Question
एक क्वांटम यांत्रिकी कण निम्नलिखित सोपान विभव पर आपतित होता है$
कण के आपतित और परावर्तित तरंगों के लिए प्रायिकता धारा घनत्व की गणना करें जब E < V0 ।
Answer :
Word Count : 432
एक क्वांटम यांत्रिकी कण जो सैद्धांतिक रूप से एक सोपान विभव (step potential) [ V(x) = \begin{cases} 0, & x<0 \ V_0 > 0, & x>0 \end{cases} ] पर आपतित होता है, उसके लिए समय-निर्भर Schrödinger समीकरण से हम स्थिर-राज्य हल निकालते हैं। यदि कण की ऊर्जा (E < V_0) है, तो हल निम्न प्रकार से होगा। क्षेत्र I ((x<0)): यहाँ विभव शून्य है, इसलिए तरंग समीकरण सरल है: [ -\frac{\hbar^2}{2m} \frac{d^2 \psi}{dx^2} = E \psi \implies \frac{d^2 \psi}{dx^2} + k^2 \psi = 0 ] जहाँ [ k = \frac{\sqrt{2 m E}}{\hbar}. ] हल: [ \psi_I(x) = A e^{i k x} + B e^{-i k x}, ] जहाँ _______ ___ ____ ___ _________ ________ ___ ______.
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एक क्वांटम यांत्रिकी कण जो सैद्धांतिक रूप से एक सोपान विभव (step potential) [ V(x) = \begin{cases} 0, & x<0 \ V_0 > 0, & x>0 \end{cases} ] पर आपतित होता है, उसके लिए समय-निर्भर Schrödinger समीकरण से हम स्थिर-राज्य हल निकालते हैं। यदि कण की ऊर्जा (E < V_0) है, तो हल निम्न प्रकार से होगा। क्षेत्र I ((x<0)): यहाँ विभव शून्य है, इसलिए तरंग समीकरण सरल है: [ -\frac{\hbar^2}{2m} \frac{d^2 \psi}{dx^2} = E \psi \implies \frac{d^2 \psi}{dx^2} + k^2 \psi = 0 ] जहाँ [ k = \frac{\sqrt{2 m E}}{\hbar}. ] हल: [ \psi_I(x) = A e^{i k x} + B e^{-i k x}, ] जहाँ _______ ___ ____ ___ _________ ________ ___ ______.
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