Podcast
Questions and Answers
Using the Kronig-Penney model, show that for $p << 1$ the energy of the lowest energy band is $E = \frac{h^2p}{mq^2}$.
Using the Kronig-Penney model, show that for $p << 1$ the energy of the lowest energy band is $E = \frac{h^2p}{mq^2}$.
To show that the energy of the lowest energy band is $E = \frac{h^2p}{mq^2}$ for $p << 1$, we can use the Kronig-Penney model to derive the energy dispersion relation and then analyze it in the limit of small potential strength, where $p <<1$ .
Study Notes
Kronig-Penney Model and Energy Band
- Using the Kronig-Penney model, for values of p much less than 1, the energy of the lowest energy band is given by the equation: E = (ћ²p²) / (mq²)
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Description
Explore the fundamentals of the Kronig-Penney model and its implications for energy bands in solid-state physics. This quiz delves into the mathematical representation of energy levels, especially for values of p much less than 1. Test your understanding of key concepts and formulas in this crucial area of physics.
