Chapter-11 DUAL NATURE OF MATTER AND RADIATION

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Transcription:

Chapter-11 DUAL NATURE OF MATTER AND RADIATION

Work function (j o ): The minimum energy required for an electron to escape from the surface of a metal i.e. The energy required for free electrons to escape from the metal surface

Work function depends on the properties of the metal and nature of the surface. Work function of alkali metals is less as compared to other metals. Work function of metals decreases with increase in atomic number.

Cesium is the best photosensitive metal as it has least Work function. Infrared radiation can not produce photoelectric effect on any metal surface

The phenomenon of emission of electrons from metal surfaces exposed to light energy of suitable frequency is called photoelectric effect.

Photoelectric current increases linearly with increase in intensity of incident light. I μa Intensity (L) i.e. the number of electrons emitted per second is directly proportional to intensity of light.

Photoelectric current increases with increase in +ve potential applied to the anode. I μa V S L L 2 2 > L 1 Saturation Current L 1 0 + Potential of A (V) When all the photoelectrons reach the plate A, current becomes maximum and is known as saturation current.

When the potential is decreased, the current decreases but does not become zero at zero potential. i.e.even in the absence of accelerating potential, a few photoelectrons manage to reach the plate on their own due to their K.E.

When ve potential is applied to the plate A with respect to cathode, photoelectric current becomes zero at a particular value of ve potential called stopping potential or cutoff potential. Intensity of incident light does not affect the stopping potential. If K max is the maximum kinetic energy of emitted electron, K max =e V o

I μa Saturation Current 2 2 > 1 1 V S 2 V S 1 0 Potential of A (V) + The saturation current is same for different frequencies of the incident lights of same intensity. Higher the frequency, higher the stopping potential. i.e. V S α

Variation of stopping potential with frequency of incident radiation V S (V) 1 ν o 1 ν o 2 2 Stopping potential 0 varies linearly with the frequency of incident radiation There exists a certain minimum cut-off frequency ν 0 for which the stopping potential is zero.

Einstein s photoelectric equation Photon energy=work function+ Max.K.E h h K. E o max

h h K. E o max hc hc KE. max o hc hc 1 mv 2 2 o 1 1 K. E hc - max o

Photon concept of electromagnetic radiation In interaction of radiation with matter, radiation behaves as if it is made up of particles called photons. Each photon has energy E (=hν) and momentum p (= h ν/c). All photons of light of a particular frequency ν, or wavelength λ, photon energy is independent of intensity of radiation.

Photons are electrically neutral and are not deflected by electric and magnetic fields. In a photon-particle collision, the total energy and total momentum are conserved. However, the number of photons may not be conserved in a collision. The photon may be absorbed or a new photon may be created.

Dual nature of matter The wave associated with material particles in motion is called matter wave or de Broglie wave. h p h mv Matter wave is neither electromagnetic nor mechanical

Relation between de-broglie wavelength and kinetic energy of the particle h 1 2mE me h 1 2mqV mqv

If electron is accelerated through a potential difference of V volt h 2meV 12.27 o A V

1.The work function of a metal is the minimum energy required 1) to remove valence electron from a metal atom 2) to remove any electron from a metal atom 3) to remove free electron from a metal surface 4) to remove any electron from a metal surface

Ans: To remove free electron from a metal surface

2.The work functions of lithium and copper are 2.1 ev and 4.3 ev respectively. Out of these, the one which is suitable for the photoelectric cell that works with the visible light is 1) Lithium 2) copper 3) both lithium and copper 4) neither lithium nor copper

Maximum energy of visible energy photon is about 3.1 ev. Hence visible light can produced photoelectric effect on Lithium but not on Copper

3.A metallic surface has a threshold wave length of 5200Å. This surface is irradiated by monochromatic light of wavelength 4500Å. Which of the following statement is true? 1) the electrons are emitted from the surface with energy between 0 to infinity 2) the electrons are emitted from the surface with energy between 0 and finite maximum value 3) the electrons are emitted from the surface all with certain finite energy 4) no electron is emitted from the surface

incident < o Ans: 2)Photoemission takes place with kinetic varying from zero to maximum

4.Blue light can cause photoelectric emission from a metal but yellow light cannot. If red light is incident on the metal then, 1) photoelectric current will increase 2) rate of emission of photoelectrons will decrease 3) no photoelectric emission will occur 4) energy of the photoelectrons will increase

Threshold wavelength is less than that of yellow light. But wavelength of red is longer than that of yellow So the red light cannot cause photoelectric emission as R > Y > B Ans: 3)

