Modern Physics-I PDF Study Material

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TG: @Chalnaayaaar PRE-MEDICAL PHYSICS ENTHUSIAST | LEADER | ACHIEVER STUDY MATERIAL Modern Physics-I ENGLISH MEDIUM  TG: @Chalnaayaaar All rights including trademark and copyrights and rights of translation etc. reserved and vested exclusively with ALLEN Career Institute Private Limited. (ALLEN) No part of this work may be copied, reproduced, adapted, abridged or translated, transcribed, transmitted, stored or distributed in any form retrieval system, computer system, photographic or other system or transmitted in any form or by any means whether electronic, magnetic, chemical or manual, mechanical, digital, optical, photocopying, recording or otherwise, or stood in any retrieval system of any nature without the written permission of the Allen Career Institute Private Limited. Any breach will entail legal action and prosecution without further notice. 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ALLEN Career Institute Private Limited is not responsible for the consequences of any action taken on the basis of this publication. ®  TG: @Chalnaayaaar Physics : Modern Physics-I Pre-Medical LOUIS VICTOR DE BROGLIE (1892-1987) French physicist who put forth revolutionary idea of wave nature of matter. ® This idea was developed by Erwin Schrodinger into a full fledged commonly known as wave mechanics. In 1929, he was awarded for his discovery of the wave nature of electrons. PHILIPP EDUARD ANTON VON LENARD (1862-1947) Was a German physicist and the winner of the Nobel Prize for Physics in 1905 for his research on cathode rays and the discovery of many of their properties. 120  Physics : Modern Physics-I TG: @Chalnaayaaar ® Pre-Medical PHOTO ELECTRIC EFFECT It is a phenomenon of ejecting electrons by falling light of suitable frequency or suitable wavelength on a metal. photon energy E=hν electrons ejected Ejected electron are called photoelectrons and current from the surface flowing due to the photoelectrons is called photoelectric current. e¯ e¯ e¯ e¯ e¯ e¯ e¯ This effect was discovered by Hertz. e – e¯ metal Laws of photo electric effect were given by Lenard. It was explained by Einstein using Quantum theory of light. ultraviolet rays DIFFERENT EXPERIMENTS Hertz Experiment cathode anode evacuated quartz tube Hertz observed that when ultraviolet rays are incident on a negative A plate of electric discharge tube then conduction takes place ® easily in the tube. 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. GOLDEN KEY POINTS  Work function :- Minimum energy required by an electron to escape from the metal surface.  For the photo electric effect the light of short wavelength (or high frequency) is more effective than the light of long wavelength (or low frequency)  Types of electron emission (i) Thermionic emission: By suitably heating, sufficient thermal energy can be imparted to the free electrons to enable them to come out of the metal. 8 –1 (ii) Field emission: By applying a very strong electric field (of the order of 10 V m ) to a metal, electrons can be pulled out of the metal, as in a spark plug. (iii) Photo-electric emission: When light of suitable frequency illuminates a metal surface, electrons are emitted from the metal surface. 1. QUANTUM THEORY 1.1 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∝ν E = hν hc hc 12400 E= ( c = νλ)   E= = eV – Å [  hc = 12400 (Å – eV)] λ λ λ –34 Here E = energy of photon, c = speed of light, h = Planck's constant (h = 6.62 × 10 J-s), e = charge of electron, ν = frequency of photon, λ = wavelength of photon 121 ®  TG: @Chalnaayaaar Physics : Modern Physics-I Pre-Medical 1.2 Linear momentum of photon E hν h Linear momentum of photon p = = = c c λ 1.3 Effective mass of photon E hc h 1 Effective mass of photon m = = 2 = i.e. effective mass m∝ c2 cλ cλ λ So mass of violet light photon is greater than the mass of red light photon. (λR > λV) Rest mass of a photon is always zero. 1.4 Intensity of light E P =I =...