1 APPLICATIONS OF SCHRÖDINGER’S EQUATION AND … · Schrödinger’s equation is extremely useful for investigating various quantum ... Schrodinger time independent wave equation

1.1 INTRODUCTIONWe have already noted that Schrödinger was influ-enced by the matter–wave postulate of de Broglie.In order to describe the behaviour of a particle as awave, Schrödinger proposed an equation in 1926 tobe known after his name as Schrödinger’s WaveEquation. The wave function ψ, which is a solu-tion of Schrödinger’s equation, is expected to de-scribe the dynamic behaviour or dynamical state ofa particle. Schrödinger’s equation has been foundto be a fundamental equation in Quantum Mechan-

ics just as Newton’s equation d pFdt

= in Classical

Mechanics or Maxwell’s equations in electromag-netism.

Schrödinger gave two types of wave equa-tions namely time dependent and time independent.The time dependent Schrödinger’s equation isgiven as

( ) ( ) ( ) ( )2

2,

, , ,2

r ti r t V r t r t

t m

ψψ ψ

∂= − ∇ +

∂(1.1)

and time independent Schrödinger’s equation is given as

( ) ( ) ( )22

2 0mr E V r rψ ψ ∇ + − = (1.2)

where the symbols carry their usual meaning.Schrödinger’s equation is extremely useful for investigating various quantum mechanical

problems; viz particle in one and three dimensional box, one dimensional harmonic oscillator,step potential and potential barrier. In the following subsections, we discuss two quantummechanical problems; particle in three dimensional box and potential barrier.

1.2 THE PARTICLE IN A THREE DIMENSIONAL BOXLet us consider a particle which is enclosed inside a rectangular box having edges a, b and c inlength as shown in Fig. 1.1. The potential function V (x, y, z) is having a constant value of zeroin the regions given below:

V (x, y, z) = 0 for 0 < x < a

1APPLICATIONS OF SCHRÖDINGER’S

EQUATION AND BAND THEORY

Erwin Rudolf Josef AlexanderSchrödinger ( 12 August 1887 – 4 Janu-ary 1961) was an Austrian physicist whoachieved fame for his contributions to quan-tum mechanics. In January 1926,Schrödinger published in the Annalen derPhysik the paper Quantisation as an Eigen-value Problem on wave mechanics andwhat is now known as the Schrödingerequation, for which he received the NobelPrize in 1933.

2 Physics–II

V (x, y, z) = 0 for 0 < y < bV (x, y, z) = 0 for 0 < z < c (1.3)

The potential outside the box is infinite. TheSchrodinger time independent wave equation insidethe box for such a particle may be written as

2 2 2

2 2 2 2

2 0m Ex y zψ ψ ψ ψ∂ ∂ ∂

+ + + =∂ ∂ ∂

(1.4)

To separate out the above equation let us put Fig. 1.1 Three dimensional box

( ), , ( ) ( ) ( )x y z X x Y y Z z XYZψ = = (1.5)Substituting the value of ψ from Eq. (1.5), we get

2 2 2

2 2 2 2

2 0X Y Z mYZ ZX XY EXYZx y z

∂ ∂ ∂+ + + =∂ ∂ ∂

Dividing this equation by XYZ , we get2 2 2

2 2 2 2

1 1 1 2 0X Y Z m EX x Y y Z z

∂ ∂ ∂+ + + =

∂ ∂ ∂

⇒2 2 2

2 2 2 2

1 1 1 2X Y Z m EX x Y y Z z

∂ ∂ ∂+ + = −∂ ∂ ∂ (1.6)

For the given energy of the particle, the term 2

2mE − is constant and each term on the left

side is a function of one variable only. If we allow only one of these (x or y or z ) to vary at atime, the sum of the three terms is still equal to the constant on the right hand side. This meansthat each of the three terms on the left is itself a constant and independent of the other variablespresent in it. Let us represent each term on left side of equation, in succession, equal to the

constants 21α− , 2

2α− and 23α− . These have a minus sign because the term on right side of

equation has minus sign. We, therefore, write2

212

1 XX x

α∂ = −∂

(1.7)

⇒2

212 0X X

xα∂ + =

∂ (1.8)

22

22

1 YY y

α∂ = −∂ (1.9)

