻌﻼﻗﺔ.
E=hw or E h
=w
ﺇﺫﺍ ﻜﺎﻨﺕ Eﻫﻲ ﻁﺎﻗﺔ ﺠﺴﻴﻡ ﺤﺭ،ﻓﺈﻨﻬﺎ ﺘﻤﺜل ﻁﺎﻗﺘﻪ ﺍﻟﺤﺭﻜﻴﺔ p2/2mﺃﻱ ﺃﻥ. ( p 2 / 2m) p 2 =w = 2mh h
)(2 − 23
ﻭﻋﻠﻴﻪ ﻓﻤﻥ ﺍﻟﻤﻤﻜﻥ ﺃﻥ ﻨﻜﺘﺏ ﻤﺘﺠﻪ ﺍﻟﻤﻭﺠﺔ ) kﺒﻨﻔﺱ ﺍﻟﻁﺭﻴﻘﺔ( ﻭﻤﻥ ﻤﻌﺎﺩﻟﺔ
ﺩﻴﺒﺭﻭﻟﻲ . λ = n p
2D 2 D p p = = = λ (h ) ( h ) h p 2D
).................(2 − 24
=k
ﻭﻟﻬﺫﺍ ﺍﻟﺴﺒﺏ ﻴﻤﻜﻨﻨﺎ ﺇﻋﺎﺩﺓ ﻜﺘﺎﺒﺔ ﺍﻟﺼﻴﻐﺔ ) (2-14ﺒﺩﻻﻟﺔ ) pﺒﺩﻻﹰ ﻤﻥ (k ψ (p ) e i (px −Et)/h
)............(2 − 25
ﺤﻴﺙ ﺃﻥ ﺍﻟﺤﺯﻤﺔ ﺍﻟﻤﻭﺠﻴﺔ
)ψ (x 1 t
ﺍﻟﺠﺯﺌﻴﺔ .partial differentiable.
1
∫ dp 2Dh
= )ψ (x 1 t
ﺘﻤﺜل ﺍﻟﺤل ﺍﻟﻌـﺎﻡ ﻟﻠﻤﻌﺎﺩﻟـﺔ ﺍﻟﺘﻔﺎﻀـﻠﻴﺔ
- ٥٧PDF created with pdfFactory Pro trial version www.pdffactory.com
iD
∂ ψ (x 1 t) ∂t
=
1
∫ dp φ (p) E e 2Dh
i (px − Et)/t
p 2 i( px − Et) /D = ∫ dp φ (p) 2m e 2Dh 1
.x ﻤﺭﺘﻴﻥ ﺒﺎﻟﻨﺴﺒﺔ ﻟـ
ψ (x 1 t)
............(2 − 26)
( ﻴﻤﻜﻥ ﺍﻟﺤﺼﻭل ﻋﻠﻴﻬﺎ ﺒﺘﻔﺎﻀل2-26) ﺤﻴﺙ ﺃﻥ
∂ ∂ψ ∂ 1 ∂ ( )= dp φ (p) e i (px − Et)/D ∫ ∂x ∂x ∂x 2Dh ∂x ∂ 1 ip = dp φ (p) . e i (px − Et)/h ∫ h ∂x 2Dh = =
1 2Dh 1 2Dh
ip ip i (px − Et)/t e h
∫ dp φ (p) h . ∫ dp φ ( p ) ( −
p 2 i (px − Et) /t )e ]................(2 − 27) D2
ﻴﻠﺯﻡ ﻀﺭﺏ،( ﻤﻥ ﺍﻟﻁﺭﻑ ﺍﻷﻴﻤﻥ2-26) ( ﺒـ2-27) ﻭﻟﻜﻲ ﺘﺘﺴﺎﻭﻯ ﻤﻌﺎﺩﻟﺔ
ﺃﻱ، − D ( ﻓﻲ2-27) ﻁﺭﻓﻲ 2
2m
−
h2 ∂2ψ 1 D2 p 2 i (px − Et)/D = dp φ (p) ( − ) ( − )e ∫ 2m ∂x 2 2m D2 2Dh 1 p 2 i (px − Et)/D ] = dp φ (p) ( )e ∫ 2m 2Dh
]
:ﻭﻋﻠﻴﻪ ﻓﺈﻥ ih
∂ψ (x 1 t) ∂t
=−
2 h 2 ∂ ψ (v1 t) 2m ∂v 2
................(2 − 28)
.ﻭﻫﺫﻩ ﺼﺤﻴﺤﺔ ﻟﺠﺴﻡ ﺤﺭ (ﻋﻼﻗﺎﺕ ﺍﻟﻼﺘﺤﺩﻴﺩ )ﺍﻟﻼﺘﻌﻴﻴﻥ
The uncertainty Relations
ﺃﺤﺩ ﺍﻟﻨﺘﺎﺌﺞ ﺍﻟﻤﻬﻤﺔ ﺍﻟﺘﻲ ﺘﻭﺼﻠﻨﺎ ﻟﻬﺎ ﻤﻥ ﻤﻨﺎﻗﺸﺘﻨﺎ ﻟﻠﺤﺯﻡ )ﺍﻟﺭﺯﻡ( ﺍﻟﻤﻭﺠﻴـﺔ .
k
ﻭ ﻓﺭﺍﻍ
v
ﻫﻲ ﺍﻟﻤﻌﻜﻭﺴﺔ ﻟﻠﻌﻼﻗﺔ ﺒﻴﻥ ﺍﻷﻋﺭﺍﺽ ﻓﻲ ﻓﺭﺍﻍ Δ k Δ v ≥1
p=h k
ﻭﺒﺎﺴﺘﻌﻤﺎل ﺍﻟﻌﻼﻗﺔ
h
- ٥٨ PDF created with pdfFactory Pro trial version www.pdffactory.com
ﺒﻀﺭﺏ ﺍﻟﻁﺭﻓﻴﻥ ﺒـ
ﺴﻨﺘﺤﺼل ﻋﻠﻰ ﻋﻼﻗﺔ ﻫﺎﻴﺯﻨﺒﺭﺝ Heisenberg ancernrainrg clarion Δ pΔ v≥h
)........(1
ﻭﺒﻤﺎ ﺃﻥ ﺍﻟﻌﺭﺽ ﻴﻤﺜل ﺍﻟﻤﻨﻁﻘﺔ ﺍﻟﺘﻲ ﻤﻥ ﺍﻟﻤﺤﺘﻤل ﺃﻥ ﻴﻜﻭﻥ ﻓـﻲ ﺍﻟﺠـﺴﻡ
)( p- , v - Iﻓﻀﺎﺀ ﻓﺈﻥ ﺍﻟﻤﻌﺎﺩﻟﺔ ) (1ﺘﺒﻴﻥ:
ﻟﻭ ﺃﺭﺩﻨﺎ ﺃﻥ ﻨﻌﻤل ﺤﺯﻤﺔ )ﺭﺯﻤﺔ( ﻤﺘﻤﺭﻜﺯﺓ )ﻤﺘﻤﻴﺯﺓ ﺠﺩﺍﹰ( ﻭﻤﻭﺠﻴﺔ ﻓـﻲ x
ﻓﻀﺎﺀ ،ﻓﺈﻨﻪ ﻤﻥ ﺍﻟﻤﺴﺘﺤﻴل ﻋﻠﻴﻨﺎ ﺃﻥ ﻨﺸﺎﺭﻙ ﺒﻬﺎ ﻜﻤﻴﺔ ﺤﺭﻜﺔ ﻤﻌﺭﻓﺔ ﺠﻴﺩﺍﹰ) .ﺍﻷﻤﺭ
ﺍﻟﺫﻱ ﻴﻌﺘﺒﺭ ﺒﺩﻴﻬﻲ ﻭﻋﻠﻴﻪ ﻓﺈﻥ ﺭﺯﻤﺔ ﻤﻭﺠﻴﺔ ﺫﺍﺕ ﻜﻤﻴﺔ ﻤﺤﺩﺩﺓ ﻤﻥ ﻜﻤﻴﺔ ﺍﻟﺤﺭﻜـﺔ
ﻴﺠﺏ ﺃﻥ ﺘﻜﻭﻥ ﻏﻴﺭ ﻤﺘﻤﻴﺯﺓ ﻓﻲ ﻓﻀﺎﺀ v-ﺃﻱ ﺃﻥ ﺘﻜﻭﻥ ﻋﺭﻴﻀﺔ ﺠﺩﺍﹰ. ﻭﻫﺫﺍ ﺍﻟﺤﺩ )ﺍﻟﺤﺼﺭ( ﻭﺍﺤﺩ ﻤﻥ ﺇﻟﺯﺍﻤﺎﺕ ﻤﻴﻜﺎﻨﻴﻜﺎ ﺍﻟﻜﻡ ﻋﻠﻰ ﻭﺼﻔﻨﺎ ﻟﻸﻨﻅﻤﺔ
ﺍﻟﻔﻴﺯﻴﺎﺌﻴﺔ ﺒﺎﻟﻤﻔﺎﻫﻴﻡ ﺍﻟﺘﻘﻠﻴﺩﻴﺔ .ﻓﻲ ﺍﻟﻔﻴﺯﻴﺎﺀ ﺍﻟﺘﻘﻠﻴﺩﻴﺔ ﻜﺎﻷﻤﺭ ) p, xﺍﻟﻤﻭﻗـﻊ ،ﻜﻤﻴـﺔ ﺍﻟﺤﺭﻜﺔ( ﻤﺴﺘﻘﻼﺕ ﻋﻥ ﺒﻌﻀﻬﻤﺎ. ﺒﻴﻨﻤﺎ ﻓﻲ ﻤﻴﻜﺎﻨﻴﻜﺎ ﺍﻟﻜﻡ p, xﻤﻜﻤﻠﻴﻥ ﻟﺒﻌﻀﻬﻤﺎ. ﻻﺤﻅ ﺃﻨﻪ ﻤﻥ ﻗﻴﻤﺔ
ﺍﻟﺼﻐﻴﺭﺓ ﻴﺘﻀﺢ ﻟﻨﺎ ﺃﻥ ﻤﻴﻜﺎﻨﻴﻜﺎ ﺍﻟﺘﻘﻠﻴﺩﻴﺔ ﺘﻔﺸل ﻓـﻲ
h
ﺤﺎﻟﺔ ﺍﻷﻨﻅﻤﺔ ﺍﻟﻤﺠﻬﺭﻴﺔ ﻓﻠﻭ ﺍﻋﺘﺒﺭﻨﺎ ﺃﻥ ﺤﺒﺔ ﻏﺒﺎﺭ ﻤﺜﺎل :ﺨﺫ ﺠﺴﻤﺎﹰ ﻜﺘﻠﺘﻪ
ﻜﻤﻴﺔ ﺤﺭﻜﺘﻪ
Δp
m 1μ g
m = 10 −4 g υ = 10 4 cn / s
ﻭﻤﻭﻗﻌﻪ ﻤﻌﺭﻭﻑ ﻟﺩﻗﺔ ، 1μ mﺍﺤﺴﺏ ﺍﻟﻼﺘﻌﻴﻴﻥ ﻓـﻲ
ﻤﻥ ﺍﻟﻤﻌﺎﺩﻟﺔ ﺍﻟﺴﺎﺒﻘﺔ. m2 s2
J = kg.
D h ≥ ⇒ Δp 2 2Δ v
≥ Δp Δv
6.63×10 −34 J.S J. S m ≥Δ p = 0.5× 10 −28 = 5 × 10 −29 k g −6 m s 4D ×1 ×10 m
ﻭﻫﺫﺍ ﻴﻁﺎﺒﻕ ﺴﺭﻋﺔ ﻤﻘﺩﺍﺭﻫﺎ
−29 m 5 ×10 u g m/s = s 1×10 −9 u g
− 20
Δν = 5 ×10
- ٥٩PDF created with pdfFactory Pro trial version www.pdffactory.com
ﻭﻫﺫﻩ ﺍﻟﻘﻴﻤﺔ ﻤﻬﻤﻠﺔ .ﻭﻋﻠﻴﻪ ﻓﺈﻥ ﻟﻸﺠﺴﺎﻡ ﺍﻟﻜﺒﻴﺭﺓ ﻋﻼﻗﺔ ﺍﻟﻼﺘﻌﻴﻴﻥ ﻟﻴﺴﺕ ﺫﺍﺕ ﺃﻫﻤﻴﺔ ﻭﻋﻠﻴﻪ ﻓﺈﻥ )ﻓﻲ ﻫﺫﻩ ﺍﻟﺤﺎﻟﺔ( ﺍﻟﻤﻴﻜﺎﻨﻴﻜﺎ ﺍﻟﺘﻘﻠﻴﺩﻴﺔ ﺘﻌﻁﻲ ﻨﻔﺱ ﺍﻟﻘﻴﻤﺔ. ﻗﻴﺎﺱ ﻤﻜﺎﻥ ﺇﻟﻜﺘﺭﻭﻥ )ﻤﻴﻜﺭﻭﺴﻜﻭﺏ ﻫﻴﺯﺒﻨﺭﺝ( )Measwrcmerg position of aneleiton (Hesibdoceg Microswgee
ﺸﻌﺎﻉ ﻤﻥ ﺍﻹﻟﻜﺘﺭﻭﻨﺎﺕ ﻟﻬﺎ ﻜﻤﻴﺔ ﺤﺭﻜﺔ ﻤﺤﺩﺩﺓ ﺠﻴﺩﺍﹰ
px
ﺘﺘﺤﺭﻙ ﻓﻲ
ﺍﺘﺠﺎﻩ t x
ﻭﺍﻟﻐﺭﺽ ﻤﻥ ﺍﻟﻤﻴﻜﺭﻭﺴﻜﻭﺏ ﺍﻟﺸﺎﺸﺔ ﻫﻲ ﺭﺅﻴﺔ ﻤﻭﻗﻊ ﺍﻹﻟﻜﺘﺭﻭﻥ ﻭﺫﻟﻙ ﺒﻤـﺸﺎﻫﺩﺓ ﺍﻟﻀﻭﺀ ﺍﻟﻤﺘﺸﺘﺕ ﻤﻥ ﺍﻹﻟﻜﺘﺭﻭﻥ.
ﺇﻥ ﺍﺴﺘﻁﺎﻡ ﺍﻟﻔﻭﺘﻭﻨﺎﺕ ﺒﺎﻹﻟﻜﺘﺭﻭﻥ ﻴﺭﺘﺩ ﺨﻼل ﺍﻟﻤﻴﻜﺭﻭﺴﻜﻭﺏ .ﻭﺇﻥ ﻗـﺩﺭﺓ ﺘﺤﻠﻴل ﺍﻟﻤﻴﻜﺭﻭﺴﻜﻭﺏ ) bastionﺃﻱ ﺍﻟﺩﻗﺔ ﺍﻟﺘﻲ ﺒﻬﺎ ﻴﻤﻜﻨﻨﺎ ﺘﺤﺩﻴﺩ ﻤﻭﻗﻊ ﺍﻹﻟﻜﺘﺭﻭﻥ ﻤﻥ ﻤﻌﻠﻭﻤﺎﺕ ﻤﻥ ﺍﻟﻀﻭﺀ(.
ﺤﻴﺙ λﻁﻭل ﻤﻭﺠﺔ ﺍﻟﻀﻭﺀ
λ sin ϕ
~ Δν
ﻴﺠﻌل λﺃﻗل ﻤﺎ ﻴﻤﻜﻥ ﻭ ϕﺃﻜﺒﺭ ﻤﺎ ﻴﻤﻜﻥ ﺒﺈﻤﻜﺎﻨﻨﺎ ﺘﺼﻐﻴﺭ
∆v
ﻟﻠﺤﺩ ﺍﻟﻤﺭﻏﻭﺏ
ﻫﺫﺍ ﻴﻤﻜﻨﻨﺎ ﻓﻌﻠﻪ ﻋﻠﻰ ﺤﺴﺎﺏ ﻤﻌﻠﻭﻤﺎﺘﻨﺎ ﻤﻥ ﺩﻗﺔ ﻜﻤﻴﺔ ﺍﻟﺤﺭﻜﺔ ﻓﻲ ﺍﺘﺠﺎﻩ .νx ﻤﻴﻜﺎﻨﻴﻜﺎ ﺍﻟﻜﻡ ﺘﺨﺒﺭﻨﺎ ﺒﺄﻥ ﺍﻹﺸﺎﺭﺍﺕ ﺍﻟﻤﺴﺠﻠﺔ ﻋﻠﻰ ﺍﻟﺸﺎﺸﺔ ﻫﻲ ﻋﺒﺎﺭﺓ ﻋـﻥ
ﻓﻭﺘﻭﻨﺎﺕ ﺘﺸﺘﺕ ﻤﻥ ﺍﻹﻟﻜﺘﺭﻭﻨﺎﺕ ﺍﺘﺠﺎﻩ ﺍﻟﻔﻭﺘﻭﻨﺎﺕ ﺒﻌﺩ ﺍﻟﺘﺸﺘﺕ ﻏﻴﺭ ﻤﺤﺩﺩ ﺨـﻼل ﺍﻟﺯﺍﻭﻴﺔ ﺍﻟﻤﻘﺎﺒﻠﺔ ﻟﻠﻔﺘﺤﺔ ﻭﻋﻠﻴﻪ ﻜﻤﻴﺔ ﺍﻟﺤﺭﻜﺔ ﺍﻟﻤﺭﺘﺩﺓ ﻟﻺﻟﻜﺘﺭﻭﻥ ﻏﻴﺭ ﻤﺤﺩﺩﺓ ﺒـ h λ c h hv λ ⇒ ϕ = 2 sin ϕ . ⇒ Δ p x ~ 2 sin φ c v c v
Δp (sin ϕ + sin ϕ ).
