Chapter 3 Quantum Mechanics (Pp 73-95)

5/28/2018 Chapter 3 Quantum Mechanics (Pp 73-95)

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CHAPTER 3

QUANTUM MECHANICS

3-1 EXPECTION VALUES, PROBABILITY,

NORMALIZATION AND ORTHOGONALITY

PROPERTIESProblem 3-1

The normalized eigenfunction of a particle in a one-

dimensional box is given by

=

L

xn

Lxn

sin

2)( for Lx =====< nL

ip

h

Problem 3-3

The eigenfunction of a particle in a one-dimensional box is

given by

=

L

xnAxn

sin)( for Lx


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Problem 3-4

The normalized eigenfunction of a particle in nth

state for a

one-dimensional box is given by

=

L

xn

Lxn

sin

2)( for Lx


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L

nm dxxx0

)()( where 0>m , 0>n and nm

( ) ( )=L

dxLxnLxmL 0

/sin/sin2

dxL

xnm

L

xnm

L

L

+

=0

)(

cos

)(

cos

1

L

Lnm

Lxnm

Lnm

Lxnm

L0

/)(}/)sin{(

/)(}/)sin{(1

+

+

=

0= since 0)sin( =k for integral values of k.

As 0)()(0

=L

nm dxxx therefore the given eigenfunctions

are orthogonal.Problem 3-6

An object in one-dimension is described by3x= for 0 < x < 1

= 0 elsewhere(a) What is the probability of finding the object within the

interval (0 , 0.5)?(b) What is average position of the object?

Solution

(a) =

==

5.0

0

5.0

0

32

5.0

0

125.03

33 x

dxxdx

(b) dxxxxdxx )3)()(3(

1

0

1

0 =

43

433

1

0

41

0

3=

== x

dxx


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Problem 3-7

Show that

=

2exp)(

224/12

0

xx

is a normalized eigenfunction.

Solution

Here we evaluate the following integral

= dxI o*0

= dx

xxI

2exp

2exp

224/12224/12

=

=

0

2222

2/12

)exp(2

)exp( dxxdxxI

Substitute xy = then dydx

1= . It may be noted that the

limits of integration remain same for new variable. Hence

=

0

2 )exp(12

dyyI

)2/1(1

)exp(21

0

2

=

=

dyyI

=

0

212 )exp(2)( dyyyn nQ

( ) 11 =

=

I = )2/1(Q

Hence the given function is a normalized eigenfunction.


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Problem 3-8

The eigenfunction for a certain object is defined as

)(cos)( 2 xAx = for22


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382

=A or38

=A

(b)The normalized eigenfunction can be written as

)(cos38

)( 2 xx

= for22


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3-2 POTENTIAL STEP

Problem 3-9

A particle of energy E approaches a potential step of heightVo. What should be the ratio (E/Vo) so that the reflectioncoefficient is 0.5? B.U. B.Sc. 2009A

SolutionThe reflection coefficient is given by

21

21

kk

kkR

+

=

0

0

0

0

(22

(22

VEE

VEE

VEmmE

VEmmE

R+

=

+

=

hh

hh

)/(

1

1

1)/()/(

1)/()/(0

00

00VExwhere

xx

xx

VEVE

VEVER =

+

=

+

=

1

15.0

+

=

xx

xx since R = 0.5

115.05.0 =+ xxxx

xx 5.015.1 = or xx =13Square both sides

xx = )1(9 or 98 =x or89

=x or 125.189

0

==

V

E

Problem 3-10

A particle of energy 8.1 eVapproaches a potential step ofheight 8 eV. Calculate the probability of reflection.Solution

The coefficient of reflection in terms of energy of incidentparticle E and height of potential step can be writtenas


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)/(1

1

1)/()/(

1)/()/(0

00

00VExwhere

xx

xx

VEVE

VEVER =

+

=

+

=

Now eVVeVE 8,1.8 0 == therefore

0125.1)8/1.8()/( 0 === VEx .