(i) At A joule watt SI UNIT : or ® 2 m −s m2 Here P = power of source, A = Area, t = time taken E = energy incident in t time = Nhν, N = number of photon incident in t time N(hν ) n(hν )  N  Intensity I = At = A...(ii)  = n = no. of photon per sec. t   P n(hν ) from equation (i) and (ii), = A A P Pλ n= =  n = (5 × 1024 J–1 m–1) P × λ hν hc 1.5 Radiation force and Radiation pressure (i) When radiations are falling normally on a perfectly reflecting surface - incident photon Let 'N' photons are falling in time t, p1= hλ Nh momentum before striking the surface (p1) = λ p2= h λ Nh momentum after striking the surface (p2) = − reflected photon λ −2Nh change in momentum of photons = p2 – p1 = λ 2Nh momentum transferred to the surface = ∆p = λ ∆p 2Nh  2h  Pλ radiation force on the surface F = = = n   but n= ∆t tλ λ  hc 2h Pλ 2P F 2P 2I  P ∴ F= × λ hc = and Radiation pressure = = = c A cA c  I = A    (ii) When radiations are falling normally on a perfectly absorbing surface - Nh −0 incident photon p − p2 λ Nh h  Pλ  p1= h Radiation force F = 1 = = F = n  n = hc  λ t t tλ λ   no reflected P F P I photon p2 = 0 F= and Pressure = = = c A Ac c 122  Physics : Modern Physics-I TG: @Chalnaayaaar ® Pre-Medical Illustrations Illustration 1. Calculate the energy of photon having λ = 4000 Å in eV and in J. Solution. 12400 E= eV = 3.1 eV  4000 –19 –19 E = 3.1 × 1.6 × 10 = 4.96 × 10 J Illustration 2. 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 nhc Pλ 60 × 10−3 × 6000 × 10−10 nhν = power or =P or n = = −34 8 photon/sec = 1.8 × 1017 photon/sec λ hc 6.62 × 10 × 3 × 10 Illustration 3. 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 wavelength of 550 nm. Solution. Power emitted P Intensity I = = = 1.388 × 103 W/m2 Area A hc 6.63 × 10 −34 × 3 × 10 8 Also Energy of a photon E = = −9 = 3.616 × 10–19 J λ 550 × 10 P/A I 1.388 × 10 3 Let n be the total number of photon/area then n = = = = 3.83 × 1021 photon/m2s E E 3.616 × 10 −19 Illustration 4. How many photons of wavelength λ = 6600 nm must strike a totally reflecting screen per second at normal incidence so as to exert a force of 1N ? [AIPMT Mains 2007] Solution. h h Momentum of the incident photon p = , Momentum after reflection = – λ λ 2h Change in momentum = ∆p = λ If n is the number of photons falling per second on the screen then force ∆p 2nh Fλ 1 × 6600 × 10−9 F= = ⇒n= = −34 = 5 × 1027 photons s–1 ∆t λ 2h 2 × 6.6 × 10 123 ®  TG: @Chalnaayaaar Physics : Modern Physics-I Pre-Medical BEGINNER'S BOX-1 1. Who discovered photo electric effect ? (1) Hertz (2) Lenard (3) Hallwach (4) Einstein 2. The energy of a photon is equal to 3 kilo eV. Calculate its linear momentum ? 3. A parallel beam of monochromatic light of wave length 500 nm is incident normally on a perfectly absorbing surface. The power through any cross-section of the beam is 10 W. Find (a) Number of photon absorbed by the surface per second. (b) The force exerted by light beam on the surface. 4. Which colour of photon has greater energy either red or violet ? 5. A TV station is operated at 100 MW with a signal frequency of 10 MHz. Calculate the number of photons radiated per second by its antenna. 6. Calculate number of photons passing through a ring of unit area in unit time if light of intensity 100 Wm–2 and of wavelength 400 nm is falling normally on the ring. 7. A special kind of light bulb emits monochromatic light of wavelength 700 nm. Electrical energy supply to it at the rate of 60 W and the bulb is 50% efficient at converting that energy to light energy. How many ® photons are emitted by the bulb during its life time of 1 day. 2. EXPERIMENTAL STUDY OF P.E.E. BY LENARD 2.1 The effect of potential difference between A and C ultraviolet rays Light of fixed frequency (υ) and intensity (I) is incident on photo – photo electrons + emissive plate C, keeping A positive w.r.t. C. If positive C(cathode) A(anode) potential of A gradually increased, it is found that first photo current increases then becomes maximum called saturation G Vac current. Now we apply negative potential to plate A w.r.t. C and on increasing it gradually then