5. Light of frequency 1.5 times the threshold frequency is incident on photo sensitive material. If the frequency is halved and intensity is doubled, the photoelectric current becomes 1)quadrupled 2) doubled 3) halved 4) zero

Ans: 4) When frequency is halved, it becomes 0.75 0. No emission of light takes place (since < 0 ) Ans: 4) zero

6. Which of the following radiation will produce photoelectrons with highest kinetic energy 1) X-rays 2) gamma rays 3) UV rays 4) visible light

Radiation with highest frequency will liberate electrons with maximum kinetic energy Ans: 2) Gamma rays

7. The work function of sodium and aluminium is 2eV and 4.2eV respectively. Then 1) Threshold of wavelength of sodium is less than that of aluminium 2) Threshold of wavelength of sodium is more than that of aluminium 3) Threshold of wavelength of sodium is equal to that of aluminium 4) None of the above

work function 1 threshold wavlength If sodium alu minium o(sodium) o(alu minium) Ans: 2) Threshold of wavelength of sodium is more than that of aluminium

1 8. The graph between and stopping potential (V) for three metals is plotted as shown. Which of the following statement is correct? 1) Ratio of work functions = 1 : 2 : 4. 2) Ratio of work functions = 4 : 2 : 1. 3) tan is equal to Planck s constant. 4) The violet colour light can eject photoelectrons from metals 2 and 3

1 o 1 1 1 : : : : 1 2 3 1 2 3 1 2 3 : : 1: 2 : 4 Ans: 1

9.According to Einstein photoelectric equation slope of plot of the stopping potential of the emitted photo electron verses frequency 1Depends on the intensity of radiation and metal used 2)Depends on the intensity of radiation 3)Depends on the metal used 4)Independent of metal used and incident radiation

Ans: (4) always the value of slope is a constant whose value is equal to plank s constant.

10. Relation between stopping potential V o and maximum velocity of photoelectron v is 1 1 1) V 2)V v v o o 2 o 2 3) V v 4) V v o

2 o max max max V K.E and K.E v V o v 2 max Ans: 3

11.Light of frequency 1.5 times the threshold frequency produce photoemission on a metal surface. If the frequency is halved and intensity is doubled, the photoelectric current becomes 1) quadrupled 2) doubled 3) halved 4) zero

When frequency is halved, it becomes 0.75 0. Since < 0 No emission of light takes place. Ans: 4) zero

12. The photoelectrons emitted from a metal surface 1) Have the same momentum 2) Have the same kinetic energy 3) Have speeds varying from zero up to a certain maximum value 4) Are all at rest

Ans: photoelectron can have kinetic varying from zero to certain maximum value. Therefore photoelectrons can have speeds varying from zero up to a certain maximum value

13. The correct curve stopping potential verses intensity of incident radiation (X-axis Intensity Y-axis Stopping potential) 1 2 3 4 Stopping potential is independent of intensity. Ans: (3)

14. The work function of a metal is 2eV. When a radiation of frequency1.6x10 15 H Z.is incident on it,then 1) There will be photoemission 2)No photoemission 3)There will be photoemission if intensity is very high 4)There will be photoemission if intensity is very less

34 15 Energy of photon E h 6.63x10 x1.6x10 J h o 6.63x10 x1.6x10 19 1.6x10 34 15 6.63eV Ans: There will be photoemission

15. A photon of energy 8 ev is incident on a metal surface of threshold frequency 1.6 10 15 Hz. The K. E. of the photo electrons emitted (in ev), (h = 6 10 34 Js) 1) 1.8 2) 6 3) 2 4) 1.2

610 34 1.610 15 o 1.610 19 work function h 6eV E max = 8 6 = 2 ev

16. Maximum velocity of photoelectron from a surface is 1.2 10 6 ms 1. Assuming the specific charge of electron to be 1.8 10 11 Ckg 1,the value of stopping potential in volt will be 1) 4V 2) 6V 3) 2V 4) 3 V

1 mv ev 2 2 max 0 1 v 2 1 (1.2x10 6 ) 2 max o 11 V x 4 e 2 2 1.8x10 m Ans: 1) 4 V

