ELEMENTS OF 1 QUANTUM MECHANICSnewagepublishers.com/samplechapter/001944.pdf · c. Since λν = c...

1

1.1 INTRODUCTION

A revolution took place in physics between 1900 and 1930. This was the eraof a new and more general scheme called quantum mechanics. This newapproach was highly successful in explaining the behaviour of atoms, mol-ecules and nuclei. Moreover, the quantum theory raduces to classical physicswhen applied to macroscopic systems.

The basic ideas of quantum theory were first introduced by Max Planck,but most of the subsequent mathematical developments and interpretationswere made by a number of distinguished physicists including Einstein, Bohr,Schrödinger, de Broglie, Heisenberg, Born and Dirac.

This chapter is simply an introduction to the underlying ideas of quan-tum theory and the wave-particle nature of matter. We shall also discusssome simple applications of quantum theory.

11111.2.2.2.2.2 WAWAWAWAWAVE-PARTICLE DUVE-PARTICLE DUVE-PARTICLE DUVE-PARTICLE DUVE-PARTICLE DUALITALITALITALITALITYYYYY

It is well recognised that the same light beam that can eject photoelectronsfrom a metallic surface can also be diffracted by a grating. In other words, thephoton concept and the wave theory of light complement each other. Thus,all forms of electromagnetic radiation can be described from two points ofview. In one extreme electromagnetic waves describe interference and dif-fraction pattern formed by a large number of photons while in other extreme,the photon description is natural when we deal with highly energetic photonof very short wavelength.

Light and other electromagnetic radiation sometimes act like waves andsometimes like particles. Interference and diffraction demonstrate wavebehaviour while emission and absorption of photons demonstrate the par-ticle behaviour. Besides waves that sometimes acts like particles, quantummechanics extend the concept of wave-particle duality to include particlesthat sometimes show wavelike behaviour. In these situations, a particle ismodelled as an inherently spread-out entity that cannot be described as apoint with a perfectly definite position and velocity.

1ELEMENTS OF

QUANTUMMECHANICS

2 LASER SYSTEMS AND APPLICATIONS

11111.3.3.3.3.3 dedededede BR BR BR BR BROGLIE WOGLIE WOGLIE WOGLIE WOGLIE WAAAAAVESVESVESVESVES

de Broglie postulated that a free particle with mass m moving with speed vshould have a wavelength λ related to its momentum exactly the same wayas for a photon. A photon of light of frequency ν has the momentum.

p = hc

.

Since λν = c so, the wavelength of a photon is specified by its momentumaccording to the relation

λ = hp

. ...(1.1)

de Broglie suggested that eqn. (1.1) is a completely general one that appliesto material particles as well as to photons. The momentum of a particle ofmass m and velocity v is p = mv and its de Broglie wavelength is accordingly.

λ = hmv

. ...(1.2)

In eqn. (1.2) m is the relativistic mass

m =m

v c02 2

.1

The wave and particle aspects of moving bodies can never be observedat the same time. In certain situations a moving body exhibits wave proper-ties and in others it exhibits particle properties. Which set of properties ismost conspicuous depends upon how its de Broglie wavelength compareswith its dimensions and the dimensions of whatever it interacts with.

de Broglie Wave Velocity

If we call the de Broglie wave velocity as W, we may writeW = νλ.

Where λ is the de Broglie wavelength and ν is the frequency. To find thefrequency, we equate the quantum expression E = hν with the relativisticformula for total energy E = mc2

hν = mc2

ν = mch

2.

The de Broglie wave velocity is therefore

W = νλ = mc h ch mv v

2 2.

Since particle velocity v must be less than the velocity of light c, de

Broglie waves always travel faster then light. In order to understand thisunexpected result, we must understand distinction between phase velocityand group velocity. Phase velocity is the velocity of wave.

ELEMENTS OF QUANTUM MECHANICS 3

11111.4.4.4.4.4 WAWAWAWAWAVE EQUVE EQUVE EQUVE EQUVE EQUAAAAATIONTIONTIONTIONTION

The equation of the wave travelling in the direction of increasing x and whosevibrations are in the y direction can be given as

y = A cos 2π ν xtW

where A is the amplitude of vibrations (that is, their maximum displacementon either side of the x-axis) and ν their frequency. The wave has some speedW and travels the distance x in time t.

Since the wave speed W is given byW = νλ.

We have wave equation

y = A cos 2π xt . ...(1.3)

Equation (1.3) is more convenient to use. The most widely used descrip-tion of a wave, however, is still another form with angular frequency andwave number. Angular frequency ω and wave number k can be defined bythe formulas

ω = 2πν, k = W

2 .

The unit of ω is the radian per second and that of k is the radian permeter. In terms of ω and k eqn. (1.3) becomes

y = A cos (ωt – kx). ...(1.4)

11111.5.5.5.5.5 PHASE AND GRPHASE AND GRPHASE AND GRPHASE AND GRPHASE AND GROUP VELOUP VELOUP VELOUP VELOUP VELOCITIESOCITIESOCITIESOCITIESOCITIES

The wave representation of a moving particle corresponds to a wave packetor wave group as shown in Fig. 1.1.

Mathematically a wave group canbe described in terms of the superpo-sition of individual waves of differentwavelengths whose interference withone another results in the variation inamplitude. If the velocities of thewaves are the same, the velocity withwhich the wave group travels is the common wave velocity. If the wavevelocity varies with wavelength, the different individual waves do not pro-ceed together and the wave group has a velocity different from those of thewaves that compose it.

Let us suppose that a wave group arises from the combination of thefollowing two waves that have the same amplitude A but differ in angularfrequency and wave number.

Fig. 1.1. Wave group


y1 = A cos (ωt – kx)y2 = A cos [(ω + dω)t – (k + dk)x].

The resultant displacement y at any time t and any position x is the sumof y1 and y2.

y = y1 + y2 = 2A cos 12

[(2ω + dω)t – (2k + dk)x]

cos 12

(dωt – dkx).

Since dω and dk are small compared with ω and k respectively.2ω + dω ≈ 2ω2k + dk ≈ 2k

y = 2A cos (ωt – kx) cos d dkt x2 2

...(1.5)

Equation (1.5) represents a wave of angular frequency ω and wave num-ber k which has superimposed upon it a wave (the process is known as

modulation) of angular frequency d2

and wave number dk

2

. The effect

of this modulation is to produce successive ‘wave groups’.The ‘phase velocity’ or wave velocity W is given by

W = νλ = k

22

while the group velocity u of the ‘group of waves’ is given by

u = ddk

.

The group velocity may be greater or less than the phase velocity. Itdepends on the manner in which phase velocity varies with wave number.For light waves in vacuum, the group and phase velocities are the same.

The angular frequency and wave number of the de Broglie waves asso-ciated with a particle of rest mass m0 moving with the velocity v are

Angular frequency of de Broglie waves

ω = 2πν = m cmc

h h v c

220

2 2

22.