ﻭﻋﻠﻴﻪ
Dv λ sin ϕ . ~ 4Dh c sin ϕ
Δ p v Δv ~ 2
- ٦٠PDF created with pdfFactory Pro trial version www.pdffactory.com
ﻟﺘﺤﺴﻴﻥ ﺩﻗﺔ ﺍﻟﻤﻭﻗﻊ ﻟﻺﻟﻜﺘﺭﻭﻥ ﻴﺠﺏ ﺃﻥ ﻨﺴﺘﺨﺩﻡ ﻀﻭﺀ ﺫﻭ λﺼﻐﻴﺭﺓ ﻟﻜﻥ ﻫﺫﺍ ﻴﻨﺘﺞ ﺨﻁﺄ ﻜﺒﻴﺭ ﺃﻭ ﺘﺸﻭﻴﺵ ﻜﺒﻴﺭ ﻓﻲ ﻜﻤﻴﺔ ﺍﻟﺤﺭﻜﺔ ﻭﺍﻟﻌﻜﺱ :ﻟﻭ ﺃﺭﺩﻨﺎ ﺃﻥ ﻨﻘﻠل ﺍﻟﺨﻁﺄ ﺒﻘﻴﺎﺴﻨﺎ ﺒﻜﻤﻴﺔ ﺍﻟﺤﺭﻜﺔ ﻴﺠﺏ ﺃﻥ ﻨﺴﺘﺨﺩﻡ ﺇﺸﻌﺎﻉ ﺫﻭ ﻁﻭل ﻤﻭﺠﻲ ﻜﺒﻴﺭ ﻭﻫﺫﺍ
ﺒﺩﻭﺭﻩ ﻴﺅﺩﻱ ﺇﻟﻰ ﺨﻁﺄ ﻜﺒﻴﺭ ﻓﻲ ﺍﻟﻤﻭﻗﻊ.
ﻭﻋﻠﻴﻪ ﻓﺈﻥ ﻤﺒﺩﺃ ﺍﻟﻼﺘﻌﻴﻴﻥ ﻫﻭ ﻨﺎﺸﺊ ﻤﻥ ﻁﺭﻴﻘﺔ ﺍﻟﻘﻴـﺎﺱ ﻨﻔـﺴﻬﺎ .ﻭﻋﻠـﻰ
ﺍﻟﻤﺴﺘﻭﻯ ﺍﻟﺫﺭﻱ ﻓﺈﻥ ﺍﻟﻘﻴﺎﺱ ﻨﻔﺴﻪ ﻴﻨﺸﺄ ﻋﻨﻪ ﺨﻁﺄ ﻭﺘﺸﻭﻴﺵ ﻤﻠﺤﻭﻅ ﻟﻠﻨﻅﺎﻡ ﻭﺫﻟـﻙ ﻟﻠﺘﻔﺎﻋل ﺒﻴﻥ ﺠﻬﺎﺯ ﺍﻟﻘﻴﺎﺱ ﺍﻟﻜﻤﻴﺔ ﺍﻟﻤﻘﺎﺴﺔ.
ﻭﻤﺒﺩﺃ ﺍﻟﻼﺘﻌﻴﻴﻥ ﻴﻘﺭﺭ ﺃﻨﻨﺎ ﻻ ﻨﺴﺘﻁﻴﻊ ﺘﺤﺩﻴﺩ ﻤﺴﺎﺭ ﺠﺴﻡ ﺒﺩﻗﺔ ﻤﺘﻨﺎﻫﻴﺔ ﻜﻤـﺎ
ﻫﻭ ﺍﻟﺤﺎل ﻓﻲ ﺍﻟﻤﻴﻜﺎﻨﻴﻜﺎ ﺍﻟﺘﻘﻠﻴﺩﻴﺔ.
ﻤﺜﺎل :ﺒﺎﻓﺘﺭﺍﺽ ﺃﻥ ﺍﻟﻼﺘﻌﻴﻴﻥ ﻓﻲ ﻤﻭﻗﻊ ﺠﺯﺉ ﻫﻴﺩﺭﻭﺠﻴﻥ )ﺍﻟﺫﻱ ﻜﺘﻠﺘﻪ
kg
−27
( 2 ×10
ﻴﻘﺩﺭ ﺒﺤﻭﺍﻟﻲ 10-10 mﺍﺤﺴﺏ ﻗﻴﻤﺔ ﺍﻟﻼﺘﻌﻴﻴﻥ ﻓﻲ ﻜﻤﻴﺔ ﺍﻟﺤﺭﻜﺔ؟ ﻭﻜﺫﻟﻙ ﺍﻟﻨـﺴﺒﺔ Δ px px 6.6 ×10 −34 h ≥ Δp x = ≥ 6.6 ×10 − 24 kg − m/s −10 Δx (2D )10
ﻜﻤﻴﺔ ﺤﺭﻜﺔ ﺍﻟﺠﺯﺉ ﺍﻟﺫﻱ ﻴﺘﺤﺭﻙ ﺒﺴﺭﻋﺔ
m s
2000
)ﺍﻟﺴﺭﻋﺔ ﺍﻟﺤﺭﺍﺭﻴﺔ ﻋﻨﺩ
ﺩﺭﺠﺔ ﺤﺭﺍﺭﺓ ﺍﻟﻐﺭﻓﺔ( m = 4 ×10 −24 kg. m/s s
Px = mc = 2 ×10 − 27 kg × 2 ×10 3
ﻭﻋﻠﻴﻪ ﻓﺈﻥ ﻨﺴﺒﺔ ﺍﻟﻼﺘﻌﻴﻴﻥ: 0.25
6.6 ×10 −24 = = 1.7 4 ×10 − 24
Δp x px
ﻭﻋﻠﻴﻪ ﻓﺈﻥ ﻜﻤﻴﺔ ﺤﺭﻜﺔ ﻫﺫﺍ ﺍﻟﺠﺯﺉ ﻻ ﻴﻤﻜﻥ ﺘﺤﺩﻴﺩﻫﺎ ﺒﺩﻗﺔ ﺃﻜﺜﺭ ﻤﻥ ﺍﻟﻘﻴﻤﺔ
ﺃﻷﺼﻠﻴﺔ.
- ٦١PDF created with pdfFactory Pro trial version www.pdffactory.com
ﻟﻜﻥ :ﻟﻭ ﺃﺨﺫﻨﺎ ﻜﺘﻠﺔ ،ﻟﻬﺎ ﺴﺭﻋﺔ ،1000 mﻤﻭﻗﻌﻬﺎ ﻤﺤﺩﺩ ﻟﺩﻗﺔ 1. mmﻫـﺫﺍ s
ﺍﻟﺠﺴﻡ ﻤﻘﺩﺍﺭ ﺍﻟﻼﺘﻌﻴﻴﻥ ﻓﻲ ﻜﻤﻴﺔ ﺤﺭﻜﺘﻪ. 6.63×10 −34 6.6 ≥ Δp = ×10 −31 kg − m/s −3 10 m 2D p = 0.005 ×1000 = 50 kg - m/s
ﻨﺴﺒﺔ ﺍﻟﻼﺘﻌﻴﻴﻥ. Δp 6.6 ×10 −31 kg − k/s = = 1.3×10 −32 p
ﻭﻫﺫﺍ ﺭﻗﻡ ﺼﻐﻴﺭ ﺠﺩﺍﹰ ﻭﻻ ﻴﻤﻜﻥ ﻗﻴﺎﺴﻪ ﺒﺄﻱ ﺠﻬﺎﺯ. ﻤﺜﺎل :ﺯﻤﻥ ﺍﻟﺤﻴﺎﺓ ﻟﻠﺤﺎﻟﺔ ﺍﻟﻤﺜﺎﺭﺓ ﻟﺫﺭﺓ ﺍﻟﻬﻴﺩﺭﻭﺠﻴﻥ
s
−8
.10ﻓﺈﻥ ﺍﻗل ﻗﻴﻤـﺔ
ﻟﻼﺘﺤﺩﻴﺩ ﻟﻠﻁﺎﻗﺔ ﻫﻭ؟ ΔE. Δt ≥ h
h 6.6 ×10 −34 = 1.0 × − 26 J = 6.5 ×10 −8 ev. = ∆t 2D ×10 −8
≥∆E
ﻭﻫﺫﺍ ﻴﺴﻤﻰ ﻋﺭﺽ ﺍﻟﻁﺎﻗﺔ ﻟﻠﺤﺎﻟﺔ ﺍﻟﻤﺜﺎﺭﺓ. ﺍﻟﻌﻼﻗﺔ ﺒﻴﻥ ﻁﻭل ﺍﻟﻤﻭﺠﺔ ﻭﺍﻟﺘﺭﺩﺩ ﻓـﻲ ﻤﻭﺠـﺔ ﺍﻟﻤﻭﺠـﺎﺕ wave Guide
ﺘﻌﻁﻲ ﺒﺎﻟﻤﻌﺎﺩﻟﺔ:
c 2
ﺃﻭﺠﺩ ﺴﺭﻋﺔ ﺍﻟﻤﺠﻤﻭﻋﺔ
vg
=λ
v 2 − v°
ﻟﻬﺫﻩ ﺍﻟﻤﻭﺠﺎﺕ. c2 2
2
v 2 − v°
= λ2
2 2 4D 2 (v − v° ) w − w° = = D2 c2 c2 2
- ٦٢PDF created with pdfFactory Pro trial version www.pdffactory.com
4D2 2D ⇒ k2 = 2 λ λ w w = 2Dv ⇒ v = 2D k=
v2 = k = 2
w2 − w° c2
w2 vD 2
2
⇒ c k = w − w° ⇒ k = 2
2
2
2
w2 − w°
2
c
w ﺒﺎﻟﺘﻔﺎﻀل ﺒﺎﻟﻨﺴﺒﺔ ﻟـ c 2 2kdk = 2 w dw ⇒
dw c 2 k = dk w 2
vg =
w dw c 2 k c 2 w l − w ° = c 1 − °2 = = dk w w w = c 1−
v°
2
v2
.1 mev = ﺤﺯﻤﺔ ﻤﻥ ﺍﻟﻨﻴﻭﺘﺭﻭﻨﺎﺕ ﺒﻁﺎﻗﺔ ﻤﺘﻭﺴﻁﺔ ﻟﻠﻨﻴﺘﺭﻭﻥ ﺍﻟﻭﺍﺤﺩ:ﻤﺜﺎل 10 km ﺍﺤﺴﺏ ﺍﺘﺴﺎﻉ ﺍﻟﺤﺯﻤﺔ ﺒﻌﺩ ﻗﻁﻊ ﻤﺴﺎﻓﺔ،1 cu ﺒﺩﺃﺕ ﺍﻟﺤﺯﻤﺔ ﺒﺎﺘﺴﺎﻉ
.ﻓﻲ ﺍﻟﻔﺭﺍﻏﺎﺕ b b Δy Δy = 1com, ⇒ Δp y ? ∴Δp y =
,
ﺍﻻﺘﺴﺎﻉ
Δp y . Δy ≥ h
6.6 ×10 −34 D = 5.3×10 −33 kgm/sec = 2DΔy 4D ×10 − 2 m
p x = 2m n E = 2 (1.67 ×10 −27 ) (10 6 en) (1.6 × 10 −19 ) = 2.3×10 − 20 ∴tomθ =
kg . m / sec
−33
5.3 ×10 2.3 × 10 − 20
R = x R 5.3× 10 −33 9 = ton θ ⇒ R = v ton θ = 10 m × = 2 ×10 −4 m − 20 x 2.3×10
.2R ﺒﻤﻘﺩﺍﺭ1am ﻭﻋﻠﻴﻪ ﻓﺈﻥ ﺍﻟﺤﺯﻤﺔ ﺴﺘﺯﺩﺍﺩ ﻤﻥ - ٦٣ PDF created with pdfFactory Pro trial version www.pdffactory.com
2 R = 4 ×10 −4 m = 4 ×10 −2 con = 0.4
ﺴﻴﺼﺒﺢ 10.4 nm
ﺱ /ﻜﻡ ﻴﺒﻠﻎ ﻋﺩﻡ ﺍﻟﺘﺤﺩﻴﺩ )ﺍﻟﻼﺘﺤﺩﻴﺩ( ﻓﻲ ﻤﻭﻀﻊ ﻓﻭﺘﻭﻥ ﻁﻭل ﻤﻭﺠﺘﻪ
°
3000 A
ﻜﺎﻨﺕ ﺩﻗﺔ ﺘﺤﺩﻴﺩ ﻁﻭل ﺍﻟﻤﻭﺠﺔ ﻴﺼل ﺇﻟﻰ ﺠﺯﺀ ﻤﻥ ﺍﻟﻤﻠﻴﻭﻥ؟
ﺇﺫﺍ
∆λ = 10 λ
ﺱ /ﺇﺫﺍ ﻜﺎﻨﺕ ﺇﻟﻜﺘﺭﻭﻥ ﻓﻲ ﺤﺎﻟﺔ ﻤﺸﺘﺎﻗﺔ ﺃﻭﻟﻲ ﻤﺩﺓ 10-6 sﺍﺤﺴﺏ ﻋﺩﺩ ﺍﻟﻠﻔﺎﺕ ﺍﻟﺘﻲ
ﺴﻴﻠﻔﻬﺎ ﻓﻲ ﻫﺫﺍ ﺍﻟﻤﺩﺍﺭ.
13.6 2D 2Dh = = ) → hw = (E 2 = 4 w hw ) 2D (1.05 × 10 −34 =T = 1.2 ×10 −15 sec 13.6 ( ) ) (1.6 ×10 −19 4 =T
~
C 10 −8 sec = =T = 8 × 10 6 tw −15 T 1.2 × 10
ﺍﺘﺴﺎﻉ ﺍﻟﺨﻁ ﺍﻟﻁﻴﻔﻲ ﺍﻟﺫﻱ ﻟﻪ ﻁﻭل ﻤﻭﺠﺔ
°
4000 A
ﻫـﻭ
ﻤﺘﻭﺴﻁ ﺯﻤﻥ ﻤﻜﻭﺙ ﺍﻟﻨﻅﺎﻡ ﺍﻟﺫﺭﻱ ﻓﻲ ﺤﺎﻟﺔ ﺍﻟﻁﺎﻗﺔ ﺍﻟﻤﻨﺎﻅﺭﺓ. °
°
0.0001A
ﺍﺤـﺴﺏ
°
D = 4000 A , Δλ = 10 −4 A = 10 −14 m. h c ΔD hc ΔD =ΔE =hΔv = 2 D D D
ﻋﺩﻡ ﺍﻟﺘﺤﺩﻴﺩ ﻓﻲ ﺍﻟﺯﻤﻥ ، ∆tﺯﻤﻥ ﻤﻜﻭﺙ ﺍﻟﺫﺭﺓ ﻓﻲ ﺍﻟﺤﺎﻟﺔ ﺍﻟﻤﺴﺘﺜﺎﺭﺓ. 4Dh 4h D 2 = C = Δt = Δ E c D ΔD `
) (4) (16) (10 −14 = 2.1× 10 −7 sec. 8 −14 ) (3×10 ) (10 = 0.21μ sec.
