1

MATTER WAVES

MATTER WAVES

1. DUAL NATURE OF RADIATION The phenomena of dispersion, interference, diffraction and polarization can only be explained on basis of wave

nature of radiation because they involve with interaction of a wave with itself. The phenomena of Compton effect, Raman effect, photoelectric effect can only be explained on basis of particle

nature of radiation because they involve with interaction of a wave with particle. The phenomena of rectilinear propagation, reflection, refraction etc. can be explained on basis of both wave

and particle nature because there is neither interaction of wave with itself nor with the particle.Thus radiation has dual nature i.e. wave and particle nature.BOHR'S COMPLIMENTARY PRINCIPLEThe complete description of radiation requires both the wave and particle nature. So still both natures are notexhibited simultaneously in one experiment.DE BROGLIE HYPOTHESISThe hypothesis is based on following observations

Nature loves symmetry The entire universe is made up of radiation and matter Since radiation has dual nature so matter must also have dual nature.

According to de-Broglie a wave can be associated with each moving particle whose wavelength is given by

h hp mv

........(1)

This wave is called de-Broglie or matter wave. The equation one is called de-Broglie relation.IMPORTANT POINTS

The de-Broglie relation connects momentum, a particle characteristic with wavelength which is characteristicof wave.

1v

, so smaller is speed of the particle, larger will be the wavelength. If v = 0 then ,so matter waves

can only be associated with moving particles.

1m

, so smaller is the mass, larger will be the wavelength

1p

, so smaller is the momentum larger will be the wavelength.

The wavelength is independent of charge and nature of particle. The matter waves can propagate in vacuum so they are not mechanical waves. These waves are not electromagnetic because they can be associated with electrically neutral moving particles. Matter waves are probability waves as they represent the probability of finding a particle in space. Matter waves propagate in the form of a wave packet moving with group velocity d/dk.

The phase velocity of these waves can be greater than c. [vp = c2/vg]ENERGY, MOMENTUM AND WAVELENGTH OF WAVES ASSOCIATED WITH PHOTON

Energy of photon E = h h = planck constant ; = frequency If m is the effective mass of photon then according to Einstein’s mass energy relation.

E = mc2 = h = hc

2

MATTER WAVES

So effective mass m = 2 2E h h

cc c

The momentum of photon p = mc = E h hc c

Photon is a chargeless, massless particle with spin h2 and travels with a velocity c.

The wavelength of wave associated with photon is h hc hp E mc

The different graphs related to photon are

2. ENERGY, MOMENTUM AND WAVELENGTH OF WAVES ASSOCIATED WITH MOVING PARTICLES

kinetic energy of a particle2

21 pE mv2 2m

momentum of a particle p mv 2mE

wavelength of associated wave h h hp mv 2mE

3

MATTER WAVES

The wave properties of macroscopic bodies cannot be observed because the associated wavelengths are verysmall ( ~ 10–34m)The wave properties of microscopic particles can be observed because the associated wavelengths are larger( ~ 10–10 m)

Ex. Determine the momentum of photon of wavelength 5000Å

Sol. According to de-Broglie relation p = h

p = 34

2710

6.6 10 1.32 10 kg m / sec5000 10

Ex. Calculate the effective mass of photon of wavelength 10Å and compare it with rest mass of electron?

Sol. effective mass of photon m = 34

10 8h 6.6 10c 10 10 3 10

= 2.2 × 10–33 kg

rest mass of electron m0 = 9.1 x 10–31 kg

0

mm

33

312.2 10 0.0249.1 10

3. DE-BROGLIE WAVELENGTH ASSOCIATED WITH CHARGED PARTICLES

The energy acquired by charged particle when accelerated by potential difference V is

E = qV = 12 mv2

so2E 2qVvm m

h h h

mv 2mE 2mqV

for electrons (me = 9.1 x 10–31 kg, qe = 1.6 x 10–19 C)

34

31 19

6.6 10 12.27 AV2 9.1 10 1.6 10 V

for proton (mp = 1.67 x 10–27 kg, qp = 1.6 x 10–19 C)

34

27 19

6.6 10 0.286 ÅV2 1.67 10 1.6 10 V

for deuterons (md = 2 x 1.67 x 10–27 kg qd = 1.6 x 10–19 C)

34

27 19

6.6 10 0.202 ÅV2 2 1.67 10 1.6 10 V

for particle (m = 4 x 1.67 x 10–27 kg, q = 2 x 1.6 x 10–19 C)

34

27 19

6.6 10 0.101ÅV2 4 1.67 10 2 1.6 10 V

4

MATTER WAVES

Ex. Find the ratio of de-Broglie wavelength of proton and particle which have been accelerated through samepotential difference?