1

Wave number of de Broglie waves

k =m vmv

h h v c02 2

222 .1

de Broglie phase velocity

W = ck v

2

which exceeds both the velocity of the particle v and the velocity of light c,since v < c.


The group velocity u of the de Broglie waves associated with the particleis

u =d dvd

dk dk dv

ddk =

m v

h v c0

2 2 3/2

2

(1 )

and dkdv

=m

h v c0

2 2 3/2

2

(1 )

so u = v.The de Broglie wave group associated with a moving particle travels

with the same velocity as the particle.The wave velocity W of de Broglie waves has no simple physical signifi-

cance.The wave velocity is a function of wavelength even in free space and

here it differs from the velocity of light which may be shown as underp = mv and total energy E = mc2

So, p = E vc2 and now using the relativistic expression for the mass of a

particle, the equation for its momentum can be given as

p =m

vvc

02

21

where m0 is the rest mass of the particle.Eliminating v between the last two equation, we get

m02 =

pEc c

22

4 2

Since we known that p = h

and E = hν.

So, m0 = hc c

2

2 21

For de Broglie wave W = νλ.

Hence m0 = h Wc c

2

2 2 21

From which W = 2 2

2021

m cc

hThis equation shows that for a particle of rest mass m0 > 0, the wave

velocity W is always greater than c and also the wave velocity of de Brogliewaves is a function of λ even in free space.


Now consider electromagnetic wave as a special case of de Broglie waveand so, these de Broglie waves are propagated with a velocity W = c. Thevelocity of the associated particle i.e., photon is also equal to c. If W = c issubstituted in the last equation given above, we find that rest mass of thephoton m0 = 0 i.e., there is no such thing as a photon at rest, photon alwaysmoves with the velocity c.

11111.6.6.6.6.6 DADADADADAVISSON AND GERMER EXPERIMENTVISSON AND GERMER EXPERIMENTVISSON AND GERMER EXPERIMENTVISSON AND GERMER EXPERIMENTVISSON AND GERMER EXPERIMENT

In 1927, Davisson and Germer predictedexperimentally the electron waves pre-dicted by de Broglie. Davisson andGermer were studying the scattering ofelectrons from a nickel target using anapparatus like that sketched in Fig. 1.2.

The energy of the electrons in theprimary beam, the angle at which theyare incident upon the target and theposition of the detector could all be var-ied. The nickel target was subjected tosuch a high temperature treatment thatthe crystal was transformed into a groupof crystals. In this case the reflection be-came anomalous and the reflected intensity showed striking maxima and minimainstead of a continuous variation of scattered electron intensity with angle.

The position of the maxima and minima observed depended upon theelectron energy. Then, they suspected that the beam of electron might bediffracted from the crystals like X-rays. This shows that electrons behave likewaves under certain circumstances. Typical polar graphs of electron intensityafter the heat treatment are shown in Fig. 1.3.

Fig. 1.3

Fig. 1.2


Fig. 1.4

The method of plotting is such that the intensity at any angle is propor-tional to the distance of the curve at that angle from the point of scattering.If the intensity were the same at all scattering angles, the curves would becircles centered on the point of scattering.

de Broglie hypothesis suggested the interpretation that the electron waveswere being diffracted by the target, much as X-rays are diffracted by theplanes of atoms in a crystal. This idea was confirmed when it was realisedthat the effect of heating a block of nickel at high temperature is to causemany small individual crystals of which it is normally composed to form intoa single large crystal all of whose atoms are arranged in a regular lattice.

To verify whether de Broglie waves are responsible for the findings ofDavisson and Germer, an analysis of the observation should be made.

For the beam of electrons falling normallyon the surface of the crystal, the current ob-served in detector is a measure of the inten-sity of the diffracted beam. Several curveswere obtained for different voltage electronswhen graphs were plotted between the co-latitude (angle between the incident beam andthe beam entering the detector) which areshown in Fig. 1.4.

It is observed that a bump begins to ap-pear in the curve at 44 volt electrons. Thisbump moves upward for 54 volts at colati-tude of 50°. Above 54 volts the bump againdiminishes. The bump at 54 volts offers the evidence for the existence ofelectron waves.

The angles of incidence and scattering relative to the family of Braggplane shown in Fig. 1.4 are both 65°. The spacing of the planes in this family,which can be measured by X-ray diffraction is 0.091 nm. The Bragg equationfor maxima in the diffraction pattern is

nλ = 2d sin θ.Here d = 0.091 nm, θ = 65°. For n = 1, the de Broglie wavelength λ of the

diffracted electrons isλ = (2) (0.091) (sin 65°) = 0.165 nm,

We use de Broglie formula to calculate the expected wavelength of theelectrons. The electron kinetic energy of 54 eV is small compared with its restenergy m0c

2 of 0.51 MeV so we can ignore relativistic considerations.

Since K = 21 .2

mv


The electron momentum mv is

mv = 31 192 (2) (9.1 10 ) (54) (1.6 10 )mk

= 4.0 × 10–24 kg m/s.

The electron wavelength is therefore

λ =34

246.63 104.0 10

hmv

= 1.66 × 10–10 m = 0.166 nm

is in excellent agreement with the observed wavelength. The Davisson-Germerexperiment thus provides direct verification of de Broglie’s hypothesis of thewave nature of moving bodies.

SOLSOLSOLSOLSOLVED EXAMPLESVED EXAMPLESVED EXAMPLESVED EXAMPLESVED EXAMPLES

EXAMPLE 1: Calculate the de Broglie wavelength associated with a proton mov-

ing with a velocity equal to 1

20 th of the velocity of light.

SOLUTION: Velocity of proton

v =83 10

20

= 1.5 × 107 m/s

Mass of the protonm = 1.67 × 10–27 kg

λ =34

27 76.6 10

1.67 10 1.5 10h

mv

= 2.634 × 10–14 m.

EXAMPLE 2: Calculate the de Broglie wavelength of neutron of energy 12.8 MeV.Given h = 6.62 × 10–34 J.sec, m = 1.67 × 10–27 kg.

SOLUTION: m0c2 =

10

191.507 10

1.6 10

= 941.87 MeV

Since 12.8 MeV is very small compared to rest mass energy hence rela-tivistic consideration may be ignored

λ =2

h h Emv mE

= Ve where V is voltage in volts

λ =34

27 6 19

6.62 10

2 1.67 10 12.8 10 1.6 10

= 8.0 × 10–5 Å.


EXAMPLE 3: Show that the de Broglie wavelength for a material particle of restmass m0 and charge q accelerated from rest through a potential difference of V voltsrelativistically is given by

λ = .