=
ﺱ /ﺍﻟﻁﺎﻗﺔ ﺍﻟﻤﺴﺘﺜﺎﺭﺓ ﻟﻬﺎ ﺇﺸﻌﺎﻉ ﻴﺴﺎﻭﻱ ،1.1 evﺍﺤﺴﺏ ﻤﺘﻭﺴﻁ ﻋﻤﺭ ﻫﺫﻩ ﺍﻟﺤﺎﻟﺔ؟ ﻤﻥ ﻋﻼﻗﺔ ﺍﻟﻼﺘﺤﺩﻴﺩ
h 2D
= Δt . ΔE
- ٦٤PDF created with pdfFactory Pro trial version www.pdffactory.com
ΔE = DΔv Δt =
6.6 × 10 −34 h = = ..............? 2D ΔE 2D ×1.1ev α∞
F (x) = ∫ dk g (k) e ikx −∞
ﺤﺯﻤﺔ ﻤﻭﺠﻴﺔ ﻤﻌﺭﻓﺔ ﺒـ ﺘﻌﻁﻲ ﺒـg ( k) ﺤﻴﺙ
g (k ) = o =N
k 〈 − k/2 − k/2 〈 k 〈 k/2 4 〈k 2
=o
ﺍﻟﺘﻲ ﻟﻬﺎN ﻗﻴﻤﺔ +∞
∫ dx f ( x )
2
=1
−∞
2N kx sin x 2 +∞ 2 kx 4N ⇒ ∫ dx 2 sin 2 = 2 x −∞
Qf( x ) =
+∞
2 4N 2 sin 2 n z z 2 du . Sin u 2kN du = ∫ k 2u 2 ∫−∞ u 2 = 2D N = 1 k2 1 ⇒ N= 2Dk
ﺍﻟﺘﻲ ﻟﻬﺎN ﻜﻴﻑ ﻫﺫﻩ ﻤﺭﺘﺒﻁﺔ ﺒﺎﺨﺘﻴﺎﺭ +∞
∫ dk g ( u )
2
−∞
+∞
=
1 2D
2 ∫ dk g ( u) = N
u/z
2
−∞
− k/z
⇒N sin
kv 2
∫ dk = N
2
k=
1 2D
1 2Dk
ﺒﻴﻥ ﻨﻘﻁﺘﻴﻥ ﺍﻟﺘـﻲ ﻋﻨـﺩﻫﺎ ﺍﻟﺩﺍﻟـﺔv ﻤﻥ ﺍﻟﻤﻨﺎﺴﺏ ﺍﺨﺘﻴﺎﺭ ﻋﺭﺽ Δu =
uD 2D ⇐ ×=± u k
- ٦٥ PDF created with pdfFactory Pro trial version www.pdffactory.com
ﻭﻫﺫﺍ ﻴﺤﺩﺙ ﻋﻨﺩ.ﺘﺘﻼﺸﻰ
uD ) u = uD u
( = ∴Δ × Δk
ﻻ ﺘﻌﺘﻤﺩ ﻋﻠﻰ .u )(1
2D dv dv dv dv = = = −D2 ) dk 2D d (1/D ) d (1/D dD
C2 2 v2 − v°
= D2
2
,
2
v 2 − v°
= vg
= QD
2
2v dv dv dv c 2 v2 v° 1 ⇒ ∴ 2 = 2 + 2 ⇒ 2 =− 2 2 = dD D 3 C c c D D c2 c2 dv 2 −D2 = (−D 2 ) . ( 3 ) = = c 1 − v ° /v 2 dD D v Dv c c c 2 v2 − v° = = = c 1 v v° (v ) 2 2 2 2 2 v − v ° /v v − v°
)(1
u g = c 1 − v ° /v 2 2
ﺍﻟﻤﻌﺎﺩﻟﺔ ﺍﻟﻤﻭﺠﻴﺔ ﻟﺸﺭﻭﺩﻨﺠﺭ ﻭﺘﻔﺴﻴﺭ ﺍﻻﺤﺘﻤﺎل The Schrödinger wave Eqn. nd the Probability Interpretation
ﻓﻲ ﻫﺫﺍ ﺍﻟﻔﺼل ﺴﻨﻨﺎﻗﺵ ﺒﻌﺽ ﺨﻭﺍﺹ ﻤﻌﺎﺩﻟﺔ ﺸﺭﻭﺩﻨﺠﺭ ﻟﻠﺠﺴﻡ ﺍﻟﺤﺭ ،ﻫﺫﻩ ﺍﻟﻤﻌﺎﺩﻟﺔ ﺍﻟﺘﻲ ﺍﺴﺘﻨﺘﺠﻨﺎﻫﺎ ﻓﻲ ﺍﻟﻔﺼل ﺍﻟﺜﺎﻨﻲ .ﻭﻜﺫﻟﻙ ﺴﻨﺴﺘﻨﺘﺞ ﺘﻔـﺴﻴﺭ ﺍﻻﺤﺘﻤـﺎل
ﻟﻠﺩﺍﻟﺔ ﺍﻟﻤﻭﺠﻴﺔ Ψﻭﻫﺫﺍ ﺒﺩﻭﺭﻩ ﻴﺅﺩﻱ ﺇﻟﻰ ﺘﻌﺭﻴﻑ ﻜﻤﻴﺔ ﺍﻟﺤﺭﻜﺔ montentumﻓﻲ
ﻤﻴﻜﺎﻨﻴﻜﺎ ﺍﻟﻜﻡ ،ﻭﻜﺫﻟﻙ ﻟﻤﻌﺎﺩﻟﺔ ﺸﺭﻭﺩﻨﺠﺭ ﺍﻟﺘﻲ ﺘﺼﻑ ﺠﺴﻡ ﻓﻲ ﺠﻬﺩ ).V (x ﻭﻨﻘﻁﺔ ﺍﻟﺒﺩﺍﻴﺔ ﻫﻲ ﺍﻟﻤﻌﺎﺩﻟﺔ ﺍﻟﺘﻔﺎﻀﻠﻴﺔ. ).............(1
)∂ ψ (x, t )h 2 ∂ 2 ψ (x, t =− ∂t 2m ∂x 2
ih
ﻭﻫﺫﻩ ﻤﻌﺎﺩﻟﺔ ﺼﺤﻴﺤﺔ ﻟﻭﺼﻑ ﺤﺭﻜﺔ ﺠﺴﻡ ﺤﺭ .ﻭﺤل ﻫﺫﻩ ﺍﻟﻤﻌﺎﺩﻟﺔ ﻴﻤﻜـﻥ ﺍﻟﺤﺼﻭل ﻋﻠﻴﻪ )ﺒﻌﻜﺱ ﺍﻟﺨﻁﻭﻁ ﺍﻟﺘﻲ ﺃﺩﺕ ﺇﻟﻴﻬﺎ( ..ﻭﻨﺘﺤﺼل ﻋﻠﻰ ﺍﻟﺤل ﺍﻟﻌﺎﻡ: i(px − Et)/h
1
∫ ap ϕ (p) e 2Dh
= )ψ (x, t
- ٦٦PDF created with pdfFactory Pro trial version www.pdffactory.com
ﻭﺤﻴﺙ ﺃﻥ Eﻟﻠﺠﺴﻡ ﺍﻟﺤﺭ ﻫﻲ p2/2m i(px − (p 2 /2m)t)lh
).............(2
1
∫ ap ϕ (p) e 2Dh
= )ψ (x, t
ﻭﻤﻥ ﺍﻟﺠﺩﻴﺭ ﺒﺎﻟﻤﻼﺤﻅﺔ :ﺃﻥ ﺍﻟﻤﻌﺎﺩﻟﺔ ) (١ﻫﻲ ﻤﻌﺎﺩﻟﺔ ﺘﻔﺎﻀﻠﻴﺔ ﻤﻥ ﺍﻟﺩﺭﺠﺔ ﺍﻷﻭﻟﻰ ﻭﻫﺫﺍ ﻴﻌﻨﻲ ﺃﻨﻨﺎ ﺒﻤﺠﺭﺩ ﺃﻥ ﻨﻌﺭﻑ ﺍﻟﻘﻴﻤﺔ ﺍﻷﻭﻟﻴﺔ ﻟـ ) (ψ
) )(ψ ( x, o
ﻋﻨﺩ ﻜل ﺍﻷﺯﻤﻨﺔ ﺍﻷﺨﺭﻯ ﻴﻤﻜﻥ ﻤﻌﺭﻓﺘﻬﺎ.
ﻓـﺈﻥ ﻗﻴﻤـﺔ
)∂ψ (x, t Δt ∂t ∂ψ (x, t) i 2h 2 ∂ψ (x, t) ih ∂ψ = = i2m h ∂xH2 2m xt 2 ∂t ih ∂ 2 ψ Δt ψ ( x, t + Δt) = ψ (x, t) + 2m ∂x 2 )ih ∂ 2 ψ (x, t ψ (x, t + Δt ) = ψ (x, t ) + )Δt .................(3 2m ∂x 2 ψ (x, t + Δt) = ψ (x, t) +
ﻭﺒﺈﻋﻁﺎﺀ ﻗﻴﻤﺔ ) ψ (x, oﺍﻟﺩﺍﻟﺔ ) ϕ (pﻴﻤﻜﻥ ﺇﻴﺠﺎﺩﻫﺎ ﻤﻥ ) (2ﻀﻊ t = oﻓـﻲ
ﺍﻟﻤﻌﺎﺩﻟﺔ ) (2ﺘﻜﺎﻤل ﻓﻭﺭﻴﺭ.
).................(4
ﻴﻤﻜﻥ ﻋﻜﺴﻪ ﻟﻠﺤﺼﻭل ﻋﻠﻰ
ipx/h
1
∫ dp φ (p) e 2xh
= ) ψ (x, o
)ϕ ( p
ﺸﺭﻭﻁ ﺍﻟﻤﻌﺎﻴﺭﺓ: ϕ (P ) = 2Dh ∫ ψ (x, o ) dx e ipx/ h
ﻭﺒﺎﻟﺘﻌﻭﻴﺽ ﻓﻲ ﺍﻟﻤﻌﺎﺩﻟﺔ ) (2ﻴﻤﻜﻨﻨﺎ ﺤﺴﺎﺏ ) ψ (x, tﻋﻨﺩ ﻜل ﻗﻴﻡ ..t ﻻﺤﻅ ﺃﻨﻪ ﻻ ﻴﻭﺠﺩ "ﻻ ﺘﻌﻴﻴﻥ" ﻓﻲ ﺍﻟﻤﻌﺎﺩﻟﺔ ﺍﻟﺘﻔﺎﻀﻠﻴﺔ: ﻁﺎﻟﻤﺎ onceﻨﻌﺭﻑ ﺍﻟﻘﻴﻡ ﺍﻷﻭﻟﻰ ﻟﻠﺤﺎﻟﺔ )ﻟﻠﺤﺯﻤﺔ ﺍﻟﻤﻭﺠﺒﺔ( ﻻ ﺘﻭﺠﺩ ﻗﻴـﻭﺩ ﻋﻠﻰ ﻗﻴﻡ ) ψ (x, oﻋﻨﺩﺌﺫ ﺍﻟﺤﺯﻤﺔ ﺍﻟﻤﻭﺠﻴﺔ ﻜﻠﻬﺎ ﻤﻌﺭﻓﺔ ﻋﻨﺩ ﻜل ﺍﻷﺯﻤﻨﺔ ﺍﻟﺘﺎﻟﻴﺔ.. ﺘﻔﺴﻴﺭ ﺍﻻﺤﺘﻤﺎل ﻟﻠﺩﺍﻟﺔ ﺍﻟﻤﻭﺠﻴﺔ
) ψ ( x, t
- ٦٧PDF created with pdfFactory Pro trial version www.pdffactory.com
ﻭﻨﺤﻥ ﺒﺼﺩﺩ ﺍﻟﺒﺤﺙ ﻋﻥ ﺘﺄﻭﻴل )ﺘﻔﺴﻴﺭ( ﻟﻠﺩﺍﻟﺔ
) ψ ( x, t
ﻴﺠﺏ ﺃﻥ ﻴﻜﻭﻥ ﻓـﻲ
ﺘﻔﻜﻴﺭﻨﺎ. ) (1ﺃﻥ ) ψ (x, tﻫﻲ ﺒﺸﻜل ﻫﺎﻡ ﺩﺍﻟﺔ ﻤﻌﻘﺩﺓ .conglex fetom ) (2ﺍﻟﺩﺍﻟﺔ
) ψ (x, t 1 424 3
ﺘﻜﻭﻥ ﻜﺒﻴﺭﺓ ﺤﻴﺙ ﻴﻔﺘﺭﺽ ﺃﻥ ﻴﻜﻭﻥ ﺍﻟﺠﺴﻡ ﻤﻭﺠـﻭﺩ ﻭﺘﻜـﻭﻥ
ﻛﻤﯿﺔﺣﻘﯿﻘﯿﺔ
ﺼﻐﻴﺭﺓ ﻓﻲ ﺃﻤﺎﻜﻥ ﻏﻴﺭ ﻫﺫﺍ ﺍﻟﻤﻜﺎﻥ. ) (3ﻭﻜﺫﻟﻙ ﻤﺼﺎﺤﺏ ﻟﻠﺩﺍﻟﺔ ﻅﺎﻫﺭﺓ ﺍﻻﻨﺒﺴﺎﻁ )) (Spreadingﺍﻟﻼﺘﺤﻴﺯ(. ﻭﻤﺒﺎﺸﺭﺓ ﺒﻌﺩ ﺍﻜﺘﺸﺎﻑ ﻤﻌﺎﺩﻟﺔ ﺸﺭﻭﺩﻨﺠﺭ )ﺍﻟﺘﻲ ﺘﺒﻌﺙ ﺍﻜﺘﺸﺎﻑ ﻤﻴﻜﺎﻨﻴﻜﺎ ﺍﻟﻜﻡ
ﺒﻭﺍﺴﻁﺔ ﻫﻴﺯﺒﻨﺭﺝ ﺴﻨﺔ (١٩٢٥ﻗﺎﻡ MaxBornﺒﺩﺭﺍﺴﺔ ﺘﺒﻌﺜﺭ )ﺘﻁﺎﻴﺭ( ﺸﻌﺎﻉ ﻤﻥ
ﺍﻹﻟﻜﺘﺭﻭﻨﺎﺕ ﺒﻬﺩﻑ ﺃﻭ ﺍﻟﺘﻲ ﺃﺭﺸﺩﺘﻪ ﺇﻟﻰ ﺍﻟﺘﺄﻭﻴل ﺍﻟﺼﺤﻴﺢ ﻟﻠﺩﺍﻟﺔ ﺍﻟﻤﻭﺠﻴﺔ .ﻭﻋﻠﻴـﻪ
ﻓﻘﺩ ﺍﻗﺘﺭﺡ ﺃﻥ.
ﺘﻜﻭﻥ ﻜﺒﻴﺭﺓ ﺤﻴﻨﻤﺎ ﻴﻔﺘﺭﺽ ﺃﻥ ﻴﻜﻭﻥ ﺍﻟﺠـﺴﻡ
⇒) ⇐ p (x, t
ﺩﺍﻟـﺔ ﺤﻘﻴﻘﻴـﺔ
ﻭﺘﻔﻠﻁﺤﻬﺎ ﻻ ﻴﻌﻨﻲ ﺃﻥ ﺍﻟﺠﺴﻡ ﻤﺘﻔﻠﻁﺢ. ﻭﻜل ﻤﺎ ﺘﻌﻨﻴﻪ ﻫﻭ ﺃﻨﻪ ﻋﻨﺩﻤﺎ ﻴﺘﻐﻴﺭ ﺍﻟﺯﻤﻥ ﻴﻜﻭﻥ ﻤﻥ ﺍﻷﻗل ﺍﺤﺘﻤﺎﻻﹰ ﺃﻥ ﻨﺠـﺩ
ﺍﻟﺠﺴﻡ ﺤﻴﻨﻤﺎ ﻜﺎﻥ ﻋﻨﺩ .t = 0
ﻭﻟﻜﻲ ﻴﻜﻭﻥ ﺍﻟﺘﺄﻭﻴل ﺼﺤﻴﺤﺎﹰ ﻴﻠﺯﻤﻨﺎ: ∞x
∫ p (x, t ) dx =1
)(6
∞
ﺸﺭﻁ ﺍﻟﻤﻌﺎﻴﺭﺓ vormalization
ﻭﺫﻟﻙ ﻷﻥ ﺍﻟﺠﺴﻡ ﻴﺠﺏ ﺃﻥ ﻴﻜﻭﻥ ﻓﻲ ﻤﻜﺎﻥ ﻤﺎ ﻭﺴﻴﺘﻀﺢ ﻟﻨﺎ ﻻﺤﻘﺎﹰ ﺃﻥ ﻴﻜﻔﻲ ﺃﻥ ﻴﻠﺯﻡ ﺃﻥ: ) (7
∞〈
2
∞x
) ∫ dx ψ (x, o
∞
- ٦٨PDF created with pdfFactory Pro trial version www.pdffactory.com
ـﺔ ـﺔ ﻤﺘﻜﺎﻤﻠـ ـﻭﻥ ﻤﺭﺒﻌـ ـﺏ ﺃﻥ ﺘﻜـ ـﺔ ) ψ (x, oﻴﺠـ ـﻲ ﺃﻥ ﺍﻟﺩﺍﻟـ ـﺫﺍ ﻴﻌﻨـ ﻭﻫـ squaredintegrabliﻭﺒﻨﻁﺎﻕ ﺘﻜﺎﻤل
∞ + ∞←−
ﻴﺠﺏ ﺃﻥ ﺘﺅﻭل ﺇﻟﻰ ﺼﻔﺭ ﺃﺴﺭﻉ ﻤﻥ ﻤﺘﺼﻠﺔ ﻓﻲ x
) ψ (x, t
ﻴﻌﻨﻲ ﺃﻥ ).ψ (x, o ﻭﻫﺫﺍ ﻴﻠـﺯﻡ ﺃﻥ ﺘﻜـﻭﻥ ﺍﻟﺩﺍﻟـﺔ
x −1 / 2
.contineon sin x.
ﺃﻫﻤﻴﺔ ﺍﻷﻁﻭﺍﺭImport on ce of the ases :
ﻴﺒﺩﻭ ﻟﻠﻭﻫﻠﺔ ﺍﻷﻭﻟﻰ ﻜﻭﻥ ﺍﻟﺩﺍﻟﺔ
2
ﺫﺍﺕ ﻤﻌﻨﻰ ﻓﻴﺯﻴـﺎﺌﻲ ،ﻴﺒـﺩﻭ ﺃﻥ
) ψ (x, t
ﺍﻟﻁﻭل ﻟﻴﺱ ﻟﻪ ﺃﻫﻤﻴﺔ .ﻭﻫﺫﺍ ﺨﻁﺄ. ﺤل.