Sol. de-Broglie wavelength h

m q V

p

p pp p

2m q V m qh 4m 2e. . 2 2h m q m e2m q V

Ex. An particle moves in circular path of radius 0.83 cm in field of 0.25 Wb/m2. Find associated de-Broglie wavelength.

Sol.2mv qvB

r or mv = qBr

de-Broglie wavelength = h h hp mv qBr

34

19 26.6 10

2 1.6 10 0.25 0.83 10

= 9.94 × 10–13 0.01 Å

Ex. The de-Broglie wavelength for a neutron is 0.1Å. Calculate its momentum and energy.

Sol. momentum of neutron P = h

= 34

2310

6.6 10 6.6 10 kg m / s0.1 10

energy E = 2 23 2

1927

p (6.6 10 ) 13.12 102m 2 1.67 10

joule = 8.2 eV

Ex. What will be the de-Broglie wavelength of electron having kinetic energy of 500eV?

Sol. de-Broglie wavelength h

2mE =

34

31 19

6.6 10

2 9.1 10 500 1.6 10

= 0.5467 Å

Ex. An electron and proton posses same kinetic energy. Which has a greater wavelength?

Sol. E = 2p

2m,

hp

so 2

2hE

2m

as per question Ee = Ep so 2 2 2

2 2 2e e e e p p

h h h2m 2m 2m

2pe

2ep

m1

m

so e > p hence electron has larger wavelength.

4. DE-BROGLIE WAVELENGTH ASSOCIATED WITH UNCHARGED PARTICLES for neutrons (mn = 1.67 x 10–27 kg)

34

27

h 6.6 10 0.286 Å2mE E(eV)2 1.67 10 E

for thermal neutron E = kT

34

27 23

h 6.6 10 30.286 Å2mkT T2 1.67 10 1.38 10 T

5

MATTER WAVES

for gas molecules rms

hmC

& 3E kT2

Crms = rms velocity of gas molecules, E is energy at T K

h

3mkT

Ex. Find the ratio of de-Broglie wavelength of molecules of hydrogen and helium which are at 27° C and 127°Crespectively.

Sol. de-Broglie wavelength h

3mkT

He He He HeH

He H HH H

3m kT m Th 4m (127 273) 8.h m T m (27 273) 33m kT

5. DAVISSON AND GERMER EXPERIMENT

The experiment demonstrates the diffraction of electron beam by crystalsurfaces

The experiment provides first experimental evidence for wave nature of thematerial particles.

The electrons are diffracted like X-rays. The Bragg's law of diffraction are

D sin = n and 2d sin = n

where D = interatomic distance and d = interplanar distance

= angle between scattering plane and incident beam

= scattering angle

2 + = 180º

The electrons are produced and accelerated into a beam by electron gun.

The energy of electrons is given as 21E mv eV2

The accelerated electron beam is made to fall on a Ni crystal. The scatteredelectrons are detected by a detector

The experimental results are shown in form of polar graphs plotted between scattering angle and intensityof scattered electron beam at different accelerating voltage. The distance of curve from point O is pro-portional to intensity of scattered electron beam

6

MATTER WAVES

IMPORTANT RESULTS Intensity of scattered electrons depends on scattering angle

The kink at = 50° is observed at all accelerating voltage

The size of kink becomes maximum at 54 volt.

For = 50° For = 180

2

= 65º

D = sin= n 2 d sin = nD = 2.15Å & n = 1 (for Ni) n = 1 and d = 0.91Å (for Ni) = 2.15 sin 50 = 1.65Å = 2 × 0.91 sin65 = 1.65Å

For V = 54 volt de-Broglie wavelength 12.27 Å 1.67Å

54

This value of is in close agreement with experimental value. Thus this experiment verifies de-Broglie's hypothesis.Ex. An electron beam of energy 10 KeV is incident on metallic foil. If the interatomic distance is 0.55Å . Find the

angle of diffraction.

Sol. = Dsin and 12.27 Å

V so 12.27 Dsin

V

10

103

12.27 10 0.55 10 sin10 10

sin = 12.27 0.22310.53 100

or 1sin (0.2231) 12.89º

6. G P THOMSON EXPERIMENTWhen a beam of electrons is made incident on thin metallic foil then they produce a diffraction pattern identicalto X-rays. The pattern consists of concentric circles around a central dark spot. The wavelength of diffractedelectron waves can be calculated by measuring the radius of rings. This shows that electrons are diffractedlike X-rays.

tan 2 ~ 2 = R/L2d sin = n sin ~ (if is small)2d = n n is order of diffraction

or RL d = n d is interplanar distance and R = radius of ring

If d and L are constant the wavelength depends on radius of ring formed in diffraction pattern.BOHR'S QUANTIZATION CONDITIONAccording to de-Broglie electrons revolve around the nucleus in form of stationary waves i.e. wave packetElectron can revolve only in those circular orbits whose circumference is an intergral multiple of de-Brogliewavelength

7

MATTER WAVES

nh2 r n h

mv

or mvrn = nh/2 which is Bohr's quantisation condition.for n = 1 1 = 2r1 thus only one wave is formed in K shellfor n = 2 22 = 2r2 thus two waves are formed in L shellThe stationary wave formed in various orbits are shown as.

