THE Kronig- Penney Model - New PDF

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Band theory of solids...

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THE KRONIG-PENNEY MODEL Q;State the drawbacks of free electron theory. Explain THE KRONIG-PENNEY MODEL. Ans: Drawbacks of free electron theory:The free electron theory of metals has the following drawbacks: 1. It could not explain why certain solid have large number of free electron and behave as conductor whereas some other have hardly any free electron and act as insulator. 2. The property of semi-conductors could not be explained by this theory. 3. This theory cannot explain the large mean free paths of electrons in metals at low temperatures. 4. This theory says that resistivity varies with temperature T as √𝑇 , whereas it is found to vary linearly with temperature. 5. This theory fails to explain molar heat capacity of metals. THE KRONIG-PENNEY MODEL: Kroning-Penny model describes all the essential features of the behavior of electrons in a periodic potential by assuming a simple one dimensional model. Kronig and Penney assumed that the potential energy of an electron has the form of a periodic array of square wells, as shown in fig. 1. The period of potential is (a+b) because, The potential energy of the electron = 0 when 0 < x < a = V0 when –b < x < 0 V(x) V0 Each of the potential energy wells is a rough approximation for the potential in the vicinity of an atom. The Schrodinger wave equation for the two regions can be expressed as:

For 0 < x < a: for –b < x < 0 :

𝑑 2 𝜓1

𝑑 𝑥2 𝑑 2 𝜓2 𝑑𝑥 2

2𝑚 + ℏ2

-b

(i) 0

a a

(a+b)

b 2(a+b)

x

+ ℏ2 𝐸𝜓1=0 2𝑚

(ii)

Fig. 1. -------------------- (1) and

(𝐸 − 𝑉0 )𝜓2=0 --------------------(2)

The solution of equation (1) is 𝜓1 (𝑥) =A𝑒 𝑖𝛼𝑥 + 𝐵𝑒 −𝑖𝛼𝑥 ----------------------------(3)

√2𝑚𝐸 -------------------------- (4) Where 𝛼= ℏ If E < V0 then equation (2) will become 𝑑 2 𝜓2 2𝑚 − ℏ2 (𝑉0 − 𝐸)𝜓2=0 -------------------------- (5) 𝑑𝑥 2 The solution of this equation is 𝜓2 (𝑥) =C𝑒 𝛽𝑥 + 𝐷𝑒 −𝛽𝑥 ------------------------------- (6)

0 Where 𝛽= -----------------------------------(7) ℏ A, B,C and D are arbitrary constants. Since the potential is periodic the wave functions represented by equations (3) and (6) must be of the Bloch form. ∴ 𝜓1 (𝑥) = 𝑢1 (𝑥)𝑒 𝑖𝑘𝑥 ----------------------------(8) and 𝜓2(𝑥) = 𝑢2 (𝑥)𝑒 𝑖𝑘𝑥 ----------------------------(9) Substituting these values in equations (3) and (6) we get 𝑢1 (𝑥) = 𝐴𝑒 𝑖(𝛼−𝑘 )𝑥 + 𝐵𝑒 −𝑖(𝛼+𝑘 )𝑥 ------------------------------ (10) And 𝑢2 (𝑥) = 𝐶𝑒 (𝛽−𝑖𝑘 )𝑥 + 𝐷𝑒 −(𝛽+𝑖𝑘)𝑥 ------------------------- (11) Now the following four boundary conditions for continuity of wave functions and their derivatives must be satisfied. 𝑑𝑢 𝑑𝑢 𝑢1 (0) = 𝑢2 (0) ( 𝑑𝑥1 ) = ( 𝑑𝑥2 )

√2𝑚(𝑉 −𝐸)

𝑢1 (𝑎) = 𝑢2 (−𝑏)

(

𝑥=0 𝑑𝑢1 ) 𝑑𝑥 𝑥=𝑎

=(

) 𝑑𝑥

𝑑𝑢2

𝑥=0

𝑥=−𝑏

Applying these boundary conditions to equations (8) and (9) we get 𝐴+𝐵 = 𝐶 +𝐷 𝐴𝑖(𝛼 − 𝑘) − 𝐵𝑖(𝛼 + 𝑘) = 𝐶 (𝛽 − 𝑖𝑘) − 𝐷(𝛽 + 𝑖𝑘) 𝐴𝑒 𝑖(𝛼−𝑘 )𝑎 + 𝐵𝑒 −𝑖(𝛼+𝑘 )𝑎 = 𝐶𝑒 −(𝛽−𝑖𝑘)𝑏 + 𝐷𝑒 (𝛽+𝑖𝑘)𝑏 𝐴𝑖(𝛼 − 𝑘)𝑒 𝑖(𝛼−𝑘 )𝑎 − 𝐵𝑖(𝛼 + 𝑘)𝑒 −𝑖(𝛼+𝑘 )𝑎 = 𝐶 (𝛽 − 𝑖𝑘)𝑒 −(𝛽−𝑖𝑘)𝑏 − 𝐷(𝛽 + 𝑖𝑘)𝑒 (𝛽+𝑖𝑘)𝑏