ﻭﺒﻤﺎ ﺃﻥ ﺍﻟﻤﻌﺎﺩﻟﺔ ) (1ﻫﻲ ﻤﻌﺎﺩﻟﺔ ﺨﻁﻴﺔ) ،ﻭﺒﺎﻗﺘﺭﺍﻥ ﺃﻥ )ψ (x, t ) = ψ1 (x, t) + ψ 2 (x, t
)(8
ﻟﻭ ﺃﻥ
iθ1
ﺤﻴﺙ ﺃﻥ
)ψ 2 (x, t ), ψ1 (x, t
ﻫﻲ
ﻫﺫﺍ ﻜﺫﻟﻙ ﺤل.
←
ψ 2 (x, t ) = R 2 e iθθ , ψ1 (x, t) = R 1 e R 1 , R 2 , θ1 , θ 2
ﻫﻲ ﺤﻘﻴﻘﺔ ،ﻋﻨﺩﻫﺎ 2
)
) i (θ 2 − θ 1
(R1 +R 2 e
iθ 1
= e
2
) ψ (x, t
) = R 1 + R 2 + 2R 1 cos (θ 1 − θ 2 2
)(9
2
ﻭﻫﺫﺍ ﻴﻅﻬﺭ ﺃﻥ ﺍﻟﻁﻭﺭ ﻤﻬﻡ ،ﺍﻟﻁﻭﺭ ﺍﻟﻜﻠﻲ ﻓﻲ ) ψ (x, tﻴﻤﻜﻥ ﺇﻫﻤﺎﻟـﻪ ،ﺃﻤـﺎ ﺍﻟﻁﻭل ﺍﻟﻨﺴﺒﻲ
) (θ 1 − θ 2
ﺒﻴﻥ ﺩﺍﻟﺘﻴﻥ ﻤﻭﺠﺒﺘﻴﻥ
ψ1
و ψ2
ﺘﻅﻬﺭ ﻓﻲ
2
ψ
.
ﻻﺤﻅ ﺃﻥ ﺍﻟﻁﺭﻑ ﺍﻷﻴﻤﻥ ﻓﻲ ﺍﻟﻤﻌﺎﺩﻟﺔ ) (9ﻫﻭ ﺍﻟﺫﻱ ﻨـﺭﺍﻩ ﻋﻨـﺩ ﺘﺭﺍﻜـﺏ
ﺍﻟﻤﻭﺠﺎﺕ .ﻭﻓﻲ ﺍﻟﺤﻘﻴﻘﺔ ﺃﻨﻬﺎ ﺍﻟﺨﻁﻭﺓ lineallyﻟﻠﺩﻭﺍل ﺍﻟﻤﻭﺠﺒﺔ ﺍﻟﺘﻲ ﺃﺩﺕ ﻟـ ﻟﺸﻜل ﺍﻟﺘﺩﺍﺨل ﺍﻟﺫﻱ ﻴﻅﻬﺭ ﻤﻥ ﺨﻼل cosineﻓﻲ ﻫﺫﻩ ﺍﻟﻤﻌﺎﺩﻟﺔ. ﻭﻋﻨﺩﻤﺎ ﻨﺘﺤﺩﺙ ﻋﻥ ﺍﻟﺴﻠﻭﻙ ﺍﻟﻤﻭﺠﻲ ﻟﻺﻟﻜﺘﺭﻭﻨﺎﺕ ،ﺍﻟﻔﻭﺘﻭﻨﺎﺕ ،ﻓﺘﺤﺩﺙ ﻤﻥ
ﺍﻟﺨﻁﻴﺔ...
- ٦٩PDF created with pdfFactory Pro trial version www.pdffactory.com
Peobalrility current ﺍﺤﺘﻤﺎل ﺍﻟﺘﻴﺎﺭ
.(1) ﺍﻟﻤﺭﺍﻓﻕ ﺍﻟﻤﻌﻘﺩ ﻟﻠﻤﻌﺎﺩﻟﺔcomplex conjugate ﻟﻨﺄﺨﺫ ∂ψ h2 ∂2ψ =− ⇒ 2m ∂t 2 ∂t ∂ψ * (x, t) h 2 ∂ 2 ψ * (x, t) ⇒ − ih =− 2m ∂t ∂t 2
ih
∂ψ ih ∂ 2 ψ = ∂t 2m ∂x 2 ih ∂ 2 ψ * ∂ψ * ⇒ =− 2m ∂x 2 ∂t
ψ 2
644744 8 ∂ ∂ p (x, t) = ( ψ (x, t) ψ x (x, t) ) ∂t ∂t ∂ψ * (x, t) ∂ψ ψ + ψ* = ∂t ∂t * ih ∂ψ * ih ∂ 2 ψ ψ = − 2 2m ∂x 2 2m ∂x 2 1 h 2 ∂ 2ψ* ∂ 2ψ * h ψ ψ − ih 2m ∂x 2 2m ∂x t ∂ ∂ p (x, t) = − ∂t ∂x ∂ ∂ j (x, t) ⇒ t (x, t) = − ∂t ∂x
=
+∞
∂ dx p (x, t) ∂x −∫∞
ﻴﺘﻌـﻭﺽ ﺒـﺎﻟﺘﻐﻴﺭ
x
+∞
= − ∫ dx −∞
∂ j (x, t) = o ∂x
= − [J (− ∞ ) − j (− ∞ ) ]= 0 , j → 0
ﺍﻟﺘﻐﻴﺭ ﻓﻲ ﺍﻟﻜﺜﺎﻓﺔ ﻓﻲ ﻤﻨﻁﻘﺔ ﻓـﻲ:ﺘﺒﻴﻥ
⇐
(12) ﻤﻌﺎﺩﻟﺔ
.ﺍﻟﺼﺎﻓﻲ ﻓﻲ ﺍﻟﻔﻴﺽ ﻓﻲ ﺘﻠﻙ ﺍﻟﻤﻨﻁﻘﺔ ∞
ﺘﺘﻼﺸﻰ ﻋﻨﺩsquare integreble ﺍﻟﺩﻭﺍل ﻤﺘﻜﺎﻤﻠﺔ ﺍﻟﺘﺭﺒﻴﻊ
Q
∞ ∂ ∂ ax p(x, t) dx j (x, t) = − ∫ ∫ ∂t ∂x −∞
= j (a, t) − j (b, t)
(1) ﻭﻗﺎﻨﻭﻥ ﺍﻟﺒﻘﺎﺀ ﺘﻜﻭﻥ ﺼﺤﻴﺤﺔ ﻟـﻭ ﺍﻟﻤﻌﺎﺩﻟـﺔ
(14)
j (x, t) , p(x, t )
- ٧٠ PDF created with pdfFactory Pro trial version www.pdffactory.com
ﻭﺘﻌﺭﻴﻑ ﻟـ
ﺘﺘﻐﻴﺭ ﻟـ
ﻭﻫﺫﻩ ﺍﻟﺘﻌﺭﻴﻔﺎﺕ ﻟـ ) d (x, t ), p (x, tﻭﻗﺎﻨﻭﻥ ﺍﻟﺤﻔﻅ )ﺍﻟﺒﻘﺎﺀ( ﺘﺒﻘـﻰ ﺼـﺤﻴﺤﺔ ﺤﺘﻰ ﻟﻭ ﻏﻴﺭﻨﺎ ﻤﻌﺎﺩﻟﺔ ﺍﻟﺠﺴﻡ ﺍﻟﺤﺭ ﻟﻠﻤﻌﺎﺩﻟﺔ ﺍﻟﺘﺎﻟﻴﺔ ﻤﻌﺎﺩﻟﺔ ﺸﺭﻭﺩﻨﺠﺭ. )(15
ﺃﺒﻌﺎﺩ.
) ∂ ψ (x, t )h 2 ∂ 2 ψ (x, t =− )+ ψ (x, t ∂t 2m ∂x
it
ﻭﻫﺫﻩ ﻤﻬﻤﺔ ﻷﻨﻬﺎ ﺘﻤﺜل ﻤﻌﺎﺩﻟﺔ ﺸﺭﻭﺩﻨﺠﺭ ﻟﺠﺴﻡ ﻓﻲ ﺠﻬﺩ ) V(xﻓﻲ ﺜﻼﺜـﺔ ψ (x, y, z, t), ) V (x, y, z ∂2 ∂2 ∂2 + + ⇒∇2 ∂x 2 ∂y 2 ∂z 2
ﻜﺫﻟﻙ ﺘﻌﻤﻴﻡ )(12 ∂ p (r, t ) + ∇ . J (r, t) = 0 ∂t 2
) p (r, t ) = ψ (r, t
)∂ψ (x, t h2 2 =− ) ∇ ψ (r, t ) + v (r ) ψ (r, t ∂t 2m h = ) j (r, t ) ψ * (r, t ) ∇ψ (x, t ) − ∇ψ * (r, t ) ψ (r, t 2im
ih
[
]
ﺍﻟﻘﻴﻡ ﺍﻟﻤﺘﻭﻗﻌﺔ ﻭﻜﻤﻴﺔ ﺤﺭﻜﺔ ﺍﻟﺠﺴﻡ Expectation Values of Canticle nonctn
ﺒﺈﻋﻁﺎﺀ ﻜﺜﺎﻓﺔ ﺍﻻﺤﺘﻤﺎل.P (x,t) . ﻓﺈﻥ ﺍﻟﻘﻴﻤﺔ ﺍﻟﻤﺘﻭﻗﻌﺔ ﻟـ ) f (xﻴﻤﻜﻥ ﺤﺴﺎﺒﻬﺎ ﺒﺸﻜل ﻋﺎﻡ. )(20
) f (x ) = ∫ ax f (x ) p (x, t
) = ∫ ax ψ * (x, t ) f (x ) ψ (x, t
ﻻﺴﺘﻨﺘﺎﺝ ﻤﺅﺜﺭ ﻜﻤﻴﺔ ﺍﻟﺤﺭﻜﺔ ﻤﻥ ﺍﻟﻤﻴﻜﺎﻨﻴﻜﺎ
- ٧١PDF created with pdfFactory Pro trial version www.pdffactory.com
p = mv = m
dx dt
(21)
d d x = m ∫ dx ψ * 9x, t) v ψ (x, t ) dt dt +∞ ∂ψ ∂ψ * ) xψ + ψ * x p (x ) = m ∫ dx ( ∂Z ∂Z −∞
p =m
+∞
p (x ) = m ∫ dx ( −∞
،ﻫﻲ ﺍﻟﺘﻲ ﺘﺘﻐﻴﺭ ﺒﺎﻟﺯﻤﻥ
ψ (x, t )
(22)
∂ψ * ∂ψ ) + ψ (x, t ) + ψ * x ∂t ∂Z
ﻓﻘﻁ.ﺘﺤﺕ ﺍﻟﺘﻜﺎﻤل
.ﻤﻊ ﺍﻟﺯﻤﻥ
x
dx dt
ﻻﺤﻅ ﺃﻨﻪ ﻻ ﻴﻭﺠﺩ
ﻭﻫﺫﺍ ﺍﻟﺘﻐﻴﺭ ﻫﻭ ﺍﻟﺫﻱ ﻴﺅﺩﻱ ﺇﻟﻰ ﺍﻟﺘﻐﻴﺭ ﻓﻲ .ﻭﺒﻤﺎ ﺃﻥ ﺒﺎﻟﺘﻌﻭﻴﺽ ﻓﻲ ﺁﺨﺭ ﻤﻌﺎﺩﻟﺔ
∂ψ h ∂ψ =− ∂t 2m ∂x 2 ∂ψ h ∂ 2ψ m =− ∂t 2i ∂x 2 ∂ψ * h ∂ 2 ψ * m = ∂t 2i ∂x 2
ih
2 h +∞ ∂ 2 ψ * * ∂ ψ xψ − ψ x 2 )dx p = ∫( 2i −∞ ∂x 2 ∂x
∂ 2 ψ* ∂ ∂ψ * ∂ψ * ∂ψ * ∂ψ xψ ( xψψ ψ x = − − x 24 x ∂x ∂x ∂x ∂4 ∂3 ∂x 2 1 ∂x * ∂ψ ψ ∂x ∂x
∂ ∂ψ * ∂ ∂ψ ( xψψ− (ψ * ψ) + ψ * ∂x ∂x ∂x ∂x 2 ∂ ∂ψ ∂ψ ∂ ψ − (ψ * x ) + ψ * + ψ*x 2 ∂x ∂x ∂x ∂x =
.ﻭﻋﻠﻴﻪ ﻓﺈﻥ ﺍﻟﺤﺩ ﺍﻟﺫﻱ ﻴﻘﻊ ﺩﺍﺨل ﺍﻟﺘﻜﺎﻤل ﻴﺼﺒﺢ 2 ∂ 2 ψ* ∂ ∂ψ * ∂ ∂ψ * ∂ ψ (xψx − ψ x = ( x ψ) − (ψ * ψ) + ψ * 2 2 ∂x ∂x ∂x ∂x ∂x ∂x ∂ ∂ψ ∂ψ − ( ψ* x ) + ψ* ∂x ∂x ∂x * ∂ ∂ψ ∂ψ ∂ψ = xψ − ψ * ψ − ψ * x + 2ψ * ∂x ∂x ∂x ∂x
- ٧٢ PDF created with pdfFactory Pro trial version www.pdffactory.com
+∞
+∞
h ∂ p = ∫ [ 2i − ∞ ∂x
]+ h ∫ 2ψ * (x, t) ∂ ψ (x, t) dx 2i −∞ ∂x
+∞
p = ∫ dx ψ * (x, t) −∞
h ∂ ψ (x, t) i ∂x
.operator ﻫﺫﻩ ﺍﻟﻤﻌﺎﺩﻟﺔ ﺘﻘﺘﺭﺡ ﺃﻥ ﻜﻤﻴﺔ ﺍﻟﺤﺭﻜﺔ ﻴﻤﻜﻥ ﺃﻥ ﺘﻤﺜل ﺒﺎﻟﻤﺅﺜﺭ p=
h ∂ i ∂x
. ﻋﻠﻰ ﺴﺒﻴل ﺍﻟﻤﺜﺎل،ﻭﻟﻬﺫﺍ p 2 = ∫ dxψ * (x, t) (−h 2
∂2 ) ψ (x, t) ∂x 2
f ( p ) = ∫ dxψ * (x, t) f ( φ ( p)
D ∂ ) ψ (x, t) t ∂x
.ﻭﺒﺸﻜل ﻋﺎﻡ
ﻭﺒﻬﺫﺍ ﺍﻟﺘﻤﺜﻴل ﻟﻜﻤﻴﺔ ﺍﻟﺤﺭﻜﺔ ﻴﻤﻜﻨﻨﺎ ﻤﻨﺎﻗﺸﺔ ﺍﻟﻤﺩﻟﻭل ﺍﻟﻔﻴﺯﻴـﺎﺌﻲ ﻟــ
.ﺍﻟﺘﻲ ﺘﻅﻬﺭ ﻓﻲ ﺍﻟﻤﻌﺎﺩﻟﺔ ﺍﻟﺩﺍﻟﺔ ﺍﻟﻤﻭﺠﻴﺔ ﻓﻲ ﻓﺭﺍﻍ ﻜﻤﻴﺔ ﺍﻟﺤﺭﻜﺔ 1
∫ 2Dh
ψ(x, t )=
ap φ ( p ) e i [px - (p/2m) t ] / h
.t ﻻ ﺘﻌﺘﻤﺩ ﻋﻠﻰ ψ (x ) =
1
∫ 2Dh
ap φ ( p ) e ipn/ h =
ﻭﺫﻟﻙ ﻷﻥt = 0 ﺩﻉ:ﺃﻭﻻﹰ
φ (p )
h ou φ (hD ) e ikv 1D ∫
ﻭﺒﺎﺴﺘﺨﺩﺍﻡ ﻤﻌﻜﻭﺱ ﺘﻜﺎﻤل ﻓﻭﺭﻴﺭ φ ( hD ) = ∫ axψ ( x) e −ikx or ⊗ φ (p) =
1 2Dh
∫
ax ψ (x) e - ipx/ h
⇒ ∫ ap φ * ( p) φ ( p) = ∫ ap φ * ( p) = ∫ ax ψ
1
1 2Dh
∫
ax ψ (x) e - ipx/ h
∫
ap φ * e - ipx/ h Dh4424443 124 ψ*
= ∫ ax ψψ * = 1
- ٧٣ PDF created with pdfFactory Pro trial version www.pdffactory.com