7. DE-BROGLIE WAVELENGTH ASSOCIATED WITH RELATIVISTIC MOTIONIf v c then motion is defined on basis of relativity

The mass varies as m = m0 / 2 21 v / c

The relation between energy and momentum is E2 = p2c2 + m02c4

The total energy (E) = kinetic energy (K) + rest mass energy (m0c2)

E = K+m0c2

pc = (E2 – m02c4)1/2 =

1/ 222 4 2 1/ 20 0 0(K m c ) m c [K(K 2m c )]

de-Broglie wavelength 2 1/ 20

h hcp [K(K 2m c )]

SOME IMPORTANT POINTS If the de-Broglie wavelength of proton and particle are equal then ratio of their velocities is

p or p p

h hm v m v

or p

p

v m 4m 4v m m 1

If de-Broglie wavelength of proton and particle is equal then ratio of potential difference is

p or p p p

h h2m q V 2m q V

p

p p

V m q 4m 2e 8V m q m e 1

If proton and particle are accelerated by same potential then ratio of their wavelengths is

pp p p

h2m q V

h

2m q V

given V = Vp

p

p p

m q 4m 2e 8m q m e 1

The ratio of de-Broglie wavelength for photon and electron both of energy E is

phhcE

eh

2mE so

2ph

e

hc 2mE 2mcE h E

8

MATTER WAVES

When a photon of wavelength is emitted from an atom then recoil energy of atom is

2 2

r 2p hE2M 2M

joule

Ex. If the momentum of proton is changed by p0 then de-Broglie wavelength is changed by 0.25%. The initial momentumof proton is

Sol.hp

so p = 2h h p

In magnitude p

p

so 0p 0.25

p 100

or 00

100Pp 400p0.25

Ex. If the kinetic energies of proton and electron are equal. Find the ratio of associated wavelength?

Sol. ee

h2m E

pp

h2m E

pe

p e

m 1836m 1

or e p: 1836 :1

DO NOT FORGET US1. Matter wave are stationary waves.2. When an electron move in an orbit, matter waves are produced due to its motion.3. Wavelength of matter waves associated with macroscopic moving particles are very small, therefore they cannot

be measured and it appears that they exhibit particle nature alone.

4.1 2

2 1

PP

5.1

2

k1

2 k

EPP E

6.1 1

2 2

P VP V

7. In davission Germer’s experiment maximum peak is obtained in I-V curve at angle of diffraction of 50º andaccelerating potential 54 volt.

8. Matter waves are probabilistic waves9. The phase velocity of matter waves can be greater than that of light.10. Matter waves are not mechanical waves11. The wavelength of matter wave does not and on the nature and charge of the particle.12. If the proton and -particle have same kinetic energy then / = 2/113. Electron microscope works on the principle of matter waves.14. Resolving power of an electron microscope is inversely proportional to wavelength and therefore can be increased

by increasing the accelerating potential.

15. The electron and photon have same energy E and mass of the electron is m then photon

e

2mCE

16. If the wavelength of de-Broglie waves associated with a proton and an -particle are same then the potential

difference are in ratio pV 8

V 1

9

MATTER WAVES

Ex.1 The de-Broglie wavelength associated with an electron having a kinetic energy of 10eV is -[1] 10 Å [2] 12.27Å [3] 3.9 Å [4] 0.10Å

Sol. The energy of 10eV means thatE = eV = 10e voltV = 10 volt

The electron was accelerated through a p.d. of 10 V

Now, 12.27 Å

V

= 12.27 Å 3.9Å

10

Ex.2 The de-broglie wavelength of the electron in the second Bohr orbit is (given the radius of the first orbit r1 = 0.53Å)Sol. The de-broglie wavelength is 2r2 = 2

= r2

r2 = 22r1 = 4r1 = 4 × 0.53 = 2.12Å= 3.14 × 2.12 = 6.66Å

Ex.3 An -particle and a singly ionized 4Be8 atom are accelerated through the same potential difference. Ratio ofde-broglie wavelength -[1] 1 : 2 [2] 2 : 1 [3] 4 : 1 [4] 1 : 1

Sol. Be Be

Be

m qm q

= 8 1 14 2

: Be = 1 : 1Ex.4 The de-broglie wavelength of an electron is 2.0Å. Calculate the potential difference required to retard it to rest -

[1] 3.76 × 10–3V [2] 3.76 × 103V [3] 3.76 × 103 eV [4] 376.5V

Sol. V = 2e

150.6 volt to determine the p.d through which it was accelerated to achieve the given de-broglie wavelength.