current decreases and becomes ip zero at a certain negative potential. Magnitude of the minimum negative potential at which ip becomes zero, called stopping potential (V0). is(Saturation current) At stopping potential most energetic electrons are stopped. It means stopping potential measures maximum kinetic –V0 Vac energy of photo electrons. i.e. Kmax. = eV0 2.2 Effect of intensity of light When intensity of given source able to produce PEE, is increased, is increases but V0 remains unchanged. Photocurrent I3 > I2 > I1 I3 I2 Photoelectric current I1 Stopping potential –V0 0 Retarding potential Collector plate potential Intensity of light 124  Physics : Modern Physics-I TG: @Chalnaayaaar ® Pre-Medical 2.3 Effect of frequency When frequency of radiations is increased, Photoelectric V0 increases but is is almost unchanged. It current ν3> ν2>ν1 potential(V0) is also observed that below a certain Saturation Stopping ν3 ν ν 2 1 frequency photoelectrons don't come out. –V03 –V02 –V01 0 ν0 Collector frequency(ν) Retarding potential plate potential The minimum frequency which can eject the photo electrons is called cut off or threshold frequency. Similarly corresponding maximum wavelength is called threshold wavelength. 2.4 Time lag : There is no time lag between the incidence of radiations and emission of electrons. It is a spontaneous process. GOLDEN KEY POINTS ®  Intensity : Energy of light passing through per unit area per unit time is known as intensity of light. 1 Intensity ∝ photon per second ∝ electron per second ∝ current ∝ (for a given point source ) d2 i 2 d1 2 i = current flows in the circuit = i 1 d 22 d = distance between the source of light and electron emitter.  Stopping potential does not depend on the distance between emitter and collector.  The photoelectric emission is an instantaneous process without any apparent time lag (~ 10–9 s or less), even when the incident radiation is made exceedingly dim.  Ultra violet light causes photo electric emission from any metal surface, while visible light causes photo emission from the alkali metals.  The work function represented the energy needed to remove the least tightly bound electrons from the surface. It depends on nature of the metal and nature of surface. 3. FAILURE OF WAVE THEORY OF LIGHT (i) According to wave theory when light incident on a surface, energy is distributed continuously over the surface. So that electron must take a time interval to accumulate sufficient energy to come out. But in experiment there is no time lag. (ii) When intensity is increased, more energetic electrons should be emitted. So that stopping potential should be intensity dependent. But it is not observed. (iii) According to wave theory, if intensity is sufficient then, at each frequency, electron emission is possible. It means there should not be existance of threshold frequency. 4. EXPLANATION BY EINSTEIN 4.1 Radiations absorbed by the surface are in the form of quanta (photon). Energy of each photon depends on frequency. One photon can interact with one electron at a time. In the interaction between photon and electron incident photon transfers its whole energy to the electron. If energy is sufficient then electron comes out without any time delay. It means photo electric effect is an instantaneous process. 125 ®  TG: @Chalnaayaaar Physics : Modern Physics-I Pre-Medical 4.2 If intensity of the given source is increased then number of photon increases. So that, more number of electrons are emitted and greater saturation current is obtained. It means saturation current depends upon intensity of the given source is ∝ I 4.3 At a time, only one photon can interact with one electron. Energy of photon used by the electron is hν = Kinetic energy of electron + Energy required to make electron free from the metal surface (φ0) + Energy lost in collision before emission (Q) If Q = 0, means there is no heat loss, then kinetic energy of electron is maximum. Now hν = (K.E.max) + φ0 It is known as Einstein's equation of P.E.E. (K.E.max) = hν – φ0 or eV0 = hν – φ0 or eV0 = hν – hν0 Here ν0 is threshold frequency for that V0 = 0 ® It means maximum K.E. and stopping potential (V0) depends on frequency. It is independent of intensity of the given source. 