XII / PHYSICS

DUAL NATURE OF MATTER & RADIATION

12/PA

DUAL NATURE OF MATTER AND RADIATION Light shows properties such as, diffraction, interference, and polarization. All such phenomena can be explained by considering light as a wave. So all such properties implies that light is a wave. But light also shows other phenomena such as the Photoelectric effect and the Compton Effect. Such phenomena can be explained by considering the particle nature of light (Photon theory of light). This means light sometimes behaves as a particle and sometimes as a wave, thus light has a dual nature. Photon theory of light assumes the particle nature of light, this was given by Planck, and hence it is also known as Planck’s Quantum or Planck’s Photon theory. MOVING FROM WAVE THEORY TO PARTICLE THEORY According to Huygen’s theory an illuminated body sends out a disturbance in the form of wave in the space. Huygens assumed the existence of hypothetical medium ether, through which waves could pass. Later on, Maxwell modified the concept by stating that light consists of electromagnetic waves requiring no material medium All the same, Maxwell’s theory of light could explained rectilinear propagation, reflection, refraction etc. in addition to phenomena like interference, diffraction, polarisation etc. Thus, the wave theory held a firm ground till spectroscopy and photoelectricity came into picture. Lines of the spectrum are possible only when an electron jumps from a higher energy level to a lower energy level, releasing the difference of energy in the form of radiation of a definite frequency in packets or discrete units called quanta. Likewise a definite quantum of light radiation, called photons, is falling o a surface to release an electron. Thus, these and similar phenomena could never be explained on the basis of wave theory. The only way we can explain these phenomena by assuming that light really consists of packets of energy called photons having a definite frequency. So, we assume the particle aspect of nature of radiation or light. PLANCK’S QUANTUM THEORY It has following postulates: 1. Light is emitted from a source not continuously but discontinuously in the form of small packets of energy, each packet is called the photon of light or a quantum. This means light emitted from a source consists of stream of large no. of photons or packets of energy traveling along straight line. 2. Energy of each photon is directly proportional to the frequency of light emitted by the source. i.e. E    E  h Where h is a constant of proportionality and is called the Planck’s constant. If there are n photons emitted by a source, then total energy of all such photons will be E  nh -34 The value of Planck’s constant is h = 6.6×10 Photons

Source

hν

PHOTON Each photon is also called the packet or particle of light. It has all the properties that a moving particle posses. i.e. it can exert impulse, it has linear momentum, it has kinetic energy etc. Following are certain properties that a photon possesses. 1. The rest mass of a photon is zero, i.e. when a photon comes to rest then its mass is lost. 2. Each photon travels in vacuum with a velocity c = 3×108 m/s, i.e. the velocity of light. visit www.TejasAcademy.in

1

XII / PHYSICS

DUAL NATURE OF MATTER & RADIATION

12/PA

3. The linear momentum associated with a photon moving with velocity c in vacuum is p = mc. 4. The total kinetic energy associated with a photon is E = mc2 [Einstein’s Equation] 5. When a photon is incident on reflecting surface is deviated from its path, i.e. it suffers reflection, in accordance with laws of reflection. PHOTOELECTRIC EFFECT We know that substances, chiefly metals, when exposed to electromagnetic radiations such as X-rays, gamma-rays, ultraviolet-rays, visible light, or in some cases even infrared rays, absorb the radiations and the energy is enough to help free electrons in the metal to come out of the surface. That emission of electrons is caused by the incident light energy. But light itself is wave, it cannot cause any particle to move, thus wave theory fails to explain this effect. This effect is better explained by the particle nature of light. Each photon is a packet of energy, when such photons interact with the free electrons on the surfaces of metal; they impart their momentum to them. Thus electrons acquire energy, and are forced to move out of surface. Thus Photoelectric effect is defined as the phenomenon of emission of free electrons from the surface of metals, when a light of sufficiently high frequency is incident on its surface Einstein’s Photoelectric Equation Photon- hν