Hence 8.010125.10125.1

10125.10125.1=+

=R The desired probability of reflection of particle from thepotential step is 0.8 or 80 percent.


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3-3 POTENTIAL BARRIER

Problem 3-11

A beam of electrons of energy 5 eVis allowed to fall on one-dimensional potential barrier of height 10 eVand width 1 .Calculate the percentage of electrons transmitted.

SolutionFor E < Vo the transmission coefficient is given by1

00

22

)]/(1)[/(4

)(sinh1

+=

VEVE

akT

Now

34

19310

2 10055.1

)10602.1)(510)(10109.9(2)(2

=

=

h

EVmk

1102 10145.1

= mk

Hence

33.0)}2/1(1){2/1(4

)]101)(10146.1[(sinh1

110102

=

+=

T 3

i.e. 33.3 % or one third of incident electrons will be transmitted.Problem 3-12

Assume that an alpha particle has energy 10 MeV andapproaches a potential barrier of height equal to 30 MeV.Determine the width of the potential barrier if the

transmission coefficient is 0.002.Solution

For E < Vo the transmission coefficient is given by1

00

22

)]/(1)[/(4)(sinh1

+=

VEVEakT

)]/(1)[/(4)(sinh

11

00

22

VEVE

ak

T +=

)]/(1)[/(4)(sinh

11

00

22

VEVE

ak

T =


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]1)/1)][(/(1)[/(4)(sinh 0022

= TVEVEak

Nowh

)(2 02

EVmk

=

34

19627

2 10055.1

)10602.1)(10)(1030)(10645.6(2

=k

11510956.1 = m and]1)/1)][(/(1)[/(4 00 TVEVE

56.4431002.01

3010

13010

4 =

=

Substitute these values in Eq(1)56.443])10956.1[(sinh 152 = a

061.2156.443])10956.1sinh[( 15 == a

741.3)061.21(sinh])10956.1[( 115 == a

ma 15

15

10913.110956.1

741.3 =

=

The desired thickness of the potential barrier isfmma 913.110913.1 15 == .

Problem 3-13

A potential barrier has a height of 8 eV and width10102 m. Calculate the lowest energy of the incident

electron in order to have 100percent transmission.Solution

The transmission coefficient for E > Vois given by1

00

32

]1)/)[(/(4)(sin1

+=VEVE

akT

It is clear that the transmission will be 100 percent if,.......3,2,3 =ak

For lowest energy electrons we have=ak3


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=

aVEm

h

)(2 0

22

20 )(2 =

h

aVEm

2

22

0 2maVE

h=

eVJVma

E 8)102)(10109.9(2

)10066.1(

2 21031

2342

02

22

+

=+=

h

eVeVE 8)10602.1()102)(10109.9(2

)10066.1(1921031

2342

+

=

eVeVeVE 41.17841.9 =+=

Problem 3-14

A proton and a deuteron (which has the same charge but

twice the mass) are incident on a barrier of thickness 10 fmand height 10 MeV. Each particle has a kinetic energy of 3.0MeV. Find the transmission probabilities for them.Solution

The approximate formula for transmission probability is

)2exp(11600

kLV

E

V

ET

whereh

)(2 0 EVmk

= . Now

36.3103

1103

1611600

=

=

V

E

V

E

For proton

h

)(2 0 EVmk

p =

34

19623

10055.1

)10620.1)(10)(310)(10673.1(2

=k


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11410806.5 = mk 612.11)1010)(10806.5(22 1514 == kL

The desired probability for proton is510043.3)612.11exp()36.3( ==T

For deuteron, we have

h

))(2(20

EVmk p

=

11414 10211.8)10806.5(2 == mk

422.16)1010)(10211.8(22 1514 == kL The desired transmission probability for deuteron is