0 20

h

qV2m qV 1

2m c

SOLUTION: Kinetic energy Ek = Vq.Ek ≠ mv2 because v is relativistic velocity and so, we cannot find momen-

tum directly from Ek. Now, we use relativistic formula

E2 = p2c2 + m02c4

E = Ek + m0c2 = Vq + m0c

2

P2c2 = E2 – m02c4 = (Vq + m0c

2)2 – m02c4

P2c2 = V2q2 + 2Vq m0c2

P2 =2 2

0 02 20

2 2 12

V q VqVq m m Vq

c m c

P = 0 20

2 12

Vqm Vq

m c

∴ de Broglie wavelength λ =

0 20

1

2 12

Vqm Vq

m c

EXAMPLE 4: Calculate the wavelength associated with an electron accelerated toa potential difference of 1.25 keV.

SOLUTION: If E is the K.E. of the electron, the de Broglie wavelength of thewave associated with the electron is

λ =34

31 3 19

6.6 102 2 9.1 10 1.25 10 1.6 10

hmE

= 3.46 × 10–11.

EXAMPLE 5: What will be the kinetic energy of an electron if its de Brogliewavelength equals the wavelength of the yellow line of sodium (5896 Å). The restmass of electron is m0 = 9.1 × 10–31 kg and h = 6.63 × 10–34 J. sec.

SOLUTION: de Broglie wavelength

λ = hmv

or v = .hm


In the absence of relativistic consideration m = m0

λ =0

hm v

, Kinetic energy K = 20

12

m v

K =2 2

0 20 0

1 .2 2

h hmm m

Putting the values of h, m0 and λ

K =34 2 25

31 10 2 19(6.63 10 ) 6.95 10

2 9.1 10 (5896 10 ) 1.6 10

= 4.34 × 10–6 eV.

EXAMPLE 6: A particle of rest mass m0 has a kinetic energy K. Show that its deBroglie wavelength is given by

λ = .( )2

0

hc

K K 2m cHence, calculate the wavelength of an electron of kinetic energy 1 MeV. What

will be the value of λ if K << m0c2?

SOLUTION: According to de Broglie’s concept of matter wave, the wave-length

λ = hmv

m = 02 2 1/2(1 )

m

v c

vc

1/22

21

=mm

0

202

1m

m= v

c

2

2

or vc

2

2 =m m

m

2 20

2

or m2v2 = (m2 – m02)c2

or mv = c(m2 – m02)1/2.

Substituting this value of mv in equation of wavelength, we get

λ = h

c m m2 20

or λ = hc

c m m2 2 20

c2(m2 – m02)1/2 = [c4(m – m0) (m + m0)]

1/2

= [c2(m – m0) {(m + m0)c2}]1/2

= [(m – m0)c2 {(m – m0)c

2 + 2m0c2}]1/2

or c2(m2 – m02)1/2 = [K(K + 2m0c

2)]1/2.


Therefore λ = hcK K m c2 1/2

0

.[ ( 2 ]

For an electron m0c2 = 9.1 × 10–31 × (3 × 108)2 joule

=15

1981.9 101.6 10

eV = 0.51 × 106 eV = 0.51 MeV

For K = 1 MeV, λ = hc hc1(1 (2 0.51)) 2.02

λ =34 8

19 66.62 10 3 102.02 1.6 10 10

m

= 8.78 × 10–13 m = 8.78 × 10–3 Å.If K << m0c

2 then K + 2m0c2 = 2m0c2

λ = hc hm Km Kc2

00

.22

EXAMPLE 7: What is the de Broglie wavelengths of any electron which has beenaccelerated from rest through a potential difference of 100 V.

SOLUTION: λ =V

12.25 Å

V = 100 volts

∴ λ = 12.25100

= 1.225 Å.

EXAMPLE 8: Can a photon and an electron of the same momentum have the samewavelength? Compare their wavelengths if the two have the same energy.

SOLUTION: Using de Broglie concept of matter wave, momentum

of the electron may be written as pe = e

h

and momentum of photon

as pph = p

h hc

.

So, if the electron and photon have same momentum, then we have λe = λp.Thus, the photon and electron of the same momentum have the same

wavelength.The de Broglie wavelength of electron is given by

λe = hmv

and mv212

= E or mv = mE2

where E is the energy of the electron

So, λe =mE1

2


The de Broglie wavelength of photon is given by

λph = hp

but E = hν = hc

= pc

λph = hcE

Now ph

e

= hc mE m mcc

E h E E

22 2 2 .

EXAMPLE 9: Find the energy of the neutron in units of electron volt whose deBroglie wavelength is 1 Å.

SOLUTION: Given mass of the neutron = 1.674 × 10–27 kgPlanck’s constant h = 6.60 × 10–34 joule/sec

We know that λ = h hmv mE2

or E = hm

2

22 where m = 1.674 × 10–27 kg

λ = 1 Å = 10–10 mh = 6.60 × 10–34 J.s.

E =

34 2

27 10 2

(6.60 10 )

2 1.674 10 (10 ) = 13.01 × 10–21 joules

=21

1913.01 101.6 10

= 8.13 × 10–2 eV.

EXAMPLE 10(a): What would be the wavelength of quantum of radiant energyemitted, if an electron transmitted into radiation and converted into one quantum?

SOLUTION: When the energy of an electron is transmitted into radiation,we use the following relations to get the value of λ

E = mc2 and E = hν = hc

So λ =h

mc

34

31 86.6 10

9.1 10 3 10

= 0.0242 × 10–10 m = 0.0242 Å.EXAMPLE 10(b): A lamp of 150 W is emitting 8% of its energy as blue light

having a mean wavelength of 4500 Å. How many photons are being emitted by thelamp per second.

SOLUTION: Power emitted by the lamp = 8150 W 12 W100

The energy emitted per second = 12 J


Energy contained in one photon = hc/λ =

34 8

106.625 10 3 10

4500 10= 44.2 × 10–20 J

No. of photons emitted per second =

1820

12 27.15 1044.2 10

.

11111.7.7.7.7.7 UNCERTUNCERTUNCERTUNCERTUNCERTAINTAINTAINTAINTAINTY PRINCIPLEY PRINCIPLEY PRINCIPLEY PRINCIPLEY PRINCIPLE

It is impossible to know both the exact position and exact momentum of anobject at the same time.

To regard a moving particle as a wave group implies that there are somefundamental limits to the accuracy with which we can measure position andmomentum of a particle.

Fig. 1.5

The narrower the wave group of the particle, its position can be preciselydetermined (Fig. 1.5a).

The wavelength and hence particle’s momentum cannot be established

because λ = hmv

and it is not well defined in a narrow packet.

On the other hand, a wide wave group such as in Fig. 1.5b, the wave-length can be precisely determined but not the position of the particle.

The relationship between the distance Δx and the wave number spreadΔk depends upon the shape of the wave group and upon how Δx and Δk aredefined. If ψ(x) representes a wave group and the function g(k) describes howthe amplitudes of the waves that contribute to ψ(x) vary with wave numberk and Δx and Δk are taken as the standard deviations of the respective func-

tions ψ(x) and g(k), then the minimum value of Δx Δk = 12

.

It is more realistic to express the relationship between Δx and Δk as

Δx Δk ≥ 1 .2

The de Broglie wavelength of a particle of momentum p is

λ = hp

.