⇒2

222 0Y Y

yα∂ + =

∂ (1.10)

22

32

1 ZZ z

α∂ = −∂

(1.11)

Y

b

O Xa

Z

c

Applications of Schrödinger’s Equation and Band Theory 3

⇒2

232 0Z Z

zα∂ + =

∂ (1.12)

On substituting equations (1.7), (1.9) and (1.11) in equation (1.6), we obtain

2 2 21 2 3 2

2mEα α α+ + = (1.13)

The solution of equations (1.8), (1.10) and (1.12) are given by

1 1 1 1cos sinX A x B xα α= + (1.14)

2 2 2 2cos sinY A y B yα α= + (1.15)

3 3 3 3cos sinZ A z B zα α= + (1.16)

In the above equations 1 2 3, ,A A A and 1 2 3, ,B B B are constants. It is possible to obtainthe values of these constants by applying the boundary conditions. As ψ vanishes at thesurfaces (boundaries) of infinite potential, it means that

ψ = 0 when x = 0, y = 0, z = 0 and when x = a, y = b, z = cIf these boundary conditions are applied, then we have

1 2 3 0A A A= = = (1.17)

Also 1 0B ≠ ∴ 1sin 0aα = ,i.e., 1 xa nα π= or 1xnaπα = (1.18)

2 0B ≠ ∴ 2sin 0bα = ,i.e., 2 yb nα π= or 2ynbπ

α = (1.19)

3 0B ≠ ∴ 3sin 0cα = ,i.e., 3 zc nα π= or 3zncπα = (1.20)

On substituting equations (1.17 –1.20) in equations (1.14), (1.15) and (1.16), weobtain

1 sin xnX B xaπ

= (1.21)

where nx = 1, 2, 3 ..............

2 sin ynY B y

bπ

= (1.22)

where ny = 1, 2, 3 .............

and 3 sin znZ B zcπ= (1.23)

where nz = 1, 2, 3 ............Thus the complete wave function is given by

( ), ,, , 1 2 3 sin sin sin

x y z

yx y z x zn n n

nn nB B B x y za b c

ππ πψ =

4 Physics–II

⇒ ( ), ,, , sin sin sin

x y z

yx y z x zn n n

nn nK x y za b c

ππ πψ = (1.24)

In the above equation, K is termed as normalization constant. It is possible to obtain thevalue of K by using the normalization condition, i.e.,

0 0 0

* 1a b c

dxdydzψψ =∫∫∫ (1.25)

⇒ 2 2 2 2

0 0 0

sin sin sin 1a b c

yx zn yn x n zK dx dy dz

a b cππ π =∫ ∫ ∫ (1.26)

But2

0 0

1 1sin cos2 2 2

a ax xn x n x adx dxa aπ π = − = ∫ ∫

Similarly 2

0

sin2

byn y bdybπ

=∫

and2

0

sin2

czn z cdzcπ =∫

Thus equation (1.26) becomes

2 . . 12 2 2a b cK =

⇒2 2Kabc

= (1.27)

Therefore, the complete wave function becomes

( ), ,, ,

2 2 sin sin sinx y z

yx y z x zn n n

n yn x n za b cabc

ππ πψ = (1.28)

From equation (1.13), we have

2 2 21 2 3 2

2mEα α α+ + = (1.29)

On substituting the value of 1 2,α α and 3α from equations (1.18 –1.20) into equation(1.29), we obtain

⇒2 22 2 2 2

2 2 2 2

2yx znn n mEa b c

ππ π+ + =

⇒22 22 2

2 2 22x y z

yx zn n n

nn nEm a b c

π = + +

(1.30)

If we consider a box that is cubical in shape such that a = b = c, energy can beexpressed as


2 2 22 2 2 2 2 2

2 22 8x y z x y zhE n n n n n n

ma maπ = + + = + +

⇒ 2 2 21 x y zE E n n n = + + (1.31)

where2 2 2

1 2 22 8hE

ma maπ= = (1.32)

1.2.1 Degeneracy of Energy LevelsIn the lowest quantum state (1,1,1), in which xn , yn and zn respectively are equal to unity, it

is seen that 2

12

3 38

hE Ema

= = . There is only one set of quantum numbers that gives this energy

state, and this level is said to be non-degenerate.