710479.2)422.16exp()36.3( ==T

Problem 3-15

When 1.0 eVelectrons are incident on a potential barrier of8.0 eV(such as the work function of a metal), what fraction

of the electrons penetrate their barrier if it is 5.0 wide?Solution

Here we can use the approximate formula for transmissioncoefficient (or probability)

)2exp(11600

kLV

E

V

ET

whereh

)(2 0 EVmk

= . Now

75.1

8

11

8

116116

00

=

=

V

E

V

E and

11034

1931

10355.110055.1

)10620.1)(18)(10109.9(2

=

= mk

55.13)105)(10355.1(22 1010 == kL 610282.2)55.13exp()75.1( ==T


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Problem 3-16

A potential barrier has a height 8 eV and thickness10102 m. What is the lowest energy incident electron may

have and be 100percent transmitted?Solution

The transmission coefficient for E > V0 is given by1

32

2

31

23

21 )(sin4

41

+= ak

kk

kkT

It is clear that the transmission coefficient will be 100 percent if nak == ..,.........3,2,3

The lowest energy electron will be that for which=ak3

=

aVEm

20 )(2

h

02

22

2 VmaE += h

eVJinE 8)102)(10109.9(2

)10055.1(21031

2342

+

=

eVeVinE 8)10602.1()102)(10109.9(2

)10055.1(1921031

2342

+

=

eVeVeVE 41.17841.9 =+=


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3-4 INFINITE POTENTIAL WELL

OR PARTICLE IN A BOX

Problem 3-17

An electron trapped in an infinite potential well of length

100 pm. What are the energies of its three lowest allowed

states? P.U. B.Sc. 2002Solution

The energy of electron in infinite potential well is given by

20

22

8 Lmhn

En = in Joules

eLm

hnEn 2

0

22

8= in eV

)10602.1()10100)(10109.9(8)10626.6(

1921231

2342

=

nEn

2

61.37 n= in eVNow ,1=n eVE 61.37)1)(61.37( 21 == ,2=n eVE 44.150)2)(61.37( 22 == ,3=n eVE 49.338)3)(61.37( 23 ==

The desired values of energy are 37.61 eV, 150.44 eV and338.49 eV.Problem 3-18

Find the lowest three energies, in MeV, of a proton trapped

in an infinite potential well of width 5 fm.

Solution

The energy of proton in infinite potential well is given by

2

22

8 Lmhn

Ep

n = in Joules

empL

hnEn 2

22

8= in eV


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)10602.1()105)(10673.1(8)10626.6(

1921527

2342

=

nEn

eVinn26 )1019.8( =

MeVnEn219.8=

Now ,1=n MeVE 19.8)1)(19.8( 21 ==

,2=n MeVE 76.32)2)(19.8( 22 == ,3=n MeVE 71.73)3)(19.8( 23 ==

The desired values of energy of the trapped proton are8.19MeV,32.76MeVand 73.71MeV.

Problem 3-19

Consider an electron trapped in an infinite well whose width

is 98.5 pm. If it is in a state with n = 15, what are (a) its

energy? (b) The uncertainty in its momentum? (c) The

uncertainty in its position?

Solution(a)The energy of electron in infinite potential well is given by

20

22

8 Lmhn

En = in Joules

eLm

hnEn 2

0

22

8= in eV

Now 15=n and mpmL 12105.985.98 == , therefore

)10602.1()105.98)(10109.9(8)10626.6()15(

1921231

2342

=E

keVeVE 72.81072.8 3 ==

(b)0

2

0

2202

0 2221

m

p

m

vmvmE ===

Emp 02 2=

Emp 02=


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keVEcmpc 4.94)72.8)(511(22 20 ===

keVMeVcm 511511.020 ==Q The direction of motion of the electron is not known. Theelectron bounces back and forth between the walls of giveninfinite potential well. Hence the uncertainty in momentum isgiven by

L

hn

L

hn

Lm

hnmEmpp

24822 2

22

20

22

00 ==

===

(c) As electron can be anywhere inside the well, there theuncertainty in position will be 98.5 pm.