Metal

Each photon of frequency  is associated with certain energy, given by, E = h . When one such photon interacts with a free electron of mass m, it imparts its energy to that electron. As a result the electron becomes energetic, and moves out of surface. Suppose it spends w0 of its total energy in just coming out 1 of surface, and remaining as its kinetic energy mv 2 then by the law of conservation of energy we have: 2 1 h  w0  mv 2 ---------------------------------------------------(1) 2 This is called Einstein’s photoelectric equation. Work Function The minimum amount of energy required by an electron to just come out of the surface of metal is called work function of metal, and is denoted by w0. Threshold Frequency It is the minimum frequency that is incident on the given metallic surface, so that an electron just comes out from the surface, without any kinetic energy. It is denoted by ν0. 1 By definition if we put ν = ν0, then mv 2 = 0, in equation (1). 2 Thus h 0  w0  0  h 0  w0 w0  h 0 Thus from equation 1, we have: visit www.TejasAcademy.in

2

XII / PHYSICS

DUAL NATURE OF MATTER & RADIATION

12/PA

1 h  h 0  mv2 ----------------------------------------------(2) 2 Graph Between K.E of ejected electron and frequency of radiation Suppose when we incident a light of frequency, v>v0 , the electrons are ejected from the surface of metal with kinetic energies, K.E. Relationship between these two is given by: h  w0  K .E

 K .E  h  w0 If we plot K.E along y-axis and v along x-axis, then above equation represents a straight line of standard form y  mx  c Having slope = h, and y-intercept = -w0. Thus following is the graph. Slope = h K.E y-int, = -w0

v

Note that the point on which line intersects the frequency axis represents the threshold frequency, v0. Experimental Study of Photoelectric Effect Consider a photocell consisting of a metallic plate, C connected to the negative terminal of battery and acting as cathode. A is the metallic plate connected to the positive terminal of the battery, and acting as anode. Light C e -

A

1 V

(A) Effect of change in intensity of light on photoelectric current. If we increase the intensity of light, more number of photoelectrons are ejected from the cathode. As a result the reading of the ammeter increases, i.e. photoelectric current increases. Thus graph between the photoelectric current (I) and intensity of light (I0) is a straight line passing through the origin, as shown in the following fig. I0

I 0

visit www.TejasAcademy.in

3

XII / PHYSICS

DUAL NATURE OF MATTER & RADIATION

12/PA

(B) Effect of change in frequency of light on photoelectric current. If we change the frequency of light, the kinetic energy of the ejected photoelectrons increases, in 1 accordance with the photoelectric equation give as: h  w0  mv 2 . But the number of photoelectrons 2 ejected remains the same, thus photoelectric current do not change. Photoelectric current is zero, when the frequency of light is less than the threshold frequency. I

ν

Conceptual Q.1: Every metal has a definite work function. Why do photoelectrons not come out al with the same energy if incident radiation is monochromatic? Why is there an intensity distribution of photoelectrons? Ans: work function merely indicates the minimum energy required for the electron in the highest level of the conduction band to get out of the metal. Not all electrons in metal belong to this level. They occupy a continuous band of levels. Consequently, for the same incident radiation, electrons knocked off from different levels come out with different energy. Conceptual Q 2: A meal emits electrons if green light falls on it but this is not true with yellow light. Will it emit electrons with red light? Will it emit electrons with blue light? Ans: No electrons will be emitted by red light. However, the blue light would cause photo=emission. It may be noted that the wavelength of red light more than that of yellow light. Also, the wavelength of blue light is less than the wavelength of yellow light. Conceptual Q 3: Which photon is more energetic: blue one or red one? Give reason. Ans: Blue photon is more energetic than red photon. This is because the frequency of blue is more than the frequency of red. Conceptual Q 4: A source of light of frequency greater than the threshold frequency is placed at a distance of 1m from the cathode of a photo cell. The stopping potential is found to be V. If the distance of the light source from the cathode is reduced explain giving reasons, what change will you observe in the (i) photoelectric current (ii) stopping potential? Ans: When the distance is reduced, the intensity is increased. (i) due to increase in intensity, photoelectric current is increased. (ii) Since the stopping potential does not depend upon intensity therefore the stopping potential remains unchanged. Stopping Potential or Cutoff Potential When the potential of the anode, is decreases with respect to cathode, it is found that the photoelectric current decreases. At certain negative potential of anode, with respect to cathode, the photoelectric current reduces to zero in any photoelectric cell; this potential of anode is called stopping potential or cutoff potential. It is denoted by V0. 1 Suppose an electron is ejected from the cathode, with kinetic energy mv 2 , and there exists stopping 2 potential on anode. Electron will come to rest on reaching to anode. By work energy theorem: Loss in kinetic energy = work done by electric field. visit www.TejasAcademy.in