If we now consider the second energy state as shown in Fig. 1.2, it is seen that there

are three sets (1,1,2), (1,2,1) and (2,1,1) of the quantum number 2 2,x yn n and 2zn that will give

the same energy level 2

2

68

hEma

= such a level is said to be degenerate, and in this particular case

the level is triply degenerate. For a cubical box, it can be concluded from the Fig. 1.2, thatvirtually all the energy levels are degenerate to some degree.

x y z

Fig. 1.2 Energy levels, degree of degeneracy and quantum numbers of a particle in a cubical box

Example 1.1 A free particle is confined in a cubical box of side a. Write the eigen values andeigen functions for an energy state represented by 4x y zn n n+ + = . What is the order ofdegeneracy in this case? [RTU B.Tech. I sem. 2007]

x y z

6 Physics–II

Solution: Given 4x y zn n n+ + =for this case, there are three possible combination of ,x yn n and zn as (2, 1, 1); (1, 2, 1);

(1, 1, 2)The eigen values and eigen functions for 3-dimensional cubical box of side a, are given

by

( )2 2

2 2 222n x y zE n n n

maπ= + +

and ( ), ,x y zψ 3

8 sin sin sinyx znn nx y z

a a a aππ π =

There are three eigen functions corresponding to a single energy value 2 2

2

62

Emaπ= .

Hence the order of degeneracy is 3.The corresponding eigen functions are

( )2,1,1ψ 3

8 2sin sin sinx y za a a a

π π π =

ψ (1,2,1) 3


π π π =

and ψ (1,1,2) 3


π π π = Ans.

Example 1.2 Answer the following questions with respect to a particular cubical box of side a.

(i) Is 1x y zn n n= = = state degenerate?

(ii) What shall happen to the degeneracies for 4x y zn n n+ + + if the box is not

cubical but rectangular parallelopiped with sides a, b and c such that ?a b c= ≠Solution: The eigen value of a particle in three dimensional box of sides a, b and c is

22 22 2

2 2 22yx z

n

nn nEm a b c

π = + +

and the eigen function for that particle is

( ) 8, , sin sin sinyx znn n

x y z x y zabc a b c

ππ πψ

=

(i) For a cubical box a = b = c and for 1,x y zn n n= = = there is only one eigen

function corresponding to eigen value 2 2

2

32maπ , so it is non-degenerate.

(ii) Given that 4x y zn n n+ + = and a b c= ≠ . Therefore, there are three states(2, 1, 1), (1, 2, 1) and (1, 1, 2).


2 2 2 2

2, 1,1 2 2 2 2 2

4 1 1 5 12 2

Em a a c m a c

π π = + + = +

2 2 2 2

1, 2, 1 2 2 2 2 2

1 4 1 5 12 2

Em a a c m a c

π π = + + = +

2 2 2 2

1, 1, 2 2 2 2 2 2

1 1 4 2 42 2

Em a a c m a c

π π = + + = +

Thus the energy 2 2

2 2

5 12m a c

π + is doubly degenerate corresponding to eigen func-

tions 2,1,1ψ and 1, 2, 1ψ while energy 2 2

2 2

2 42m a c

π + is non-degenerate.

1.3 THE POTENTIAL BARRIER (TUNNEL EFFECT)Let us consider the motion of a particle in the x-direction, under the action of a potential whichis zero every where except over a certain region AB where it has a constant value 0V . Such aregion AB is called a square potential barrier (Fig. 1.3).

According to classical mechanics a particletravelling in region I (on the left of the barrier) willovercome the barrier if its initial kinetic energy E isgreater than the barrier height 0V . If its energy issmaller than 0V , it will be totally reflected and theparticle can never be found is region III (on the rightof the barrier). Quantum mechanical calculationspredict that the particle can penetrate through thebarrier even when its energy is less than 0V . Thisphenomenon is known as tunnelling and the effectis known as ‘Tunnel Effect’.

Consider a particle of mass m and energy E in the region I. The potential field can be represented as

( ) 0V x = 0x < Region I

0( )V x V= 0 x a≤ ≤ Region II

( ) 0V x = x a> Region III (1.33)The situation represents the encounter between a free particle and a potential barrier.

