11.7 PARTICLE NATURE OF LIGHT: THE PHOTON Photoelectric effect thus gave evidence to the strange fact that light in interaction with matter behaved as if it was made of quanta or packets of energy, each of energy h ν. Is the light quantum of energy to be associated with a particle? Einstein arrived at the important result, that the light quantum can also be associated with momentum (h ν/c). A definite value of energy as well as momentum is a strong sign that the light quantum can be associated with a particle. This particle was later named photon. The particle-like behaviour of light was further confirmed, in 1924, by the experiment of A.H. Compton (1892-1962) on scattering of X-rays from electrons. In 1921, Einstein was awarded the Nobel Prize in Physics for his contribution to theoretical physics and the photoelectric effect. In 1923, Millikan was awarded the Nobel Prize in physics for his work on the elementary charge of electricity and on the photoelectric effect. We can summarise the photon picture of electromagnetic radiation as follows: (i) In interaction of radiation with matter, radiation behaves as if it is made up of particles called photons. (ii) Each photon has energy E (=hν) and momentum p (= h ν/c), and speed c, the speed of light. (iii) All photons of light of a particular frequency ν, or wavelength λ, have the same energy E (=hν = hc/λ) and momentum p (= hν/c= h/λ), whatever the intensity of radiation may be. By increasing the intensity of light of given wavelength, there is only an increase in the number of photons per second crossing a given area, with each photon having the same energy. Thus, photon energy is independent of intensity of radiation. (iv) Photons are electrically neutral and are not deflected by electric and magnetic fields. (v) In a photon-particle collision (such as photon-electron 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 283 absorbed or a new photon may be created. Reprint 2025-26 Physics Example 11.1 Monochromatic light of frequency 6.0 ´1014 Hz is produced by a laser. The power emitted is 2.0 ´10–3 W. (a) What is the energy of a photon in the light beam? (b) How many photons per second, on an average, are emitted by the source? Solution (a) Each photon has an energy E = h n = ( 6.63 ´10–34 J s) (6.0 ´1014 Hz) = 3.98 ´ 10–19 J (b) If N is the number of photons emitted by the source per second, the power P transmitted in the beam equals N times the energy 11.1 per photon E, so that P = N E. Then P 2.0 × 10 −3 W N = = − 19 E 3.98 × 10 J EXAMPLE = 5.0 ´1015 photons per second. Example 11.2 The work function of caesium is 2.14 eV. Find (a) the threshold frequency for caesium, and (b) the wavelength of the incident light if the photocurrent is brought to zero by a stopping potential of 0.60 V. Solution (a) For the cut-off or threshold frequency, the energy h n0 of the incident radiation must be equal to work function f0, so that 2.14eV φ0 n0 = = h 6.63 × 10 −34 J s 2.14 × 1.6 × 10 − 19 J 14 = −34 = 5.16 × 10 Hz 6.63 × 10 J s Thus, for frequencies less than this threshold frequency, no photoelectrons are ejected. (b) Photocurrent reduces to zero, when maximum kinetic energy of the emitted photoelectrons equals the potential energy e V0 by the retarding potential V0. Einstein’s Photoelectric equation is hc eV0 = hn – f0 = – f0 λ or, l = hc/(eV0 + f0) (6.63 × 10 −34 Js) × (3 × 10 8 m/s) = 11.2 (0.60eV−26 + 2.14eV) 19.89 × 10 J m = (2.74eV) 19.89 × 10 − 26 J m EXAMPLE λ = − 19 = 454 nm 2.74 × 1.6 × 10 J 11.8 WAVE NATURE OF MATTER The dual (wave-particle) nature of light (electromagnetic radiation, in general) comes out clearly from what we have learnt in this and the preceding chapters. The wave nature of light shows up in the phenomena of interference, diffraction and polarisation. On the other hand, in Reprint 2025-26 Dual Nature of Radiation and Matter photoelectric effect and Compton effect which involve energy and momentum transfer, radiation behaves as if it is made up of a bunch of particles – the photons. Whether a particle or wave description is best suited for understanding an experiment depends on the nature of the experiment. For example, in the familiar phenomenonof seeing an object by our eye, both descriptions are 1987) important. The gathering and focussing mechanism of – light by the eye-lens is well described in the wave picture. But its absorption by the rods and cones (of the retina)requires the photon picture of light. (1892 A natural question arises: If radiation has a dual (wave- particle) nature, might not the particles of nature (the electrons, protons, etc.) also exhibit wave-like character? In 1924, the French physicist Louis Victor de Broglie Louis Victor de Broglie (1892 – 1987) French(pronounced as de Broy) (1892-1987) put forward the BROGLIE physicist who put forth bold hypothesis that moving particles of matter should revolutionary idea of wavedisplay wave-like properties under suitable conditions. DE nature of matter. This idea He