4.4 Kinetic energy cannot be negative so that, hν > φ0  hc 12400  hν > hν0 Here φ0 = hν0 = , φ0 = eV − A   λ0 λ0  ν > ν0 It means if frequency is less than 'ν0' , electron does not come out. Graph between (K.E.)max. and frequency metal A metal B (K)max (K)max. = hν – φ0 [ Y = mx – c] θ θ slope = m = tanθ = h (same for all metals) (υ0)A (υ0)Bυ (φ0)A frequency (φ0)B > (φ0)A (φ0)B Graph between stopping potential (V0) and frequency (ν) metal A metal B  eV0 = hν – φ0 V0 h φ  θ V0 =   ν −  0  θ e   e (υ0)A (υ0)B υ (φ0)A frequency slope = m = tanθ = h (same for all metals) (φ0)B e e e 4.5 Quantum efficiency number of electrons emitted per second n Quantum efficiency = = e total number of photons incident per second n ph ne x=...(i) n ph x 24 –1 –1 If quantum efficiency is x% then ne = n [from equation (i)] [Here nph = (5 × 10 J m ) Pλ] 100 ph 4.6 Photoelectric current charge Q –19 Photoelectric current Ie = = = ne e = 1.6 × 10 ne time t 126   Physics : Modern Physics-I TG: @Chalnaayaaar ® Pre-Medical GOLDEN KEY POINTS  Einstein's Photo Electric equation is based on conservation of energy.  Einstein explained P.E.E. on the basis of quantum theory, for which he was awarded nobel prize.  According to Einstein one photon can eject one e– only. But here the energy of incident photon should be greater than work function (threshold energy) to bring out the electron.  Particle nature of light is also supported by Compton effect and Raman effect. Compton effect : The reduction in the energy (hence increase in wavelength) of high-energy (X-ray or gamma ray) photons when they are scattered by free electrons, or loosely bound electrons which thereby gain energy. Raman effect : It is the inelastic scattering of monochromatic light as it passes through a transparent medium due to interaction of photons with the molecules of medium. This results in wavelengths being increased or decreased. ® Illustrations Illustration 5. The threshold wavelength of a metal is 400 nm. Photo electrons have kinetic energy maximum 1.5 eV. Find the wavelength of incident photon. Solution. 12400 eVÅ 12400 eVÅ λ0 = 400 nm = 4000 Å K.E.max. = − λ λ0 12400 eVÅ 12400 eV 12400 eVÅ 1.5 eV = −  1.5 eV = − 3.1 eV λ 4000 λ 12400 eVÅ 12400Å (1.5 + 3.1) eV =   λ=   λ = 2696 Å λ 4.6 Illustration 6. The work function of a metal is 2.3 eV and the wavelength of incident photon is 4.8 × 10–7m. Find maximum kinetic energy of photo electrons. Solution. φ = 2.3 eV λ = 4.8 × 10 m = 4800 Å –7 and hc (6.62 × 10−34 Js)(3 × 108 ms −1 ) 12400 eVÅ = K.E.max. = −φ −φ = −φ λ λ λ 12400  K.E.max. =  − 2.3  eV  0.28 eV  4800  Illustration 7. Light quanta with an energy 4.9 eV eject photoelectrons from metal with work function 4.5 eV. Find the maximum impulse transmitted to the surface of the metal when each electron flies out. Solution. According to Einstein's photoelectric equation 1 Kmax = mv2max = hν – φ0 = 4.9 – 4.5 = 0.4 eV p1= – E → ^ 2 ci M  E E  change of momentum = impulse ∴ impulse = mv –  −  = mv + E/c  c T A e– (p2 – p1) L → E p2= mv ^i Maximum impulse = 2mK max + c 4.9 × 1.6 × 10−19 = 2 × 0.4 × 1.6 × 10 −19 × 9.1 × 10 −31 + = 3.43 × 10–25 kg m/s 3 × 108 127 ®  TG: @Chalnaayaaar Physics : Modern Physics-I Pre-Medical Illustration 8. The stopping potential for the photoelectrons emitted from a metal surface of work function 1.7 eV is 10.4V. Find the wavelength of the radiation used. Also identify the energy levels in hydrogen atom which will emit this wavelength. Solution. Energy of radiation hν = KEmax + φ0 = eV0 + φ0 = 10.4 eV + 1.7 eV = 12.1 eV hc 12400 = Wavelength corresponding to this energy λ= eV−Å = 1024 Å E 12.1eV 1 1  As ∆E = 13.6  2 − 2  = 12.1 eV 1 3  so this radiation will be emitted by transition n=3→n=1 ® Illustration 9. A metallic surface is illuminated alternatively with light of wavelength 3000 Å and 6000 Å respectively. It is observed that the maximum