Title MODERN PHYSICS.pdf ✅

 Digital www.allendigital.in [ 181 ] Dual Nature of Light Experimental phenomena of light reflection, refraction, interference, diffraction are explained only on the basis of wave theory of light. These phenomena verify the wave nature of light. Experimental phenomena of light photoelectric effect and Compton effect, pair production and pair annihilation can be explained only on the basis of the particle nature of light. These phenomena verify the particle nature of light. it is inferred that light does not have any definite nature, rather its nature depends on its experimental phenomenon. This is known as the dual nature of light. The wave nature and particle nature both cannot be possible simultaneously. Electron Emission ➢ Emission of electron from a metal surface, when energy is incident upon the surface. ➢ Minimum energy required to bring out the electron from metal surface is called work function (φ0). ➢ It depends on nature of material and nature of surface. Types of Electron Emission 1. Thermionic emission 2. Field emission 3. Photoelectric emission 1. Thermionic emission : The phenomenon of emission of electrons from the metal surface when it is heated is called thermionic emission. 2. Field emission : In this process strong electric field ( 108 V/m) is applied to a metal, and the electrons are emitted from the surface. 3. Photoelectric emission: When light of suitable frequency illuminates a metal surface, electrons are emitted from the metal surface. This phenomenon is called photo electric emission. Different Experiments Hertz Experiment: Hertz observed that when ultraviolet rays are incident on a negative plate of electric discharge tube then conduction takes place easily in the tube. Modern Physics-1 [Dual Nature of Radiation and Matter] V/m(E) 102 104 106 108 e e e e 03 Modern Physics
NEET : Physics [ 182 ] www.allendigital.in  Digital Hallwach Experiment Hallwach observed that if negatively charged Zn plate of electroscope is illuminated by ultra violet light, its negative charge decreases and becomes neutral and after some time, it gains positive charge. It indicates that under the action of ultra violet light, some negative charged particles are emitted from the metal. Lenard Experiment He told that when ultraviolet rays are incident on cathode, electrons are ejected, these electrons are attracted by anode and due to complete path of photo electrons, photo current flows. When ultra violet rays are incident on anode, electrons are ejected but current does not flow. Work Function () It is the minimum energy required by an electron to escape from metal surface. Work function depends on the nature of metal and its surface. Minimum : Cs = 2.14 eV ; Maximum : Pt = 5.65 eV (1eV = 1.6 ×10–19 J) Einstein's Quantum theory of Light ➢ Light behaves as Quanta, when it interacts with the matter. These energy quanta are called photons, having definite Energy and Momentum. ➢ Photons always travel with speed of light in vacuum. (c = 3 × 108 m/s) ➢ Photons are electrically neutral and are not affected by electric and magnetic fields. Energy of photon Energy radiated from a source propagates (microscopically) in the form of small packets and these are known as photons. according to Planck the energy of a photon is directly proportional to the frequency of the radiation. E    E = h hc E =  ( c = )  E = hc  = 12400  eV – Å [ hc = 12400 (Å – eV)] Here; E = energy of photon, c = speed of light, h = Planck's constant (h = 6.62 × 10–34 J-s), e = charge of electron,  = frequency of photon, l = wavelength of photon Linear momentum of photon Linear momentum of photon p = E h h c c  = =  Effective mass of photon Effective mass of photon m = 2 2 E hc h c c c = =   i.e. effective mass 1 m   So, mass of violet light photon is greater than the mass of red light photon. ( R > V) Rest mass of a photon is always zero. evacuated quartz tube cathode anode ultraviolet rays – + A
Modern Physics  Digital www.allendigital.in [ 183 ] Intensity of light E P I At A = = ...(i) SI Unit : 2 joule m s − or 2 watt m Here; P = power of source, A = Area, t = time taken E = energy incident in t time = Nh, N = number of photon incident in t time Intensity  = N(h ) At  = n(h ) A  ...(ii) N n no. of photon per sec. t   = =     from equation (i) and (ii), P A = n(h ) A  n = P h = P hc   n = (5 × 1024 J –1 m–1) P ×  Illustration 1: Power of a radiator is 100 watt and its  is 400 nm. Calculate No. of photon emitted in 10 hrs. Solution: n = (5 × 1024 J –1 m–1) P ×  n = 5 × 1024 × 100 × 4 × 10–7 = 2 × 1020 N = n × t = 2 × 1020 × 10 × 3600 = 7.2 × 1024 Illustration 2: A source S1 is producing 1015 photon/s of wavelength 5000 Å. Another source S2 is producing 1.02 × 1015 photon/s of wavelength 5100 Å. Calculate ratio of power of S2 to power of S1. Solution: P nhc n or P hc  = =  15 2 2 15 1 2 1 P n 1.02 10 5000 1 P n 1 5100 10    =  = =   Illustration 3: In an photo-electric experiment frequency of incident light is doubled and its intensity is trippled. How much time the photon current is obtained? Solution:  = nh ; 3 n 2 = times i  n So, i = 3 2 times Illustration 4: The energy flux of sunlight reaching the Earth's surface is 1.388 × 103 Wm–2. How many photons (nearly) per square meter are incident on the earth per second? Assume that the photons in the sunlight have an average wavelength of 1550 nm. Solution: IA IA n E hc  = =
NEET : Physics [ 184 ] www.allendigital.in  Digital Illustration 5: The power of a bulb is 60 milliwatt and the wavelength of light is 6000 Å. Calculate the number of photons/second emitted by the bulb? Solution: Energy released per second nh = power  nhc = P   P n hc  = = 3 10 34 8 60 10 6000 10 6.62 10 3 10 − − −       photon/sec = 1.8 × 1017 photon/sec Illustration 6: The energy flux of sunlight reaching on the Earth's surface is 1.388 × 103 W/m2. How many photons (nearly) per square metre are incident on the Earth per second? Assume that the photons in the sunlight have an average length of 550 nm. Solution: Intensity,  = Power emitted Area = P A = 1.388 × 103 W/m2 Also, Energy of a photon, E = –34 8 9 hc 6.63 10 3 10 550 10−    =   = 3.616 × 10–19 J Let n be the total number of photon/area then, n = 3 19 P/ A I 1.388 10 E E 3.616 10−  = =  = 3.83 × 1021 photon/m2s Radiation Force and Pressure (i) When radiations are falling normally on a perfectly reflecting surface - Let 'N' photons are falling in time t, momentum before striking the surface 1 Nh (p )=  momentum after striking the surface 2 Nh (p )= −  change in momentum of photons = 2 1 2Nh p p − − =  momentum transferred to the surface = 2Nh  = p  radiation force on the surface p 2Nh 2h F n t t    = = =        but P n hc  =  2h P 2P F hc c  =  =  and Radiation pressure = F 2P 2I P I A cA c A   = = =     incident photon reflected photon