reasoned that nature was symmetrical and that the was developed by Erwin two basic physical entities – matter and energy, must have Schródinger into a full- symmetrical character. If radiation shows dual aspects, fledged theory of quantum so should matter. De Broglie proposed that the wave mechanics commonly VICTOR length l associated with a particle of momentum p is known as wave mechanics. given as In 1929, he was awarded the Nobel Prize in Physics for his h h discovery of the wave nature l = = (11.5) LOUIS p m v of electrons. where m is the mass of the particle and v its speed. Equation (11.5) is known as the de Broglie relation and the wavelength l of the matter wave is called de Broglie wavelength. The dual aspect of matter is evident in the de Broglie relation. On the left hand side of Eq. (11.5), l is the attribute of a wave while on the right hand side the momentum p is a typical attribute of a particle. Planck’s constant h relates the two attributes. Equation (11.5) for a material particle is basically a hypothesis whose validity can be tested only by experiment. However, it is interesting to see that it is satisfied also by a photon. For a photon, as we have seen, p = hn /c (11.6) Therefore, h c = = λ (11.7) p ν That is, the de Broglie wavelength of a photon given by Eq. (11.5) equals the wavelength of electromagnetic radiation of which the photon is a quantum of energy and momentum. Clearly, from Eq. (11.5 ), l is smaller for a heavier particle (large m) or more energetic particle (large v). For example, the de Broglie wavelength of a ball of mass 0.12 kg moving with a speed of 20 m s–1 is easily calculated: 285 Reprint 2025-26 Physics p = m v = 0.12 kg × 20 m s–1 = 2.40 kg m s–1 h 6.63 × 10 − 34 J s l = = −1 = 2.76 × 10–34 m p 2.40 kg m s This wavelength is so small that it is beyond any measurement. This is the reason why macroscopic objects in our daily life do not show wave- like properties. On the other hand, in the sub-atomic domain, the wave character of particles is significant and measurable. Example 11.3 What is the de Broglie wavelength associated with (a) an electron moving with a speed of 5.4´106 m/s, and (b) a ball of mass 150 g travelling at 30.0 m/s? Solution (a) For the electron: Mass m = 9.11´10–31 kg, speed v = 5.4´106 m/s. Then, momentum p = m v = 9.11´10–31 (kg) ´ 5.4 ´ 106 (m/s) p = 4.92 ´ 10–24 kg m/s de Broglie wavelength, l = h/p 6. 63 × 10 –34 J s = –24 4 . 92 × 10 kg m/s l = 0.135 nm (b) For the ball: Mass m ’ = 0.150 kg, speed v ’ = 30.0 m/s. Then momentum p’ = m’ v’ = 0.150 (kg) ´ 30.0 (m/s) p ’= 4.50 kg m/s de Broglie wavelength l’ = h/p’. 6. 63 × 10 – 34 Js = 4. 50 × kg m/s 11.3 l’= 1.47 ´10–34 m The de Broglie wavelength of electron is comparable with X-ray wavelengths. However, for the ball it is about 10–19 times the size of EXAMPLE the proton, quite beyond experimental measurement. SUMMARY 1. The minimum energy needed by an electron to come out from a metal surface is called the work function of the metal. Energy (greater than the work function (fo) required for electron emission from the metal surface can be supplied by suitably heating or applying strong electric field or irradiating it by light of suitable frequency. 2. Photoelectric effect is the phenomenon of emission of electrons by metals when illuminated by light of suitable frequency. Certain metals respond to ultraviolet light while others are sensitive even to the visible light. Photoelectric effect involves conversion of light energy into electrical energy. It follows the law of conservation of energy. The photoelectric emission is an instantaneous process and possesses certain special features. Reprint 2025-26 Dual Nature of Radiation and Matter 3. Photoelectric current depends on (i) the intensity of incident light, (ii) the potential difference applied between the two electrodes, and (iii) the nature of the emitter material. 4. The stopping potential (Vo) depends on (i) the frequency of incident light, and (ii) the nature of the emitter material. For a given frequency of incident light, it is independent of its intensity. The stopping potential is directly related to the maximum kinetic energy of electrons emitted: e V0 = (1/2) m v2max = Kmax. 5. Below a certain frequency (threshold frequency) n 0, characteristic of the metal, no photoelectric emission takes place, no matter how large the intensity may be. 6. The classical wave theory could not explain the main features of photoelectric effect. Its picture of continuous absorption of energy from radiation could not explain the independence of Kmax on intensity, the existence of no and the instantaneous nature of the process. Einstein explained these features on the basis of photon picture of light. According to this, light is composed of discrete packets of energy called quanta or photons. Each photon carries an energy E (= h n) and momentum p (= h/l), which depend on the frequency (n ) of incident light and not on its intensity. Photoelectric emission from the metal surface occurs due to absorption of a photon by an electron. 