17. When a metal surface is illuminated with a monochromatic radiation of wavelength λ the stopping potential is 3V. When it is illuminated with radiation of wavelength 2λ,stopping potential reduces to V. Threshold wavelength for the metal is 1) 8 λ 2) 6 λ 3) 4 λ 4) 4 λ/3

hc hc K.E= - λ λ o hc hc 3Ve= -...(1) λ λ o hc hc Ve= -...(2) 2λ λ o Ans:(3) o 4

18. The kinetic energies of photoelectrons emitted from a metal are K 1 and K 2 when it is irradiated with lights of wavelength 1 and 2 respectively. The work function of the metal is K λ 1) λ K λ λ 1 1 2 2 2 1 K λ 2) λ K λ λ 1 1 2 2 2 1 3) K λ λ K λ λ 1 2 2 1 2 1 4) K λ λ K λ λ 1 2 2 1 2 1

h c λ 1 h c λ 2 W + K h c W λ + K λ...(1) 1 1 1 1 W + K h c W λ + K λ...(2) 2 2 2 2 From (1) and (2) W λ + K λ W λ + K λ or W λ W λ K λ K λ 2 1 1 1 2 2 K 1 λ1 K 2 λ2 W λ λ 2 1 Answer: (1) 2 2 2 1 1 1

19. When ultraviolet light is incident on a photocell, its stopping potential is V o and the maximum momentum of electron is P o.when X- ray is incident on the same cell 1) Both V o and P o will increase. 2) Both V o and P o will decrease 3) V o increase and P o will decrease 4) V o remain same and P o will increase.

Energy of X-ray photon is greater than ultraviolet photon Therefore Kinetic energy of emitted electron and stopping potential w ill be more for ultraviolet light In turn the momentum of the emitted electron Ans: 1)Both V o and P o will increase.

20. In an experiment with photoelectric effect, the slope of stopping potential verses frequency of incident light is found to be 4.130x10-15 Vs. The value of plank s constant from the graph is 1)6.591x10-34 Js 3) 6.616x10-34 Js 2)3 6.650x10-34 Js 4) 6.631x10-34 Js

Slope of V o verses = h/e h =slope x e=4.130x10-15 x1.6x10-19 = 6.616x10-34 Ans: (2)

21. When a monochromatic point source of light at a distance of 0.2m from a photocell. The cut off voltage and saturation current is 0.6V and 18mA. If the same source is placed 0.6m away from the cell, the cut off voltage and saturation current is 1) 0.2V and 2mA 2) 1.8V and 6mA 3) 0.6V and 6mA 4) 0.6V and 2mA

As the source is moved away,the intensity of light falling on the surface reduces according to the relation 1 intensity α distance Therefore when the distance is increased by 3 times intensity reduces by 9 time and hence the current reduces to 2mA Frequency remains same and hence the cut off voltage remains same (0.6V) 2

Ans: 4) 0.6V and 2mA

22. The maximum kinetic energy of photoelectrons is 1eV and 3eV for incident 3 photon frequency and respectively. 2 Maximum kinetic energy of photoelectrons for 9 the incident frequency 4 1) 8eV 2) 6eV 3) 45eV 4) 3eV

Let Energy of light freuency E ; 3 3 Energy of light freuency E 2 2 9 9 Energy of light freuency E 4 4 From Einstein 's equation 1E1 3 E1 4, 3 3 E1 2 9 K.E E1 6eV 4 1 1 1

Ans: 2) 6eV

23. A photosensitive metal is first incident with the radiation of wavelength 400 nm and then with radiations of wavelength 800 nm. The change in the maximum kinetic energy of the photoelectron is 1) 0.55 ev 2) 1.55 ev 3) 2.0 ev 4) 1.0 ev

1 1 E hc 1 2 20x10 1.56eV 26 1 1 4x10 8x10 7 7 Ans: (2)

24. Maximum velocity of photoelectrons emitted by a photo emitter is 2X10 6 m/s. If e/m=1.8x10 11 for electrons, the stopping potential of the emitter is 1) 11.1V 2) 20V 3) 40V 4) 2V

1 mv ev 2 2 max 0 1 v 2 1 (2x10 6 ) 2 max o 11 V x 11.1V e 2 2 1.8x10 m Ans: 2) 11.1V

25. Light photons of energies 1 ev and 2.5 ev are successively incident on a metal surface of work function 0.5 ev, then the ratio maximum velocities of the emitted photo-electrons will be 1) 1 : 5 2) 1 : 4 3) 1 : 3 4) 1 : 2

26. When UV light of wavelength 100 nm is incident on a silver surface of work function 4.7eV, a negative potential of 7.7V is required to stop the photo-electrons from reaching the collector plate. The potential which is required to stop the photoelectrons when light of wavelength 200 nm is incident on it will be 1) 1.5V 2) 3V 3) 4.5V 4) 6V

E = 7.7 ev, W=4.7eV K1 Energy of photon: E = W + E 1 K1 = 4.7 + 7.7 = 12.4 ev E λ 2 1 100 nm 1 = = = E λ 200 nm 2 1 2 E E = 1 6.2eV 2 2