The wave number corresponding to this wavelength is

k =p

h22 .


Hence an uncertainty Δk in the wave number of the de Broglie wavesassociated with the particle results in an uncertainty Δp in the particle’smomentum according to the formula

Δp = h k .2

Since Δx Δk ≥ 12

, Δk ≥ x

12( )

Δx Δp ≥ h .4

...(1.6)

Equation (1.6) states that the product of the uncertainty Δx in the positionof an object at some instant and the uncertainty Δp in its momentum compo-

nent in the x direction at the same instant is equal to or greater than h .4

For

a narrow wave group Δx is small and then Δp will be large and in a broadwave group Δp is small and then Δx will be large.

Generally h2

is abbreviated as (h bar)

= h2

= 1.054 × 10–34 J.s.

Thus, in terms of , the uncertainty principle becomes

Δx Δp ≥ .2

In the above discussion we have considered wave nature of particlewhile uncertainty principle can be arrived considering particle nature ofwaves.

It is worth mentioning that the lower limit of 2 for Δx Δp is rearely

attained; more usually Δx Δp ≈ or even Δx Δp ≈ h.Another form of the uncertainty principle is sometimes useful. Let en-

ergy emitted E be measured during the time interval Δt in an atomic process.The limited time available imposes restriction on the accuracy with which wecan determine the frequency ν of the waves.

The uncertainty Δν in our frequency measurement is

Δν ≥t

1 .

Because the minimum uncertainty in the number of waves we count ina wave group is one wave and to get uncertainty Δν, it is divided by the timeinverval.

The corresponding energy uncertainty isΔE = hΔν

and so ΔE ≥ ht

or ΔE Δt ≥ h.

A more precise calculation changes this result to

ΔE Δt ≥ .2



EXAMPLE 11: Calculate the smallest possible uncertainty in the position of an

electron moving with velocity 3 × 107 m/s.

SOLUTION: (Δx)min (Δp)max = h2

(Δp)max = p (momentum of the electron)

mv =m v

v c0

2 2(1 )

(Δx)min =h v c

m v

2 2

0

12

h2

= 1.05 × 10–34 J.s., v = 3 × 107 m/s, m0 = 9 × 10–31 kg, c = 3 × 108 m/s

(Δx)min =

7 234

8 2

31 7

(3 10 )1.05 10 1

(3 10 )

9 10 3 10

= 3.8 × 10–12 m = 0.038 Å.

EXAMPLE 12: An electron is confined to a box of length 10–8 meter; calculate the

minimum uncertainty in its velocity. Given m = 9 × 10–31 kg; = 1.05 × 10–34 joulesecond.

SOLUTION: (Δx)max = 10–8 meter

(Δp)min =x

34

8max

1.05 10( ) 10

kg-m/s

= 1.05 × 10–26 kg-m/s

(Δp)min = m(Δv)min = 1.05 × 10–26

(Δv)min = m

26 26

311.05 10 1.05 10

9 10

= 1.17 × 104 m/s.

EXAMPLE 13: Find the uncertainty in the momentum of a particle when itsposition is determined within 0.01 cm. Find also the uncertainty in the velocity ofan electron and an α-particle respectively when they are located within 5 × 10–8 cm.

SOLUTION: According to uncertainty principle

(Δx) (Δp) = h2

or Δp = hx2


h2

= 1.05 × 10–34 J.s Δx = 0.01 × 10–2 meter

Δp =34

21.05 100.01 10

= 1.05 × 10–30 kg m/s

Δp = mΔv

∴ Δv = hm x

.2

Uncertainty in the velocity of electron

m = 9 × 10–31 kg

Δx = 5 × 10–10 m

Δv =34

31 101.05 10

9 10 5 10

= 2.33 × 105 m/sec.

Uncertainty in the velocity of α-particle, mass of α-particle = 4 × mass ofproton

= 4 × 1.67 × 10–27 = 6.68 × 10–27 kg

Δx = 5 × 10–10 m

Δv =34

27 101.05 10

6.68 10 5 10

= 31.4 m/sec.

EXAMPLE 14: An electron has a speed 4 × 105 m/s within the accuracy of 0.01per cent. Calculate the uncertainty in the position of the electron. Given h = 6.625 ×10–34 J.s. Mass of electron = 9.11 × 10–31 kg.

SOLUTION: Uncertainty in velocity = 0.01100

× 4 × 105 = 40 m/s

(Δx) (Δp) = h2

(Δx) = hp m v

341.055 102 ( )

=34

311.055 10

9.11 10 40

= 2.895 × 10–6 m.

EXAMPLE 15: An excited atom has an average lifetime of 10–8 sec. That is duringthis period it emits a photon and returns to the ground state. What is the minimumuncertainty in the frequency of this photon?

SOLUTION: ΔE Δt ≥ h2

E = hν


or ΔE = hΔν

hΔν Δt ≥ h2

or Δν ≥t

12

Δt = 10–8 sec

Δν ≥ 81

2 3.14 10 = 15.92 × 106 sec–1.

EXAMPLE 16: If an excited state of hydrogen atom has a lifetime of 2.5 × 10–14

sec. What is the minimum error with which the energy of this state can be measured?

SOLUTION: The uncertainty in the energy of the photon is equal to theuncertainty in the energy of the excited state of the atom due to energyconservation.

According to uncertainty principle

ΔE Δt ≥ h2

ΔE ≥ ht

34

146.63 10

2 2 3.14 (2.5 10 )

= 4.22 × 10–21 J

=21

194.22 101.6 10

= 0.026 eV.

EXAMPLE 17: Using the uncertainty relation ΔE.Δt = h2

, calculate the time

required for the atomic system to retain rotation energy for a line of wavelength 6000Å and width 10– 4 Å.

SOLUTION: We know that E = hν = hc

ΔE = hc2

where Δλ is the width of the spectral lines. Here Δλ = 10–4 × 10–10 m.Using uncertainty relation

ΔE . Δt = h2

, we have

Δt = h hE chc

2

22 2

2

Putting the value we get Δt =

7 2

8 14

(6 10 )

2 3.14 3 10 10

= 1.9 × 10– 8 second.


EXAMPLE 18: A nucleon is confined to a nucleus of diameter 5 × 10–14 m.Calculate the minimum uncertainty in the momentum of the nucleon. Also calculatethe minimum kinetic energy of the nucleon.

SOLUTION: We have the relation

(Δp)min (Δx)max =2h

(Δp)min =

34

14max

6.626 102 ( ) 5 10 2 3.14

hx

= 0.21 × 10–20 kg m/sec.Since p cannot be less than (Δp)min

Emin =2

min( )2

Pm

Putting the values ofPmin = 0.21 × 10–20 kg . m/sec and m = 1.675 × 10–27 kg.

Emin =

20 2

27(0.21 10 )

2 1.675 10 = 0.063 × 10–13 J

=

13

190.063 101.6 10

= 0.039 MeV.