Then the same p.d. will retard it to rest. Thus,

V = 150.6 volt

0.2 0.2, V = 3765 volt = 3.76 kV

Ex.5 A hydrogen atom moving at a speed v absorbs a photon of wavelength 122nm and stops. Find the value of v.(mass of hydrogen atom = 1.67 × 10–27 kg)

Sol. The linear momentum of the photon = h

= 34

279

6.63 10 kg m5.43 10s122 10

p = mv v = pm v =

27

27

5.43 10 3.25 m/ s1.67 10

Ex.6 An electron microscope uses 40KeV electrons. Find its resolving limit on the assumption that it is equal to thewavelength of the electron -[1] 0.61Å [2] 0.6Å [3] 0.06Å [4] 0.061Å

SOLVED EXAMPLE

10

MATTER WAVES

Sol. Wavelength of electrons is =150 ÅV

Now, electron have energy of 40KeV, therefore they are accelerated

through a potential difference of 40 × 103 volt.

3150 0.061Å

40 10

Resolving limit of electron microscope = 0.061ÅEx.7 From rest a electron is accelerated between two such points which has potential 20 & 40 volts respectively.

Associated Debroglie wavelength of electron is[1] 0.75Å [2] 7.5Å [3] 2.75Å [4] 2.75m

Sol.12.27 Å

V

V = 40 – 20 = 20 volt

= 12.27 Å 2.75Å

20

Ex.8 De-broglie wavelength of a rotating electron around a nucleus of hydrogen atom at the fundamental energy levelis -[1] 0.3Å [2] 3.3Å [3] 6.62Å [4] 10Å

Sol.1

h2mE

E1 = –13.6 eVE1 = –13.6 × 1.6 × 10–19J, h = 6.62 × 10–34 J-secm = 9.1 × 10–31 kg

34

31 19

6.62 102 9.1 10 13.6 1.6 10

= 3.3 × 10–10m = 3.3Å

Ex.9 Associated de-broglie wavelength of a electron in nth bohr’s orbit is -

[1] 2 r Ån

[2] 2nR [3] 1 Ån [4] nÅ

Sol. 2r = n = 2 r Ån

Ex.10 Velocity of a proton is C20 . Associated de-broglie wavelength is

[1] 2.64 × 10–24 mm [2] 2.64 × 10–24 cm [3] 2.64 × 10–24 Å [4] 2.64 × 10–14m

Sol.h

mv

v = 8

7C 3 10 1.5 10 m/sec20 20

h = 6.62 × 10–34 J-s, m = 1.67 × 10–27 kg

= 34

27 7

6.62 101.6 10 1.5 10

= 2.64 × 10–14 m

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MATTER WAVES

Ex.11 Energy of a -particle, having debroglie wavelength of 0.004 Å.[1] 1275eV [2] 1200 KeV [3] 1200MeV [4] 1200GeV

Sol.0.101 0.101Å, V

0.004V

V 25.25V, V 637V

E q v 1275eV

Ex.12 Ratio of wavelength of deuteron & proton accelerated by equal potential

[1] 12 [2]

21

[3] 12 [4]

21

Sol. p dp p d d

h h2m e V 2m e V

p pd

p d d

m em e

md = 2mp, ed = ep p pd

p p p

m e 12m e 2

Ex.13 One electron & one proton have equal energies then ratio of associated de-broglie wavelength will be

[1] 1 : (1836)2 [2] 1836 : 1 [3] 1836 : 1 [4] (1836)2 : 1

Sol.pe

p e

m1 1836,m 1m

Ex.14 De-broglie wavelength of a electron is 10Å then velocity will be[1] 7.2 × 107 m/s [2] 7.2 × 106 m/s [3] 7.2 × 105 m/s [4] 7.2 × 104 m/s

345

31 10

h h 6.6 10V , V 7.2 10 m/smv m 9.1 10 10 10

Ex.15 On electron & one proton is accelerated by equal potential . Ratio in their de-broglie wavelength is

[1] p

e

mm [2]

e

p

mm [3]

p

e

mm [4] 1

Sol. e pe p

1 1 1,m m m

pe

p e

mm

12

MATTER WAVES

EXERCISE # 1Q.1 The ratio of deBroglie wavelength of a proton and an -particle accelerated through the same potential

difference is

[1] 2 [2] 2 2 [3] 22/1 [4] 2

Q.2 The ratio of deBroglie wavelengths of a proton and an alpha particle of same energy is .