7. Einstein’s photoelectric equation is in accordance with the energy conservation law as applied to the photon absorption by an electron in the metal. The maximum kinetic energy (1/2)m v2max is equal to the photon energy (hn ) minus the work function f0 (= hn0) of the target metal: 1 m v2max = V0 e = hn – f0 = h (n – n0) 2 This photoelectric equation explains all the features of the photoelectric effect. Millikan’s first precise measurements confirmed the Einstein’s photoelectric equation and obtained an accurate value of Planck’s constant h . This led to the acceptance of particle or photon description (nature) of electromagnetic radiation, introduced by Einstein. 8. Radiation has dual nature: wave and particle. The nature of experiment determines whether a wave or particle description is best suited for understanding the experimental result. Reasoning that radiation and matter should be symmetrical in nature, Louis Victor de Broglie attributed a wave-like character to matter (material particles). The waves associated with the moving material particles are called matter waves or de Broglie waves. 9. The de Broglie wavelength (l) associated with a moving particle is related to its momentum p as: l = h/p. The dualism of matter is inherent in the de Broglie relation which contains a wave concept (l) and a particle concept (p). The de Broglie wavelength is independent of the charge and nature of the material particle. It is significantly measurable (of the order of the atomic-planes spacing in crystals) only in case of sub-atomic particles like electrons, protons, etc. (due to smallness of their masses and hence, momenta). However, it is indeed very small, quite beyond measurement, in case of macroscopic objects, commonly encountered in everyday life. 287 Reprint 2025-26 Physics Physical Symbol Dimensions Unit Remarks Quantity Planck’s h [ML2T –1] J s E = hn constant Stopping V0 [ML2T –3A–1] V e V0= Kmax potential Work f0 [ML2T –2] J; eV Kmax = E –f0 function Threshold n0 [T –1] Hz n0 = f0/h frequency de Broglie l [L] m = h/p wavelength POINTS TO PONDER 1. Free electrons in a metal are free in the sense that they move inside the metal in a constant potential (This is only an approximation). They are not free to move out of the metal. They need additional energy to get out of the metal. 2. Free electrons in a metal do not all have the same energy. Like molecules in a gas jar, the electrons have a certain energy distribution at a given temperature. This distribution is different from the usual Maxwell’s distribution that you have learnt in the study of kinetic theory of gases. You will learn about it in later courses, but the difference has to do with the fact that electrons obey Pauli’s exclusion principle. 3. Because of the energy distribution of free electrons in a metal, the energy required by an electron to come out of the metal is different for different electrons. Electrons with higher energy require less additional energy to come out of the metal than those with lower energies. Work function is the least energy required by an electron to come out of the metal. 4. Observations on photoelectric effect imply that in the event of matter- light interaction, absorption of energy takes place in discrete units of hn. This is not quite the same as saying that light consists of particles, each of energy hn. 5. Observations on the stopping potential (its independence of intensity and dependence on frequency) are the crucial discriminator between the wave-picture and photon-picture of photoelectric effect. h 6. The wavelength of a matter wave given by λ = has physical p significance; its phase velocity vp has no physical significance. However, the group velocity of the matter wave is physically meaningful and equals the velocity of the particle. EXERCISES 11.1 Find the (a) maximum frequency, and 288 (b) minimum wavelength of X-rays produced by 30 kV electrons. Reprint 2025-26 Dual Nature of Radiation and Matter

13.5 The Q value of a nuclear reaction A + b ® C + d is defined by Q = [ mA + mb – mC – md]c2 where the masses refer to the respective nuclei. Determine from the given data the Q-value of the following reactions and state whether the reactions are exothermic or endothermic. (i) 11 H+13 H →12 H+12 H (ii) 126 C+126 C →1020 Ne+ 24 He Atomic masses are given to be m ( 12 H ) = 2.014102 u m ( 13 H) = 3.016049 u m ( 126 C ) = 12.000000 u m ( 1020 Ne ) = 19.992439 u

11.4 Monochromatic light of wavelength 632.8 nm is produced by a helium-neon laser. The power emitted is 9.42 mW. (a) Find the energy and momentum of each photon in the light beam, (b) How many photons per second, on the average, arrive at a target irradiated by this beam? (Assume the beam to have uniform cross-section which is less than the target area), and (c) How fast does a hydrogen atom have to travel in order to have the same momentum as that of the photon?