Su'aalo Tusaale ah oo Ka Hadlaya Ifafaalaha Quantum
Ifafaalaha Quantum-ka, ama ifafaalaha ay xukumaan farsamoyaqaannada quantum-ka, waxay ka kooban yihiin fikrado iyo mabaadi'yo kala duwan oo u baahan faham qoto dheer iyo kakanaanta xisaabta. Farsamooyinka Quantum-ka waa laan fiisikis ah oo sharraxaysa dhaqanka walxaha subatomic-ka ah, sida elektaroonada iyo fotonnada, kuwaas oo aan lagu sharxi karin fiisikiska caadiga ah. Maqaalkan, waxaan ku sahamin doonnaa dhowr dhibaato oo tusaale ah iyo xalalkooda la xiriira ifafaalaha quantum-ka si aan uga caawinno fahamka mabaadi'da aasaasiga ah ee farsamoyaqaannada quantum-ka.
Su'aal Tusaale 1aad: Mabda'a Hubanti la'aanta ee Heisenberg
Su'aal:
Waa la ogyahay in booska elektaroonada ee atamka lagu cabbiro saxnaanta \( \Delta x = 0.1 \text{ nm} \). Go'aami hubanti la'aanta ugu yar marka la cabbirayo dhaqdhaqaaqa elektarooniga (\( \Delta p \)) adoo isticmaalaya mabda'a hubanti la'aanta Heisenberg.
Jawaab:
Mabda'a hubanti la'aanta ee Heisenberg wuxuu leeyahay:
\[ \Delta x \cdot \Delta p \geq \frac{\hbar}{2} \]
halkaas oo \( \hbar \) uu yahay joogtada Planck ee la dhimay, oo leh qiimaha \( \hbar \qiyaastii 1.054 \times 10^{-34} \text{ Js} \).
Beddel \( \Delta x = 0.1 \qoraal{ nm} = 0.1 \times 10^{-9} \qoraal{ m} \):
\[ \Delta p \geq \frac{\hbar}{2 \Delta x} \]
\[ \Delta p \geq \frac{1.054 \times 10^{-34}}{2 \times 0.1 \times 10^{-9}} \]
\[ \Delta p \geq \frac{1.054 \times 10^{-34}}{2 \times 10^{-10}} \]
\[ \Delta p \geq \frac{1.054 \times 10^{-34}}{2 \times 10^{-10}} = 5.27 \times 10^{-25} \text{ kg m/s} \]
Markaa hubanti la'aanta ugu yar ee lagu cabbiro dhaqdhaqaaqa elektarooniga waa \( 5.27 \times 10^{-25} \text{ kg m/s} \).
Tusaale Su'aal 2aad: Tamarta Suurtagalka ah ee ku jirta Sanduuq (Qayb ka mid ah Sanduuqa)
Su'aal:
Walax leh cuf m ayaa ku xayiran sanduuq hal-cabbir ah oo dhererkiisu yahay L. Waa maxay tamarta aasaasiga ah (tamarta xaaladda dhulka) ee walaxda?
Jawaab:
Tamarta aasaasiga ah (tamarta xaaladda dhulka) ee walaxda ku jirta sanduuq hal-cabbir ah waxaa lagu bixiyaa isla'egta:
\[ E_n = \frac{n^2 h^2}{8mL^2} \]
Xaaladda dhulka (\( n=1 \)):
\[ E_1 = \frac{h^2}{8mL^2} \]
halkaas oo \( h \) uu yahay joogtada Planck \( (h \qiyaastii 6.626 \times 10^{-34} \text{ Js}) \).
Ka soo qaad \( m = 9.109 \times 10^{-31} \text{ kg} \) (cufka elektarooniga) iyo \( L = 1 \times 10^{-9} \text{ m} \):
\[ E_1 = \frac{(6.626 \jeer 10^{-34})^2}{8 \jeer 9.109 \jeer 10^{-31} \jeer (1 \jeer 10^{-9})^2} \]
\[ E_1 = \frac{4.39 \jeer 10^{-67}}{7.287 \jeer 10^{-50}} \]
\[ E_1 = 6.02 \jeer 10^{-18} \qoraal{ J} \]
Markaa tamarta aasaasiga ah ee walaxda waa \( 6.02 \times 10^{-18} \text{ J} \).