A non-zero solution of these equation of these equations exists only if the determinant of the co-efficients of AB,C and D vanishes, i.e., 1 1 𝑖(𝛼 − 𝑘) −𝑖(𝛼 + 𝑘) | 𝑒 𝑖(𝛼−𝑘 )𝑎 𝑒 −𝑖(𝛼+𝑘 )𝑎 (𝛼−𝑘 )𝑎 𝑖 𝑖(𝛼 − 𝑘)𝑒 −𝑖(𝛼 + 𝑘)𝑒 −𝑖(𝛼+𝑘 )𝑎 On solving this determinant we get

𝛽 2 −𝛼2 2𝛼𝛽

1 ( 𝛽 − 𝑖𝑘)

1 −(𝛽 + 𝑖𝑘) |=0 𝑒 (𝛽+𝑖𝑘)𝑏 (𝛽+𝑖𝑘)𝑏 −(𝛽 + 𝑖𝑘)𝑒

𝑒 −(𝛽−𝑖𝑘)𝑏 (𝛽 − 𝑖𝑘)𝑒 −(𝛽−𝑖𝑘)𝑏

∙ 𝑠𝑖𝑛ℎ𝛽𝑏 ∙ 𝑠𝑖𝑛𝛼𝑎 + 𝑐𝑜𝑠ℎ𝛽𝑏 𝑐𝑜𝑠𝛼𝑎 = 𝑐𝑜𝑠𝑘(𝑎 + 𝑏) ------------------------------------- (12)

Kroning and Penny simplified the equation by assuming that 𝑉0 → ∞ and 𝑏 → 0 but the product 𝑉0 𝑏 remains finite. So the potential become Dirac delta function. Under these conditions we get 𝑠𝑖𝑛ℎ𝛽𝑏 → 𝛽𝑏, 𝑐𝑜𝑠ℎ𝛽𝑏 → 1 and 𝛽2 − 𝛼 2 → 𝛽2 So equation (12) becomes 𝛽2

2𝛼𝛽

∙ 𝛽𝑏 ∙ 𝑠𝑖𝑛𝛼𝑎 + 𝑐𝑜𝑠𝛼𝑎 = 𝑐𝑜𝑠𝑘𝑎

⇒ 𝛽2

𝑎𝑏 2

𝑠𝑖𝑛𝛼𝑎 𝛼𝑎

+ 𝑐𝑜𝑠𝛼𝑎 = 𝑐𝑜𝑠𝑘𝑎 ---------------------- (13)

= =a measure of area of potential well. We define a quantity 𝑃 as 𝑃 = If P is increased the electron is more strongly bound to a particular potential well.. 𝑠𝑖𝑛𝛼𝑎 Equation (13) becomes 𝑃 𝛼𝑎 + 𝑐𝑜𝑠𝛼𝑎 = 𝑐𝑜𝑠𝑘𝑎 --------------------------------- (14) Band Gap: The plot of L.H.S of equation (14) as a function of 𝛼𝑎 is shown In figure 2. 𝑎𝑏 𝛽2 2

𝑚𝑉0 𝑎𝑏 ℏ2

∙

Fig 2

Since −1 ≤ 𝑐𝑜𝑠𝑘𝑎 ≤ +1, only those value of 𝛼𝑎 are allowed which make L.H.S between +1 𝑎𝑛𝑑 − 1 . Since 𝛼 is related to ‘E’, it means that electron may possess energies within certain bands but not outside them. The allowed and forbidden bands of energy are shown by shaded and un-shaded regions. Discussion: The following conclusions may be drawn from the above diagram 1. The energy spectrum of electrons consists of alternate allowed and forbidden energy bands. 2. The width of the allowed energy bands increases with 𝛼𝑎 or with the energy. 3. For large value of P, the L.H.S of equation (14) will cross +1 to -1 region at a steep angle. Thus the allowed band will be narrow. 4. When 𝑃 → ∞, equation (14) gives 𝑠𝑖𝑛𝛼𝑎 = 0 ⇒ 𝛼𝑎 = ± 𝑛𝜋, 𝑛 = 1,2,3,….. ∴ th

where En= energy of n level. In this case allowed regions become extremely narrow.

5. When 𝑃 → 0, equation (14) gives 𝑐𝑜𝑠𝛼𝑎 = 𝑐𝑜𝑠𝑘𝑎 ⇒ 𝛼 = 𝑘 ⇒ 6.

This is the energy of free electron. All the boundaries of allowed bands is defined as 𝑐𝑜𝑠𝑘𝑎 = ±1 = 𝑐𝑜𝑠𝑛𝜋, 𝑛 = 1,2,3, … … ⇒ 𝑘 =

𝑛𝜋 𝑎

2𝑚𝐸 = ℏ2

√2𝑚𝐸𝑛 ℏ

𝑘2 ⇒ 𝐸 =

𝑎 = ±𝑛𝜋 ⇒ 𝐸𝑛 =

ℏ2 𝑘 2 2𝑚

These values of k define the boundaries of the brillouine zone. These results are shown in figure3. On the extreme left (P=0) the energy spectrum is almost continuous. On the extreme right (P =∞)a line spectrum is resulted. For any other value of P, the position and width of the allowed and forbidden bands are obtained by drawing a vertical line.

𝑛2 𝜋2 ℏ2 2𝑚𝑎2

,...