[1] 1 [2] 2 [3] 4 [4] 0.25

Q.3 The ratio of de broglie wavelengths of a proton and an alpha particle moving with the same velocity is

[1] 1 [2] 2 [3] 4 [4] 0.25

Q.4 The ratio of de Broglie wavelengths of a proton and a neutron moving with the same velocity is nearly

[1] 1 [2] 2 [3] 2/1 [4] none of the above

Q.5 In a TV tube the electron are accelerated by a potential difference of 10 kV. Then, their deBroglie wavelengthis nearly

[1] 1.2 Å [2] 0.12 Å [3] 12 Å [4] 0.01 Å

Q.6 Electron microscope works on the principle of

[1] wave nature of light [2] particle nature of light [3] wave nature of electron [4] particle nature of electron

Q.7 The group velocity of the de Broglie wave packet associated with a particle moving with velocity v is

[1] equal to v [2] less than v [3] greater than v [4] equal to speed of light

Q.8 The de Broglie waves are associated with moving particles. These particle may be

[1] electrons [2] He+, Li2+ ions [3] Cricket ball [4] all of the above

Q.9 The de Broglie wavelength of a particle is the same as the wavelength of a photon. Then, the photon'senergy is

[1] equal to the kinetic energy of the particle [2] less than the kinetic energy of the particle

[3] greater than the kinetic energy of the particle [4] nothing can be specified

Q.10 Determine the kinetic energy of a neutron having de Broglie wavelength 1 Å (mass of neutron is 1.67 x 10–24 gm,h = 6.63 x 10–27 erg.)

[1] 1.2564 eV [2] 0.8888 eV [3] 0.0825 eV [4] 0.7825 eV

Q.11 What voltage must be applied to an electron microscope to produce electrons of . = 1.0 Å

[1] 190 volt [2] 180 volt [3] 160 volt [4] 150 volt

Q.12 An electron microscope uses 40 KeV electrons. find its resolving limit on the assumption that it is equalto the wavelength of the electrons

[1] 0.061 Å [2] 1.061 Å [3] 0.182 Å [4] 0.29 Å

Q.13 Calculate the de Broglie wavelength of an electron whose kinetic energy is 50 eV, h = 6.62 x 10–34 joule-sec,m0 9.1 x 10–31 kg.

[1] 3.73 Å [2] 2.73 Å [3] 1.73 Å [4] None of these

13

MATTER WAVES

Q.14 An -particle moves along a circular path of radius 0.83 cm in a magnetic field of 0.25 Wb/m2. The de-Brogliewavelength associated with it will be

[1] 10 Å [2] 1 Å [3] 0.1 Å [4] 0.01 Å

Q.15 Momentum of -ray photon of energy 3 keV in kg-m/s will be

[1] 1.6 x 10–19 [2] 1.6 x 10–21 [3] 1.6 x 10–24 [4] 1.6 x 10–27

Q.16 In Davisson-Germer experiment an electron beam of energy 60 eV falls normally on the surface of a crystal. Ifthe maximum intensity is obtained at an angle of 60° to the direction of incident beam, then the inter-atomicdistance in the lattice plane of the crystal will be

[1] 18 Å [2] 3.6 Å [3] 1.8 Å [4] 0.18 Å

Q.17 The mass of a particle is m. If its kinetic energy is increased to four times, then de-Broglie wavelength will

[1] Remain unchanged. [2] Become half [3] Become double [4] Become 21

times.

Q.18 Wrong statement in connection with Davisson-Germer experiment is

[1] The inter-atomic distance in nickel crystal is of the order of the de-Broglie wavelength.

[2] Electrons of constant energy are obtained by the electron gun.

[3] Nickel crystal acts as a three dimensional diffracting grating.

[4] Davisson-Germer experiment is an interference experiment.

Q.19 Mean kinetic energy of a free electron at high temperatures is 23

kT. At what temperature will be mean de-Broglie wavelength of the electron be 0.5 nm

[1] 466 K [2] 4660 K [3] 46600 K [4] 46.6 K

Q.20 In Davisson-Germer experiment the relation between the angle of diffraction and the grazing angle is

[1] 90º2

[2] 90º2

[3] = 90º – [4] = (90º – )/2

Q.21 An electron beam is incident on a crystal. The distance between atomic planes of the crystal is b. For whichde-Broglie wavelength electron beam will be directly reflected

[1] nb

[2] nb2

[3] nb3

[4] nb4

Q.22 The wavelengths of an electron and a photon are same and equal to 1Å. The ratio of their energies is

[1] 12 : 1 [2] 1.2 : 1 [3] 1.2 : 10 [4] 1.2 : 102

Q.23 In Davisson-Germer experiment an electron beam accelerated with 54 volt is diffracted at an angle of 50°by a nickel crystal and produces first diffraction maxima. The interatomic distance in Nickel crystal is