ﻭﻫﺫﻩ ﺍﻟﻨﺘﻴﺠﺔ ﺘﺴﻤﻰ ﺒﻨﻅﺭﻴﺔ ﺒﺎﺭﺴﻴﻔﺎل Parsevali
ﻭﻤﻨﻁﻭﻗﻬﺎ :ﻟﻭ ﺃﻥ ﺩﺍﻟﺔ ﻤﻌﺎﻴﺭﺓ ﻟـ :1ﻓﺈﻥ ﺘﺤﻭﻴل ﻓﻭﺭﻴﺭ ﻟﻠﺩﺍﻟﺔ ﻴﻜﻭﻥ ﻤﻌﺎﻴﺭ ﻜﺫﻟﻙ If a function is normalized to 1, 50 is its Fourier Transform
) ﻋﻨﺩﻨﺎ ﺘﻤﺜﻴﻼﻥ ﻟﻠﺩﻭﺍل :ﺘﻤﺜﻴل ،xﺘﻤﺜﻴل .(regasrauney p ﺨﺫ: h d )ψ (x i dx h d 1 ) = ∫ ax ψ * ( x ap φ (p) e ipx/h ∫ i dx 2Dh 1 = ∫ ap φ ( p) p ax ψ * (x) e ipx/ h ∫ 2Dh
) p = ∫ ax ψ * ( x
)p = ∫ ap φ ( p ) p φ * (p
ﻭﻫﺫﻩ ﺍﻟﻨﺘﻴﺠﺔ ﺘﻌﻨﻲ ﺃﻥ
ﺍﻟﺤﺭﻜﺔ ﺒـ
2
) φ( p
) φ (p
ﺘﻔﺴﻴﺭ ﺒﺄﻨﻬﺎ ﺍﻟﺩﺍﻟﺔ ﺍﻟﻤﻭﺠﻴﺔ ﻓﻲ ﻓـﺭﺍﻍ ﻜﻤﻴـﺔ
ﺘﻌﻁﻴﻨﺎ ﻜﺜﺎﻓﺔ ﺍﺤﺘﻤﺎل ﺇﻴﺠﺎﺩ ﺍﻟﺠﺴﻡ ﺍﻟﺫﻱ ﻜﻤﻴﺔ ﺤﺭﻜﺘﻪ .p
ﻓﻲ ﻓﺭﺍﻍ ﻜﻤﻴﺔ ﺍﻟﺤﺭﻜﺔ xﻟﻬﺎ ﺍﻟﺘﻤﺜﻴل
∂ ∂p
x = ih
ﺴﺘﺠﺩ ﻓﻴﻤﺎ ﺒﻌﺩ ﺃﻥ ﺍﻟﻤﺅﺜﺭﺍﺕ
ogenatrsﺘﻠﻌﺏ ﺩﻭﺭﺍﹰ ﻤﻬﻤﺎﹰ ﻓﻲ ﻤﻴﻜﺎﻨﻴﻜﺎ ﺍﻟﻜﻡ. ﻭﺍﻵﻥ ﺴﻨﻭﻀﺢ ﺒﻌﺽ ﺨﺼﺎﺌﺼﻬﺎ. ﻋﻠﻰ ﻋﻜﺱ ﺍﻷﺭﻗﺎﻡ ﺍﻟﻌﺎﺩﻴﺔ ﺍﻟﻤﺅﺜﺭﺍﺕ ﻟﻴﺴﺕ ﺒﻨﺩ ﺩﺍﺌﻤﺎﹰ ،ﻟﻭ ﻋﺭﻓﻨﺎ
[ A B ] = AB − BA 1
∂ )D∂ ψ (x, t x ψ (x, t ) − x i ∂x i ∂x D ∂ψ ∂ψ = (x +ψ ( x, t ) − x ) i ∂x ∂x ) [ p, x ] ψ (x, t) = h ψ (x, t i
[ p, x ] ψ (x, t) = h
ﺤﻴﻨﺌﺫ:
ﻭﻋﻠﻴﻪ ﻨﻘﻭل ﺃﻨﻨﺎ ﻋﻨﺩﻨﺎ ﻋﻼﻗﺔ ﺍﻟﺘﺒﺩﻴل Commutation relatively
- ٧٤PDF created with pdfFactory Pro trial version www.pdffactory.com
[ p, x ] = h i
normalized ware fumets ﺨﺫ ﺠﺴﻡ ﺍﻟﺫﻱ ﺩﺍﻟﺘﻪ ﺍﻟﻤﻌﺎﻴﺭﺓ:ﻤﺜﺎل ψ (x ) = 2 α α v e − αv v > 0 =
v0
= a/h e −pa/ h p2 p
2
= =
2a h
∫
∞
∫
p 2 e − pa/h ap
0
φ * p 2 φ ap
ap =z h p
2
p>0
p2 = z2
2a h 3 ( ) = h a = 2(
p =0 Δ p = 2h/a
∞
∫
h2 a2
z 2 e − z az
0
3
h h ) Γ (3) = 4 ( ) 2 a a Γ (n + 1 ) = n ! Γ (2 + 1) 2! Δp =
p2
Δx =
Δx
- ٧٩ PDF created with pdfFactory Pro trial version www.pdffactory.com
.ﺃﺜﺒﺕ ﺃﻥ ﺍﻟﻤﺅﺜﺭ
ﻟﻭ ﺃﻥ ﺍﻟﺩﺍﻟﺔ ﺍﻟﻤﻭﺠﻴﺔ ﺘﺤﻘﻕ ﺍﻟﺸﺭﻁ/ﺱ
ψ (D ) = ψ ( − D )
∠=
h d i dθ
ﻤﺅﺜﺭ ﻫﻴﺭﻤﻴﺘﻲ ∠ − ∠ D
∫
∠ =
dθ ψ * ( θ )
−D D
∠
*
=−
∫
dθ ψ
−D
∠ − ∠ = *
*
ﻹﺜﺒﺎﺕ ﻫﺫﺍ ﻴﺠﺏ ﺃﻥ
=0
h d ψ i dθ
h d * ψ i dθ
d d D h * dθ dθ + dθ ( ψ ( θ ) ψ ) (ψ ψ * ) ∫ i −D
=
h i
D
∫
dθ
−D
d ( ψ* ψ ) dθ
[
h * = ψ ( D ) ψ ( D ) − ψ * ( − D ) ψ (D ) i ψ (D ) =ψ ( − D ) = 0
]
∠ = ∠ ∞
3.c.
∫
k/2
dk N 2 = N 2
−∞
∫
dk = N 2 k =
− k/2
1 ⇒ 2D
N=
*
∴ ﺍﻟﻤﺅﺜﺭ
1 2D k
. ﻓﻲ ﺍﻟﺤﺎﻟﺘﻴﻥ ﻤﺘﺴﺎﻭﻴﺔN ﻭﻋﻠﻴﻪ ﻓﺈﻥ ﻗﻴﻡ ( ﻴﻤﻜﻥ ﺘﻘﺩﻴﺭﻩ ﺒﺎﻟﻤﺴﺎﻓﺔ ﺒﻴﻥ ﻨﻘﻁﺘـﻴﻥ ﻋﻨـﺩﻫﺎp) ﻋﺭﺽ ﺍﻟﺩﺍﻟﺔ ﻤﻥ ﺍﻟﺠﺯﺀ : ﻭﻫﺫﺍ ﻴﺤﺩﺙ ﻋﻨﺩ3.d ﺘﺨﺘﻔﻲ sin
.k ﻭﻫﻲ ﻨﺘﻴﺠﺔ ﻻ ﺘﻌﺘﻤﺩ ﻋﻠﻰ
2D kx x = + k =0 2D 2 x = − k
⇒ Δx=
4D k
⇒ Δ x. Δ k =
, 4D . k
- ٨٠ PDF created with pdfFactory Pro trial version www.pdffactory.com
k u Δk = − (− ) = k 2 2
k = 4D
sin
kx 2
ﺍﻟﺩﻭﺍل ﺍﻟﺨﺎﺼﺔ ﻭﺍﻟﻘﻴﻡ ﺍﻟﺨﺎﺼﺔEigenfunctions and Eigenvalues :
ﻟﻭ ﺃﺨﺫﻨﺎ ﻤﻌﺎﺩﻟﺔ ﺸﺭﻭﺩﻨﺠﺭ ﺍﻟﺘﻲ ﺘﻌﺘﻤﺩ ﻋﻠﻰ ﺍﻟﺯﻤﻥ: )∂ψ (x, t )h 2 ∂ 2 ψ(x, t =− )+ v (x) ψ (x, t ∂t 2m ∂x 2
)(1
ih
ﻭﻟﺤل ﻫﺫﻩ ﺍﻟﻤﻌﺎﺩﻟﺔ ﻨﻘﺼﻬﺎ ﻟﻤﻌﺎﺩﻟﺘﻴﻥ ﻨﻕ )ψ (x, t) , T(t) u (x )dT (t) h 2 d 2 u (x = − + v (x) u (x) 2 dt 2m dx
)⇒ ih u (x
ﺒﺎﻟﻘﺴﻤﺔ ﻋﻠﻰ )u (x) T (t )2m) (a 2 u (x) / dx 2 ) + V (x)u (x )u (x
(sin 1
dT (t)dt ⇒ ih = )T (t
ﻭﻫﺫﺍ ﻜﻤﻴﺔ ﻴﻤﻜﻥ ﺘﺤﻘﻴﻘﻪ ﻓﻘﻁ ﺍﻓﺘﺭﺍﺽ ﺃﻥ ﻜﻼ ﺍﻟﻁﺭﻓﻴﻥ ﻴﺴﺎﻭﻱ ﺜﺎﺒﺕ ﻭﺍﻟﺫﻱ ﻨﺴﻤﻴﻪ E )dT (E i1 i = ∫ − E dt = − Et h )T (E ih
∫
)d T (t ⇒ )= E T (t dt
⇒ ih
i
− ET i lm T (t) = - Et ⇒ T (t) = Ce t h
ﻭﺍﻟﻤﻌﺎﺩﻟﺔ ﺍﻟﺜﺎﻨﻴﺔ: )h 2 d 2 u (x )+ v (x) u (x) = Eu (u 2m dx 2
−
ﻤﻌﺎﺩﻟﺔ ﺸﺭﻭﺩﻨﺠﺭ ﺍﻟﻐﻴﺭ ﻤﻌﺘﻤﺩﺓ ﻋﻠﻰ ﺍﻟﺯﻤﻥ time indgader S-E :ﻭﻫـﺫﻩ
ﺍﻟﻤﻌﺎﺩﻟﺔ ﺘﻔﺭﻕ ﻋﻥ ) (1ﻭﺍﻟﺘﻲ ﺘﺼﻑ ﺍﻟﺘﻐﻴﺭ ﺍﻟﺯﻤﻨﻲ ﻟﻠﺩﺍﻟﺔ
)ψ (x, t
ﻤﻌﺎﺩﻟﺔ ﻫﻲ ﺩﺍﻟﺔ
ﺨﺎﺼﺔ .Ligea valu equation .ﻭﻟﻔﻬﻡ ﻤﺎﺫﺍ ﻴﻌﻨﻲ ﻫﺫﺍ ﺍﻟﻜﻼﻡ ﺩﻋﻨﺎ ﻨﻌـﺩﺩ ﻤـﺭﺓ ﺃﺨﺭﻯ ﻟﻤﻔﻬﻭﻡ ﺍﻟﻤﺅﺜﺭﺓ.
ﺒﺸﻜل ﻋﺎﻡ ﻤﺅﺜﺭ ﻤﺎ :ﻴﺅﺜﺭ ﻋﻠﻰ ﺩﺍﻟﺔ ﻭﻴﺄﺨﺫﻫﺎ ﻟﺩﺍﻟﺔ ﺃﺨﺭﻯ ،ﻤﺜﺎل ﻋﻠﻰ ﺫﻟﻙ:
- ٨١PDF created with pdfFactory Pro trial version www.pdffactory.com
of (x) = f (x) + x 2
1.
] )of (x) = [f (x
2.
)of (x) = f (3 x 2 + 1
3.
2
3
d f (x) 4. of (x) = dx )5. of (x) = df / dx − 2f (x )6. of (x) = λf (x
ﻜل ﻫﺫﻩ ﺍﻷﻤﺜﻠﺔ ﺘﺅﺜﺭ ﻓﻲ ﺍﻟﺨﺎﺼﻴﺔ ﻫﻲ ﺒﺈﻋﻁﺎﺀ ﺩﺍﻟﺔ ) f (xﺘﻭﺠﺩ ﻗﺎﻋﺩﺓ ﺍﻟﺘﻲ ﺘﺤﺩﺩ ) of (xﻟﻨﺎ. ﻴﻭﺠﺩ ﻨﻭﻉ ﺨﺎﺹ ﻤﻥ ﺍﻟﻤﺅﺜﺭﺍﺕ ﻴﺴﻤﻰ ﺍﻟﻤﺅﺜﺭﺍﺕ ﺍﻟﺨﻁﻴﺔ. ﻭﻫﺫﻩ ﺍﻟﻤﺅﺜﺭﺍﺕ ﻟﻪ ﺍﻟﺨﺎﺼﻴﺔ: ] )∠ [f 1 (x) + f 2 (x) ] = ∠ [f 1 (x) + ∠ f 2 (x
ﻭﺒﻭﺠﻭﺩ cﻜﺭﻗﻡ ﻤﻭﺠﺏ ﺍﺨﺘﻴﺎﺭﻱ. ﻤﻼﺤﻅﺔ :ﻴﻭﺠﺩ ﻤﺅﺜﺭﺍﺕ ﻋﻜﺴﻴﺔ ﺍﻟﺨﻁﻴﺔ ﻭﻟﻬﺎ )∠ cf (x) = c * ∠ f (x
ﺍﻟﺨﻁﻴﺎﻥ.
)∠cf(x) = c * ∠ f(x
ﻭﻋﻠﻴﻪ ﻓﺈﻨﻪ ﻓﻲ ﺍﻟﻘﺎﺌﻤﺔ ﺍﻟﺴﺎﺒﻘﺔ ﻓﺈﻥ ﺍﻟﻤﺅﺜﺭﻴﻥ ﺍﻷﺨﻴﺭﻴﻥ ﻫﻤـﺎ
ﺍﻟﻤﺅﺜﺭ ﺍﻟﺨﻁﻲ ﺴﻴﻨﻘل ﺩﺍﻟﺔ ﻤﺎ ﻟﺩﺍﻟﺔ ﺃﺨﺭﻯ. )df (x )− 2 f (x ax
= )∠ f (x
ﻭﻴﻤﻜﻨﻨﺎ ﺘﺸﺒﻪ ﺍﻟﺩﻭﺍل ﻜﻤﺘﺠﻬﺎﺕ ﻓﻲ ﺜﻼﺜﺔ ﺃﺒﻌﺎﺩ ﻭﺘﺄﺜﻴﺭ ﺍﻟﻤﺅﺜﺭﺍﺕ ﻴﻜﻭﻥ ﺒﻨﻘل )ﺘﺤﻭﻴل( ﺍﻟﻤﺘﺠﻪ ﻟﻤﺘﺠﻪ ﺁﺨﺭ ،ﻭﻴﺄﺨﺫ ﺍﻟﻤﺘﺠﻪ ﻜﻭﺤﺩﺓ ) (amitvedreﻭﻤـﺅﺜﺭ ﻴﻨﻘـل
ﻨﻘﻁﺔ ﻋﻠﻰ unit sphereﻟﻨﻘﻁﺔ ﺃﺨﺭﻯ ﻭﻤﺅﺜﺭ ﻤﺎ ﻴﻤﻜﻥ ﺍﻋﺘﺒـﺎﺭﻩ ﺒﻘﺎﺒـل ﺩﻭﺭﺍﻥ ﺍﻟﻨﻘﻁﺔ ﺤﻭل ﻤﺤﻭﺭ zﺒﺜﻼﺜﻴﻥ . 30 °
ﺜﻼﺙ ﺤﺎﻻﺕ:
A → A′
- ٨٢PDF created with pdfFactory Pro trial version www.pdffactory.com
B → B′ C → C′
ﻻﺤﻅ ﺃﻥ ﺍﻟﻨﻘﻁﺔ ﻋﻠﻰ ﺍﻟﻘﻁﺏ ﺘﻨﻘل ﻋﻠﻰ ﻨﻔﺴﻬﺎ )ﻭﻜﺫﻟﻙ ﺍﻟﻨﻘﻁﺔ ﺍﻟﺘﻲ ﻋﻠـﻰ
ﺍﻟﻘﻁﺏ ﺍﻟﺠﻨﻭﺒﻲ( ﻭﻫﺫﻩ ﺍﻟﺤﺎﻟﺔ ﺍﻟﺨﺎﺼﺔ ﻟﻤﺜﺎل ﻤﺅﺜﺭ ﻤﺜل ﻤﻌﺎﺩﻟﺔ ) ( 4 – 6ﻭﺍﻟـﺫﻱ ﻴﻤﻜﻥ ﻜﺘﺎﺒﺘﻪ ﻜﻤﺎ ﻴﻠﻲ: )(4-10
)u E (x
H u E (x) = E
ﻤﻨﻁﻭﻕ ﺍﻟﻤﻌﺎﺩﻟﺔ ،H :ﻤﺅﺜﺭ ﺍﻟﻬﺎﻤﻠﺘﻴﻭﻨﻴﺎﻥ ﻴﺅﺜﺭ ﻋﻠﻰ ﻨﻭﻑ ﻤﻌﻴﻥ ﻤﻥ ﺍﻟﺩﻭﺍل
ﺴﻴﻌﻁﻴﻨﺎ ﻤﻥ ﺠﺩﻴﺩ ﺍﻟﺩﺍﻟﺔ ﺍﻟﺘﻲ ﻴﺅﺜﺭ ﻋﻠﻴﻬﺎ ﻤﻀﺭﻭﺒﺔ ﻓﻲ ﺜﺎﺒﺕ.
