Fundamentals of modern physics, the de-Broglie hypothesis

The presentation uploaded here educates about the failure of classical physics to explain Blackbody radiation and the success of quantum theory to explain the Blackbody radiation spectrum and other phenomena, the de-Broglie hypothesis and its significance, nature of de-broglie waves and the representation. Numerical problems are given at the end.

1. Modern Physics 2021 Prof Praveen Vaidya Department of Physics SDMCET, Dharwad . https://youtu.be/eXXEOuyPhcY https://www.youtube.com/watch?v=eXXEOuyPhcY&feature=youtu.be https://www.youtube.com/watch?v=6qWsmv3Y3rk

2. Classical Physics Physics of Macroscopic matter. To find the Kinetic energy of car at time. To find potential energy of water in a dam. Conversion fuel into kinetic energy of car. Quantum Physics Physics of Microscopic matter. Kinetic energy of electron under potential. Nuclear energy stored in a nucleus Change in energy of electron to convert into light energy. Engineering Physics Engineering Physics - Study of matter and energy related to Engineering and technology. Deals with the characteristic properties of in Microscopic level.

3. Important equations Energy of ……. E =mv 2 /2 E = mc 2 E = eV E = h υ = hc / λ Where υ = c/ λ Momentom p = mv p = hk

4. Classical physics have limita tions Black body radiation:

5. Explanation by classical Physics Wien’s law of energy Distribution W. Wien Derived an equation to explain BBRS given by Drawbacks The results are in good agreement with the smaller wavelengths (higher frequencies) . At large wavelengths values deviate away from experimental values. Rayleigh Jean’s law Rayleigh Jean separately proved that, BBRS given by Drawbacks The results are in good agreement with the larger wavelengths (lower frequencies) . At smaller Energy reaches to infinity called Ultraviolet catastrophe.

6. Birth of Quantum theory Plank’s Quantum theory The law successfully Explained the BBRS The energy emitted by the blackbody per unit volume in the range of wavelength from λ to λ+dλ is given by Where c = velocity of light = 3 × 10 8 m s -1 This equation is suits correctly with the experimental results for both lower and higher wave lengths According to Plank’s theory, Radiation contains stream of particle called photon Energy of each particle is equal to quanta. One Quanta of energy is given by value E = h ν , Energy of n number of photon E = nh ν , where n = 1, 2, 3….. Hence the energy has discrete nature.

7. According classical physics Light or Energy has wave nature. Interference of light, Diffraction of light and Polarization of light are successfully explained by wave nature of light. According Plank’s theory Light has particle nature and particle is called as the photon. The blackbody radiation, Photoelectric effect, Compton effect are successfully explained by the particle nature of light Energy has dual nature..? Compton effect

8. Exercises: The device which absorb all the radiations incident on it is called____________ ___________ emits all the radiations when maintained at__________________ The radiation (spectrum) emitted by_________ is called Blackbody spectrum. Sun is blackbody because_____________________ Temperature of a body at which it behaves like blackbody is called___________ Intensity of blackbody radiation reduces considerably at__________ The lower wavelength side BBR is explained by_____________law The higher wavelength side BBR is explained by_____________law The Quantum theory is explained ___________side of blackbody spectrum The Quantum theory assumed light has ______________nature. According to Quantum theory of radiation energy of a photon is__________ Photons have a wavelength of 500 nm. (The symbol nm is defined as a nanometer = 10 -9 m) What is the energy of this photon, and what is the type of the electromagnetic wave in this case? A beam of electromagnetic rays green colour has energy 10eV. Determine the total number of photons emitted.

9. De-Broglie hypothesis Fight continued till a young researcher giving a hypothetical solution called In 1924, Louis de Broglie a French physicist, proposed a hypothesis to explain the wave nature of particles. According to this, under some circumstances light behaves like particle so, as nature likes symmetry, the entities those have particle nature basically are also behave like waves under special circumstances.

10. Explanation of de-Broglie Hypothesis If light has wave nature then, E = h υ ----- 1 According to mass energy Relation We have E = mc 2 -------- 2 Equating equation 1 and 2 h υ = mc 2 or hc / λ = mc 2 λ = h/mc Taking c = v in case of typical particle, λ = h/ mv It is called as de-Broglie wavelength or wavelength of matter waves.