The particle can be turned back by the barrier or it can penetrate through . Thus in the region Ithe desired wave function has to represent a particle moving to the right (incident particle) anda particle moving to the left (reflected particle). In the region III the wave function has torepresent only a particle moving to the right (transmitted particle).

In the force free, region a particle moving in a definite direction (x-direction) with adefinite energy E has a definite momentum xP . However its position is completely uncertainand it can be represented by a travelling wave. In the region I and III where ( ) 0V x = , the waveequation is given by

Fig. 1.3 Potential barrier

VV(x)

III III

Ba

xOA

0

8 Physics–II

2

2 2

2 0m Exψ ψ∂ + =

∂

⇒2

22 0k

xψ ψ∂ + =

∂ (1.34)

where 22

2mEk =

The solutions of this equation are

( )1ikx ikxx Ae Beψ −= + for 0x < (Region I) (1.35)

and ( )3ikxx Ceψ = for x a> (Region III) (1.36)

For x a> , ikxe− term has been omitted as there is no wave travelling in - x direction.The physical meaning of the coefficients A, B and C can be understood in terms of

probability current density. If v is the speed of a particle with propagation constant k, then theflux of particles in the incident beam is given by

f1 = probability density of particles in the incident beam × particle velocity.

⇒ ( )1 *ikx ikxf Ae A e v−=

( ) 2*AA v A v= =Similarly the flux of particles in the reflected beam and transmitted beam may be

written as 22f B v= and 2

3f C v= , respectively. Therefore reflection coefficient (R) and

transmission coefficient (T) are defined as 2 2

2

B v BRAA v

= = and 2 2

2

C v CTAA v

= = respec-

tively. Thus the absolute values of A, B and C are not important only their ratios are of moresignificance.

In the region II ( )( )00 and xx a V V≤ ≤ = , the solution of Schrödinger wave equation will de-

pend on the energy of particle, i.e., 0E V< or 0E V> . Therefore both cases are discussedseparately.Case I When 0E V< (Tunnel Effect)

In this case the wave equation is

( )2

20 22 2

2 0m V Exψ ψ∂ − − =

∂

⇒2

2222 0

xψ α ψ∂ − =

∂ (1.37)

where( )

12

02

2m V Eα

−=


The solution of this wave equation is

( )2x xx Fe Geα αψ −= + (1.38)

Now we will use the boundary condition to calculate the reflection coefficient andtransmission coefficient.

At x = 0, 1 2ψ ψ= and 1 2

x xψ ψ∂ ∂=∂ ∂

(1.39)

Applying this boundary condition on equations (1.35) and (1.38)A + B = F + G (1.40)i k (A – B) = α (F – G) (1.41)

Similarly = x a, 2 3ψ ψ= , and 32

x xψψ ∂∂ =

∂ ∂ (1.42)

using this boundary condition on equations (1.36) and (1.38)ika a aCe Fe Geα α−= + (1.43)

( )ika a aikCe Fe Geα αα −= − (1.44)From equations (1.40) and (1.41)

2 1 1 1 1i iA F G F Gik ik k kα α α α = + + − = − + + (1.45)

and 2 1 1 1 1i iB F G F Gik ik k kα α α α = − + + = + + −

(1.46)

⇒( ) ( )( ) ( )k i F G k iB

A k i F G k iα αα α

+ + −=

− + +

( ) ( )

( ) ( )

Gk i k iFGk i k iF

α α

α α

+ + −=

− + + (1.47)

Now dividing equation (1.44) by equation (1.43)

( )a a

a a

Fe Geik

Fe Ge

α α

α α

α −

−

−=

+

⇒ a a a aikFe ikGe Fe Geα α α αα α− −+ = −

⇒ ( ) ( )a aik Fe ik Geα αα α −− = − +

⇒ 2 aG ik eF ik

ααα

− = − +

2 ak i e

k iαα

α− − = − − +

10 Physics–II

2 aG k i eF k i

ααα

+ = − − (1.48)

Put this value in equation (1.47), we get

( ) ( )

( ) ( )

2

2

a

a

k ik i k i ek iB

A k ik i k i ek i

α

α

αα αααα αα

+ + + − − − = + − + + − −

( ) ( )( ) ( )

2

2 2 2

1 a

a

k i e k i

k i k i e

α

α

α α

α α

+ − − =− − +

( )