[1] 1 Å [2] 2 Å [3] 2.15 Å [4] 3.12 Å

Q.24 The energy of an electron moving in the second orbit of Bohr atom is –3.4 eV. The radius of this orbit forits stability will be

[1] 2.08 Å [2] 1.04 Å [3] 0.52 Å [4] 0.26 Å

14

MATTER WAVES

Q.25 The wavelength of de-Broglie waves associated with neutrons at room temperature T is

[1] ÅT82.1

[2] ÅT82.1

[3] ÅT7.30

[4] ÅT

7.30

Q.26 The mean thermal energy of oxygen molecules of mass m at temperature T is 23

kT where k is Boltzmann

constant. The average value of de-Broglie wavelength will be

[1] mkT2h

[2] mkT3h

[3] mkT3h2

[4] mkT3h

Q.27 The wavelength of electrons moving with a very large velocity (v c) will be

[1] ph

[2] mE2h

[3] mE2

h22 [4]

22

0c/v1

vmh

Q.28 The energy of a neutron of de-Broglie wavelength 1 Å will be

[1] 6.6 x 10–13 eV [2] 4 x 10–8 eV [3] 8.13 x 10–2 eV [4] 1.6 x 10–19 eV

Q.29 An electron and a positron are having same kinetic energy. The ratio of their de-Broglie wavelengths willbe

[1] 2 : 1 [2] 1 : 2 [3] 1 : 1 [4] 2:1

Q.30 De-Broglie wavelength of an electron is 10–10 m. Its energy will be

[1] 1.51 x 102 eV [2] 3 x 104 eV [3] 1.6 x 10–10 eV [4] 10 x 105 eV

Q.31 A photon of wavelength 1Å is emitted by an atom. Due to emission of photon the recoil energy of the atomwill be- (mass of the atom = 1 amu)

[1] 1.304 x 10–20 J [2] 1.304 eV [3] 1.532 x 10–19 J [4] 1.532 eV

Q.32 In Davisson-Germer experiment an electron beam of wavelength 1.5 Å is incident normally on a crystal havingatomic spacing of 3Å. The angle at which the first maximum if formed is

[1] 30° [2] 60° [3] 90° [4] 180°

ANSWER KEY EXERCISE # 1

Que . 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15Ans. 2 2 3 1 2 3 1 4 3 3 4 1 3 4 3Que . 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30Ans. 3 2 4 3 2 2 4 3 1 3 2 4 3 3 1Que . 31 32Ans. 1 1

15

MATTER WAVES

Q.1 The de-Broglie wavelengths of a photon, an electron and a uranium nucleus are equal. Which of these willhave highest total energy[1] Photon [2] Electron[3] Uranium nucleus [4] It will depend on wavelength

Q.2 The wavelength of a photon and the deBroglie wavelength of an electron and uranium atom are identical.Which one of them will have highest kinetic energy

[1] Photon [2] electron [3] U-atom [4] nothing can be predicted

Q.3 Consider the statements given below

(A) The wave and the particle aspects are both necessary for a complete description of light

(B) The wave and particle aspects cannot be revealed simultaneously in a single experiment. The

[1] A is true, B is false [2] A is false, b is true [3] Both A and B are false [4] Both A and B are trueQ.4 Two particles have identical charges. If they are accelerated through identical potential differences, then the

ratio of their deBroglie wavelength would be[1] 1 : 2 = 1 : 1 [2] 1 : 2 = m2 : m1

[3] 1 : 2 = 12 m:m [4] 1 : 2 = 21 m:m

Q.5 Which of the following figure, represents the variation of particle momentum and associated de-Broglie wave-length

[1]

P

O

[2]

P

O

[3]

P

O

[4]

P

O Q.6 In Bohr's atom the number of de Broglie waves associated with an electron moving in nth permitted orbit is

[1] n [2] 2n [3] n/2 [4] n2

Q.7 The increase in the energy of an electron for changing its de-Broglie wavelength from 1 Å to 0.5 Å will be[1] 1.5 keV [2] 0.45 keV [3] 0.25 keV [4] 100 eV

Q.8 If the velocity of a moving particle is reduced to half, then percentage change in its wavelength will be[1] 100% decrease [2] 100% increase [3] 50% decrease [4] 50% increase

Q.9 Linear momenta of a proton and an electron are equal. Relative to an electron[1] Kinetic energy of proton is more. [2] De-Broglie wavelength of proton is more.[3] De-Broglie wavelength of proton is less. [4] De-Broglie wavelength of proton and electron are equal.

Q.10 Helium atom emits a photon of wavelength 0.1 A. The recoil energy of the atom due to the emission of photonwill be[1] 2.04 eV [2] 4.91 eV [3] 1.67 eV [4] 9.10 eV

Q.11 The resolving power of an optical instrument depends upon the wavelength. In electron microscope electronbeam is used. Its resolving power can be increased if we[1] Increase the accelerating voltage. [2] Decrease the accelerating voltage.[3] Use protons in place of electrons. [4] Increase intensity of electron beam.