EXAMPLE 19: An electron is confined to a box of length 1.1 × 10–8 m. Calculate

the minimum uncertainty in its velocity. Given m = 9.1 × 10–31 kg and h

2 = 1.05

× 10–34 J.s.

SOLUTION: We know from uncertainty principle that

Δx . Δp ≥2h Δp =

2h

x

Let Δv be the uncertainty in the velocity of a particle of mass m, so wehave

Δp = mΔv or Δv = pm

Δx = 1.1 × 10–8 m, 2h = 1.05 × 10–34 J.s. and m = 9.1 × 10–31 kg

Δv =

34

31 81.05 10

2 9.1 10 1.1 10hm x

= 1.06 × 104 m/sec.


EXAMPLE 20(a): What is the minimum uncertainty in the frequency of a photonwhose life time is about 10–8 sec?

SOLUTION: From uncertainty principle, we have

ΔE Δt ≥2h But E = hν or ΔE = h Δν

h Δν Δt ≥2h or Δν ≥

1

2 t

Putting Δt = 10–8 sec, Δν = 81

2 3.14 10 = 15.92 × 106/sec.

EXAMPLE 20(b): A typical atomic nucleus is about 5 × 10–15 m in radius. Usethe uncertainty principle to place a lower limit on the energy an electron must haveif it is to be part of a nucleus.

SOLUTION: Let Δx = 5 × 10–15 m

Δp ≥

34

151.054 10

2 2 5 10x

≥ 1.1 × 10–20 kg . m/sIf this is the uncertainty in a nuclear electron’s momentum, the momen-

tum p itself must be at least comparable in magnitude. An electron with sucha momentum has kinetic energy many times greater than its rest energy m0c

2.Thus we have E = pc

E = pc ≥ (1.1 × 10–20) (3 × 108)≥ 3.3 × 10–12 J

Since 1 eV = 1.6 × 10–19 J, the energy of an electron must exceed 20 MeVif it is to be on nuclear constituent. Experiments indicate that even the elec-trons associated with unstable atoms never have more than a fraction of thisenergy and we conclude that electrons are not present within nuclear.

11111.8.8.8.8.8 WAWAWAWAWAVE FUNCTION AND WAVE FUNCTION AND WAVE FUNCTION AND WAVE FUNCTION AND WAVE FUNCTION AND WAVE EQUVE EQUVE EQUVE EQUVE EQUAAAAATIONTIONTIONTIONTION

The quantity with which quantum mechanics is concerned is the wave func-tion Ψ of a body. The square of its absolute magnitude | Ψ |2 evaluated at aparticular place at a particular time is proportional to the probability offinding the body there at that time. The momentum, angular momentum andenergy of the body are other quantities that can be established from Ψ. Inquantum mechanics we determine Ψ for a body when its freedom of motionsis limited by the action of external forces.

If the wave function Ψ is complex with both real and imaginary parts, theprobability density | Ψ |2 is written as ΨΨ*. Where Ψ* is complex conjugate ofΨ. A complex wave function Ψ can be written in the form

Ψ = A + iB where A and B are real functionsΨ* = A – iB

Ψ*Ψ = A2 + B2 [i2 = –1].


Thus ΨΨ* is always a positive real quantity. The condition for an accept-able wave function is that the integral of | Ψ |2 over all space must be finite,if | Ψ |2 is equal to P, then it must be true that

2 dV = 1 since

PdV = 1

is the mathematical statement that the particle exists somewhere at all times.A wave function that obeys above equation is said to be normalised. Everyacceptable wave function can be normalised by multiplying by an appropri-ate constant.

Besides being normalisable, Ψ must be single valued, since P can haveonly one value at a particular place and time and continuous. Momentum

considerations require that the partial derivatives

, ,x y z

be finite con-

tinuous and single valued. The wave functions with all these properties canyield physically meaningful results when used in calculations.

For a particle restricted to motion in the x direction, the probability offinding it between x1 and x2 is given by

Probability = 2

1

2x

xdx

Schrödinger’s equation which is the fundamental equation of quantummechanics is a wave equation in the variable Ψ.

1.9 SCSCSCSCSCHRHRHRHRHRÖDINGER’S EQUÖDINGER’S EQUÖDINGER’S EQUÖDINGER’S EQUÖDINGER’S EQUAAAAATION: TIME-DEPENDENTTION: TIME-DEPENDENTTION: TIME-DEPENDENTTION: TIME-DEPENDENTTION: TIME-DEPENDENT

The wave function of a freely moving particle moving in x direction can berepresented by a wave function Ψ as

Ψ =

xi tAe ...(1.15)

Replacing ω in the above formula by 2πν and v = νλ, we get

Ψ = Ae–2πi(νt – x/λ) ...(1.16)Which is convenient since we know ν in terms of the total energy E and

momentum p of the particle being described by Ψ.

Since E = hν = 2π ν and λ = 2hp p

we have Ψ = Ae–(i/)(Et – px). ...(1.17)Equation (1.17) is a mathematical description of the wave equivalent of

an unrestricted particle of total energy E and momentum p moving in the+ x direction. The expression for the wave function Ψ given by eqn. (1.17) iscorrect only for freely moving particles.


Schrödinger’s equation which is the fundamental equation of quantummechanics is a wave equation in the variable Ψ. Differentiating eqn. (1.17)twice with respect to x, we get

2

2x=

2

2p

...(1.18)

and differentiating eqn. (1.17) once w.r.t. t, we get

x

=

.iE ...(1.19)

At speeds small compared with that of light, the total energy E of aparticle is the sum of its kinetic energy p2/2m and its potential energy V,where V is in general a function of position x and time t.

E = 2

.2p

Vm

...(1.20)

Multiplying both sides of eqn. (6) by the wave function Ψ gives

E Ψ =

2

.2p

Vm

...(1.21)

From eqns. (1.18) and (1.19), we see that

E Ψ =

i t

and p2 Ψ =

2

22x

...(1.22)

Substituting these expressions for E Ψ and p2 Ψ into eqn. (1.21), we get

it

=

2 2

2 .2

Vm x

...(1.23)

Equation (1.23) is the time dependent form of Schrödinger’s equation.In three dimensions the time-dependent form of Schrödinger’s equa-

tion is

it

=

2 2 2 2

2 2 2.

2V

m x y z...(1.24)

Where the particle’s potential energy V is some function of x, y, zand t.

The restrictions on particle’s motion affect the potential energy funcitonV. Once V is known Schrödinger’s equation may be solved for the wavefunction Ψ of particle from which its probability density | Ψ |2 may be deter-mined for a specified x, y, z, t. The extension of Schrödinger’s equation fromthe special case of freely moving particle (V = constant) to the general caseof a particle subject to arbitrary forces that vary in time and space [V = V(x,y, z, t)] is entirely plausible.