ﻫﺫﺍ ﺍﻟﺜﺎﺒﺕ ﻴﺴﻤﻰ eigeulalwﻭﻤﻥ ﺍﻟﻤﻌﺎﺩﻟﺔ ﻴﻌﺘﻤﺩ ﻋﻠﻰ Eﻭﺤل ﺍﻟﻤﻌﺎﺩﻟـﺔ ﻴﻌﺘﻤﺩ ﻋﻠﻰ .E ﻭﺤل ﺍﻟﻤﻌﺎﺩﻟﺔ
)u E (x
ﻟﻠﻤﺅﺜﺭ .H
ﻴﺴﻤﻰ eigeufactionﺍﻟﺫﻱ ﻴﻨﺎﻅﺭ ﺍﻟﻘﻴﻤﺔ ﺍﻟﺨﺎﺼﺔ E
ﻤﺜﺎل :ﺤل ﻤﺴﺄﻟﺔ ﺍﻟﻘﻴﻤﺔ ﺍﻟﺨﺎﺼﺔ
)∠ f (x) = λ f (x
ﺤﻴﺙ: )h df (x )− β x f (x i ax
ﻓﻲ ﺍﻟﻤﻨﻁﻘﺔ
a ≤x≤a
= )∠ f (x
ﻭﻟﻬﺎ ﺸﺭﻁ ﺍﻷﻁﺭﺍﻑ )ﺍﻟﺸﺭﻭﻁ ﺍﻟﺤﺩﻴﺔ( .bouday )f (a) = f ( − a
- ٨٣PDF created with pdfFactory Pro trial version www.pdffactory.com
:ﺍﻟﺤل ∠ f (x) = λ f (x) h a f (x) − β v f (x) = λ f (x) i ax df (x) i i = [β + t (x) + λ f (x)]= [β x + λ ]+ (x) dx h h df i x = (β + λ) ax f h i x2 h f = (β + λx )+c h 2 x 2 iβ + iλλxh 2 f (x) = Ae
energy eigenvahes ﺘﺴﻤﻰ ﻗﻴﻡ ﺍﻟﻁﺎﻗﺔ ﺍﻟﺨﺎﺼﺔH ﺍﻟﻘﻴﻤﺔ ﺍﻟﺨﺎﺼﺔ ﻟﻠﻤﺅﺜﺭ H=
p 2 op + V (x) 2m
the Eigenvalne Poeem for a ﻤﺴﺄﻟﺔ ﺍﻟﻘﻴﻤﺔ ﺍﻟﺨﺎﺼﺔ ﻟﺠﺴﻡ ﻓﻲ ﺼﻨﺩﻭﻕ
.Particle in box
h 2 a 2 u (x) − + V (x) u (x) = Eu (x) 2m ax 2
(4 – 6) ﺒﺎﻋﺘﺒﺎﺭ
:ﺤﻴﺙ V (x) = ∞ = 0 =∞
x N
~ am + 2
ﻭﻫﺫﺍ ﻴﻌﻨﻲ ﺃﻥ ﺍﻟﺤﻠﻭل ﺘﻘﺭﻴﺒﺎﹰ )ﺒﺎﺨﺘﻴﺎﺭ ﺍﻟﺤﻠﻭل ﺍﻟﺯﻭﺠﻴﺔ( .ﻭﻫﺫﺍ )h ( y ) = (atolyromialiny
2 22 23 + a N y N + y N+2 + yN+4 + y N + 6 + .......... N ) N (N + 2 ) N (N + 2 ) (N + 4
ﻭﺒﺎﻟﺘﺒﺴﻴﻁ N N −1 +1 N 2 2 2 2 2 2 N (y ) (y ) (y ) + + + .......... a N y 2 ( −1 )! N N 2 ! ) ( N − 1 !) ( !)( + 1 2 2 2
ﺒﺎﺨﺘﻴﺎﺭ
N = 2k
)ﻟﻠﻤﻼﺌﻤﺔ( ﻓﺈﻥ ﺍﻟﻤﺘﺴﻠﺴﻠﺔ ﺘﺄﺨﺫ ﺍﻟﺸﻜل.
(y 2 ) k − 1 (y 2 ) k (y 2 ) k + 1 y 2 (k − 1 )! + + + .......... !k ! )(k + 1 ! ) (k − 1
- ١١٤PDF created with pdfFactory Pro trial version www.pdffactory.com
2 (y 2 ) k −1 (y 2 ) 2 = y 2 (k − 1) ! e y − {1 + y 2 + + .................. + (k − 2)! !2
ﻭﻫﺫﺍ ﺸﻜل ﻤﺘﻌﺩﺩ ﺍﻟﺤﺩﻭﺩ +ﺜﺎﺒﺕ x
y2 ey
ﻭﺒﺎﻟﺘﻌﻭﻴﺽ ﻫﺫﻩ ﺍﻟﻤﻌﺎﺩﻟﺔ ﻓﻲ ﺍﻟﻤﻌﺎﺩﻟﺔ
2
−y2
ﻓﺈﻨﻨﺎ ﺴﻨﺘﺤﺼل ﻋﻠﻰ ﺤل ﻻ ﻴﺘﻼﺸﻰ ﻋﻨﺩ
4 (y ) = h (y ) e ∞
ﻟﻭ ﺃﻥ ﻤﻌﺎﺩﻟﺔ ﺍﻻﺴﺘﻁﺭﺍﺩ ﺘﻨﺘﻬﻲ ،ﺃﻱ ﺃﻥ ﻟﻭ ﺃﻥ.
ﻭﺍﻟﺤل ﺍﻟﻤﻘﺒﻭل ﻴﻤﻜﻥ ﺃﻥ ﻴﻭﺠﺩ
( N + 1 ) ( N + 2 ) a n + 2 = 2N − ∈+ 1 a n = 0 1424 3 =0
⇒ ∈ = 2N + 1 2E = 2N + 1 hw
)(17
ﻭﺍﻟﻨﺘﺎﺌﺞ ﻫﻲ: -1ﺃﻨﻪ ﻴﻭﺠﺩ ﻗﻴﻡ ﺫﺍﺘﻴﺔ ﻤﻨﻔﺼﻠﺔ )ﻏﻴﺭ ﻤﺘﺼﻠﺔ( ﻤﺘﺴﺎﻭﻴﺔ ﺍﻟﻤﺴﺎﻓﺎﺕ ﻤﻥ ﺍﻟﻤﻌﺎﺩﻟـﺔ
) (4ﻭﺍﻟﻤﻌﺎﺩﻟﺔ ) (17ﻴﻤﻜﻨﻨﺎ ﺍﻟﺤﺼﻭل ﻋﻠﻰ ﺍﻟﻤﻌﺎﺩﻟﺔ ﺍﻟﺘﺎﻟﻴﺔ. 1 1 hw (2m + 1) = (n + ) h w 2 2
)(18
=E
ﻭﻫﺫﻩ ﻗﺭﻴﺒﺔ ﻤﻥ ﻤﻌﺎﺩﻟﺔ Planckﻟﻺﺸﻌﺎﻉ. 1 E = (n + ) hw 2
-2ﻤﺎ ﻋﺩﺍ ﺜﻭﺍﺒﺕ ﺍﻟﻤﻌﺎﻴﺭﺓ ،ﻓﺈﻥ ﻤﺘﻌﺩﺩ ﺍﻟﺤﺩﻭﺩ ﻫﻭ ﻤﺘﻌﺩﺩ ﺍﻟﺤﺩﻭﺩ
ﺍﻟﻬﻴﺭﻤﻴﺘﻲ)Hn (y
ﻭﺍﻟﺫﻱ ﻴﻤﻜﻥ ﺇﻴﺠﺎﺩ ﺨﻭﺍﺼﻪ ﺒﺴﻬﻭﻟﺔ ﻓﻲ ﺃﻱ ﻜﺘﺎﺏ ﻓﻲ ﺍﻟﻔﻴﺯﻴﺎﺀ ﺍﻟﺭﻴﺎﻀﻴﺔ. ﻭﻟﻭ ﻗﺼﺭﻨﺎ ﺃﻨﻔﺴﻨﺎ ﻟﻠﺨﻭﺍﺹ ﺍﻟﺘﺎﻟﻴﺔ ﻤﻨﻬﺎ: )Hn (y
ﺘﺤﻘﻕ ﺍﻟﻤﻌﺎﺩﻟﺔ ﺍﻟﺘﻔﺎﻀﻠﻴﺔ. ﺘﺤﺕ ﺸﺭﻁ
∈ = 2 n +1
+ 2nH n (y) = 0
)dH n (y dy
− 2y
) d 2 H n (y 2
dy
- ١١٥PDF created with pdfFactory Pro trial version www.pdffactory.com
ecussionedations ﻭﻫﺫﻩ ﺘﺤﻘﻕ ﺍﻟﻌﻼﻗﺎﺕ ﺍﻻﺴﺘﻁﺭﺍﺩﻴﺔ ﺍﻟﺘﺎﻟﻴﺔ H n +1 − 2y H n + 2n H n −1 = 0 H n +1 +
∑ n
dH n dy
− 2y H n + 2n H n = 0
ﻭﻜﺫﻟﻙ
2 Zn H n (y) = e 2zy − z n!
H n (y) = (−1) n e y
2
ﻭ
d n − y2 e dy n
ﻭﺍﻟﺘﻁﺒﻴﻊ ﻟﻜﺜﻴﺭﺍﺕ ﺤﺩﻭﺩ ﻫﻴﺭﻤﻴﺕ ﻴﻜﻭﻥ ﺒﺤﻴﺙ ∞
∫
2
dy e − y H n (y) 2 = d n n! D
−∞
.ﻫﺫﻩ ﻗﺎﺌﻤﺔ ﺒﺒﻌﺽ ﻋﺩﻴﺩﺍﺕ ﺤﺩﻭﺩ ﻫﻴﺭﻤﻴﺕ H 0 (y) = 1 H1 (y) = 2 y H 2 (y) = 4 y 2 − 2 H 3 (y) = 8 y 2 − 12 y H 4 (y) = 16 y 4 − 48 y 2 + 12 H 5 (y) = 32 y 5 − 160 y 3 + 129 y
(ﻭﻤﻌﺎﺩﻻﺕ ﺍﻟﻘﻴﻡ ﺍﻟﺫﺍﺘﻴﺔ )ﺍﻟﺨﺎﺼﺔ d2u n dx
2
=
mk 2 2mEn x un − un 2 h h2 و
d2u* e mk 2mEn * u e = 2 x 2 u* e − 2 dx h h2
ﺍﻟﻤﻌﺎﺩﻟﺔ ﺍﻟﺜﺎﻨﻴﺔ،ﺒﺎﻟﻤﻌﺎﺩﻟﺘﻴﻥ ﺍﻟﺴﺎﺒﻘﺘﻴﻥ ﻋﻠﻰ ﺍﻟﺘﻭﺍﻟﻲ
un
ﻭ
u*e
ﻭﻋﻨﺩ ﻀﺭﺏ .ﺘﻁﺭﺡ ﻤﻥ ﺍﻷﻭﻟﻰ
* 2m d * dun du x ux − u x = 2 (E x − E n ) u *x u n h dx dx dx
- ١١٦ PDF created with pdfFactory Pro trial version www.pdffactory.com
ﻭﺒﺘﻜﺎﻤل ﻫﺫﻩ ﺍﻟﻤﻌﺎﺩﻟﺔ ﻋﻠﻰ xﻤﻥ
∞−
∞+
←
ﻓﺈﻥ ﺍﻟﺠﺎﻨـﺏ ﺍﻷﻴـﺴﺭ
ﻴﺘﻼﺸﻰ ،ﺒﻤﺎ ﺃﻥ ﺍﻟﻤﻌﺎﺩﻻﺕ ﺍﻟﺨﺎﺼﺔ ﻭﺘﻔﺎﻀﻼﺘﻬﺎ ﺘﺘﻼﺸﻰ ﻋﻨﺩ dx u *e (x) u n (x) = 0
∞
∫
) − En
x
∞ x= ±
ﻭﻫﻜﺫﺍ.
(E
∞−
ﻭﺍﻟﺫﻱ ﻴﻌﻨﻲ ﺃﻥ ﺍﻟﺩﻭﺍل ﺍﻟﺨﺎﺼﺔ ﺍﻟﺘﻲ ﻟﻬﺎ -3ﺍﻟﻤﻌﺎﺩﻟﺔ
1 ) E = h w (n + 2
Ee ± En
ﻫﻲ ﻤﺘﻌﺎﻤﺩﺓ.
ﺘﻭﻀﺢ ﺃﻥ ﺃﻗل ﺤﺎﻟﺔ ﻟﻬﺎ ﻁﺎﻗﺔ ﻭﻫﺫﻩ ﻫﻲ ﻁﺎﻗﺔ ﻨﻘﻁـﺔ
ﺍﻟﺼﻔﺭ .Zero – point evergy ﺇﻥ ﻭﺠﻭﺩﻫﺎ ﻫﻭ ﺘﺄﺜﻴﺭ ﻜﻤﻲ ﺒﺤﺕ ،ﻭﻴﻤﻜﻥ ﺘﺄﻭﻴﻠﻪ ﺒﺩﻻﻟﺔ ﻤﺒﺩﺃ ﺍﻟﻼﺘﻌﻴﻴﻥ .ﺇﻨﻬﺎ ﻁﺎﻗﺔ ﻨﻘﻁﺔ ﺍﻟﺼﻔﺭ ﺍﻟﻤﺴﺌﻭﻟﺔ ﻋﻥ ﺍﻟﺤﻘﻴﻘﺔ ﺃﻥ ﺍﻟﻬﻴﻠﻴﻭﻡ ﻻ ﻴﺘﺠﻤـﺩ ﻋﻨـﺩ ﺩﺭﺠـﺎﺕ ﺍﻟﺤﺭﺍﺭﺓ ﺍﻟﻘﻠﻴﻠﺔ ،ﻭﻟﻜﻨﻪ ﻴﺒﻘﻰ ﻓﻲ ﺍﻟﺤﺎﻟﺔ ﺍﻟﺴﺎﺌﻠﺔ ﺤﺘﻰ ﺩﺭﺠﺎﺕ ﺤـﺭﺍﺭﺓ ﻤﻘـﺩﺍﺭﻫﺎ 10-3kﻭﺫﻟﻙ ﻋﻨﺩ ﺍﻟﻀﻐﻁ ﺍﻟﻌﺎﺩﻱ .ﺍﻟﺘﺭﺩﺩ ﻟﻬﺎ ﻴﻜﻭﻥ ﻜﺒﻴﺭﺍﹰ ﻟﻠﺫﺭﺍﺕ ﺍﻟﺨﻔﻴﻔﺔ .ﻭﺍﻟﺫﻱ
ﻫﻭ ﺍﻟﺴﺒﺏ ﻓﻲ ﺃﻥ ﺍﻟﻨﻴﺘﺭﻭﺠﻴﻥ ﻻ ﻴﻅﻬﺭ ﻫﺫﺍ ﺍﻟﺘﺄﺜﻴﺭ ﻤﺜل ﺍﻟﻬﻴﻠﻴﻭﻡ .ﺇﻨﻪ ﻜﺫﻟﻙ ﻴﻌﺘﻤﺩ
ﻋﻠﻰ ﺍﻟﺨﻭﺍﺹ ﺍﻟﺘﻔﻀﻴﻠﻴﺔ ﻟﻠﻘﻭﻯ ﺍﻟﻔﺎﻋﻠﺔ ﺒﻴﻥ ﺍﻟﺠﺯﻴﺌﺎﺕ ﻭﺍﻟﺘﻲ ﺘﺸﺭﺡ ﺍﻟﺴﺒﺏ ﻓﻲ ﺃﻥ ﺍﻟﻬﻴﺩﺭﻭﺠﻴﻥ ﻴﺘﺠﻤﺩ.