Q.12 Which of the following statements is wrong[1] De-Broglie waves are probability waves and there is no physical existence of these.[2] De-Broglie wavelength of a moving particle is inversely proportional to its momentum.[3] Wave nature is associated with atomic particles only.[4] In general wave nature of matter is not observed.

EXERCISE # 2

16

MATTER WAVES

Q.13 If the momentum of an electron is changed by p, then the de-Broglie wavelength associated with it changesby 0.50%. The initial momentum of the electron will be

[1] 200p

[2] 199p

[3] 199 p [4] 400 p

Q.14 The interatomic distance in the surface of a crystal is 1.227 Å. For electrons which are accelerated by 10 kV,the maximum order of diffraction will be

[1] 1 [2] 10 [3] 100 [4] 1000

Q.15 If a proton and an -particle are accelerated by same potential difference, then the ratio of wavelengths ofmatter waves associated with them will be

[1] 2:1 [2] 1:2 [3] 8:1 [4] 1:8

Q.16 A photon and an electron have equal energy E. photon/electron is proportional to -

[1] E [2] 1/ E [3] 1/E [4] Does not depend upon E

Q.17 The ratio of de-Brolie wavelength of a -particle to that of a proton being subjected to the same magnetic field sothat the radii of their path are equal to each other assuming the field induction vector B

is perpendicular to thevelocity vectors of the -particle and the proton is -

[1] 1 [2] 14 [3]

12 [4] 2

Q.18 The potential energy of a particle of mass m is given by 0E ; 0 x 1U(x)

0; x 1

, 1 and 2 are the de-broglie

wavelengths of the particle, when 0 x 1 and x > 1 respectively . If the total energy of particle is 2E0, the ratio

1

2

will be

[1] 2 [2] 1 [3] 2 [4] 12

Q.19 A proton, a deuteron and an -particle having the same momentum, enters a region of uniform electron fieldbetween the parallel plate of a capacitor. The electric field is perpendicular to the initial path of the particle. Thenthe ratio of deflections suffered by them is -

p d

screen+

-

[1] 1 : 2 : 8 [2] 1 : 2 : 4 [3] 1 : 1 : 2 [4] None of these

ANSWER KEY EXERCISE # 2Que . 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15Ans. 3 1 4 3 4 1 2 2 4 1 1 3 3 2 4Que . 16 17 18 19Ans. 2 3 3 1

17

MATTER WAVES

Q.1 The energy of electron with de-Broglie wavelength of 10–10 meter, in [ev] is - [RPMT- 88][1] 13.6 [2] 12.27 [3] 1.27 [4] 150.6

Q.2 The de-Broglie wavelength of a particle accelerated with 150 volt potential is 10–10 m : If it is acceleratedby 600 volts p.d. its wavelength will be - [RPET-88][1] 0.25 Å [2] 0.5 Å [3] 1.5 Å [4] 2 Å

Q.3 If the momentum of photon is P then its frequency will be [MPPMT-89][1] Ph/c [2] Pc/h [3] mh/c [4] mc/h

Q.4 The value of momentum of an electron which is accelerated a volt potential difference V is [RPMT- 89]

[1] me V [2] meV [3] meV2 [4] m/eV2Q.5 An electron [charge = 1.6 x 10–19 coulomb] is accelerated by potential of 1,00,000 volt. The energy acquired

by electron will be - [RPET- 89][1] 1.6 x 10–24 Joule [2] 1.6 x 10–14 erg [3] 0.53 x 10–17 Joule [4] 1.6 x 10–14 Joule

Q.6 The de-Brolie wavelength of a moving particle is [RPET- 89]

[1] h/P [2] h/ P [3] P/h [4] hPQ.7 The de-Broglie wavelength of an atom at Tº K is [RPMT-89]

[1] mKT3 [2] mKT3h [3] mKT3h

[4] hmKT3

Q.8 The deacceleration voltage for an electron, whose de-Broglie wavelength is 1Å will be [RPMT-89][1] 12.27 V [2] 1/12.27 V [3] 150.6 V [4] 1/150.5 V

Q.9 An electron is accelerated through a potential difference of 104 volts, it’s de-Broglie wavelength will be -[RPMT 97, PET 89]

[1] 12.27 Å [2] 1.227 Å [3] 0.1227 Å [4] .001227 Å

Q.10 Electron volt is unit of [RPMT-87, PET-89][1] Charge [2] Momentum [3] Energy [4] Potential difference

Q.11 The energy of an electron is E, then the associated wavelength is [MPPET-90]

[1] mE2/h [2] 2h/mE [3] 2mhE [4] h/Em22Q.12 In Davisson-Germer experiment, the filament emits [RPET-90]