4

XII / PHYSICS

DUAL NATURE OF MATTER & RADIATION

12/PA

1 2 mv = eV0 -----------------------------------------------(3) 2 From equation (2) and (3) we have: h  h 0  eV0 -------------------------------------------(4) Conceptual Q 5: Define the terms threshold frequency and stopping potential in relation to the phenomenon of photoelectric effect. How is the photoelectric current affected on increasing the (i) frequency (ii) intensity of the incident radiations and why? Ans: (i) The minimum frequency of incident radiation which can eject electrons from a metal is called threshold frequency. Below threshold frequency, there is no photoelectric emission. The minimum negative potential given to the anode of a photocell for which the photoelectric current becomes zero is called stopping potential. (i) Increase in frequency of incident radiation has no effect on the photoelectric current. (ii) The photoelectric current increases proportionally with the increase in intensity of incident radiation.

Graph between stopping potential and frequency of incident light From equation 4 we have, eV0  h  h 0 h h  V0     0 -------------------------------------------------(5) e e Above equation in variables, V0 and ν represents a straight, line of the form y = mx + c It has slope h/e and y-intercept = -h/e. V0

ν

(C) Effect of change in intensity of light on stopping potential: Stopping potential is directly proportional to the kinetic energy of ejected electrons in accordance with equation 3. When we increase intensity of light number of photoelectrons ejected increase, however their kinetic energy does not change. Thus stopping potential remains constant. Thus there is no effect of change in intensity on stopping potential of a photocell. PHOTOCELL A photocell is a device which is used to convert light energy to electric energy. It consists of inverted glass tube consisting of a metallic cathode, and positively charged anode. Light incident on cathode causes emission of photoelectrons. These electrons are captured by anode, and hence photoelectric current is produced.

visit www.TejasAcademy.in

5

XII / PHYSICS

DUAL NATURE OF MATTER & RADIATION

12/PA

Following are applications of photocell: 1. Used to generate electric energy from light energy. 2. Used in burglar alarm. 3. Used to count no. of persons entering in a hall. 4. Used in cameras 5. Used to find the thickness of paper. DUAL NATURE OF MATTER It is a fact that radiation has a dual aspect behaving either as waves or as particles under suitable circumstances. A particle having a finite mass has been supposed to be associated with definite amount energy, given by Einstein in his mass-energy relationship. E = mc2, where E is the energy associated with a particle of mass m, c being the velocity of light. On this basis, a wave has a particle aspect. In 1924, it was argued by de-Broglie that if radiation has a dual aspect why not think of particles of matter also as having this dual aspect? Nature loves symmetry and if mass-energy symmetry is present for waves, then the same symmetry should be available for material particles. In simple words what we know as material particles like electrons, protons, neutrons etc. should also behave like waves of a definite wavelength, under suitable conditions. Thus particles may also behave as waves. This is the dual nature of matter. The waves which are associated with matter are called matter waves or de Broglie waves after the name of the French scientist who gave us this idea. DE BROGLIE’S HYPOTHESIS It states that every moving particle is associated with wave like characteristics, so it should have wavelength. The wavelength of moving object is given by: h  p Where p is the linear momentum of particle. And h is the Planck’s constant. PROOF: We will prove de-Broglie’s hypothesis for a moving photon. Consider a photon of mass m and having frequency v. By Planck’s quantum theory: E = hv --------------------------------------------------(1) By Einstein’s hypothesis: E = mc2 ------------------------------------------------(2) Equating 1 and 2 we get: hv = mc2 c  h  mc2



 

h mc

Here m is the linear momentum of a photon. h p DE-BROGLIE’S WAVELENGTH OF AN ELECTRON Consider an electron accelerated by applying a potential difference V. Electron is starting from rest, and acquires velocity v, thus Gain in kinetic energy = work done by electric field.  

visit www.TejasAcademy.in

6

XII / PHYSICS

DUAL NATURE OF MATTER & RADIATION

12/PA

1 2 mv  eV 2 2eV ---------------------------------------------------(1) v m

By de-Broglie’s hypothesis we have:  

h mv

Putting value from 1 we have: h 2eV m m h   2meV

 

Putting values of constants.

