Question

Figure 41-21 shows three leveled levels in a band and also the Fermi level for the material. The temperature is 0K. Rank the three levels according to the probability of occupation, greatest first if the temperature is (a) 0K and (b) 1000K. (c) At the latter temperature, rank the levels according to the density of states N(E) there, greatest first.

Short answer

1. The ranking of the three levels according to the probability of occupation at T = 0 K is $level3>level2=level1$.
2. The ranking of the three levels according to the probability of occupation at $T=1000K$is $level3>level2>level1$.
3. At the latter temperature, the ranking of the levels according to the density of states is $\mathrm{level}3>\mathrm{level}2<\mathrm{level}1$.

Step-by-step solution

The given data

1. The levels in a band and also the Fermi level for the material is shown.
2. Temperature, T = 0k and T = 1000 k

Understanding the concept of band levels

Using the occupancy probability and the energy at the three levels, we can get the required probabilities of occupying the band at the two different temperatures. Again, the density of states equation indicates that the density value is directly proportional to the square root of energy. Hence, it determines the value of density of all three leveled energies.

Formulae:

The occupancy probability of the energy level (E) , according to Fermi-Dirac Statistics, having Fermi energy $\left(E_F\right)$ and absolute temperature (T) , is given as-

$P\left(E\right)=\frac{1}{e^{(E-E_F)/kT}+1}$

The density of states associated with the conduction electrons of a metal, is given as-

$N\left(E\right)=\frac{8\sqrt{2}{\mathrm{\pi m}}^{3/2}}{h^3}{E}^{1/2}$

$whereh=6.63\times {10}^{-34}J.s$and $m=9.1x{10}^{-31}kg$ is the mass of the electron.

a) Calculation of the rank of the levels according to the probability at T=0K

At absolute zero temperature ( T = 0 K ), the probability using equation (i) is 1 for all energies less than the Fermi energy and zero for energies greater than the Fermi energy.

Thus, energy at level 3 is 1, while at the other two levels is zero as their energies are greater than Fermi energy.

Hence, according to the diagram, according to occupancy probability the levels are ranked as

level 3 > level 2 = level 1.

b) Calculation of the rank of the levels according to the probability at T=1000K

At temperature, T = 1000 K a few electrons whose energies were less than that of the Fermi energy at T = 0 K move up to states with energies slightly greater than the Fermi energy. That means probability at Fermi energy is $P\left(E_F\right)=0.5$.

Thus, the occupancy probability of the levels below the Fermi energy level is higher as compared to the ones above the Fermi level. Also, for the levels present above the fermi level, the higher the level, the less is the occupancy probability.

Hence, at 1000 K , the three levels are ranked as-

level 3>level 2>level 1.

c) Calculation of ranking of the levels according to the number density

From equation (ii), we get that the density of states is proportional to square root of energy

$\mathrm{N}\left(E\right)\infty E^{1/2}$.

According the diagram, the ranking of the energies of the three levels is $E_3<E_2<E_1$.