speeds of the photoelectrons under these illuminations are in the ratio 3 : 1. –34 8 Calculate the work function of the surface in eV. (h = 6.62 × 10 J-s, c = 3 × 10 m/s) Solution 1 hc Maximum kinetic energy of photo electrons Kmax. = mv2max = – φ0 2 λ Now let 3000 Å = λ then 6000 Å = 2λ hc − φ0 7 × 6.62 × 10 −34 × 3 × 10 8 2 (v max ) 1 9 7hc ∴ λ= = ⇒ φ = = = 1.81 eV hc 2 1 0 16λ 16 × 3000 × 10 −10 × 1.6 × 10 −19 − φ 0 (v max ) 2 2λ Illustration 10. A light of wavelength 1240 Å incident on a metal having threshold frequency 4.8 × 1014 Hz. What is 2 maximum kinetic energy of photo electron ? If light has intensity 100 W/cm then calculate number of 2 incident photons per m per second. Solution Maximum kinetic energy of photo electrons hc hc Kmax = –φ= – hν0 λ λ 6.62 × 10 −34 × 3 × 10 8 6.62 × 10 −34 × 4.8 × 10 14 Kmax = −10 −19 eV – eV 1240 × 10 × 1.6 × 10 1.6 × 10 −19 Kmax = (10 – 2) eV = 8 eV = 12.8 × 10–19 J Pλ Iλ Number of photons per unit area per unit time n = = hcA hc 100 × 1240 × 10 −10 n= m–2s–1 = 62.43 × 1022 m–2 s–1 10 −4 × 6.62 × 10 −34 × 3 × 10 8 128  Physics : Modern Physics-I TG: @Chalnaayaaar ® Pre-Medical Illustration 11. When a metal is irradiated by monochromatic light, the maximum kinetic energy of the photo–electrons is 1.2eV. If frequency of the light is increased 50% then maximum kinetic energy of photo–electron is 3.6 eV. Evaluate the work function of the metal. Solution Einstein's equation of photo electric effect is (KE)max = hν – φ0 ⇒ 1.2eV = hν – φ0 When frequency of light is increased 50% then 3.6eV = 1.5 hν – φ0 from above equation 3.6eV = 1.5 (1.2eV + φ0) – φ0 ⇒ 3.6eV = 1.8 eV + 0.5 φ0 ⇒ φ0 = 3.6 eV Illustration 12. A light beam of 2mW power and 6000Å wavelength incident on a photo–cell. If threshold wavelength of emitter is 7000Å and 2% incident photons eject the photo–electron then find out value of saturation current in the photo cell. ® Solution Let n photons per second incident on the emitter nhc Pλ 2 × 10 −3 × 6 × 10 −7 then power P = ⇒n = = = 6 × 1015 λ hc 6.62 × 10 −34 × 3 × 10 8 If per second number of emitted electrons is ne then 2 2 ne = ×n = × 6 × 1015 = 12 × 1013 100 100 saturated current is = (ne) e = 12 × 1013 × 1.6 ×10–19 = 19.2 × 10–6 A Illustration 13. The strength of magnetic field required to bend photoelectrons of maximum energy in a circle of radius 50 cm when light of wavelength 3300 Å is incident on a barium emitter is 6.7 × 10–6 T. What value of charge on the photoelectrons is obtained from this data ? (Given : Work function of barium = 2.5 eV; mass of the electron = 9 × 10–31 kg) [AIPMT Mains 2007] Solution 1 hc Maximum KE of photoelectron mv 2max = −φ 2 λ 2  hc  2  6.6 × 10−34 × 3 × 108  ⇒ vmax = − φ =  − 2.5 × 1.6 × 10−19  m  λ  −31  9 × 10  3300 × 10 −10   4 2 × 1012 = × 106 ms –1 = 9 3 Mv 2max Now Bevmax = R max −31 2 6 Mv max 9 × 10 × 3 × 10 ⇒e= = −6 = 1.8 × 10–19C BR max 6.7 × 10 × 0.5 129 ®  TG: @Chalnaayaaar Physics : Modern Physics-I Pre-Medical Illustration 14. 1.656 The graph between the stopping potential and frequency of the V0(in volt) → incident radiation is shown in figure. Calculate. [AIPMT Mains 2008] (i) Planck's constant (ii) Work function 1 5 14 Solution v(1 × 10 Hz)→ (i) According to Einstein's equation of photo electric effect eV0 = hν – hν0 eV0 1.6 × 10−19 × 1.656 1.6 × 1.656 ⇒h= = = × 10−33 = 6.624 × 10−34 J-s ( ν − ν0 ) ( 5 − 1) × 1014 4 (ii) Work function φ0 = hν0 = 6.624 × 10 –34 14 × 1 × 10 = 6.624 × 10 –20 J ® 6.624 × 10−20 = eV = 0.414eV 1.6 × 10−19 BEGINNER'S BOX-2 1. The work function of a metal is 4 eV if 5000Å wavelength of light is incident on the metal. Is there any photo electric effect ? 2. In a photo cell 4 unit photo electric current is flowing, the distance between source and cathode is 4 unit. Now distance between source and cathode becomes 1 unit. What will be photo electric current now ? 3. The wavelength of photons in two cases are 4000 Å and 3600 Å respectively what is difference in stopping potential for these two ? 