Ngā Tauira Pātai e Matapaki ana i ngā Āhuatanga Irahiko
Ko ngā āhuatanga matūriki, arā, ngā āhuatanga e whakahaerehia ana e te miihini matūriki, e kapi ana i te whānuitanga o ngā ariā me ngā mātāpono e hiahia ana kia mārama hōhonu me te uaua o te pāngarau. Ko te miihini matūriki he peka o te ahupūngao e whakaahua ana i te whanonga o ngā matūriki iti-atomika, pērā i ngā irahiko me ngā photon, kāore e taea te whakamārama e te ahupūngao matarohia. I roto i tēnei tuhinga, ka tūhuratia e mātou ētahi tauira raruraru me ā rātou otinga e pā ana ki ngā āhuatanga matūriki hei āwhina i te mārama ki ngā mātāpono taketake o te miihini matūriki.
Tauira Pātai 1: Te Mātāpono Korekore o Heisenberg
Pātai:
E mōhiotia ana ko te tūranga o te irahiko i roto i te ngota ka inehia me te tika o \( \Delta x = 0.1 \text{ nm} \). Whakatauhia te iti rawa o te koretake i te ine i te nekehanga irahiko (\( \Delta p \)) mā te whakamahi i te mātāpono koretake a Heisenberg.
Whakautu:
E mea ana te mātāpono o Heisenberg mō te koretake:
\[ \Delta x \cdot \Delta p \geq \frac{\hbar}{2} \]
ko \( \hbar \) te pūmau Planck kua whakaitihia, me te uara \( \hbar \approx 1.054 \times 10^{-34} \text{ Js} \).
Whakakapia \( \Delta x = 0.1 \text{ nm} = 0.1 \times 10^{-9} \text{ m} \):
\[ \Delta p \geq \frac{\hbar}{2 \Delta x} \]
\[ \Delta p \geq \frac{1.054 \times 10^{-34}}{2 \times 0.1 \times 10^{-9}} \]
\[ \Delta p \geq \frac{1.054 \times 10^{-34}}{2 \times 10^{-10}} \]
\[ \Delta p \geq \frac{1.054 \times 10^{-34}}{2 \times 10^{-10}} = 5.27 \times 10^{-25} \text{ kg m/s} \]
Nō reira, ko te iti rawa o te koretake i te ine i te nekehanga irahiko ko \( 5.27 \times 10^{-25} \text{ kg m/s} \).
Tauira Pātai 2: Pūngao Pūmanawa i roto i te Pouaka (Matūriki i roto i te Pouaka)
Pātai:
Kei roto i tētahi pouaka kotahi-ahu te matūriki, ko tōna papatipu he m, ko te roa he L. He aha te pūngao taketake (pūngao āhua whenua) o te matūriki?
Whakautu:
Ko te pūngao taketake (pūngao āhua whenua) o tētahi matūriki i roto i tētahi pouaka kotahi-ahu ka hoatuhia e te whārite:
\[ E_n = \frac{n^2 h^2}{8mL^2} \]
Mō te āhua whenua (\( n=1 \)):
\[ E_1 = \frac{h^2}{8mL^2} \]
ko \( h \) te pūmau a Planck \( (h \approx 6.626 \times 10^{-34} \text{ Js}) \).
Me kī ko \( m = 9.109 \times 10^{-31} \text{ kg} \) (te papatipu o te irahiko) me \( L = 1 \times 10^{-9} \text{ m} \):
\[ E_1 = \frac{(6.626 \times 10^{-34})^2}{8 \times 9.109 \times 10^{-31} \times (1 \times 10^{-9})^2} \]
\[ E_1 = \frac{4.39 \times 10^{-67}}{7.287 \times 10^{-50}} \]
\[ E_1 = 6.02 \times 10^{-18} \text{ J} \]
Nō reira, ko te pūngao taketake o te matūriki ko \( 6.02 \times 10^{-18} \text{ J} \).
