# How to draw different zone schemes in the Kronig-Penney model?

Kronig-Penney model does not give us an equation of how the electron energy $$E$$ depends on its wavenumber $$k$$. Then how do they draw the $$E-k$$ curves in different zone schemes such as extended, periodic and reduced? If one can explain the method of drawing the extended zone scheme that will solve for my purpose.

• "Kronig-Penney model does not give us an equation of how the electron energy 𝐸 depends on its wavenumber 𝑘." -- Why not? Isn't that the point - that it is a nice simple solvable model? Commented Feb 29, 2020 at 22:40
• Here is a reference to the KP model en.wikipedia.org/wiki/… Could you tell me what is $E$ as a function of $k$? Commented Feb 29, 2020 at 22:46
• It seems to be all there, just after "For energy values inside the well (E < 0), we get: " Commented Feb 29, 2020 at 22:51

$$\cos(ka) = \cos(\alpha (a-b))\cos(\beta b) – \frac{\alpha ^2 + \beta^2}{2\alpha \beta}\sin(\alpha (a-b))\sin(βb)$$
$$\alpha ^2 = \frac{2mE}{\hbar^2}$$ and $$\beta^2 = \frac{2m(E + V_0 )}{\hbar^2}$$
When the electron energy increases, values of $$\alpha$$ e $$\beta$$ come close. When $$\alpha \approx \beta$$ the expression: $$\frac{\alpha ^2 + \beta^2}{2\alpha \beta} \approx 1$$. And the formula tends to: $$\cos(ka) = \cos(\alpha (a-b))\cos(\beta b)–\sin(\alpha (a-b))\sin(\beta b)$$ But the right side of that equation is $$\cos(\alpha(a-b)+\beta b)$$.
One simple solution is: $$ka = \alpha (a-b) + \beta b$$. When $$\alpha \approx \beta$$, $$k \approx \alpha$$, and how energy is proportional to $$\alpha^2$$, graphics $$E$$ x $$k$$ is approximately a parabola.
But according to Bloch theorem, $$k$$ is restricted to the Brillouin zone. This can be achieved by $$k = \frac{\cos^{-1}[\cos(\alpha (a - b)+\beta b)] }{a}$$ No matter how energy grows, the values of $$k$$ oscillates in the interval $$\left[- \frac{\pi}{a+b} , \frac{\pi}{a+b}\right]$$, what is the Brillouin zone of that periodic frame.