( ) ( )2 2 2

2 2 2 2 2

1

2 2

a

a

k e

k i k k i k e

α

α

α

α α α α

+ − =− − − + −

( )

( )2 2 2

2 2 2 2

1

1 2 1

a

a a

k e

k e i k e

α

α α

α

α α

+ − = − − − +

( )( )

2 2 2

2 2 2 2

1

1 2 1

a

a a

k eBA k e i k e

α

α α

α

α α

+ − = − − + +

Since sinh2

x xe ex−−= and cosh

2

x xe ex−+=

∴( )

( )2 2

2 2 2

a a

a a a a

k e eBA k e e i k e e

α α

α α α α

α

α α

−

− −

+ − = − − + +

( )( )

2 2

2 2

sinh

sinh 2 cosh

k aBA k a i k a

α α

α α α α

+=

− + (1.49)

Now equation (1.43) dividing by equation (1.45), we get

2 1 1

ika a aCe Fe Gei iA F Gk k

α α

α α

−+= − + +

⇒ ( ) ( )

2 a aGk e eC F

GA k i k iF

α α

α α

− + =− + +


Put the value of GF

using equation (1.48)

( ) ( )

2

2

2 a a a ika

a

k ik e e e ek iC

A k ik i k i ek i

α α α

α

αα

αα αα

− − + + − − = + − + + − −

( ) ( )

( ) ( )2 2 2

2 a

a

ke k i k i

k i k i e

α

α

α α

α α

− − + =− − +

( ) ( )2 2 2 2 2

42 2

a ika

a

i ke ek i k i e

α

α

αα α α α

−

= −− − − − +

( )( ) ( )2 2 2 2

41 2 1

a ika

a a

i ke ek e i k e

α

α α

αα α

−−=− − − +

( )( ) ( )2 2 2 2

41 2 1

a ika

a a

i ke ek e i k e

α

α α

αα α

−

=− − + +

( )2 2

2

22 2

ika

a a a a

i kee e e ek i kα α α α

α

α α

−

− −=

− +− +

( )2 2

2sinh 2 cosh

ikaC i keA k a i k a

αα α α α

−

=− + (1.50)

Now Reflection coefficient 2 *B B BR

A A A = =

From equation (1.49)

( )( ) ( )

22 2 2

2 2 2 2

sinh

sinh 2 cosh sinh 2 cosh

k aR

k a i k a k a i k a

α α

α α α α α α α α

+= − + − −

( )

( )22 2 2

4 4 2 2 2 2 2 2

sinh

2 sinh 4 cosh

k a

k k a k a

α α

α α α α α

+=

+ − +

12 Physics–II

( )( ) ( )

22 2 2

4 4 2 2 2 2 2 2

sinh

2 sinh 4 1 sinh

k a

k k a k a

α α

α α α α α

+=

+ − + +

( )( )

22 2 2

22 2 2 2 2

sinh

sinh 4

k aR

k a k

α α

α α α

+=

+ + (1.51)

( )2 2

22 2 2

1

41sinh

Rk

k aα

α α

= + +

(1.52)

As we know that 22

2mEk = and ( )202

2m V Eα = −

( )( )

1

02 2

0

41

sinh

E V ER

V aα

− − = +

(1.53)

and Transmission coefficient 2 *C C CT

A A A = =

From equation (1.50)

( )2 2

22 2 2 2 2

4

sinh 4

kTk a k

α

α α α=

+ + (1.54)

( )

122 2 2

2 2

sinh1

4

k ak

α αα

− + = +

( )( )

12 20

0

sinh1

4V a

TE V E

α−

= +

− (1.55)

Thus 0T ≠ , this implies that the particle has finite probability of penetrating the poten-tial barrier. This penetration is due to the wave nature of matter and is known as tunnel effect orquantum tunneling. This effect provides explanation for the following phenomena:

(i) the electrical breakdown of insulators(ii) the switching action of a tunnel diode(iii) the emission of α-particles from a radioactive nuclei(iv) the emission of electrons from a cold metallic surface, i.e., electron emission from a

metal under the action of an electric field.(v) the reverse breakdown of semiconductor diodes

From equations (1.51) and (1.54) it can be proved that R + T = 1. (1.56)


The transmission coefficient is maximum when E approaches 0V . So aα is very small, i.e.,

0E V→ , 0aα → ⇒ sinh a aα α→Using equation (1.55)

( )( )0

0

122 2 2

014

E V

V aCTA E V E

α−

→

= = + −

( ) ( )

( )0

00

12 20

2

21

4E V

V m V E a

E V E

−

→

− = + −

( )

0

12 20

212

E V

m V aE

−

→

= +

( ) 120

212

m V aT

−

= +

(1.57)

From equation (1.57) it is clear that the width a or the height of potential barrier ( )0V or bothincreases, the transmission probability of particle decreases.