[1] Photons [2] Protons [3] X-rays [4] Electrons

Q.13 The energy of a Proton and an -particle is same. The ratio of associated de-Broglie wavelength will be -

[1] 1: 2 [2] 2 : 1 [3] 1: 4 [4] 4: 1 [RPET- 91]

Q.14 In Davisson-Germer experiment, Nickel crystal acts as [RPET- 91][1] Perfect reflector [2] Three dimensional diffraction grating[3] Ideal absorber [4] Two dimensional diffraction grating

Q.15 The charge on an electron is [RPMT 93][1] 1.6 x 10–19 coulomb [2] –1.6 x 10–19 coulomb [3] – 9.1 x 10–31 coulomb [4] 6.6 x 10–34 Joule

Q.16 Which experiment explains the wave nature of electron [RPET-93][1] Michelson’s experiment [2] Davisson Germer experiment[3] Roentgen experiment [4] Rutherford experiment

Q.17 The diffracted waves in the Davisson -Germer experiment are [RPMT-93][1] Electrons [2] X-rays [3] Photons [4] Protons

EXERCISE # 3

18

MATTER WAVES

Q.18 In Davisson-Germer experiment when electron strikes the Ni-crystal which of the following is produced -

[1] X-rays [2]-rays [3] Electron [4] photon [RPMT-93]Q.19 In an electron gun electron are accelerated through a potential difference V. If e = charge of electron and

m = mass of electron then maximum electron velocity will be [MPPET, CPMT-93, RPMT 93]

[1] 2eV/m [2] m/eV2 [3] eV/m2 [4] V2 /2em

Q.20 The rest mass of a photon is [MP PET 84, RPMT-94][1] 0 [2] [3] Between 0 and [4] Equal to the rest mass of electron

Q.21 The magnitude of de-Broglie wavelength for electron in first Bohr orbit of an atom is [RPMT- 95][1] Equal to the circumference of orbit [2] Half to the circumference of orbit[3] One fourth to the circumference of orbit [4] None of the above

Q.22 The momentum of Photon having energy E is [RPET-88, RPMT-95][1] E/C [2] 1/E [3] E/C2 [4] None of the above

Q.23 The wavelength of x-ray photon is 0.01 Å, its momentum in Kg m/sec is [RPMT-95]

[1] 6.6 x 10–22 [2] 6.6 x 10–20 [3] 6.6 x 10–46 [4] 6.6 x 10–27

Q.24 The accelerating voltage of an electron gun is 50,000 volt. De-Broglie wavelength of the electron will be-

[1] 0.55 Å [2] 0.055 Å [3] 0.077 Å [4] 0.095 Å [RPMT- 95]

Q.25 If wave length of a proton and photon are identical, which of the following quantities will be identical- [CPMT-95, RPET-95]

[1] Velocity [2] Kinetic energy [3] Mass [4] MomentumQ.26 If particles are moving with same velocity, then maximum de-Broglie wavelength is for [CPMT- 2002]

[1] Proton [2] -particle [3] Neutron [4] -particleQ.27 The wave nature of electron was confirmed experimentally by [RPET- 2002]

[1] Davisson-germer [2] de-Broglie [3] Rutherford [4] Milikan-oil dropQ.28 If the kinetic energy of a free electron doubles, its deBroglie wavelength changes by the factor -

[1] 21

[2] 2 [3] 21

[4] 2 [AIEEE- 2005]

Q.29 When the speed of electron beam used in Young’s double slit experiment is increases, then which among thefollowing statement is correct? [IIT- 2005][1] interference pattern will not be observed in case of electrons[2] distance between consecutive fringes increases[3] distance between consecutive fringes decreases[4] distance between consecutive fringes remain same.

Q.30 The de Broglie wavelength of a tennis ball of mass 60 g moving with a velocity of 10 metres per second isapproximately. (Planck’s constant, h = 6.63 × 10–34Js)- [AIEEE - 2003][1] 10–33 metre [2] 10–31 metre [3] 10–16 metre [4] 10–25 metre

Q.31 The time taken by a photoelectron to come out after the photon strikes is approximately [AIEEE - 2006][1] 10–10 sec [2] 10–16 sec [3] 10–1 sec [4] 10–4 sec

Q.32 The momentum of a photon of energy 1 MeV in kg m/s, will be - [CPMT- 2006][1] 7 × 10–24 [2] 10–22 [3] 5 × 10–22 [4] 0.33 × 106

ANSWER KEY EXERCISE # 3Que. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15Ans. 4 2 2 3 4 1 3 3 3 3 1 4 2 2 2Que. 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30Ans. 2 1 3 2 1 1 1 1 2 4 4 1 1 3 1Que. 31 32Ans. 1 3