Tusaale 3: Hawlgallada Hawl-wadeenka Hamiltonian ee Hawlaha Mowjadaha
Su'aal:
Shaqada hirarka ee walaxda ku jirta sanduuq hal-cabbir ah waa \( \psi(x) = \sqrt{\frac{2}{L}} \sin\left(\frac{n\pi x}{L}\right) \) ee \( n=1,2,3,\ldots \). Go'aami tamarta walaxda adoo isticmaalaya hawlwadeenka Hamiltonian \( \hat{H} \).
Jawaab:
Hawlwadeenka Hamiltonian hal dhinac waa:
\[ \hat{H} = -\frac{\hbar^2}{2m} \frac{d^2}{dx^2} \]
Waa inaan hawlwadeenka Hamiltonian ku dabaqnaa shaqada hirarka \( \psi(x) \):
\[ \hat{H} \psi(x) = -\frac{\hbar^2}{2m} \frac{d^2}{dx^2} \left( \sqrt{\frac{2}{L}} \sin\left( \frac{n\pi x}{L} \right) \right) \]
Qodobkii ugu horreeyay ee ka soo jeeda \( \psi(x) \):
\[ \frac{d}{dx} \left( \sqrt{\frac{2}{L}} \sin\left( \frac{n\pi x}{L} \right) \right) = \sqrt{\frac{2}{L}} \left( \frac{n\pi}{L} \cos\left( \frac{n\pi x}{L} \right) \right) \]
Nooca labaad ee la soo xulay:
\[ \frac{d^2}{dx^2} \left( \sqrt{\frac{2}{L}} \sin\left( \frac{n\pi x}{L} \right) \right) = \sqrt{\frac{2}{L}} \left( -\left( \frac{n\pi}{L} \right)^2 \sin\left( \frac{n\pi x}{L} \right) \right) \]
\[ \frac{d^2}{dx^2} \left( \sqrt{\frac{2}{L}} \sin\left( \frac{n\pi x}{L} \right) \right) = -\frac{n^2 \pi^2}{L^2} \sqrt{\frac{2}{L}} \sin\left( \frac{n\pi x}{L} \right) \]
Hadda, natiijada dib ugu celi hawlwadeenka Hamiltonian:
\[ \hat{H} \psi(x) = -\frac{\hbar^2}{2m} \left( -\frac{n^2 \pi^2}{L^2} \sqrt{\frac{2}{L}} \sin\left( \frac{n\pi x}{L} \right) \right) \]
\[ \hat{H} \psi(x) = \frac{\hbar^2 n^2 \pi^2}{2m L^2} \sqrt{\frac{2}{L}} \sin\left( \frac{n\pi x}{L} \right) \]
Laga bilaabo halkan, waxaan aragnaa taas:
\[ \hat{H} \psi(x) = \frac{\hbar^2 n^2 \pi^2}{2m L^2} \psi(x) \]
Sidaa darteed, tamarta walxaha waa:
\[ E_n = \frac{\hbar^2 n^2 \pi^2}{2m L^2} \]
Ka soo qaad inaan rabno inaan helno tamarta loogu talagalay \( n=1 \):
\[ E_1 = \frac{\hbar^2 \pi^2}{2m L^2} \]
Gabagabo
Xalinta dhibaatooyinka la xiriira ifafaalaha kuantumka waxay u baahan tahay faham adag oo ku saabsan mabaadi'da aasaasiga ah ee makaanikada kuantumka, sida mabda'a hubin la'aanta Heisenberg iyo tamarta walxaha ku jira sanduuq suurtagal ah. Iyada oo loo marayo dhowr dhibaato oo tusaale ah iyo doodahooda, waxaan rajeyneynaa inaan gacan ka geysanno xoojinta fikradaha aasaasiga ah ee makaanikada kuantumka iyo codsigeeda xaaladaha fiisigiska ee kala duwan. Inkasta oo makaanikada kuantumka ay u ekaan karaan kuwo adag, dhibaatooyinka ku dhaqanka iyo fahamka fikradeed waxay si weyn uga caawin doonaan barashada agabkan aasaasiga ah.