4. Threshold frequency of a surface is ν0. It is illuminated by 3 ν0 frequency, then maximum speed of photo electrons is V m/sec. What will be maximum speed if incident frequency is 9 ν0 ? 5. When incident wavelength is λ, stopping potential is 3 V0. If incident wavelength is 2λ then stopping potential is V0. Find out threshold wavelength in terms of λ. 6. A light beam of power 1.5 mW and 400 nm wavelength incident on a cathode. If quantum efficiency is 0.1% then, find out obtained photo current and number of photoelectron per second. 7. A metalic surface of work function hν is illuminated by a radiation beam of frequency 5ν. Stopping potential observed is X. What will be stopping potential if the surface is illuminated by radiations of 7ν frequency? 8. The kinetic energy of the fastest moving photo electron from a metal of work function 2.8 eV is 2eV. If the frequency of light is doubled, then find the maximum kinetic energy of photo electron. 9. A monochromatic light incident on metal ‘A’ having threshold frequency ν0. It emits photo electrons of maximum kinetic energy K. Now incident frequency is made three times and fall on a metal ‘B’ having threshold frequency 2ν0. What will maximum kinetic energy of photo electrons emitted by metal ‘B’ ? 10. If light of wavelength 4000 Å falls on a metal which has a stopping potential 1·4 volt against photoelectric –34 emission then what is the work function of the metal. [Take h = 6·6 × 10 Js and c = 3 × 108 ms–1] 130  Physics : Modern Physics-I TG: @Chalnaayaaar ® Pre-Medical 5. PHOTO CELL A photo cell is a practical application of the phenomenon of photo electric effect, It converts light energy into electrical energy. Construction A photo cell consists of an evacuated sealed glass tube containing anode and a concave cathode of suitable emitting material such as Cesium (Cs). Working When light of frequency greater than the threshold frequency of cathode material falls on the cathode, emitted photoelectrons are collected by the anode and an electric current starts flowing in the external circuit. The current increases with the increase in the intensity of light. The current would stop, if the light does not fall on the cathode. ® mA + anode + e¯ e¯ light V e¯ e¯ glass tube e¯ cathode - key Application (i) In television camera. (ii) In automatic door (iii) Burglar’s alarm (iv) Automatic switching of street light and traffic signals. 131 ®  TG: @Chalnaayaaar Physics : Modern Physics-I Pre-Medical MATTER WAVES THEORY 1. 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 can not be possible simultaneously. 2. de-Broglie HYPOTHESIS ® De Broglie imagined that as light possess both wave and particle nature, similarly matter must also posses both nature, particle as well as wave. De Broglie imagined that despite particle nature of matter, waves must also be associated with material particles. Wave associated with material particles, are defined as matter waves. 2.1 de-Broglie wavelength associated with moving particles If a particle of mass m moving with velocity v 1 p2 = Kinetic energy of the particle E = mv 2 2 2m 2E Velocity of the particle v = m Momentum of the particle p = mv = 2mE h h h The wave length associated with the particles is λ= = = p mv 2mE –24 The order of magnitude of wave lengths associated with macroscopic particles is 10 Å. The smallest wavelength whose measurement is possible is that of γ - rays (λ ~ 10–5 Å). This is the reason why the wave nature of macroscopic particles is not observable. The wavelength of matter waves associated with the microscopic particles like electron, proton, neutron, α - particle, atom, molecule etc. is of the order of 10–10 m,it is equal to the wavelength of X-rays, which is within the limit of measurement. Hence the wave nature of these particles is observable. 