11. E xperimental proof of wave nature of particles E – Electron gun (filament ), S – Circular carriage with scale, G – Galvanometer , C – Nickel target , Ф – Angle of satterin g Experimental setup: Electron gun, which emit the electrons on application of voltage. Unoxidized nickel target: electrons emitted from electron gun diffract from it. Detector: Fitted with Galvanometer, collects the diffracted electrons at various angles and gives the intensity. Applying a voltage to electron gun, the intensity of the diffracted electrons from target is measured at various angles of diffraction using detector.

12. The experiment is repeated for different voltage of electron gun and draw the graph angle of diffraction versus Intensity each time. It is found that Intensity is highest at voltage is 54V at 50 o Below and above 54V the intensity reduces. Measure the de- broglie wavelength using equation The stream of electrons considered as electron beam as they diffracted like wave. So, wavelength of electrons also found using from Bragg’s X-ray diffraction theory where is glancing angle

13. Proof of Dual nature of matter or wave nature of electron by Division- Germer’s experimental data: The maximum ionization current was observed for the grid voltage of V = 54V, the de-Broglie wavelength for corresponding electron is given by Diffraction is the property of wave nature of light, using wave property Bragg gave formula to find wavelength of X ray diffracted from a crystal, If above formula applied to diffracted electron beam, taking angle of diffraction φ = 50 o , the Bragg’s angle becomes

14. = 2 x 0.91 x 10 -10 x Sin 65 o = 1.65 x10 -10 m d = inter planer spacing of the nickel crystal = 0.91 x 10-10m m = 1, for first order diffraction The de Broglie formula thus gives a value of wavelength, for the electron that is in excellent agreement with their experimentally measured wavelength. Therefore, Davisson- Germer’s experiment provides an evidence for the Dual nature of matter or wave nature of electron.

15. Exercises According to plank’s quantum theory, one quanta of energy is equals to=__________ An electron just emits out of metal surface, the energy of photon is______________. The scientific phenomenon successfully explained by Quantum theory are________. The optical phenomena explained only by wave theory of light are______________. Light has been regarded as dual nature because ______________________________. According de-Broglie, basically light has ________________ nature, but under some circumstances it behaves like ____________. The de-Broglie proposes, nature like symmetry hence _____also behave like waves. The special circumstances under which de broglie hypothesis is valid are_________. Davison Germer experiments used to prove __________________________________. _____________________is used to accelerate the electron towards the nickel crystal. When electrons interact with Nickel crystal they produce __________________ like the electro magnetic wave. The electrons _______________ from nickel target are collected at various angles by device called____________. Electron detector measures the _______________ using a galvanometer. The value of accelerating voltage and angle for max. Intensity are _____ and ___ resp. The wavelength of electron beam is calculated using _________ and ________ equations respectively. The Davison Germer experiment proved that, the electrons behave like wave, as wavelength of scattered electrons found by ______ and ______ theory are same.

16. Phase velocity: The velocity with which a point marked on the wave, which represents certain phase of a wave at that point, is traveling is called as phase velocity. If matter behaves like waves then what type of wave? It converts into single wave or group of waves? If de-Broglie waves are single wave then, Monochromatic in nature, so not stationary. Velocity of wave is phase velocity. The de-Broglie waves are not continues but localized: So are not monochromatic in nature.

17. Group Velocity: Matter waves is regarded as the single resultant wave of the superposition of large number waves of different wavelengths and with different velocities. This single resultant wave forms the wave packet or wave group (figure below). The velocity with which this group of waves travels is called group velocity. Particle velocity: Velocity possessed by the particle in motion is called particle velocity. Particle velocity is given by where p is momentum of a particle.

18. The velocity with which the group of waves which represents the particle is called group velocity and it is given by Where, angular frequency Also, the wave number 3 2 Relation between particle velocity ( v particle ) and Group velocity ( v group ) E = h υ or υ = E/ h λ = h/p or 1/ λ = p/h

19. On substituting equations (2) and (3) in equation (1), Hence p = mv particle

20. Expression for de Broglie wavelength using group velocity concept The velocity with which this group of waves which represents the particle is called group velocity and it is given by Where, angular frequency Also, the wave number On substituting equations (2) & (3) in equation (1)

21. ---------- (5) ---------- (6) Substitute Eqn 5 in 4

22. Relation between velocity of light (c), group velocity (v g ) and phase velocity ( v p ­ )

23. important properties of matter waves: 1. Matter waves are waves associated with the moving particle 2. They are not electromagnetic in nature 3. Matter waves are localized and not continuous in the medium. 4. The amplitude of the matter waves at the given point determines the probability of finding the particle at that point at the given instant of time. 5. The wave nature of matter is explained only by group velocity. 6. The de Broglie matter waves are suitable only for microscopic matters like, electrons, photons and other inter atomic particles.