- ١١٧PDF created with pdfFactory Pro trial version www.pdffactory.com
ﺍﻟﺒﻨﺎﺀ ﺍﻟﻌﺎﻡ ﻟﻤﻴﻜﺎﻨﻴﻜﺎ ﺍﻟﻜﻡ )ﺍﻟﻤﻴﻜﺎﻨﻴﻜﺎ ﺍﻟﻤﻭﺠﻴﺔ(: ﺴﻨﺘﻌﺭﺽ ﻓﻲ ﻫﺫﺍ ﺍﻟﻔﺼل ﻟﺩﺭﺍﺴﺔ ﺍﻟﻤﺒﺎﺩﺉ ﺍﻷﺴﺎﺴﻴﺔ ﺍﻟﺘﻲ ﺫﻜﺭﺕ ﺁﻨﻔﺎﹰ ﻓـﻲ
ﺴﻴﺎﻕ ﺤل ﺒﻌﺽ ﺍﻟﻤﺴﺎﺌل ﺍﻟﻔﻴﺯﻴﺎﺌﻴﺔ ﻤﻥ ﻫﺫﻩ ﻨﻅﺭﻴﺔ ﺍﻟﻤﻔﻜﻭﻙ ﻭﻤﻌﻨﺎﻫﺎ ﺍﻟﻔﻴﺯﻴـﺎﺌﻲ ﻓﻲ ﺍﻟﻤﺅﺜﺭﺍﺕ ﻤﺘﺠﻬﺎﺕ ﺍﻟﺤﺎﻟﺔ ،ﺍﻟﺘﻜﺭﺍﺭ ﻭﺍﻟﻨﻬﺎﻴﺔ ﺍﻟﺘﻘﻠﻴﺩﻴﺔ .ﻭﺴﻨﻘﺩﻡ ﺘﺭﻤﻴﺯ ﺩﻴـﺭﺍﻙ
ﺒﺎﺨﺘﺼﺎﺭ.
ﺍﻟﺩﻭﺍل ﺍﻟﺨﺎﺼﺔ ﻭﺍﻟﻘﻴﻡ ﺍﻟﺨﺎﺼﺔ: "ﻤﺅﺜﺭ ﺍﻟﻬﺎﻤﻠﺘﻭﻨﻴﺎﻥ" ﺇﻥ ﺤﺎﻟﺔ ﺃﻱ ﻨﻅﺎﻡ ﻓﻴﺯﻴﺎﺌﻲ ﻴﻭﺼﻑ ﺒﺎﻟﺩﺍﻟﺔ ﺍﻟﻤﻭﺠﻴﺔ ﻭﺍﻟﺘﻲ ﺘﺤﺘﻭﻱ ﻋﻠﻰ ﻜل
ﺍﻟﻤﻌﻠﻭﻤﺎﺕ ﻋﻥ ﻫﺫﺍ ﺍﻟﻨﻅﺎﻡ .ﺇﻥ ﺍﻟﺩﺍﻟﺔ ﺍﻟﻤﻭﺠﻴﺔ ﺘﻌﺘﻤﺩ ﻋﻠﻰ ﺍﻟﺯﻤﻥ ،ﻭﺘﻁﻭﺭﻫﺎ ﻤـﻊ
ﺍﻟﺯﻤﻥ ﻴﻌﻁﻲ ﺒـ:
)(1
∂ )ψ (x, t) = H ψ (x, t xt
ih
ﺇﻥ ﺍﻟﺩﺍﻟﺔ ﺍﻟﻤﻭﺠﻴﺔ ﺁﺜﺭﻨﺎ ﻋﻠﻴﺎ ﺒﻤﺅﺜﺭ ،Hﺍﻟﻬﺎﻤﻠﺘﻭﻨﻴﺎﻥ ﻭﺍﻟـﺫﻱ ﻴﻠﻌـﺏ ﺩﻭﺭﺍﹰ
ﻜﺒﻴﺭﺍﹰ ﻓﻲ ﻤﻴﻜﺎﻨﻴﻜﺎ ﺍﻟﻜﻡ .ﺍﻟﻤﺅﺜﺭ Hﻟﻨﻅﺎﻡ ﻤﻥ ﺠﺴﻡ ﻭﺍﺤﺩ ﻓﻲ ﺠﻬﺩ Vﻟﻪ ﺍﻟﺸﻜل. )+ V (x
p sp2 2m
=H
ﻭﺇﺫﺍ ﻜﺎﻨﺕ ) V (xﻟﻴﺱ ﻟﻬﺎ ﺍﻋﺘﻤﺎﺩ ﻭﺍﻀﺢ ﻋﻠﻰ ﺍﻟﺯﻤﻥ ﻓﺈﻥ ﺍﻟﻤﻌﺎﺩﻟـﺔ )(١
ﻴﻤﻜﻥ ﺤﻠﻬﺎ ﺒـ:
ψ (x, t) = u E (x) e − Et/h
ﺤﻴﺙ: )H u E (x) = E u E (x
- ١١٨PDF created with pdfFactory Pro trial version www.pdffactory.com
ﺇﻥ ﺤﻠﻭل ﻫﺫﻩ ﺍﻟﻤﻌﺎﺩﻟﺔ
)u E (x
ﻫﻲ ﺍﻟﺩﻭﺍل ﺍﻟﺨﺎﺼﺔ ﻟﻠﻬﺎﻤﻠﺘﻭﻨﻴﺎﻥ ﻭ Eﻫـﻲ ﺍﻟﻘـﻴﻡ
ﺍﻟﺨﺎﺼﺔ ﻭﻓﻲ ﻫﺫﺍ ﺍﻟﻔﺼل ﺴﻨﺫﻜﺭ ﻋﻠﻰ ﺨﺎﺼﻴﺘﻥ ﻤﻬﻤﺘﻴﻥ ﻟﻠﺩﻭﺍل ﺍﻟﺨﺎﺼﺔ ﻟـ .H -1ﺍﻟﺩﻭﺍل ﺍﻟﺨﺎﺼﺔ ﺍﻟﻤﻁﺎﺒﻘﺔ ﻟﻘﻴﻡ ﺨﺎﺼﺔ ﻤﺨﺘﻠﻔﺔ )ﺃﻱ ﻗﻴﻡ ﻤﺨﺘﻠﻔﺔ ﻟـ (Eﺘﻜـﻭﻥ
ﻫﺫﻩ ﺍﻟﺩﻭﺍل ﻤﺘﻌﺎﻤﺩﺓ ﺃﻱ:
E ≠ E′
(x) = 0
E′
dx u *E (x) u
∫
-2ﺇﻥ ﺍﻟﺩﻭﺍل ﺍﻟﺨﺎﺼﺔ ﺘﻜﻭﻥ ﻓﺌﺔ ﺍﻟﻬﺎﻤﻠﺔ – ﺃﻱ ﺃﻥ – ﺩﺍﻟﺔ ﺍﺨﺘﻴﺎﺭﻴﺔ
ﻤﺭﺒﻌﺔ ﺍﻟﺘﻜﺎﻤل ﻟﻜﻲ.
∞
)(6
0
ﻟﺩﻴﻨﺎ ﺍﻟﺤﻠﻭل ﻭﺍﻟﺘﻲ ﺘﻜﻭﻥ ﻤﻌﺎﻴﺭﺓ ﻓﻲ ﺩﺍﺨل ﺼﻨﺩﻭﻕ. )(55
ﺍﻟﺤل ﻋﻨﺩﻤﺎ ﺘﻜﻭﻥ ﻤﻥ
e ikr
ﻭ
− ikr
γ
2μ E =k2 2 h d 2u −u2 =0 2 dr
ﻜﺒﻴﺭﺓ ﺒﺤﻴﺙ ﺃﻥ ) V(rﻤﻬﻤﻠﺔ ﻴﻜﻭﻥ ﻫﻭ ﺍﺘﺤﺎﺩ ﺨﻁـﻲ
.e
- ١٤٥PDF created with pdfFactory Pro trial version www.pdffactory.com
ﺫﺭﺓ ﺍﻟﻬﻴﺩﺭﻭﺠﻴﻥ
The Hydrogen Atom
ﺫﺭﺓ ﺍﻟﻬﻴﺩﺭﻭﺠﻴﻥ ﺃﺒﺴﻁ ﺫﺭﺓ ﻭﺫﻟﻙ ﻷﻨﻬﺎ ﺘﺤﺘﻭﻱ ﻋﻠﻰ ﺇﻟﻜﺘﺭﻭﻥ ﻭﺍﺤﺩ .ﻭﻟﻬﺫﺍ
ﺍﻟﺴﺒﺏ ﺘﺼﺒﺢ ﻤﻌﺎﺩﻟﺔ ﺸﺭﻭﺩﻨﺠﺭ ﻤﻌﺎﺩﻟﺔ ﺠﺴﻡ ﻭﺍﺤﺩ ﻭﺫﻟﻙ ﺒﻌﺩ ﻓﺼل ﺤﺭﻜﺔ ﻤﺭﻜﺯ ﺍﻟﻜﺘﻠﺔ ﻭﺴﻭﻑ ﻨﺘﻌﺎﻤل ﻤﻊ ﺫﺭﺍﺕ ﻤﺸﺎﺒﻬﺔ ﻟﻠﻬﻴﺩﺭﻭﺠﻴﻥ ،ﺃﻱ ﺃﻥ ﺍﻟﺫﺭﺍﺕ ﺍﻟﺘﻲ ﺘﺤﺘﻭﻱ ﺘﺤﺘﻭﻱ ﻋﻠﻰ ﺃﻜﺜﺭ ﻤـﻥ ﺒﺭﻭﺘـﻭﻥ
ﻋﻠﻰ ﺇﻟﻜﺘﺭﻭﻥ ﻭﺍﺤﺩ ﻓﻘﻁ ،ﻭﻟﻜﻥ ﺴﻨﺴﻤﺢ ﻭﺍﺤﺩ .ﻭﺍﻟﺠﻬﺩ ﻓﻲ ﻫﺫﻩ ﺍﻟﺤﺎﻟﺔ ﻴﻜﻭﻥ
Ze2 r
)(1
V(r ) = −
ﻭﻤﻌﺎﺩﻟﺔ ﺸﺭﻭﺩﻨﺠﺭ ﺍﻟﻨﺼﻑ ﻗﻁﺭﻴﺔ ﺘﻜﻭﻥ d2 2 d 2μ Ze 2 l (l + 1) h 2 2 + R + 2 E + − R =0 r dr r 2μ r 2 h dr
)(2
ﻭﺴﻨﺭﻜﺯ ﺍﻟﺩﺭﺍﺴﺔ ﻋﻠﻰ ﺍﻟﺤﺎﻻﺕ ﺍﻟﻤﻘﻴﺩﺓ ،ﺃﻱ ﺤﻠﻭل
E < 0
ﺃﻥ ﻨﻌﻤل ﺘﻐﻴﻴﺭ ﻓﻲ ﺍﻟﻤﺘﻐﻴﺭﺍﺕ.
2
)(3
1
r
ﺇﻨﻪ ﻤﻥ ﺍﻟﻤﻨﺎﺴﺏ
8µ E δ = 2 h
ﻓﻲ ﻫﺫﻩ ﺍﻟﺤﺎﻟﺔ ﺘﺼﺒﺢ ﺍﻟﻤﻌﺎﺩﻟﺔ ) (2ﻋﻠﻰ ﺍﻟﺼﻭﺭﺓ ﺍﻟﺘﺎﻟﻴﺔ: )d 2 R 2 dR l ( l + 1 λ 1 + − R+ − R = 0 2 2 δ dδ dδ δ δ 4
)(4
ﻭﻗﺩ ﺍﺴﺘﺤﺩﺜﻨﺎ ﺍﻟﻤﻌﺎﻤل )ﺒﺩﻭﻥ ﻭﺤﺩﺓ( )(5
2
1
μ C2 = Zα 2 E
ﺍﻟﻤﻌﺎﺩﻟﺔ ) (4ﻤﻥ ﺍﻟﺴﻬل ﺤﻠﻬﺎ .ﻭﺫﻟﻙ
2
1
1 137
μ 2 E
Z e2 =λ h
= ، αﻭﺍﻟﻁﺎﻗﺔ ﻤﻌﺒﺭ ﻋﻨﻬﺎ ﺒﻭﺤﺩﺍﺕ
ﺍﻟﻜﺘﻠﺔ ﻋﻨﺩ ﺍﻟﺴﻜﻭﻥ . rest – mass
- ١٤٦PDF created with pdfFactory Pro trial version www.pdffactory.com
ﻁﻴﻑ ﺍﻟﻁﺎﻗﺔ
The Energy Spectrum
ﺴﻨﺤﺎﻭل ﺤل ﻤﻌﺎﺩﻟﺔ ) (4ﺒﻁﺭﻴﻘﺔ ﻤﻌﺭﻭﻓﺔ ﻟﺩﻴﻨﺎ ﺍﻵﻥ .ﺃﻭﻻﹰ ﺴـﻨﺠﺩ ﺴـﻠﻭﻙ
ﺍﻟﺩﺍﻟﺔ ﻋﻨﺩﻤﺎ ﺘﻜﻭﻥ δﻜﺒﻴﺭﺓ .ﻭﻋﻨﺩﻤﺎ ﺘﻜﻭﻥ
δ
ﻜﺒﻴﺭﺓ ﻓﺎﻟﺤﺩﻭﺩ ﺍﻟﻤﺘﺒﻘﻴﺔ ﻓﻲ ﺍﻟﻤﻌﺎﺩﻟﺔ.
d2R 1 − R=0 4 dδ 2
) (6
ﻭﺍﻟﺤل ﺍﻟﻤﻘﺒﻭل ﻋﻨﺩ ﻤﺎ ﻻ ﻨﻬﺎﻴﺔ 2
+δ
⇐ R ~e
ﻋﻨﺩ 2
−δ
∞ → ∞ ← R ⇐δ
R~e
ﻭﻜﻤﺎ ﻋﺎﻟﺠﻨﺎ ﺍﻟﻤﺘﺫﺒﺫﺏ ﺍﻟﺘﻭﺍﻓﻘﻲ ) G (δ
) (7
δ 2
−
R= e
ﻭﺒﺎﻟﺘﻌﻭﻴﺽ ﻓﻲ ﺍﻟﻤﻌﺎﺩﻟﺔ )(4
ﺴﻨﺘﺤﺼل ﻋﻠﻰ ﻤﻌﺎﺩﻟﺔ ﻟـ ):G (s d 2 G 2 dG λ − 1 l (l + ) G =0 − 1 − + − dδ 2 δ dδ δ δ 2
)(8
ﻭﺍﻵﻥ ﻨﻜﺘﺏ ﻤﻔﻜﻭﻙ ﺍﻟﻘﻭﺓ ﻟـ ) G (δﻭﺍﻟﺘﻲ ﺘﺄﺨﺫ ﺍﻟﺸﻜل: )(9
δn
∞
n
∑a
n =ο
G (δ ) = δ l
ﻭﺒﺘﻌﻭﻴﺽ ﺍﻟﻤﻌﺎﺩﻟﺔ ) (aﻓﻲ ﺍﻟﻤﻌﺎﺩﻟﺔ ﺍﻟﺘﻔﺎﻀـﻠﻴﺔ ،ﺴـﻨﺠﺩ ﺍﻟﻌﻼﻗـﺔ ﺒـﻴﻥ
ﺍﻟﻤﻌﺎﻤﻼﺕ ﺍﻟﻤﺨﺘﻠﻔﺔ
an
ﻭﻤﻌﺎﺩﻟﺔ ﺍﻻﺴﺘﻁﺭﺍﺩ ﻴﻤﻜﻥ ﺍﻟﺤﺼﻭل ﻋﻠﻴﻬﺎ ﻤﻥ ﺍﻟﻤﻌﺎﺩﻟـﺔ
ﺍﻟﺘﻔﺎﻀﻠﻴﺔ ﺍﻟﺨﺎﻀﻌﺔ ﻟﻠﻌﻼﻗﺔ
)(10
∞
H (δ ) = ∑ a n δ n n =ο
- ١٤٧PDF created with pdfFactory Pro trial version www.pdffactory.com
ﻭﺍﻟﺘﻲ ﺘﻜﻭﻥ d 2 H 2 l + 2 d H λ −1− l H=0 + + − 1 δ dδ 2 δ dδ
)(11
ﻭﻫﺫﻩ ﺍﻟﻤﻌﺎﺩﻟـﺔ ﻴﻤﻜﻨﻨـﺎ ﺍﻟﺤـﺼﻭل ﻋﻠﻴﻬـﺎ ﺒـﺴﻬﻭﻟﺔ ﻭﺫﻟـﻙ ﺒﺘﻌـﻭﻴﺽ ) G (δ ) = δ l H ( p
ﻓﻲ ﻤﻌﺎﺩﻟﺔ )(8
ﻓﻲ ﻫﺫﻩ ﺍﻟﺤﺎﻟﺔ: ) (12
2l + 2 δ n − 2 + n a n δ n −1 − 1 + (λ − 1 − l ) a n δ n −1 = 0 δ
]
}
+ (2 l + 2 ) a n + 1 + ( λ − 1 − l − n ) a n δ n −1 = 0
∑ n ( n − 1 ) a ∞
n
n=0
∑ {(n + 1)[n a ∞
n +1
n=0
ﻭﺒﻤﺎ ﺃﻥ ﻫﺫﻩ ﺍﻟﻤﻌﺎﺩﻟﺔ ﺘﺘﻼﺸﻰ ﺤﺩ ﺒﺤﺩ ،ﻭﺴﻨﺘﺤﺼل ﻋﻠﻰ ﻤﻌﺎﺩﻟﺔ ﺍﻻﺴﺘﻁﺭﺍﺩ. n + l +1− λ ) ( n + 1 )(n + 2l + 2
)(13
=
a n +1 an
ﻭﻟﻘﻴﻡ ﻜﺒﻴﺭﺓ ﻟﻘﻴﻤﺔ nﻓﺈﻥ ﺍﻟﻨﺴﺒﺔ ﺘﺼﺒﺢ ) (14
1 n
≈
a n +1 an
ﻭﻜﻤﺎ ﻓﻌﻠﻨﺎ ﻓﻲ ﺤﺎﻟﺔ ﺍﻟﻤﺘﺫﺒﺫﺏ ﺍﻟﺘﻭﺍﻓﻘﻲ ،ﻓﺈﻨﻨﺎ ﻟﻥ ﻨﺘﺤﺼل ﻋﻠﻰ ﺴﻠﻭﻙ ﻤﻘﺒﻭل
ﻟﻠﺩﺍﻟﺔ ) R ( pﻋﻨﺩ
∞
ﺇﻻ ﺇﺫﺍ ﺍﻨﺘﻬﺕ ﺍﻟﺴﻠﺴﻠﺔ ) (aﺃﻱ ﺃﻨﻪ ﻟﻘﻴﻤﺔ ﻤﻌﻁﺎﺓ ﻟـ
l
ﻭ
n = nr
ﻴﺠﺏ ﺃﻥ ﻨﺘﺤﺼل ﻋﻠﻰ. ﺩﻋﻨﺎ ﺍﻵﻥ ﻨﺴﺘﺤﺩﺙ ﺍﻟﺭﻗﻡ ﺍﻟﻜﻤﻲ ﺍﻷﺴﺎﺴﻲ Nﺍﻟﻤﻌﺭﻑ ﺒـ )(16
ﻭﻴﻨﺘﺞ ﻤﻥ ﺍﻟﺤﻘﻴﻘﺔ ﺃﻥ
n = nr + l + 1
⇐ nr ≥ 0
1. n ≥ l +1 2. n is au int eger
- ١٤٨PDF created with pdfFactory Pro trial version www.pdffactory.com
) (Zα 1 E = − µ c2 2 n2 E = hw
2
)(14
ﻭﻫﺫﻩ ﺍﻟﻨﺘﻴﺠﺔ ﺸﺒﻴﻬﺔ ﺒﻨﻤﻭﺫﺝ ﺒﻭﺭ .ﻻﺤﻅ ﺃﻥ ﺍﻟﻜﺘﻠﺔ ﺍﻟﻤﺨﺘﺯﻟﺔ
µ
ﻓﻲ ﺍﻟﻤﻌﺎﺩﻟﺔ .ﺇﻥ ﻭﺠﻭﺩ ﺍﻟﻜﺘﻠﺔ ﺍﻟﻤﺨﺘﺯﻟﺔ. )(18
mµ m+ M
ﺍﻟﺘﻲ ﺘﻅﻬﺭ
=µ
ﺤﻴﺙ mﻜﺘﻠﺔ ﺍﻹﻟﻜﺘﺭﻭﻥ ،ﻭ Mﻜﺘﻠﺔ ﺍﻟﻨﻭﺍﺓ ،ﻴﻌﻨﻲ ﺃﻥ ﺍﻟﺘﺭﺩﺩﺍﺕ. )(19
Ei − Ej mC 2 /2h (Zα )2 12 − 12 = m h n ∫ n ∫ 1+ M
= ωiy
ﻭﺍﻟﺫﻱ ﻴﺨﺘﻠﻑ ﻗﻠﻴﻼﹰ ﻻﺨﺘﻼﻑ ﺫﺭﺍﺕ ﺍﻟﻬﻴﺩﺭﻭﺠﻴﻥ ﺍﻟﺸﺒﻴﻬﺔ ﻋﻠـﻰ ﺍﻷﺨـﺹ
ﺍﻟﻔﺭﻕ ﺒﻴﻥ ﻁﻴﻑ ﺫﺭﺓ ﺍﻟﻬﻴﺩﺭﻭﺠﻴﻥ ﻭﺫﺭﺓ ﺍﻟﺩﻴﺘﻭﻴﺭﻭﻡ.