2.2 de-Broglie wavelength associated with the charged particles Let a charged particle having charge q is accelerated by potential difference V 1 = Kinetic energy of this particle E = mv 2 qV 2 Momentum of particle= = p mv 2mE = 2mqV h h h h The De Broglie wavelength associated with charged particle λ= = = = p mv 2mE 2mqV 132  Physics : Modern Physics-I TG: @Chalnaayaaar ® Pre-Medical e.g. de-Broglie wavelength for an electron –31 –19 –34 me = 9.1 × 10 kg, q = 1.6 × 10 C, h = 6.62 × 10 J-s 6.62 × 10−34 De Broglie wavelength associated with electron λ = m 2 × 9.1 × 10−31 × 1.6 × 10−19 V 12.27 × 10−10 12.27  1 λ= meter volt  λ = A volt so λ∝ V V V 150.6 Potential difference required to stop an electron of wavelength λ is V = 2 volt (Å). λ2 2.3 de-Broglie Wavelength Associated with Uncharged Particles 3 If an uncharged particle has thermal kinetic energy E = kT then its de-Broglie wavelength 2 ® h h λ= λ= 3  3mkT 2m  kT  2  where k = Boltzmann's constants & T = temperature in K scale. 3. DAVISSON GERMER EXPERIMENT The experimental arrangement is shown in figure. There are three main parts of this experiment electron gun nickel electron crystal beam V electron scale detector accelerating voltage G (i) Electron gun Electrons of desired energy are produced in it by the process of thermionic emission. (ii) Nickle crystal diffracts the electrons beam obtained from electron gun.Nickle crystal behaves like a three dimensional diffraction grating (iii) Detector (Ionisation chamber) It detects the electron beam diffracted by the nickle crystal. (iv) Conclusion and results Curve between the intensity (I) of diffracted electrons and diffracting angle (φ) I v = 54 volt φ φ=50° 133 ®  TG: @Chalnaayaaar Physics : Modern Physics-I Pre-Medical Graph I versus φ for different accelerating potential V electrons of 54eV φ = 50° φ = 50° Nickel crystal V < 54V V = 54V V > 54V In this experiment, on drawing different I–φ curve intensity maxima is obtained at an angle of diffraction of 500 and accelerating potential 54 volt. From the differaction measurements wavelength associated with electron is obtained as 1.65 Å whereas according to de Broglie theory this wavelength comes out to be 1.66 Å. Because both the results are same, therefore we can say wave nature is associated with the moving electrons. diffraction from crystal is studied using two equations called Bragg's equations. This are - 2d sinθ = nλ or D sin φ = nλ ® Where d = distance between two consequtive crystal plane incident electron or interplanar distance. beam D = Distance between two atoms in the same lattice plane diffracted θ φ electron n = order of diffraction beam λ = De broglie wavelength associated with electron. θ θ = glancing angle φ = angle of diffraction atomic lattice φ Relation between θ and φ : φ = 180° – 2θ or θ = 90° – 2 4. EXPLANATION OF BOHR QUANTISATION CONDITION According to De Broglie electron revolves round the nucleus in the form of stationary waves (i.e. wave packet) in the similar fashion as stationary λ waves in a vibarting string. Electron revolves in those circular orbits 6th Bohr orbit whose circumference is an integral multiple of de–Broglie wavelength associated with the electron, 2πr = nλ h  λ= and 2πr = nλ mv nh ∴ mvr = equivalent straightened orbit 2π This is the Bohr quantisation condition. GOLDEN KEY POINTS Quantum view of probable waves  h  If a particle has definite momentum p, (∆p = 0) then it has definite wavelength  λ =  , which extends in  p  entire space. By Max Born it means particle can be find in whole universe. up to ∞ (means ∆x = ∞) (single value of λ) 134  Physics : Modern Physics-I TG: @Chalnaayaaar ® Pre-Medical But for a good description, particle should be in a finite region (means ∆x = finite). So, it is rather good idea to consider a particle have group of wave (multiple value of λ) with central wavelength ‘h/p’. In this case superposition of constituent wave takes place and results out as wavepacket. (∆x = finite, ∆p = finite) ∆x Above concepts are consistent with uncertainty principle.  Wave function of matter wave is an imaginary function ψ. If amplitude of matter wave is ψ then ψ (∆V) = probability of finding in ∆V volume. 2   Truely satisfactory physics of dual nature of matter has not developed so far. Quantum wave theory is still subject of research.  