2 nd case : Energy of photon : E 6.2 ev 2 E E - W 6.2-4.7 1.5 ev K2 2 Hence the stopping potential is 1.5V Ans:1

27. The ratio of linear momentum of an electron and alpha particle when both are accelerated through a potential difference of 100 volt 1) 1 2) 2m e m a me 3) m a 4) me 2m a

1 p mq & q 2q pe meqe me p m q 2m e Ans : 4) me 2m a

28. An X-ray tube operates at 10kV. The ratio of X-ray wavelength to that of de-broglie is 1)10:1 2) 1:10 3) 1:100 4) 100:1

de Broglie wavelength Wavelength of X ray x d hc ev h 2meV d x d 2m c ev : 10 :1 x 10 Ans: 1)10:1

29. If the kinetic energy of the particle is increased by 16 times,the percentage change in the de-broglie wavelength is 1)25% 2)75% 3)60% 4)50%

1 E E E 1 E 16E 4 2 1 1 1 2 1 1 2 1 0.25 1 4 change in wavelength 0.75 or 75% Ans: 2)75%

30. The de-broglie wavelength associated with proton changes by 0.25% if its momentum changes by p. The initial momentum is 1) 100 p 2) p/400 3) 401 p 4) p/100

h p (1) 0.25 h 1 (2) 100 p p solving above equations p 401p Ans: 3) 401 p

31. The de-broglie wavelength of electron in ground state of hydrogen atom is o o o o 1) 0.5A 2)1.06A 3) 1.67A 4)3.33A

The circumference of ground state orbit is equal to the wavelength of corresponding de-broglie wave i.e. 2r ( 0.53A) o o 3.33A r o Ans :4)3.33A o o

32. The maximum kinetic energy of photoelectrons from the surface of silver of threshold wavelength λ o due to a radiation of wavelength λ (λ< λ o ) is given by 1) hc( ) o hc 2) ( ) 0 1 1 3) h c o o 4) hc o

KE. max hc hc o hc o o Ans: (4)

33. Of the following the graph which represents the variation of momentum (p) of a particle with the wavelength () of matter wave is p p p p

h p 1 p Hence the plot is a rectangular hyperbola Ans:3

34.If the velocity of a particle is reduced to one-third then the percentage increase in its de-broglie wavelength is 1) 50% 2) 100% 3) 200% 4) 300%

h h h λ = and λ' = = 3 = 3λ m v v m v m 3 Change in de - Broglie wavelength = 2λ 2λ Percentage change = 100% = 200% λ Answer is (3)

35. The ratio of de-broglie wavelength associated with proton and particle accelerated through same potential difference 1) 1 2)2 3) 8 4) 1 8

h 1 For a charged particle : λ = 2 m q V m q 1 For a proton : λp m e 1 1 For α - particle : λ α = (4m) (2e) 8 m e λ λ p α 8 m e = = 8 m e Answer is (3)

36. If the momentum of two particles of mass m and 2m are equal, then the ratio of wavelength of the matter wave is 1) 1 : 2 2) 2 : 1 3) 1 : 1 4) 1 : 3

h p 1 2 p p 2 1'; 1 1 : 2 = 1 : 1 Ans: (3)

37. An X ray photon has a wavelength of 0.02 Å. Its momentum is 1) 3.3 10 22 kg m/s 2) 6.6 10 21 kg m/s 3) 6.6 10 24 kg m/s 4) 1.65 10 22 kg m/s

h Momentum: p= λ -34-34 6.6 10 = 6.6 10 = 0.02 10 2 10 =3.3 10-10 -12-22 Answer: (1)

38. If E is the energy, de-broglie wavelength is proportional to 1) E 1 for both photons and particles 2) E 1 for photons and E 1/2 particles 3) E 1/2 for both photons and particles 4) E 1/2 for photons and E 1 for particles

h c For photon: E = λ 1 λ λ E 1 E h For a particle: λ= 2mE 1 1 λ λ E 2 E Answer is (2)

39. For given kinetic energy which of the following has the longest de-broglie wavelength? 1) Electron 2)Proton 3)Neutron 4) alpha particle

λ = h or λ 1, 2mE m > (m < m < m < m ) e p n e p n α Answer: (1)

40. Electrons used in an EM are accelerated by a voltage of 25kV. If the voltage is increased to 100kV then the de-broglie wavelength associated with the electrons would 1) Increases to 2 times 2) increases to 4 times 3)decreases by 2 times 4)decreases by 4 times

λ λ V 1 2 1 25 kv 1 V λ V 100 kv 2 1 2 λ 1 λ 2 2 Ans: 3