Thus, Schrödinger’s equation is postulated, solve it for a variety of physi-cal situations and compare the results of the calculations with the results ofexperiments. If they agree, the postulate is valid and if disagree it maybe discarded and some other approach would have to be explored.Equation (1.24) can be used for non-relativistic problems and in practice ithas turned out to be accurate in predicting the results of experiments.

11111.....1111100000 SCSCSCSCSCHRHRHRHRHRÖDINGER’S EQUÖDINGER’S EQUÖDINGER’S EQUÖDINGER’S EQUÖDINGER’S EQUAAAAATION: STEADTION: STEADTION: STEADTION: STEADTION: STEADYYYYY-ST-ST-ST-ST-STAAAAATETETETETEFORMFORMFORMFORMFORM

It has been observed that in many situations, the potential energy of a par-ticle does not depend on time and vary with the position of the particle only.When this is true, Schrödinger’s equation may be simplified by removingtime.

The one-dimensional wave function Ψ of an unrestricted particles may bewritten as

Ψ = Ae–(i/)(Et – px) = Ae–(iE/)t e+(ip/)x ...(1.25)Thus, Ψ is the product of a time-dependent function e–(iE/)t and a posi-

tion dependent function Ψ. (Ψ = Ae(ip/)x)Substituting the Ψ of eqn. (1.25) into the time-dependent form of

Schrödinger’s equation, we obtain

E ψ e–(iE/)t =

22( / ) ( / )

22iE t iE te V e

m xDividing throughout by the common exponential factor

2

2 22 ( )m E V

x= 0. ...(1.26)

Equation (1.26) is the steady-state form of Schrödinger’s equation. Inthree-dimensions it is

2 2 2

2 2 2 22 ( )m E V

x y z= 0. ...(1.27)

Schrödinger’s steady-state can be solved only for certain values of theenergy E. To solve Schrödinger’s equation we obtain a wave function ψ whichfulfils the requirements for an acceptable wave function, i.e., its derivativesbe continuous, finite and single valued. If there is no such wave function, thesystem cannot exist in a steady-state.

The energy quantisation in wave mechanics is a natural element of thetheory and is a characteristic of all stable systems.

Particle in a Box

The simplest quantum-mechanical problem is that of a particle trapped in abox with infinitely hard walls. We may specify the particle’s motion along the


x-axis between x = 0 and x = L. A particle does not lose energy when collideswith infinitely hard walls so that its total energy remains constant. From theformal point of view of quantum mechanics, the potential energy V of theparticle is infinite on both sides of the box while V is constant say 0 forconvenience on the inside. The particle cannot have infinite amount of en-ergy. So, the wave function Ψ is zero for x ≤ 0 and x ≥ L. Thus, our aim is tofind Ψ between x = 0 and x = L (Fig. 1.6).

Schrödinger’s equation for the condition specified (x = 0 and x = L)becomes

2

2 22m E

x = 0 [V = 0] ...(1.28)

Equation (1.28) has the solution

ψ = 2 2sin cosmE mEA x B x

where A and B are constants to be evaluated.Applying the boundary conditions that

Ψ = 0 for x = 0 and for x = L. Since cos θ =1, the second term cannot describe the par-ticle because it does not vanish at x = 0.Hence, B = 0. Since sin θ = 0, the sine termalways yields ψ = 0 at x = 0 but ψ will be 0 at x = L only when

2mE L = nπ, n = 1, 2, 3 ... ...(1.29)

This is true because the sines of the angles π, 2π, 3π, ... are all 0.From eqn. (1.29) it is clear that the energy of the particle can have only

certain values known as eigenvalues. These eigenvalues constituting the energylevels of the system are

En =2 2 2

22n

mL where n = 1, 2, 3 ...(1.30)

The wave function of a particle in a box whose energy is En is

ψn =2

sin nmEA x ...(1.31)

Substituting eqn. (1.30) for En, we get

ψn = sin .n xAL ...(1.32)

where ψn is the eigenfunctions corresponding to the energy eigenvalues En.

It can be verified that these eigenfunctions meet all the requirements. Foreach quantum number n, ψn is a single valued function of x and ψn and ∂ψn/∂x are continuous. The integral of | ψn |2 over all space is finite. By integrating| ψn |2 dx from x = 0 and x = L.

Fig. 1.6. Potential well whichcorresponds to a box with

infinitely hard walls


2n dx

= 2 2 2

0 0sin

L L

nn xdx A dx

L

= 2 .2LA ...(1.33)

To normalise ψ we assign a value of A such that | ψn |2 dx is equal to theprobability P dx of finding the particle between x and x + dx, rather thanmerely proportional to P. If | ψn |2 dx is equal to P dx, then it must be true that

2n dx

= 1 since P dx

= 1. ...(1.34)

is the mathematical way of stating that the particle exists somewhere at alltimes. Comparing eqns. (1.33) and (1.34), we find that the wave function ofa particle in a box are normalised if

2

2LA = 1, A = 2 .

LThe normalised wave functions of the particle are therefore

ψn = 2 sin n xL L

n = 1, 2, 3.

The normalised wave functions ψ1, ψ2 and ψ3 together with the probabil-ity densities | ψ1 |2, | ψ2 |2 and | ψ3 |2 are plotted in Fig. 1.6(a). While ψn maybe negative as well as positive, | ψn |2 is always positive and, since ψn isnormalised, its value at a given x is equal to the probability density P offinding the particle there. In every case | ψn |2 = 0 at x = 0 and x = L, theboundaries of the box.

Fig. 1.6(a). Wave functions and probability densities of a particleconfined to a box with rigid walls

At a particular place in the box the probability of the particle beingpresent may be very different for different quantum numbers. For instance| ψ1 |2 has its maximum value of 2/L in the middle of the box while | ψ2 |2 =0. A particle in the lowest energy level of n = 1 is most likely to be in themiddle of the box, while a particle in the next higher state of n = 2 is neverthere. Classical physics suggests the same probability for the particle beinganywhere in the box.



EXAMPLE 21: Show that ψ(x) = eicx where c is some finite constant is acceptableeigenfunctions. Also normalise it over the region –a ≤ x ≤ a.

SOLUTION: The wave function ψ(x) can be written asψ(x) = eicx = cos cx + i sin cx.

Its derivative is given by

( )d xdx = ic eicx = ic (cos cx + i sin cx)

= – c sin cx + ic cos cx.The following points may be observed:(i) sin cx and cos cx are periodic functions with maximum value 1 and

c is finite constant. Thus ψ(x) and ( )d xdx

are finite for all values of

x.

(ii) The function ψ(x) and ( )d xdx

are single-valued because cos cx and

sin cx are also continuous for all values of x.Hence ψ(x) is an acceptable form of the eigenfunction. To normalise, the

wave function we may write ψ(x) asψ(x) = Aeicx.