If 1aα >> , then 1sinh2

aa eαα ≈

Put this value in equation (1.55), the transmission coefficient becomes

( )( )0

12 2

0116

aV eT

E V E

α−

≈ + −

⇒( )( )

0 22

0

16a

E V ET e

Vα−

−≈ (1.58)

⇒ As aα increases, T decreases exponentially.

In the case of 0

E V< , the wave function will be of the form as depicted in Fig. (1.4).

14 Physics–II

II IIITransmitted Wave

E

x=0 x=a

V0

ψI

Fig. 1.4 Barrier penetration

Case II When 0

E V> the Schrödinger equations for all three regions

21

12 2

2 0m Exψ ψ∂ + =

∂

or2

2112 0k

xψ ψ∂ + =

∂x < 0 (1.59)

222

22 ' 0kxψ ψ∂ + =

∂0 < x < a (1.60)

and2

2332 0k

xψ ψ∂

+ =∂

x > a (1.61)

where 22

2mEk = and ( )0

22

2' mk E V= − (1.62)

As we know that ( )0

22

2m V Eα = − in previous case 0

E V< .

Therefore ( )0

22

2m E Vα = − −

2 2 2i kα =⇒ 'ikα = ∴ sinh sin 'a k aα = (1.63)Thus the value of reflection coefficient (R) and transmission coefficient (T) can be calculatedby putting 'ikα = in equations (1.52) and (1.54).

( )

12 2

2 2 2

4 '1' sin 'k kR

k k k a

− = +

− and

( ) 12 2 2

2 2

' sin '1

4 'k k k a

Tk k

− − = +

(1.64)

Put the values of k and k’ then R and T become

0


( )( )

0

1

2 20

41

sin '

E E VR

V K a

− − = +

(1.65)

and( )

( )

12 20

0

sin '1

4V K a

TE E V

−

= + −

(1.66)

From equation (1.64) it can be proved that R + T = 1. Keeping 0E V> , as E approaches its

lower limit 0V , 'K becomes small so that ( )02 2 2 2

2

2sin ' '

m E VK a K a a

−≈ = .

In this limiting condition

( ) ( )( )

0

122 20 0

20

21

4E V

V m E V aCTA E E V

−

→

−= = +

−

120

212

mV aT−

= +

(1.67)

i.e., the transmission probability of wave packet is less than one, even when the energy of theparticle is greater than the height of potential barrier. When ' ,2 ,3 ,......k a nπ π π π= = then

sin ' 0k a = and T = 1, i.e., there is perfect transmission.

Since 2

'kπλ′ =

∴ 2 a nπ πλ

=′

⇒ 2na λ′

=

Hence when the barrier thickness is equal to anintegral multiple of half wavelength, the barrierdoes not obstruct the motion of particle, i.e., T = 1. Fig. 1.5 Oscillations of transmission

coefficient

As E increases (with 0E V> ) the transmission coefficient oscillates and for large values of E itapproaches unity (Fig. 1.5).

Example 1.3 The potential barrier problem is a good approximation to the problem of anelectron trapped inside but near the surface of metal. Calculate the probability of transmissionthat a 1.0 eV electron will penetrate a potential barrier of 4.0 eV when the barrier width

is 2.0 Å .

T

O 1 2 3 4

(EN )0

16 Physics–II

Solution: The transmission coefficient is given by

( ) ( )002

0

16 2exp 2E V E aT m V E

V− − ≈ −

Here E = 1 eV, V0 = 4 eV, 102Å 2 10 ma −= = × ,

341.054 10−= × J-s, 319.1 10 kgm −= ×

( ) ( ) 190 4 1 eV 3 1.6 10 JV E −− = − = × ×

∴1 116 1 exp4 4

T = − 10 31 19

34

2 2 10 2 9.1 10 3 1.6 101.054 10

− − −

−

− × × × × × × × ×

× ~ 0.084 8.4%= Ans.

