Energy gap between brillouin zones?

[unscientific]

Homework Statement

Consider a monovalent 2D crystal with a rectangular lattice constants $a$ and $b$. Find expressions for the fermi energy and fermi wavevector in 2D. Show that the fermi surface extends beyond first zone if $2a > b\pi$. If the crystal is now divalent, estimate the energy gap between the first and second brillouin zone.

Homework Equations

The Attempt at a Solution

Part(a)
I found the density of states to be $g(E) dE= \frac{1}{\pi} \frac{m}{\hbar^2} dE$. Fermi energy is then found to be $E_F = \frac{n\pi \hbar^2}{m}$. Wavevector is also found to be $k_F = (2n\pi)^{\frac{1}{2}}$. Since it atom is monovalent, $n = \frac{1}{ab}$. The fermi energy and wavevector thus becomes $E_F = \frac{\pi \hbar^2}{mab}$ and $k_F = \left( \frac{2\pi}{ab} \right)^{\frac{1}{2}}$.

How do I continue and show the relation $2a > b\pi$?

 

[theodoros.mihos]

When $a=b$ then $1 < \pi/2$.

 

[unscientific]

When $a=b$ then $1 < \pi/2$.

How did you get that?

 

[theodoros.mihos]

1st zone is between $\pm \pi/2$.

 

[unscientific]

1st zone is between $\pm \pi/2$.

I just solved the first part. The brillouin zone exceeds when
$$k > \frac{\pi}{a}$$
$$\sqrt{\frac{2\pi}{ab}} = \frac{\pi}{a}$$

For the second part,
is the gap simply
$$\frac{\hbar^2}{2m}( k_x^2 - k_y^2) = \frac{\hbar^2 \pi^2}{2m} \left( \frac{1}{a^2} - \frac{1}{b}^2 \right)$$

 

[theodoros.mihos]

Ok is the same. I use numerics. Check values of $a,b$ for left part be smaller than right.

 

[unscientific]

Ok is the same. I use numerics. Check values of $a,b$ for left part be smaller than right.

It's given that $a>b$. Is my part (b) right?

 

[theodoros.mihos]

inverce a,b roles. the answer is indepented to which is a an which is b