Problem 3-20

What must be the width of an infinite well such that a

trapped electron in the n = 3 state has an energy of 4.70 eV.

Solution

The energy of the electron in an infinite potential well is given

by2

22

8mL

hnEn =

For n = 3,2

2

3 89

Lm

hE = or

3

22

89

Em

hL =

or)10602.170.4)(10109.9(8

)10626.6(3

8

31931

34

3

==

Em

hL

mL1910486.8 = or 8.486

Problem 3-21The ground-state energy of an electron in infinite potential

well is 2.6 eV. What will be the ground-state energy if the

width of the well is doubled?

Solution

The energy of the electron in an infinite potential well is givenby


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2

22

8mLhn

En = or 21

LEn

If the width of the well is doubled, then its energy will becomeone-fourth of the first value i.e.

eVE

E 65.046.2

41

1 ===

Problem 3-22An electron, trapped in an infinite well of width 253 pm, is

in the ground (n = 1) state. How much energy must it absorb

to jump up to the third excited (n = 4) state?

Solution

The energy absorbed by the electron will be

2

2

2

2

2

2

14 8

15

88

16

mL

h

mL

h

mL

hEE ==

21231

234

)10253)(10109.9(8)10626.6(15

=

keVorJ 74.810399.1 15=

Problem 3-23

(a) Calculate the smallest allowed energy of an electron

confined to an infinitely deep well with a width equal to the

diameter of an atomic nucleus (about14104.1 m).

(b) Repeat for a neutron.

Solution

The energy of the electron in an infinite potential well is given

by 2

22

8mLhn

En =

. The value of lowest energy is obtained bysubstituting n = 1 in above equation i.e.

mmmL

hE

40

214

234

2

2

1

10800.2

)104.1(8

)10626.6(

8

=

==

(a)For electron emm = and


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GeVJE 918.110074.310109.9

10800.2 1031

40

1 ==

=

Q JeV 1910602.11 = (b) For proton pmm = and

MeVJE 045.110674.1

10673.1

10800.2 1327

40

1 ==

=

Problem 3-24

A proton confined in a one-dimensional box has energy 400keVin its first excited state. How wide is the box?Solution

The energy of the particle in one-dimensional box is given by

2

22

8mLhn

En =

For first excited state, we substitute n = 2

2

2

2

22

2 28

)2(

mL

h

mL

h

E ==

2

22

2mEh

L = or22mE

hL =

Now kgmmsJh p2734 10673.1,10626.6 === and

JJkeVE 14196 10408.6)10602.1)(10)(400(400 === therefore

mL 141427

34

10525.4)10408.6)(10673.1(2

10626.6

=

=


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CONCEPTUAL QUESTIONS

(1) The wavefunction associated with a physical system is, in

general, a complex quantity but does contain all relevant

information needed to describe the system. How will you get

any significant result from this wavefunction?

Answer: - By measuring | |2 =which is always real.(2)What are the units of one dimensional wavefunction?Answer: - Probability per unit distance or m-1.(3) What is equivalent operator for linear momentum?

Answer: -dx

dip h

(4) What will be the effect on transmission coefficient if the

width of the potential barrier is increased?

Answer: - The transmission coefficient T will decrease.(5) Name an application of potential barrier.

Answer: - Alpha decay.

(6) What will be the effect on the number of bound states ofa potential well if it is made deeper and / or broader?

Answer: - The number of bound states will increase.


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ADDITIONAL PROBLEMS

(1)A particle is defined in the space


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Answers

(1) (a)4/1

2

=

A (b) )2exp(

2 22/1

x

(c) 1

(2) (a)54

3

a (b) 0 (3) (49/48) (4) 7109.4

(5) 1910056.2 (6) 904.5)/( 12 =TT

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Chapter 3 Quantum Mechanics (Pp 73-95)