Now we have to determine the value of A and we may write

* ( ) ( )a

ax x dx

= 1

2a

icx icx

aA e e dx

= 1

2 aaA x

= 1

A2(2a) = 1 or A = .2a

Hence normalised wave function is

ψ(x) = .2

icxea

EXAMPLE 22: A particle is moving in one-dimensional potential box (of infiniteheight) of width 25 Å. Calculate the probability of finding the particle within aninterval of 5 Å at the centres of the box when it is in its state of least energy.

SOLUTION: The wave function of the particle can be written as

ψ(x) = 2 sin n xa a


For the particle in the least energy state n = 1 and hence

ψ(x) = 2 sin xa a

At the centre of the box x = 2a , the probability of finding the particle in

the interval Δx is given asP = | ψ(x) |2 Δx

| ψ(x) |2 =2

2( 2)2 2 2sin sin2

aa a a a

P =10

102 5 102

25 10x

a

= 25 Å0.45 Å

ax

EXAMPLE 23: A particle is in motion along a line between x = 0 and x = a withzero potential energy. At points for which x < 0 and x > a, the potential energy isinfinite. The wave function for the particle in the nth state is given by

ψn = .n xA sina

Find the expression for the normalised wave funciton.

SOLUTION: The probability of the particle between x and x + dx for the nthstate is given as

| ψn(x) |2 dx = 2 2sin .n xA dxa

Since the particle is found in the region x = 0 and x = a for all times, wehave

2

0

a

n dx = 1

2 2

0sin

a n xA dxa = 1

2

0

1 21 cos2

a nxA dxa

= 1

2

0

2sin2 2

aA a nxxn a

= 1

2

2A a = 1 or A = 2 .

aNow the normalised wave funciton is

ψn(x) = 2 sin .n xa a


EXAMPLE 24: Find the energy of an electron moving in one dimension in aninfinitely high potential box of width 1 Å, given mass of the electron 9.11 × 10–31

kg and h = 6.63 × 10–34 Js.

SOLUTION: Since we know that E = 2 2

28n hma

(n = 1, 2, 3, ...)

For least energy of the particle n = 1.

Now2

28hma

=34 2

31 10 2(6.63 10 )

8 9.11 10 (10 )

joules

= 9.1 × 10–19 joules

=19

199.1 10

1.602 10

eV = 5.68 eV.

EXAMPLE 25: Show that Ψ = ψe–iωt is a wave function of a stationary state.

SOLUTION: If ψ is the wave functions of a stationary state, then the value

of 2 at each point must be constant, independent of time. To find 2 ,we first take the complex conjugate of Ψ = ψe–iωt which is Ψ* = ψ* e+iωt. Then

2 = Ψ*Ψ = (ψ* e+iωt) (ψ e–iωt)

= ψ*ψ e0 = 2

where ψ is not a function of time, so 2 is also independent of time. Now,

it has been shown that 2 = 2 , so 2 is independent of time and

Ψ = ψe–iωt is a wave function of a stationary state.

EXAMPLE 26: Find the probability that a particle trapped in a box L wide canbe found between 0.45 L and 0.55 L for the ground and first excited states.

SOLUTION: This part of the box is one tenth of the box’s width and iscentered on the middle of the box. Classically, we could expect the particleto be in this region 10 per cent of the time. The quantum mechanics givesquite different predictions that depend on the quantum number of the particle’sstate. The probability of finding the particle between x1 and x2 when it is inthe nth state is

P =2 2

1 1

2 22 sinx x

nx x

n xdx dxL L

=

2

1

1 2sin2

x

x

x n xL n L

x1 = 0.45 L and x2 = 0.55 LFor ground state which corresponds to n = 1,we have P = 0.198 = 19.8%


For the first excited state n = 2P = 0.0065 = 065%

This low figure is consistent with the probability density of 2

n = 0 atx = 0.50 L.

EXAMPLE 27: (a) Show that y = Ax + B where A and B are constants, is asolution to the Schrödinger equation for an E = 0 energy level of a particle in a box.

(b) Show, however, that the probability of finding a particle with this wavefunction is zero.

SOLUTION: (a) The Schrödinger equation for a particle in a box is

2

2 2

2d mE

dx= 0

Differentiating ψ = Ax + B twice with respect to x gives 2

2

d

dx

= 0 for

the left side of the Schrödinger equation. Also E = 0 gives zero of theright side. Since 0 = 0, ψ = Ax + B is a solution to this Schrödingerequation for E = 0.

(b) The wave function ψ equals zero outside the box (x < 0 and x > L). Inorder that ψ = Ax + B may be continuous at x = 0, it must be true thaty = 0 at x = 0. So A(0) + B = 0 or B = 0.Similarly, in order that ψ may be continuous at x = L, it must be truethat ψ = 0 at x = L, so A (L) + 0 = 0 or A = 0. With both A and B equalto zero, ψ = Ax + B = 0. Thus the wave function equals zero insidethe box as well as outside the box and the probability of finding theparticle anywhere with this wave function is zero.

EXEREXEREXEREXEREXERCISESCISESCISESCISESCISES

1. Discuss the dual nature of matter and waves.2. Find de Broglie wavelength of an electron of energy MeV.3. What are de Broglie matter waves?4. Calculate the de Broglie wavelength associated with a proton moving

with a velocity equal to 1/20 velocity of light.5. Derive time dependent and time-independent Schrödinger wave

equation.6. What is uncertainty principle? Apply this to prove the non-existence

of the electron in the nucleus.7. Calculate the smallest possible uncertainty in the position of electron

moving with velocity 3 × 107 m/sec. [Ans. 0.038 Å]8. An electron has a speed of 1.05 × 104 m/sec within the accuracy of

0.01 per cent. Calculate the uncertainty in the position of the electron.[Ans. 1.10–4 m]


9. Show that the de Broglie wavelength for a material particle of restmass m0 and charge q, accelerated from rest through a potential dif-ference of V volts relativistically is given by λ = h/{2m0qV [1 + qV/2m0c

2]}1/2.10. Find the uncertainty in the momentum of a particle when its position

is determined within 0.01 cm. Find also the uncertainty in the veloc-ity of an electron and α-particle respectively when they are locatedwithin 5 × 10–8 cm.

[Ans. 1.05 × 10–30 kg m/s, 2.33 × 105 m/s, 31.4 m/s]11. An electron is confined to a box of length 1.1 × 10–8 meter. Calculate

the minimum uncertainty in its velocity. Given m = 9.1 × 10–31 kg, = 1.05 × 10–34 J. sec. [Ans. 1.06 × 104 m/s]

12. Calculate the kinetic energy of an electron if the wavelength of elec-tron equals the yellow line of Sodium. [Ans. 4.34 × 10–6 eV]

13. Calculate the de Broglie wavelength of neutron of energy 28.8 eV.Given h = 6.62 × 10–34 J.sec, m = 1.67 × 10–27 kg. [Ans. 5.3 × 10–2 Å]

14. Derive the formula for de Broglie wavelength of particle in terms ofkinetic energy and its rest energy m0c

2.15. Describe Davisson-Germer experiment to demonstrate the wave na-

ture of particle.16. Deduce an expression for de Broglie wavelength of helium atom

having energy at temperature T°K.17. What are de Broglie waves and how do they help in the interpreta-

tion of the Bohr’s quantisation rule?18. Show that the phase velocity of de Broglie wave is greater than the

velocity of light; but the group velocity is equal to the velocity of theparticle with which the waves are associated.