12.27 A0 V Thus λ varies inversely as the square root of applied potential difference.  

DAVISSON AND GERMER EXPERIMENT Davisson and Germer performed an experiment to analyze the wave nature of a moving electrosn. Experimental setup: A beam of electrons emitted by the electron gun is allowed to fall on nickel crystal cut along cubical axis at a particular angle. The scattered beam of electrons is received by the detector which can be positioned at any angle by rotating it about the point of incidence. The intensity of scattered beam of electrons is measured as the function of the angle of scattering (φ). Theory: According to classical physics, the intensity of scattered beam of electrons at all scattering angles should be same.

visit www.TejasAcademy.in

7

XII / PHYSICS

DUAL NATURE OF MATTER & RADIATION

12/PA

But when the graphs between the intensity of scattered beam of electrons at different accelerating potential were plotted by Davisson and Germer, they found that the intensity pattern of scattered beam of electrons from nickel crystal was not the same but different at different angles of scattering. They found that when the incident beam of electron accelerated through 54V is allowed to incident on the nickel crystal, the intensity of scattered beam was maximum at an angle of 500. θ + φ + θ = 1800. 2θ = 180 – 50. θ = 650 According to Bragg’s equation for maxima in diffraction pattern, 2dsinθ = nλ For nickel crystal, d = 0.91A0 Hence 2× 0.91×sin650 = λ λ = 1.65A0 Now by De-Broglie’s hypothesis, we have: 12.27 A0    1.67 A0 54 We observe that wavelength of an electron comes out to be nearly same as was produced by Bragg’s equation. So De-Broglie hypothesis for an electron is verified. Conceptual Q 6: What is the basic principle of electron microscope? Ans: It is based on de-Broglie hypothesis. A beam of accelerated electrons behaves like wave. This wave can be handled by electric and magnetic fields in exactly the same way as electromagnetic waves can be handled by lenses. Conceptual Q 7: What is the basic difference between light waves and matter waves? Ans: The velocity of light waves in vacuum is a constant quantity. On the other hand, the velocity of matter waves in vacuum depends upon their wavelength. Conceptual Q 8: Why a fast neutron beam needs to be thermalised with the environment, before it can be used for neutron diffraction experiments? Ans: The wavelength of fast neutrons is very small as compared to inter atomic spacing. So these are not useful for diffraction experiments. On the other hand the wavelength of thermal neutrons is useful for diffraction experiments. Thus, a fast neutron beam needs to be thermalised with the environment before it can be used for neutron diffraction experiments.

