FAQs & their solutions for Module 4: Particle in a three ... - Nptel

FAQs & their solutions for Module 4:

Particle in a three-dimensional box

Question1: For a free particle inside a cube, the potential energy variation of the form

else everywhere

0

; 0

; 0

for )

, ,

(x y z x L y L z L

V (1)

Solve the Schrödinger equation and obtain the energy eigenvalues and eigenfunctions.

Solution1: For a free particle inside the box, the Schrödinger equation is given by

L z

L y

L E x

0 0 0 2 0

2 2

 (2)

The boundary condition is that should vanish everywhere on the surface of the cube. We use the method of separation of variables and write = X(x) Y(y) Z(z) to obtain

2 2

2 2

2 2

2 1 1 2

1

 E dz

Z d Z dy

Y d Y dx

X d

X (3)

The first term is a function of x alone, the second term of y alone, etc., so that each term has to be set equal to a constant. We write

2 2

1 2

kx

dx X d

X (4)

and similar equations for Y(y) and Z(z) with

2 2 2

2 2

 k E k

k_{x y z} (5)

We have set each term equal to a negative constant; otherwise the boundary conditions cannot be satisfied. The solution of Eq. (3) is

x k B x k A x

X( ) sin _x cos _x

and since has to vanish on all points on the surface x = 0 we must have B = 0. Further, for to vanish on all points on the surface x = L, we must have

0 sink_xL or

...

, 2 , 1 with _x

x n

L

k n (6)

Similarly for k_y and k_z. Using Eq.(5) we get the following expression for energy eigenvalues ,..

3 , 2 , 1 , ,

; ) 2 (

2 2 2 2 2 2

z y x z y

x n n n n n

L n

E 

(7) The corresponding normalized wave functions are

L z y n L x n L n z L

y

x 8 sin ^x sin ^y sin ^z )

, , (

2 / 1

3 (8)

Question2: For a free particle inside the box, the energy eigenvalues are given by ,..

3 , 2 , 1 , ,

; ) 2 (

2 2 2 2 2 2

z y x z y

x n n n n n

L n

E 

(9) Calculate the density of states g E where g E dE represents the number of states whose energy lies between E and E dE

Solution2: Let g E dE represents the number of states whose energy lies between E and E dE. If N E represents the total number of states whose energies are less than E , then

E

dE E g E N

0

) ( )

( (10)

and therefore

dE E E dN

g ( )

)

( (11)

Now,

(say)

2 2

2 2

2 2

2

2 L E R

n n n_{x y z}

 (12)

Thus N E will be the number of sets of integers whose sum of square is less than R². In the , ,

x y z

n n n space each point corresponds to a unit volume and if we draw a sphere of radius R then the volume of the positive octant will approximately represent¹ N E ; we have to take the positive octant because n n and _x, _y n_z take positive values. Thus

2 / 3 3 2

3 2 / 3 3

3 ) 2 ( 3

4 8 2 1 )

( L E

R E

N  (13)

where an additional factor of 2 has been introduced as a state can be occupied by two electrons (corresponding to the two spin states). Using Eq. (9) we get

2 / 1 3 2

2 / 3

2 ) 2 ) (

( V E

E

g  (14)

where V( = L³) represents the volume of the box.

1 If the reader finds it difficult to understand he may first try to make the corresponding two-dimensional calculations in which one is interested in finding the number of sets of integers such that n_x² + n_y² < R ². If one takes a graph paper then each corner corresponds to a set of integers and each point can be associated with a unit area.

Thus the number of sets of integers would be R ²/ 4 where the factor ¼ is because of the fact that we are interested only in the positive quadrant.

Question3: In a metal (like Na, Cu etc.) electrons are assumed to be free and at absolute zero (i.e., atT 0) all states below a certain level

F0

E are assumed to be occupied; the energy

F0

E is referred to as the Fermi energy at absolute zero. Using the results of the previous problem, derive an expression for

F0

E .

Solution3: In a metal like Na, electrons are assumed to be free and at absolute zero (i.e., at T = 0) all states below a certain level

F0

E are assumed to be occupied. Thus

0 0

0 0

3/ 2 1/ 2

2 3 0

3/ 2 3/ 2

2 3

(2 ) 2 (2 ) 3

EF

EF

F

N g E dE

V E dE

V E





(15)

Simple manipulations give us

0

2

2 2 / 3

(3 )

F 2

E  n

(16)

where n(= N/V) represents the number of free electrons per unit volume.

Substituting the numerical values of various constants

= 9.1093897 10-31 kg

me and 1.05457266 10 ³⁴ Js

2

 h we get

eV 10

65 . 3

erg 10

84 . 5

3 / 2 15

3 / 2 27

0

n n E_F

(17) where n is measured in cm^-3. For sodium with one free electron per atom and having a density of 0.97 g/cm³ we get

3 22

23

cm electrons/

10 54 . 23 2

97 . 0 10 023 . n 6 Thus

eV 2 .

0 3 EF

Now at T 300 K^ , kT 0.025 eV. Thus

F0

E  kT

Question4: For sodium with one free electron per atom and having a density of 0.97 g/cm³ calculate the value of

F0

E and show that at room temperatures (T 300^o K), the electron gas is almost completely degenerate i.e.,

F0

E  kT.

Solution4: For copper we assume one free electron per atom. Its density is about 8.94 g/cm³ and its atomic mass is 63.5. Thus

23

22 3

6.023 10 8.94

8.48 10 electrons/cm n 63.5

Using the formula derived above we get

F0

E 7.0 eV