19. Calculate de Broglie wavelength associated with nitrogen at 3.0 at-mospheric pressure and 27°C temperature.

20. Can a photon and an electron of the same momentum have the samewavelength? Compare their wavelengths if the two have the sameenergy.

21. X-rays of wavelength 0.91 Å fall on a metal plate having work func-tion 2.0 eV. Find the wavelength associated with the emitted photo-electrons. [Ans. 8.7 Å]

22. Calculate kinetic energy of a neutron having de Broglie wavelength1 Å. [Ans. 8.13 × 10–2 eV]

23. What is the minimum uncertainty in the frequency of a photon whoselifetime is about 10–8 sec? [Ans. 15.92 × 106 per sec.]

24. A certain excited state of hydrogen atom is known to have a lifetimeof 2.5 × 10–4 sec. What is the minimum error, with which energy ofthe excited state can be measured. [Ans. 0.026 eV]


25. What is de Broglie wavelength of an electron accelerated from restthrough a potential difference of 100 volts. [Ans. 1.23 Å]

26. Find the expression for the energy state of a particle in one-dimen-sional box.

OBJECTIVE QUESOBJECTIVE QUESOBJECTIVE QUESOBJECTIVE QUESOBJECTIVE QUESTTTTTIONSIONSIONSIONSIONS

1. A particle possesses a kinetic energy E and mass m, then its de Brogliewavelength is

(a)2h

mE(b) 2h mE

(c)2mE

h(d)

12h mE

2. An electron is accelerated through a potential difference of V volts.The de Broglie wavelength of the electron is

(a)12.27

V (b)12.27

V

(c) 12.27 V (d) 12.27 V

3. Read the statements (A and B) and choose the answer:(A) The de Broglie wavelength of a moving particle is inversely

proportional to its momentum(B) Only a charged particle in motion is associated with matter waves

(a) A and B are correct(b) A and B are wrong(c) A is wrong and B is correct(d) A is correct and B is wrong

4. If the momentum of a particle is doubled then its de Broglie wave-length(a) halves (b) doubles(c) quadruples (d) remains the same

5. The masses of neutron and electron are mn and me respectively. If theyhave the same de Broglie wavelength, then their velocities should bein the ratio

(a) 1 : 1 (b) e

n

mm

(c) n

e

mm

(d)2

2e

n

m

m

6. An α-particle has a mass 4 mp and a proton mp. If they possess thesame kinetic energy, then the ratio of the de Broglie wavelengths is


(a) 1 : 4 (b) 4 : 1(c) 2 : 1 (d) 1 : 2

7. The characteristic of wave function ψ are(a) real function, finite and discontinuous(b) complex, single valued, finite and continuous function(c) complex, infinite and discontinuous function(d) complex single valued and infinite

8. The wave function for the motion of a particle in a potential well of

width a is given as ψn = sinn x

Ba

, then B is

(a)1a (b)

1a

(c)2a (d)

2a

9. When the wave function is normalized then(a) ψ ψ* = 1 (b) ψ ψ* dx = 1

(c)1

1

* 1xd

(d) * 1a

x

a

d

10. Increasing the potential difference between anode and filament in

Davisson-Germer experiment(a) causes an increase in the wavelength of the electron-waves(b) causes a decrease in the wave velocity(c) causes a decrease in the wavelength of the electron-waves(d) causes a decrease in the momentum of the electron

11. For a particle of mass m confined to a cubical box of side L, theallowed values of energy E are given as

(a)2 2

2,

2n h

nmL

= 1, 2, 3 (b)2 2

2,

8n h

nmL

= 1, 2, 3

(c)2 2

2,

2h L

nmn

= 1, 2, 3 (d)2 2

,n h

nmL

= 1, 2, 3

12. An electron is confined to a potential box of infinite height and width10 Å. The probability of finding the particle in a small interval Δx atthe centre of the box for the energy state immediately above theground state is(a) zero (b) 0.5(c) 0.9 (d) 1


13. If a charged particle of mass m is accelerated through a potentialdifference of V volts, the de Broglie wavelength is proportional to

(a)12V (b) V

(c)

12V (d) V2

14. The de Broglie hypothesis is associated with(a) wave nature of α-particles only(b) wave nature of radiations(c) wave nature of electrons only(d) wave nature of all material particles

15. Which of the following phenomena cannot be expressed by wavenature of light?(a) Interference (b) Diffraction(c) Polarization (d) Photoelectric effect

16. Matter waves(a) are latitudinal (b) show diffraction(c) are electromagnetic (d) always travel with speed of light

17. Of the following moving with the same velocity, the one which haslargest wavelength is(a) a photon (b) an electron(c) an α-particle (d) a neutron

18. The uncertainty principle holds for(a) macroscopic particles only(b) microscopic particles only(c) macroscopic and microscopic particles both(d) neither macroscopic nor microscopic particles

19. The uncertainty principle cannot hold for the following pairs:(a) Energy and time(b) Position and momentum(c) Angular momentum and angle(d) Linear momentum and angle

20. Matter waves were first experimentally observed by(a) Frank-Hertz (b) Davisson and Germer(c) Ster-Gerlach (d) de Broglie

21. If the momentum of a particle is increased to four times, then the deBroglie wavelength will become(a) half (b) four times(c) twice (d) one-fourth

22. The Schrödinger time independent wave equation for free particle is

(a) 2

2 ( ) 02

E vm

(b) 2

2 ( ) 02

v Em


(c) 2

2 02

Em

(d) 2

2 02

Em

23. Schrödinger time dependent wave equation for free particle is

(a)

2 2

22i

m tx(b)

2 2

22i

m tx

(c)

2

2 2

2m itx

(d)

2

2 2

2mi

tx24. An α-particle and a proton have the same kinetic energy. The ratio of

their wavelength is (m∝ = 4mp)(a) 1 : 2 (b) 2 : 1(c) 1 : 4 (d) 4 : 1

25. In Davisson-Germer experiment, nickel crystal acts as(a) perfect absorber(b) perfect reflector(c) two dimensional grating(d) three dimensional diffraction grating

26. Momentum of a photon of energy hν is(a) hν (b) hνc(c) hν/c (d) has no momentum

ANSWERSANSWERSANSWERSANSWERSANSWERS

1. (a) 2. (b) 3. (d) 4. (a) 5. (b)6. (d) 7. (b) 8. (c) 9. (d) 10. (c)

11. (b) 12. (a) 13. (c) 14. (d) 15. (d)16. (b) 17. (b) 18. (b) 19. (d) 20. (b)21. (d) 22. (a) 23. (b) 24. (c) 25. (d)26. (c)

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