Tauira 3: Ngā Mahi Whakahaere Hamiltonian mō ngā Mahi Ngaru
Pātai:
Ko te mahi ngaru o tētahi matūriki i roto i tētahi pouaka kotahi-ahu ko \( \psi(x) = \sqrt{\frac{2}{L}} \sin\left(\frac{n\pi x}{L}\right) \) mō \( n=1,2,3,\ldots \). Whakatauhia te pūngao o te matūriki mā te whakamahi i te kaiwhakahaere Hamiltonian \( \hat{H} \).
Whakautu:
Ko te kaiwhakahaere Hamiltonian i te taha kotahi ko:
\[ \hat{H} = -\frac{\hbar^2}{2m} \frac{d^2}{dx^2} \]
Me whakamahi tātou i te kaiwhakahaere Hamiltonian ki te mahi ngaru \( \psi(x) \):
\[ \hat{H} \psi(x) = -\frac{\hbar^2}{2m} \frac{d^2}{dx^2} \left( \sqrt{\frac{2}{L}} \sin\left( \frac{n\pi x}{L} \right) \right) \]
Te taupū tuatahi o \( \psi(x) \):
\[ \frac{d}{dx} \left( \sqrt{\frac{2}{L}} \sin\left( \frac{n\pi x}{L} \right) \right) = \sqrt{\frac{2}{L}} \left( \frac{n\pi}{L} \cos\left( \frac{n\pi x}{L} \right) \right) \]
Te tuarua o ngā pānga:
\[ \frac{d^2}{dx^2} \left( \sqrt{\frac{2}{L}} \sin\left( \frac{n\pi x}{L} \right) \right) = \sqrt{\frac{2}{L}} \left( -\left( \frac{n\pi}{L} \right)^2 \sin\left( \frac{n\pi x}{L} \right) \right) \]
\[ \frac{d^2}{dx^2} \left( \sqrt{\frac{2}{L}} \sin\left( \frac{n\pi x}{L} \right) \right) = -\frac{n^2 \pi^2}{L^2} \sqrt{\frac{2}{L}} \sin\left( \frac{n\pi x}{L} \right) \]
Inaianei, whakakapia te hua ki te kaiwhakahaere Hamiltonian:
\[ \hat{H} \psi(x) = -\frac{\hbar^2}{2m} \left( -\frac{n^2 \pi^2}{L^2} \sqrt{\frac{2}{L}} \sin\left( \frac{n\pi x}{L} \right) \right) \]
\[ \hat{H} \psi(x) = \frac{\hbar^2 n^2 \pi^2}{2m L^2} \sqrt{\frac{2}{L}} \sin\left( \frac{n\pi x}{L} \right) \]
Mai i konei, ka kite tātou:
\[ \hat{H} \psi(x) = \frac{\hbar^2 n^2 \pi^2}{2m L^2} \psi(x) \]
Nō reira, ko te pūngao matūriki ko:
\[ E_n = \frac{\hbar^2 n^2 \pi^2}{2m L^2} \]
Me kī tātou e hiahia ana ki te kimi i te pūngao mō \( n=1 \):
\[ E_1 = \frac{\hbar^2 \pi^2}{2m L^2} \]
Whakamutunga
Ko te whakaoti rapanga e pā ana ki ngā āhuatanga matūriki me mārama pai ki ngā mātāpono taketake o te miihini matūriki, pērā i te mātāpono koretake o Heisenberg me te kaha o ngā matūriki i roto i te pouaka pūmanawa. Mā roto i ētahi tauira rapanga me ā rātou matapakinga, ko te tumanako ka āwhina i te whakapakari i ngā ariā taketake o te miihini matūriki me ōna tono i roto i ngā āhuatanga ahupūngao rerekē. Ahakoa te āhua uaua o te miihini matūriki, ko ngā rapanga mahi me te māramatanga ariā ka tino āwhina i te mōhio ki tēnei rauemi taketake.