visit www.TejasAcademy.in

8

XII / PHYSICS

DUAL NATURE OF MATTER & RADIATION

12/PA

ASSIGNMENT 1. The wavelength of electromagnetic radiation is doubled. What will happen to the energy of the photons? 2. If the intensity of the incident radiation in a photocell increased how does the stopping potential vary? 3. The frequency of incident radiation is greater than threshold frequency in a photocell. How will the stopping potential vary if frequency is increased keeping other factors constant? 4. How does the maximum kinetic energy of electrons emitted vary with the work function of the metal? 5. On what factor does the energy carried by a photon of light depend? 6. State two applications of photo cell. 7. Two metals A and B have work functions 2eV and 4eV respectively. Which metal has a lower threshold wavelength for photoelectric effect? 8. If the intensity of incident radiation on metal is doubled, what will happen to the energy of the electrons emitted? 9. If the maximum kinetic energy of electrons emitted in a photocell is 5 eV, what is the stopping potential? 10. If the maximum kinetic energy of electrons emitted y a photocell is 4 eV, what is the stopping potential? 11. Obtain the energy in joules acquired by an electron beam when accelerated through a potential difference of 2000 V. [3.2 ×10-16 ] 12. What is the energy associated in joules with a photon of wavelength 4000 A? [4.98 ×10-19 J] 13. Two metals A and B have work function 2eV and 4 eV. Which of the two metals has a smaller threshold wavelength? 14. What is the charge on metal in the photoelectric experiment? 15. Can photo electrons be emitted by a surface whose work function is 4.4 eV, when illuminated with visible light? 16. The photoelectric cutoff voltage in a certain experiment is 1.5 V. What is the maximum kinetic energy of photoelectrons emitted? [2.4 ×10-19 J] [NCERT] 17. Find the (a) maximum frequency and (b) minimum wavelength of X-ray produced by 30kV electrons. [7.24 ×1018 Hz, 0.041 m] [NCERT] 18. What is the de-broglie wavelength of a 3 kg object moving with a speed of 2 m/s ? [1.1 ×10-34 m] 19. Calculate the frequency associated with a photon of energy 3.3 ×10-10 J. [5 ×1013 Hz] 20. Light of frequency 1.5 times the threshold frequency is incident on a photosensitive material. If the frequency is halved and intensity is doubled, what happens to photoelectric current? 21. The work function of aluminum is 4.2 eV. IF two photons each of energy 2.5 eV strike an electron of aluminium, will the emission of electrons be possible? 22. Are matter waves electromagnetic? Write the de Broglie wave equation. 23. What is the effect on the velocity of the emitted photoelectrons if the wavelength of the incident light is decreased? -----------NUMERICALS------------24. Ultraviolet light of wavelength 2271 A from a 100 W mercury source radiates a photo cell made of molybdenum metal. If the stopping potential is -1.3 V, estimate the work function of the metal. How would the photo cell respond to a high intensity red light of wavelength 6328 A produced by a He-Ne laser?[4.2 eV] [NCERT] 25. A monoenergetic electron beam with electron speed of 5.20 ×106 m/s is subject to magnetic field of 1.30 ×10-4 T normal to the beam velocity. What is the radius of the circle traced by the beam? Given e/m for electron equals 1.76 ×1011 C/kg. [22.7 cm] [NCERT] visit www.TejasAcademy.in

9

XII / PHYSICS

DUAL NATURE OF MATTER & RADIATION

12/PA

26. The work function for the following metals is given: Na : 2.75 eV ; K : 2.30 eV ; Mo : 4.17 eV ; Ni : 5.15 eV. Which of these metals will not give photoelectric emission for a radiation of wavelength 3300 A from a He-Cd laser placed 1 m away from the photocell? What happens if the laser is brought nearer and placed 50 cm away? 27. What is the acceleration a n electron in an electric field of magnitude 40 V/cm ? Given that its e/m = 1.76 ×1011 C/kg [5.28 ×1014 m/s2] 28. When radiation of wavelength 3000 A and 4000 A falls in certain metal surface the photoelectrons emitted have maximum kinetic energies of 2 eV and 1 eV respectively. Calculate the maximum wavelength of the incident radiation for which there will be photoelectron emission from the same surface. [5824 A ; 5893 A] 29. What is the wavelength of a photon whose energy is 1 eV ? Which part of the electromagnetic spectrum does it belong to ? [12375 A] 30. What is the frequency of a photon whose wavelength is 1 A? Which part of the electromagnetic spectrum does it belong to? [3 ×1018 Hz] 31. Calculate the energy and momentum of a photon of wavelength 6600 A. [3 ×10-19 J ; 10-27 kgm/s] 32. A particle is moving three times as fast as an electron. The ration of the de broglie wavelength of the particle to that of the electron is 1.813 ×10-4 . Calculate the particle’s mass and identify the particle. [1.67 ×10-27 kg, neutron] [NCERT] 33. Calculate the momentum and de Broglie wavelength of the electrons accelerated through a potential difference of 56 V. [4.04 ×10-24 kgm/s ; 0.164 nm] [NCERT] 34. Light of wavelength 488 nm is produced by an argon laser which is used in the photoelectric effect. When light from this spectral line is incident on the cathode the stopping potential of photoelectron is 0.38 V. Fine the work function of the material from which the cathode is made. [2.16 eV] [NCERT] 35. Light of frequency 7.21 ×1014 Hz is incident on metal surface. Electrons wit a maximum speed of 6 ×105 m/s is ejected from the surface. What is the threshold frequency for photoelectric emission of electrons? [4.73 ×1014 Hz] [NCERT] 36. Work functions of three elements A,B and C are as given below: A : 5.0 eV B : 3.8 eV C : 2.8 eV A radiation of wavelength 4125 A is made to be encident on each of these elements. By appropriate calculations show in which case photoelectrons will not be emitted. 37. If a photoemessive surface has a threshold frequency of 4.6 ×1014 Hz, calculate the energy of the photon in eV. [1.00 eV] 38. Two metals X and Y have functions 2 eV and 5 eV respectively. Which metal will emit electrons, when it is radiated with light of wavelength 400 nm and why? 39. An alpha particle and a proton are accelerated through the same potential difference. Calculate the ration of velocities acquired by the two particles [1: 21/2 ] 40. A radio power transmitter operates at a frequency of 880 kHz and power of 10 kW. Find the no of photons emitted per sec.? [1.72 ×1031 ] 41. Work function of sodium is 2.3 eV. Does the sodium show photoelectric emission for arranged light? Given that the wavelength of the light is 8600 A. 42. Define the terms (i) work function (ii) threshold frequency and (iii) stopping potential with reference to photoelectric effect. Calculate the maximum kinetic energy of electrons emitted from a photosensitive surface of work function 3.2 eV, for the incident radiation of wavelength 300 nm. -----------CONCEPTUALS----------43. Derive and expression for the de Broglie wavelength of the fast moving electrons. visit www.TejasAcademy.in

