![Now let us consider the energy level 𝐸1 which is below the Fermi level.
The probability that the level 𝐸1 is not occupied.
1 − 𝑓 𝐸1 = [1 − 𝑓 𝐸 𝐹 − ∆𝐸 ]
= 1 −
1
1+𝑒(𝐸 𝐹−∆𝐸−𝐸 𝐹)/𝑘𝑇
= 1 −
1
1+𝑒−∆𝐸/𝑘𝑇
=
𝑒−∆𝐸/𝑘𝑇
1+𝑒−∆𝐸/𝑘𝑇
=
1
1+𝑒∆𝐸/𝑘𝑇 …(2)
From (1) and (2) we have
𝑓 𝐸2 = 1 − 𝑓 𝐸1
This means that the probability of an energy level [𝐸 𝐹 + ∆𝐸] (∆𝐸
above the Fermi level) being occupied is the same as the probability of
energy level [𝐸 𝐹 − ∆𝐸] (∆𝐸 below the Fermi level)being vacant.](https://image.slidesharecdn.com/problemsonfermi-160725110949/85/Problems-on-fermi-part-2-7-320.jpg)
Problems on fermi part 2
- 1. A Presentation on Problems on Fermi -Dirac distribution function By Ms.Chetana Magadum Smt. Kasturbai Walchand College,Sangli
- 2. Problems on Fermi –Dirac distribution function:- 1. The Fermi level for potassium is 2.1eV. Calculate the velocity of the electron at the Fermi level. Solution:- We have the formula, 𝐸 𝐹 = 1 2 𝑚𝑣 𝐹 2 Therefore, 𝑣 𝐹 = 2𝐸 𝐹 𝑚 1 2 = 2 × 2.1𝑒𝑉 × 1.602 × 10−19 𝐽/𝑒𝑉 9.11 × 10−31 𝑘𝑔 1 2 𝑣 𝐹 = 8.6 × 10 5 𝑚/𝑠.
- 3. 2. In a solid, consider the energy level lying 0.01 eV below Fermi level. What was is the probability of this level not being occupied by an electron? Given : Energy Difference 𝐸 𝐹 − 𝐸 = 0.01𝑒𝑉 Thermal energy at room temperature, kT=0.026eV Solution:- We have the formula, 𝑓 𝐸 = 1 1 + 𝑒−(𝐸 𝐹−𝐸)/𝑘𝑇 𝑓 𝐸 = 1 1 + 𝑒−0.01𝑒𝑉/0.026𝑒𝑉 = 1 1 + 𝑒−0.3846 = 0.595 Therefore, 𝑝 = 1 − 𝑓 𝐸 = 1 − 0.595 = 0.405
- 4. 3.Calculate the probabilities for an electronic state to be occupied at 20o C, if the energy of these states lies 0.11eV above and 0.11eV below the Fermi level. Solution:- Probability of occupying an energy level E, 𝑓 𝐸 = 1 1 + 𝑒(𝐸−𝐸 𝐹)/𝑘𝑇 Probability of occupying an energy level 0.11eV above Fermi level 𝑓 𝐸 = 1 1 + 𝑒0.11𝑒𝑉/𝑘𝑇 = 1 1 + 𝑒4.2307 = 0.0126 Probability of occupying an energy level 0.11eV below Fermi level 𝑓 𝐸 = 1 1 + 𝑒−0.11𝑒𝑉/𝑘𝑇 = 1 1 + 𝑒−4.2307 = 0.987
- 5. 4. Evaluate the Fermi function for energy kT above the Fermi energy. Given: Fermi function 𝑓 𝐸 =? Boltzmann’s con stant 𝑘 = 1.38 × 10−23 𝐽/𝐾 Kelvin’s temperature T=300K Formula: 𝑓 𝐸 = 1 1+𝑒(𝐸−𝐸 𝐹)/𝑘𝑇 Solution:- We have the formula, 𝑓 𝐸 = 1 1+𝑒(𝐸−𝐸 𝐹)/𝑘𝑇 Here, 𝐸 − 𝐸 𝐹 = 𝑘𝑇 then 𝑓 𝐸 = 1 1 + 𝑒1 = 1 1 + 2.78 = 1 3.78 = 0.269
- 6. 5) Show that the probability of finding an electron of energy ∆𝑬 above the Fermi level is same as the probability of not finding as electron at energy ∆𝑬 below the Fermi level. Solution: Let us consider an energy level 𝐸2 above the Fermi energy level by an amount of energy ∆𝐸 .The probability that the energy level 𝐸2 occupied is 𝑓 𝐸2 = 𝑓 𝐸 𝐹 + ∆𝐸 = 1 1+𝑒(𝐸2−𝐸 𝐹)/𝑘𝑇 = 1 1+𝑒(𝐸 𝐹+∆𝐸−𝐸 𝐹)/𝑘𝑇 = 1 1+𝑒∆𝐸/𝑘𝑇 …(1)
- 7. Now let us consider the energy level 𝐸1 which is below the Fermi level. The probability that the level 𝐸1 is not occupied. 1 − 𝑓 𝐸1 = [1 − 𝑓 𝐸 𝐹 − ∆𝐸 ] = 1 − 1 1+𝑒(𝐸 𝐹−∆𝐸−𝐸 𝐹)/𝑘𝑇 = 1 − 1 1+𝑒−∆𝐸/𝑘𝑇 = 𝑒−∆𝐸/𝑘𝑇 1+𝑒−∆𝐸/𝑘𝑇 = 1 1+𝑒∆𝐸/𝑘𝑇 …(2) From (1) and (2) we have 𝑓 𝐸2 = 1 − 𝑓 𝐸1 This means that the probability of an energy level [𝐸 𝐹 + ∆𝐸] (∆𝐸 above the Fermi level) being occupied is the same as the probability of energy level [𝐸 𝐹 − ∆𝐸] (∆𝐸 below the Fermi level)being vacant.
- 8. Thank you.