- ١٤٩PDF created with pdfFactory Pro trial version www.pdffactory.com
ﻅﺎﻫﺭﺓ ﺘﺤﻠل ﺍﻟﻁﻴﻑ
The Degeneracy of the spectrum
ﺍﻵﻥ ﺴﻨﻨﺎﻗﺵ ﺘﺤﻠل ﻁﻴﻑ ﺍﻟﻁﺎﻗﺔ ﻟﻘﻴﻤﺔ ، λ =1ﺍﻟﺤﺎﻟﺔ ﺍﻟﺩﻨﻴﺎ ،ﻴﺠﺏ ﺃﻥ ﻴﻜـﻭﻥ )ﻤـﻥ ﺍﻟﻤﻌﺎﺩﻟـﺔ = o, l = 0
ﺃﺨﺫﻨﺎ
r
λ = nr + l +1
ﻭ
. nﻫﺫﺍ ﻴﻌﻨﻲ ﺃﻨﻪ ﺘﻭﺠﺩ ﺤﺎﻟﺔ ﻭﺤﻴﺩﺓ ﻓﺭﻴﺩﺓ ﻟﻠﺤﺎﻟﺔ ﺍﻟﺩﻨﻴﺎ ground staleﺇﺫﺍ
⇐ λ=2
ﻴﻜﻭﻥ ﻋﻨﺩﻨﺎ ﺍﺤﺘﻤﺎﻟﻴﻴﻥ
(i ) n r =1, l = 0 (ii ) n r = 0 , l =1
ﺒﺎﻟﻨﺴﺒﺔ ﻟﻠﺤﺎﻟﺔ ﺍﻷﻭﻟﻰ
=1, l = 0
r
(i ) nﻓﺈﻥ ﺍﻟﻤﻌﺎﺩﻟﺔ ) ،(13ﺘﻜﺘﺏ ﺒﺎﻟﺸﻜل
n + l + 1− nr − l −1 n + l +1 − λ = )(n + 1)(n + 2l + 2 ) (n + 1)(n + 2 l + 2 n − nr
) (20
=
) (n + 1)(n + 2 l + 2
a n +1 an
=
ﻭﻫﺫﺍ ﻴﺒﻴﻥ ﺃﻥ: a −1 −1 = ⇒ a1 = − ο (ο + ο + 2) 1 × 2 2
=
a1 aο
∞
⇒ H (δ ) = ∑ a n δ n = a ο δ ο + a 1 δ1 n=ο
δ
)(21
a = a ο + − ο 2 δ = a ο 1 − 2
) H (δ
ﺒﻴﻨﻤﺎ ﺍﻟﺘﻭﺯﻴﻊ ﺍﻟﺯﺍﻭﻱ ﻤﺘﻤﺎﺜل ﻜﺭﻭﻴﺎﹰ ﻟﻨﺄﺨﺫ ﺍﻟﺤﺎﻟﺔ ﺍﻟﺜﺎﻨﻴﺔ(i i ) n r = 0, l = 1 .
ﻫﻨﺎ ﺍﻟﺩﺍﻟﺔ ﺍﻟﻨﺼﻑ ﻗﻁﺭﻴﺔ ﺜﺎﺒﺘﺔ H (δ ) = a ο
ﻟﻜﻥ ﺍﻟﺠﺯﺀ ﺍﻟﺯﺍﻭﻱ ﻤﻥ ﺍﻟﺩﺍﻟﺔ ﺍﻟﻤﻭﺠﻴﺔ ﻴﺘﻀﻤﻥ ) . Y (ο , ϕ 1m
- ١٥٠PDF created with pdfFactory Pro trial version www.pdffactory.com
ﺍﻟﺘﺤﻠل ﻴﻜﻭﻥ ) (2 l +1ﻭﻋﻠﻴﻪ ﻴﻭﺠﺩ ﺜﻼﺙ ﻤﻥ ﻤﺜل ﻫﺫﻩ ﺍﻟﺤﺎﻻﺕ. ﺍﻟﺘﺤﻠل ﺍﻟﻜﻠﻲ ﻟﻠﺤﺎﻟﺔ
ﺍﻻﺤﺘﻤﺎﻻﺕ ﺍﻟﺘﺎﻟﻴﺔ:
λ =n=2
ﻴﻜﻭﻥ
3 +1= 4 = 2 2
ﻓﻲ ﺤﺎﻟﺔ
ﻴﻭﺠﺩ ﻟـﺩﻴﻨﺎ
λ =3
n r = 2, l = 0
ﻫﻨﺎ ﻴﻭﺠﺩ ﺤﺎﻟﺔ ﻭﺍﺤﺩﺓ ﻭﺍﻟﺘﻲ ﻟﻬﺎ
1 6
=−
a2 a1
ﻭ
= −1
a1 aο
ﻭﻋﻠﻴـﻪ ﻓـﺈﻥ ﺍﻟﺩﺍﻟـﺔ
ﺍﻟﻨﺼﻑ ﻗﻁﺭﻴﺔ ﺘﺘﻀﻤﻥ. 1 H (δ ) = a ο 1 − δ + δ 2 6
)(22 nr = 1 , l = 1
ﺒﻴﻨﻤﺎ
ﻟﻬﺎ ﺜﻼﺙ ﺤﺎﻻﺕ ،ﺒـ
nr = 0 , l = 2
ﻟﻬﺎ
ﻭﻫﻜﺫﺍ ﻴﻭﺠﺩ λ =n=3
)5 = (2l +1
δ 1 − 4
ο
ﺤﺎﻻﺕ ﺒـ
1+ 3 + 5 = 9 = 32
، H (δ ) = a H (δ ) = a ο
ﺘﺤﻠل ﻟﻠﺤﺎﻻﺕ ﺍﻟﺘﻲ ﻟﻬﺎ ﻗﻴﻤﺔ ﺨﺎﺼـﺔ )ﺫﺍﺘﻴـﺔ(
ﻭﺒﺸﻜل ﻋﺎﻡ ،ﺍﻟﺘﺤﻠل ﻓﻲ ﺤﺎﻟﺔ
ﻴﻜﻭﻥ
λ=n
1 + 3 + 5 + ... + [2 (n − 1) + 1]= n 2
ﻭﻤﺴﺒﻘﺎﹰ ﻨﺤﻥ ﺘﻭﻗﻌﻨﺎ ﺘﺤﻠل ) (2l +1ﻟﻠﺠﻬـﺩ ﺍﻟﻨـﺼﻑ ﻗﻁـﺭﻱ ،ﻭﺫﻟـﻙ ﻷﻥ
ﺍﻟﻬﺎﻤﻠﺘﻭﻨﻴﺎﻥ ﺍﻟﻨﺼﻑ ﻗﻁﺭﻱ ﻻ ﻴﻌﺘﻤﺩ ﻋﻠﻰ
z
،ﻭﻟﻜﻥ ﻴﻌﺘﻤﺩ ﻋﻠﻰ
.
2
ﻫﻨﺎ ﻴﻭﺠﺩ ﺘﺤﻠل ﺇﻀﺎﻓﻲ .ﻭﺍﻟﺘﺤﻠل ﺍﻟﺨﺎﺹ ﺼﻔﺔ ﻤﻤﻴﺯﺓ ﻟﻠﺠﻬﺩ . 1 r
- ١٥١PDF created with pdfFactory Pro trial version www.pdffactory.com
ﺍﻟﺩﻭﺍل ﺍﻟﺫﺍﺘﻴﺔ ﺍﻟﻨﺼﻑ ﻗﻁﺭﻴﺔ The Radial Eigen functions
ﺒﺎﻟﺭﺠﻭﻉ ﻟﻠﺩﺍﻟﺔ ﺍﻟﺘﻔﺎﻀﻠﻴﺔ .ﻟﻭ ﻋﻭﻀﻨﺎ
λ=n
ﺒﺤﻴﺙ:
ﻓﻲ ﻋﻼﻗـﺔ ﺍﻻﺴـﺘﻁﺭﺍﺩ )(13
k + l +1− n a (k + 1)(k + 2l + 2 ) k
) (26
= a k +1
ﺴﻨﺠﺩ ﺃﻥ: )n − (k + l + 1 ) n − (k + l . )(k + 1)(k + 2l + 2) k (k + 2l +1 )n − ( l + 1 a ......... 1. ( 2l + 2) ο
) (27
k +1
)a k + 1 = (− 1
ﺒﻤﺴﺎﻋﺩﺓ ﻫﺫﻩ ﺍﻟﻌﻼﻗﺔ ﺒﺈﻤﻜﺎﻨﻨﺎ ﺍﻟﺤﺼﻭل ﻋﻠﻰ ﻤﻔﻜﻭﻙ ﺴﻠﺴﻠﺔ ﺍﻟﻘﻭﺱ ﻟﻠﺩﺍﻟـﺔ ) . H (δﻭﻤﺎ ﻴﻜﺎﻓﺊ ﻫﺫﺍ.ﻨﺠﺩ ﺃﻥ ﺍﻟﻤﻌﺎﺩﻟﺔ ﻟـ ) H (δﺘﻜﻭﻥ ﻟﻤﻌﻘﺩﺓ ﺍﻟﺤﺩﻭﺩ ) (δ
)(28
= ) H (δ
)( 2 l + 1 n − l −1
ﻭﺒﻌﺩﻤﺎ ﻨﹸﺤﻭل ﺒﺎﻟﻌﻜﺱ ﻟﻠﻭﺼﻭل ﻟﻺﺤـﺩﺍﺜﻴﺎﺕ ﺍﻟﻨـﺼﻑ ﻗﻁﺭﻴـﺔ rﻭﺒﻌـﺩ ﺍﻟﻤﻌﺎﻴﺭﺓ ﻨﺠﺩ ﺃﻥ ﺍﻟﺩﻭﺍل ﺍﻟﻨﺼﻑ ﻗﻁﺭﻴﺔ ﺍﻷﻭﻟﻰ ﻴﻤﻜﻥ ﻜﺘﺎﺒﺘﻬﺎ ﻋﻠﻰ ﺍﻟﺼﻭﺭﺓ 12 − 30
h = a ο µ c α
Rnl (r ): zr aο
− 2 aο e zr
−
2
e
1 − z r 2a ο
2 2 z r 2 ( z r ) − 3 aο + 1− e 3 aο 27 a ο2 zr
2
2
3
2 2 z r 2 ( z r ) − 3 aο + 1− e 3 aο 27 a ο2 zr
3
3
2
z R10 (r ) = 2 aο
z R20 (r ) = 2 2 aο
z 3 2 a ο
1
3
= ) R21 (r
z R30 (r ) = 2 3 aο
- ١٥٢PDF created with pdfFactory Pro trial version www.pdffactory.com
ﺭﺴﻡ ﻜﺜﺎﻓﺔ ﺍﻻﺤﺘﻤﺎل ﺍﻟﻨﺼﻑ ﻗﻁﺭﻱ
ﻟﻭﺠﻭﺩ ﺍﻹﻟﻜﺘﺭﻭﻥ ﻋﻨﺩ ﻤﺴﺎﻓﺔ r
) P (r
ﻤﻥ ﻨﻘﻁﺔ ﺍﻷﺼل ﻴﻤﻜﻥ ﺇﻴﺠﺎﺩﻩ ﻤﻥ ﻫﺫﻩ ﺍﻟﺩﻭﺍل ﺍﻟﻤﻭﺠﻴﺔ ﻜﻤﺎ ﻫﻭ ﻤﻭﻀﺢ ﺒﺎﻟـﺸﻜل
) (12-2ﻴﺠﺏ ﺃﻥ ﻨﺘﺫﻜﺭ ﺃﻥ ﺍﻟﺩﺍﻟﺔ ﺍﻟﻤﻭﺠﻴﺔ ﻟﻬﺎ ﺠﺯﺀ ﺯﺍﻭﻱ ،ﺍﻟﺫﻱ ﻤﺭﺒﻌﻪ ﺍﻟﻤﻁﻠﻕ ﻴﻜﻭﻥ
2
) Pem (cosθ
ﺍﻟﺭﺴﻭﻤﺎﺕ ﻟﺩﻭﺍل ﻟﻴﺠﻨﺩﺭ ﺍﻟﻤﺸﺎﺭﻜﺔ
) Pem (cosθ
2
ﻤﻭﻀﺤﺔ ﺒﺎﻟﺭﺴﻡ
).(12-3 ﻭﺒﻤﻌﺭﻓﺔ ﺍﻟﺩﻭﺍل ﺍﻟﻤﻭﺠﻴﺔ ﺒﺈﻤﻜﺎﻨﻨﺎ ﺤﺴﺎﺏ ﺍﻟﻘﻴﻤﺔ ﺍﻟﻤﺘﻭﻗﻌﺔ. ∞
]) = ∫ dr r 2 + k [ Rnl (r
2
k
r
ο
ﻭﻫﺫﻩ ﺒﻌﺽ ﺍﻟﻘﻴﻡ ﺍﻟﻤﺘﻭﻗﻌﺔ ﺍﻟﻤﻔﻴﺩﺓ. 2 2 a n = ο 2 5 n 2 + 1 − 3 l (l + 1) 2z ) (12 − 36 z = aο n 2 2 z = 1 2 3 aο n l + 2
]
)− l (l + 1
]
2
[
[3n 2z aο
= r r2 1 r 1 r2
- ١٥٣PDF created with pdfFactory Pro trial version www.pdffactory.com
![Quantum Mechanics ميكانيكا الكم [PDF]](https://pdfs.asia/img/200x200/quantum-mechanics.jpg)