24. Application of de-Broglie wavelength, Electron Microscope Electron microscopes use the electron beam to get the magnified image of an object. In any microscope , higher the wavelength lower is the resolving power, according to above equation.

25. Resolving Power of electron Microscope If the accelerated electron beam is used in microscope then average wavelength deBrolie waves of electrons is 0.5Å (average). If the visible light is used in place of electron beam let us take for example it is 5000Å. keeping refractive index constant( μ ), For a given microscope of given half angle of cone( θ ), we have, The resolving power, Using this equation if the resolving power of electron beam compared with that of visible light then, Therefore using deBroglie hypothesis, electron microscope gives the magnification 10000 times of visible light

26. Some Questions: Explain the dual nature of energy and matter. State and explain the de-Broglie hypothesis. Set up an experiment to prove the wave nature of electrons and draw the conclusion. Outline the Phase velocity and Group velocity concepts applied to matter waves. Derive the relation between Group velocity and Particle velocity. Derive equation of de-Broglie wavelength by group velocity concept. Evaluate the significance of de-Broglie hypothesis in the design of high resolution electron microscope.

27. Find the de Broglie wavelength of a moving electron of energy 100 keV . Make a similar exercise for a proton of mass 1.67 x 10 -27 kg. What must be the potential difference through which an electron should be accelerated to attain its de Broglie wavelength to 0.01 nm. Calculate the de Broglie wavelength (in nm) of a 1361 kg automobile moving at 90 km/hr. Radiation of wavelength 1800Å is incident on the surface of photo-sensitive plate of work-function 4.3 eV ; find the de-Broglie wavelength of emitted photoelectron. An electron has a de Broglie wavelength of 2pm. Find it’s kinetic energy, phase and group velocities of the de Broglie waves.{Rest mass of electron=511keV} Calculate the de Broglie wavelength in the following cases A cricket ball of mass 0.3 kg moving with a speed of 120 km/hr. 9 keV electrons of a typical TV picture tube. Compare the results and draw the conclusions

28. Numericals : An X-ray photon of wavelength 20 Ǻ collides with an electron assumed to be in rest and diffracts, through an angle 45 o . Calculate, a) The energy of scattered x-ray photon and b) Maximum kinetic energy of recoiled electron. De-Broglie wavelength of an electron in motion is 1.2 Ǻ. Determine the phase velocity group velocity and kinetic energy of electron in eV . An electron has a de-Broglie wavelength of 2pm. Find the kinetic energy, phase velocity and group velocity of its de-Broglie waves. Rest mass energy of electron is = 511eV. Calculate the de-Broglie wavelength of an electron accelerated with a potential of 1k Find the de Broglie wavelength of an electron possessing a kinetic energy of 24.6 eV . An electron accelerated a potential acquire a speed of 6.2x10 7 . Find its kinetic energy and group velocity Electrons are accelerated to a potential of 100 V. Calculate their energy, group velocity and phase velocity. Compare the de-Broglie wavelength of electron and a rocket part mass 10gm moving around the earth with a velocity of 320km/s.

29. Quantum Mechanics Superposition of number of waves results in localization of waves. More the number of waves more the localization and better identification of particle and poor the identification of momentum and wavelength (wavelength is average wavelength of all the waves). . Lesser the number of waves less the localization and better to find the momentum. Radiation of wavelength 1800Å is incident on the surface of photo-sensitive plate of work-function 4.3 eV ; find the de-Broglie wavelength of emitted photoelectron. An electron has a de Broglie wavelength of 2pm. Find it’s kinetic energy, phase and group velocities of the de Broglie waves.{Rest mass of electron=511keV} Calculate the de Broglie wavelength in the following cases A cricket ball of mass 0.3 kg moving with a speed of 120 km/hr. 9 keV electrons of a typical TV picture tube. Compare the results and draw the conclusions