Thus, only about 8 electrons of energy 1.0 eV out of every hundred, penetrate thebarrier.

Example 1.4 Consider an electron whose total energy is 5 eV approaching a barrier whoseheight is 6 eV and width is 7Å. Find out the de Broglie wavelength of incident electron and

probability of transmission through the barrier. (Mass of electron 319.1 10 kg.,m −= × Planck’s

constant 346.6 10−= × J-s). [Raj. B.E.I.2000]Solution: The de Broglie wavelength of an electron is given by

12.27 ÅV

λ =

Here V = 5 Volt (as electron energy is 5 eV)

∴ 12.27 5.485

λ = = Å Ans.

The transmission coefficient is given by

( ) ( )002

0


V− − = −

Here 0V = 6 eV, 107 10 ma −= × , 341.054 10−= × J-sE = 5 eV

∴ ( ) 1031 19

34

16 5 6 5 2 7 10exp 2 9.1 10 1 1.6 1036 1.054 10

T−

− −−

× × − × ×= × × × × × × 31.69 10−≈ ×

= 0.169 % Ans.

Example 1.5 Electrons of energy 2 eV are incident on a potential barrier of height 5 eV andwidth 5 Å. Find tunnel probability for these electrons. [RTU B.Tech. I sem. 2008]


Solution: Given E = 2 eV 192 1.6 10−= × × J

V0 = 5 eV 195 1.6 10−= × × J and a = 5 Å = 5× 10–10 mThe tunnel probability is given by

T ( )0 22

0

16 aE V Ee

Vα−−

=

where α ( )1

2

02

2m V E = −

( )( )

12

31 19

234

2 9.1 10 5 2 1.6 10

1.054 10

− −

−

× × × − × × = ×

12

929.12 3 101.11

× = ×

( )1 9226.23 3 10= × ×

95.12 3 10= × ×95.12 1.732 10= × ×

98.8628 10= ×

Now T( ) 9 1016 2 5 2

exp 2 8.867 10 5 1025

−× × − = − × × × ×

[ ]96 exp 8.86725

= −

( )3.84 exp 8.867= −43.84 1.42 10−= × ×

T 45.452 10−= × = 0. 054 % Ans.

Example 1.6 A beam of electrons is incident on a potential barrier of height 5 eV and width 0.3

nm. What should be the energy of the electrons so that 13

of them are able to penetrate

through the barrier ?Solution: Given V0 = 6 eV and a = 0.3 nm = 3×10–10 m

The probability of transmission is given by

( )0 22

0

16 aE V ET e

Vα−−

= 20

aT e α−=

18 Physics–II

where ( )00 2

0

16E V ET

V−

= maximum transmission probability

and ( )02

2m V Eα = −

Given 0

3TT =

∴20

03aT T e α−=

⇒ 2 3ae α =⇒ 2 ln 3aα =

⇒ 2(2 ) (ln 3)aα =

2 202

2 ( )4. (ln 3)m V E a− =

2 2

0 2

(ln 3)( )8

V Ema

− =

34 2 2

0 31 10 2

(1.054 10 ) (1.098)( )8 9.1 10 (3 10 )

V E−

− −

×− =× × × ×

190 0.201 10 JV E −− = ×

19

0 19

0.201 10 eV 0.125eV1.6 10

V E−

−

×− = =×

⇒ 0( 0.125)eV 6 0.125E V= − = −

5.875eVE = Ans.

1.4 THEORY OF α – DECAYThe theory of α-decay was explained by George Gamow in 1928 on the basis of quantummechanical barrier penetration. There are certain nuclei which emit α–particles and get con-verted into new nuclei with atomic number less by two and mass number less by four. Thisprocess is known as α -decay and is depicted as follows:

4 42 HeA A

Z Z zX Y −−→ +

When anα -particle is present inside the nucleus, the energy of α -particle can not begreater than the height of the potential barrier which is existing at the surface of the nucleus. Itimplies that according to classical mechanics it is never possible for the α -particle to come outof the nucleus.