Observations on photoelectric effect imply that in the event of matter-light interaction, absorption of energy ® takes place in discrete units of hv. This is not quite the same as saying that light consists of particles, each of energy hυ.  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  The wavelength of a matter wave given by λ = has physical significance; its phase velocity vp has no p physical significance. However, the group velocity of the matter wave is physically meaningful and equals the velocity of the particle. Illustrations Illustration 1. Find the initial momentum of electron if the momentum of electron is changed by Pm and the De Broglie wavelength associated with it changes by 0.50% Solution. dλ dλ 0.5 1 × 100 = 0.5  = = and ∆P = Pm λ λ 100 200 h dp h h 1 p |dp| dλ  p= differentiating =– 2 =– × = −  = λ dλ λ λ λ λ p λ Pm 1 ∴ =  p = 200 Pm p 200 Illustration 2. A deutron is accelerated through a potential of 500 volts. Find the potential through which a singly ionised helium ion is to be accelerated for the same De Broglie wavelength. Solution h λ= or mV = constant V = P.d., q is same 2 mqV mHe × VHe = mdVd or 4VHe = 2 × 500   VHe = 250 V 135 ®  TG: @Chalnaayaaar Physics : Modern Physics-I Pre-Medical Illustration 3. An α-particle moves in circular path of radius 0.83 cm. in the presence of a magnetic field of 2 0.25 Wb/m. Find the De Broglie wavelength associated with the particle. Solution. h h 6.62 ×10−34 λ= = = meter p qBr 2 ×1.6 ×10−19 × 0.25 × 83 ×10−4  mv 2   λ = 0.01 Å  = qvB   r  Illustration 4. For what kinetic energy of a neutron will the associated de-Broglie wavelength be 1.40 × 10–10 m ? Mass of neutron is 1.675 × 10–27 kg. (h = 6.6 × 10–34 J-s). Solution. ® h h 6.6 × 10−34 de-Broglie wavelength λ = ∴ v= = ms –1 mv mλ 1.675 × 10−27 × 1.40 × 10−10 2 1 1  6.6 × 10−34  ∴ K.E. of neutron is mv 2 = × (1.675 × 10−27 )  −27 −10  2 2 1.675 × 10 × 1.40 × 10  (6.6 × 10−34 )2 = = 6.69 × 10–21 J 2 × 1.675 × 10−27 × 1.4 × 1.4 × 10−20 Illustration 5. An electron and a photon have got the same de–Broglie wavelength. Prove that total energy of electron is greater than energy of photon. Solution. h h Total energy of electron Ee = mc2 and λ = ⇒m= mv λv hc2 ∴ Ee = λv hc E e hc2 λ c For energy of photon EP = therefore = × =  c>v So Ee > Ep λ EP λv hc v Illustration 6. If λe and λp denote the de–Broglie wavelength of electron and proton after they are accelerated from rest through potential difference V0 then find the relation between λe and λp. Solution. h h  De-Broglie wavelength λ = = 2mE 2mqV0 1 Here v0 is constant so λ ∝  mp > me ∴ λe > λp m OR h h λe mp  λ= = ∴ = 2mE 2mqV0 λp me 136  Physics : Modern Physics-I TG: @Chalnaayaaar ® Pre-Medical Illustration 7. Find out ratio of wavelength for proton, deutron and α-particle if they are accelerated through same potential difference. Solution. h h 1 1 1 1 =  λ = ∴ λ∝ ⇒ λp : λd : λα = : : 2mE 2mqV mq mp q p md q d mα q α 1 1 1 1 1 = : : =1: : mp q p 2m p q p 4m p 2q p 2 2 2 BEGINNER'S BOX-3 1. de-Broglie wavelength of an electron, accelereted by potential V is λ. What will be de-Broglie wavelength of the electron which is accelerated by 4V potential ? ® 2. Find the ratio of de Broglie wavelength of molecules of hydrogen and helium which are at temperatures 27°C and 127°C respectively. 3. A particle of mass M at rest decays into two particles of masses m1 and m2 having non zero velocities. Find out the ratio of the de–Broglie wavelengths of the two particles. 4. Find out velocity of an electron so that its momentum is equal to that of photon with a wavelength of λ = 5200Å. 5. Find out voltage applied to an electron microscope to produce electron of wavelength 0.6Å. 6. What is the effective mass of a photon having wavelength λ ? [AIPMT 2006] ANSWER'S KEY BEGINNER BOX-1 BEGINNER BOX-3 1. Hertz 2. 1.6 × 10–24 kg-m/s λ 1. 2. λ H2 : λ He = 8 : 3 2 3. (a) 2.5 × 1019 (b) 3.33 × 10–8 N 3. 1 : 1 4. 1400 m/s 4. Violet 5. 1.5 × 1034 h 5. 416.6 volt 6. m = λc 6. 2 × 1020 7. 9 × 1024 BEGINNER BOX-2 1. No 2. 64 unit 3. 0.34 V 12 4. 2V 5. 4λ 6. 0.48 µA, 3 × 10 7. 1.5 X 8. 6.8 eV 9. 3K + hν0 10. 1.7 eV 137