10

XII / PHYSICS

DUAL NATURE OF MATTER & RADIATION

12/PA

44. What is photoelectric effect? Explain the effect of increase of (i) frequency (ii) intensity of the incident radiation on photoelectrons emitted by a photo tube. 45. If the frequency of incident light on metal surface is doubled for the same intensity what change would you observe in: (i) kinetic energy of photoelectrons emitted (ii) photo electric current and (iii) stopping potential? Justify your answer in each case. 46. State the dependence of work function on the kinetic energy of electrons emitted in photocell. If the intensity of incident radiation doubled, what changes occur in the stopping potential and (ii) photoelectric current? 47. A source of light of frequency greater than its threshold frequency of metal surface is place at 2 m from the cathode of photo cell. The stopping potential is found to be V. If the distance of the light source halved, state with reason what changes occur I (i) stopping potential (ii) photoelectric current and (iii) maximum velocity of photoelectrons emitted. 48. A proton and an alpha particle having the same kinetic energy are in turn allowed to pass through a uniform magnetic field perpendicular to their direction of motion. Compare the radii of the paths of protons and alpha particle [1:1] 49. What is meant by the work function of a metal? Discuss how the value of work function influences the kinetic energy of the electrons liberated by photoelectric emission. 50. For the photoelectric effect in sodium the figure shows the plot of cut off voltage versus frequency of (i) the threshold frequency (ii) the work function for sodium. ------------NUMERICALS------------51. What is the frequency of a photon whose energy is 75 eV? [Ans: 18×1015Hz] 52. Calculate the energy of a photon whose (i) frequency is 1000 kilocycles/s (radio waves ), (ii) wavelength is 6000 A0 (yellow light) and (iii) wavelength is 0.6 A0. (X-rays). [Ans:6.63×10-28J, 3.31×10-19J, 3.31×10-15J] 53. Calculate the energy and momentum of a photon of wavelength 6600A0. [Ans:3×10-19J, 10-27 kgms-1] 54. Find the number of photons emitted per second by a 25 watt source of monochromatic light of wavelength 6000A0. [Ans: 7.54×1019] 55. Light of wavelength 5000 A0 falls on a sensitive plate with photoelectric work function 1.90 eV. Find (i) energy of the photon in eV (ii) kinetic energy of the photoelectrons emitted and (iii) stopping potential. Given h = 6.62×10-34Js and 1eV = 1.6×10-19J. [Ans: 2.48 eV, 0.5825 eV, 0.5825 eV] 56. Light of wavelength 3500 A0 in incident on two metals A and B. Which metal will yield photoelectrons, if their work functions are 4.2 eV and 1.9eV respectively? [Ans: Metal B will emit photoelectrons] 57. Find the frequency of light, which ejects electrons from a metal surface fully stopped by a retarding potential of 3V,. The photoelectric effect begins in this metal at a frequency of 6×10 14Sec-1. Find the work function of this metal. [Ans:2.48eV] 58. Determine the region of the electron spectrum which liberates photoelectrons from potassium. Electron work function

visit www.TejasAcademy.in

11