The first difficulty involves the energy of α -particle in the field of the nucleus. It wasconcluded from Rutherford’s scattering experiments that the force experienced by α -particules


is a Coulomb force which acts at very small distance ( )1410 m−= from the centre of the nucleus.After this direct distance, Coulomb’s law breaks down and the boundary of the nucleus starts.Thus, there is only Coulomb potential outside the nucleus which is depicted in Fig. 1.6. Assoon as the α -particle enters the nucleus (0 < r < a), it will be under the influence of verystrong attractive nuclear forces. These forces are represented by a potential well as depicted inFig. 1.6.

If the radius of the radium nucleus is159.1 10 m−× , the Coulomb potential outside the

nucleus is 27.8 MeV. However, the energy ofα-particle emitted by a radium nucleus is 4.88MeV only. Now the question arises : How cana α -particle of 4.88 MeV energy go througha potential barrier of 27.8 MeV ?

The second difficulty is concernedwith the law of causality. According to the ra-dioactive decay law, the amount of radium leftafter 1620 years should be half. After severalhalf lives, a small fraction of radium should beleft. Now the question arises. Why are only someradium atoms decaying in the first few yearsand why are some others surviving for thousands of years ?

When quantum mechanics was applied to the problem of α -decay, the difficulties ofclassical mechanics as described above disappear. According to quantum mechanics, everyparticle is assigned a wave aspect. If this fact is taken into consideration, then the electrostaticpotential barrier, although very high, can not completely rule out the passage of wave throughit. There always exists certain probability of the particle to penetrate through the barrier, howsmall the energy might be. It is possible to obtain some quantitative features of the theory fromthe expression of the transmission coefficient.

( ) ( )002

0


V− − = − (1.68)

Now an attempt is being made to calculate the half life time in case of uranium nucleus

which is having radius of about 1410 m− . There is an evidence that the α -particle moves back

and forth freely with an average speed of 710 m/s. Thus the α -particle will strike the barrier7 14 2110 /10 10− = times per second. The probability that it penetrates the barrier is equal to the

transmittance (T).Hence the probability (P) that α -particle leaks out from nucleus in one second, which

is called decay constant ( λ ), i.e.,

λ = 1021 T per sec

( ) ( )021

020

16 210 exp 2E V E a m V E

V− − = × − (1.69)

Fig. 1.6 Potential for the ααααα-particle in the nucleus

~

V

0V27.8 MeV

E = 4.88 Meα

2(z–2)e2

4 rπε 00V =

0r rO

V

20 Physics–II

For uranium, E = 4.2 MeV and 0V = 30 MeV.

Mass of α -particle 274 1.67 10 kg−= × × .341.054 10−= × J-s, a 142 10 m−= ×

Thus, ( ) ( )( )

( )2102

4.2MeV 30 42 MeV 210 16 exp 230MeV

a m V Eλ× − − = × × −

⇒

( )27 13 1421

34

2 2 4 1.67 10 25.8 1.6 10 2 1016 4.2 25.810 exp30 30 1.054 10

λ− − −

−

× × × × × × × ×× × = × × ×

( )8921 21 391.9 10 1.9 2.22 10 10e ≈− −= × = × × × 184.218 10−≈ × per sec

101.328 10−= × per yearTherefore half-life time of nucleus is

91

2

0.693 5.21 10Tλ

= = × years (1.70)

But experimentally, 12

T had been found to be 1910 years. The discrepancy has been

ascribed to the fact that the barrier is not rectangular. This result explains why most energeticparticles are decaying in first few years and why least energetic particles are surviving forthousands of years.

Example 1.7 Find the height of the potential barrier for α-particles emitted from radon ( )22286 Rn

assuming that the effective nuclear radius is given by 15 1 30 1.5 10 mr A−= × .

Solution: ( ) 2

00 0

2 24Z e

Vrπε

−=

Here Z = 86, 191.6 10 ce −= ×

9

0

1 9 104πε

= × , ( )1 3150 1.5 10 222r −= ×

∴( ) ( )

( )

219 9

0 1/ 315

2 86 2 1.6 10 9 10

1.5 10 222V

−

−

× − × × × ×=

× ×

626.62 10 eV= × 26.62 MeV= Ans.

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