# Energy band structure (3 questions)

1) Three Ge atoms are brought close one to each other up until a distance comparable with the distance between atoms in crystal. Draw a schematics of the valence band, conduction band and of the forbidden band. How many allowed energy states are contained into the valence band? What about the conduction band? Where can be found the valence electrons?

2) A Ge crystal has a mass $m=100 g$. Knowing that the molecular mass is $M =78.52 kg/kmol$ and the Avogadro number is $N_A$, find the number of allowed states $N_c$ from the conduction band.

3) The conduction band of a semiconductor crystal has a width of $Delta E_c=1.8 eV$ and is made by the splitting of a $4p$ level. If the crystal contains $n=6*10^{22} atoms$, find the energy difference $Delta epsilon$ between two adjacent energy levels from the conduction band. Compare this value with the $K_BT$, for $T =290K$.

1. The 4 valence electrons of a Ge atom are found in the states $4s^2$, $4p^2$. If the 3 Ge atoms are brought together up until the distance from the Ge crystal, then the $4s$ and $4p$ levels split themselves in 2 allowed energy bands, each band containing 6 energy levels. Thus, the valence band will have 6 energy levels and each energy level can be occupied by 2 electrons with opposite spin.

The same, the conduction band will have also 6 allowed energy levels. Since 3 Ge atoms have together a total of 12 valence electrons, this means that all the possible states from the valence band will be occupied

2. The total number of atoms from the Ge crystal is

$n =(m*N_A)/M$

Since each atom contributes with 4 possible states (2 opposite states on each energy level), the number of allowed states of the conduction band will be:

$N_c =4n =(4mN_A)/M =3.07*10^{24} states$

3. Since the conduction band is made from the “split” of a $p$ level, this means it contains $3n$ energy levels. Therefore

$Delta epsilon =(Delta E_c)/(3n) =10^{-23} eV$

For $T =290 K$ the value of $K_BT =4*10^{-21} J =0.025 eV$ . Thus

($Delta epsilon)/(K_BT) =4*10^{-22}$