Quantum Mechanics Reference: Concepts of Modern Physics “A. Beiser” https://www.youtube.com/watch?v=cz27jq4ATL4 Reference: Concepts of Modern Physics “A. Beiser” 1 1

De- Broglie Hypothesis of matter waves: Wave theory of light ----------- interference and diffraction. Quantum (Planck’s) theory of light ----- Photoelectric effect and Compton effect, etc. Thus Light possesses a dual character, behaving as a wave as well as a particle (Photon). In analogy with dual character of light, de-Broglie in 1924 introduced the hypothesis that all material particles in motion possess a wave character also. Accordingly, particles such as electrons, protons, etc. have waves associated with them. These waves are called ‘matter waves’ or ‘de-Broglie waves’

MATTER WAVES de Broglie relation In 1924 Prince Louis de Broglie postulated that ordinary matter can have wave-like properties, with the wavelength λ related to momentum p in the same way as for light In analogy with photon, wavelength of matter waves de Broglie relation Planck’s constant de Broglie wavelength Where ‘h’ is planck’s constant, ‘p=mv’ is momentum If v =0, then λ= infinite It means that “waves are associated with moving particles only”. Note: Wavelength depends on momentum, not on the physical size of the particle Prediction: We must see diffraction and interference of matter waves.

De Broglie’s Hypothesis Louis de Broglie postulated that the dual nature of the light must be “expanded” to ALL matter In other words, all material particles possess wave-like properties, characterized by the wavelength, λB, related to the momentum p of the particle in the same way as for light Planck’s Constant Momentum of the particle de Broglie wavelength of the particle

Frequency of a Particle In an analogy with photons, de Broglie postulated that a particle would also have a frequency associated with it These equations present the dual nature of matter Particle nature, p and E Wave nature, λ and ƒ ( and k)

Different forms of the expression Matter waves are sound waves or electromagnetic waves or new kind of waves? Like electromagnetic (EM) waves, matter waves can also be propagated through vacuum. Hence they are not mechanical waves as the sound waves. Unlike EM waves, they are also associated with the motion of electrically neutral body. Hence they are not electromagnetic waves but they are new kind of waves called matter waves. Different forms of the expression de-Broglie wavelength of a particle of mass m and velocity v is λ=h/mv, where h is Planck’s constant. The greater the particle’s momentum, the shorter its wavelength. Here in above expression m is the relativistic mass.

Larger the momentum of the particle, shorter is the wavelength. Discussion Thus matter waves are associated with material particles only if they are in motion. Wavelength of particle should decrease as its velocity increases. For a given velocity, heavier particles should have shorter wavelength than lighter particles. In case of macroscopic objects such as bullet or a ball, the de-Broglie wavelength is so small that it can not be measured directly by any known means. Larger the momentum of the particle, shorter is the wavelength.

2. If Ek is the kinetic energy of the material particle, then 3. For a material particle , like neutron at temperature T having kinetic energy Ek=(3/2) kT , where k is Boltzmann constant 4. If a charged particle of charge q is accelerated through a potential difference of V volts, then

4a. If protons are accelerated through a potential difference of V volts, then 4b. If α-particles (doubly ionized helium atom) are accelerated through a potential difference of V volts, then 4c. If deuterons ( ionized deuterium atom 1H2) are accelerated through a potential difference of V volts, then

Now Let us test de-Broglie hypothesis….. Example: de-Broglie wavelength of associated matter wave with (a) a ball of 1 gm moving with velocity 30 m/s. Ans: 2.2 x 10-32 m. (b) Electron moving with velocity 1007 m/s. Ans: 0.7 Å. (c) Alpha particle (2 proton+2 neutron) moving with same velocity. Ans: 9910 Å (d) Electron accelerated by 50 volts (or having kinetic energy 50 eV). Ans: 1.7 Å (e) Wavelength of Nitrogen molecule at room temperature. Mass of Nitrogen atom = 4.65X10E-26 Kg. Ans: 0.2754 Å Q: A proton and electron have equal kinetic energies. Compare their De-Broglie wavelengths. Now Let us test de-Broglie hypothesis…..

Example: Find the Kinetic energy of a proton whose de Broglie wavelength is 1fm , which is roughly the proton diameter. Solution : A relativistic calculation is needed when pc>Eo . In case of proton rest mass energy Eo=0.938GeV. pc=(mv)c=hc/=1.2410GeV Since pc>Eo a relativistic calculation is required.

Derive the expression for de-Broglie wavelength of electron in each case. Non relativistic (m=m0): Relativistic Case: ( K.E (eV) is comparable to rest mass energy.) Where K is Kinetic energy and m0 is rest mass of electron

Note: For electron; Kinetic energy is eV.

Matter Waves Vs EM Waves Matter waves are associated with moving particles irrespective of whether the particles are charged or not. EM waves are produced only by charged accelerated particles. Matter waves are obtained by charged particles are associated with electric and magnetic fields. EM waves are associated with electric and magnetic fields perpendicular to each other as well as to the direction of propagation of wave. Matter waves are neither emitted by the particles nor radiated into space. These are simply associated with the particles. EM waves can be radiated into space. The velocity of matter waves depends upon the velocity of the material particles. Velocity of EM waves is constant in a given medium. The velocity of matter waves are generally less than the velocity of light. The velocity of EM waves is equal to the velocity of light. The mater waves have shorter wavelengths given by de-Broglie equation. =h/p The wavelength of EM waves are given by relation: =c/

1.Calculate the velocity and kinetic energy of neutron having de-Broglie wavelength 1Ao Velocity= 3.96103 m/sec Kinetic Energy= 0.082eV 2. An electron and a photon each have a wavelength of 2.0 Ao. Compare their (a) momentum, (b) total energies, ( c) ratio of their kinetic energies. (a) pe=3.3110-24 kg m/s pp=3.3110-24 kg m/s (b)Ke=37.6eV , Eo=0.5MeVE=0.51 MeV Kp=6.21keV, Eo=ZERO E=6.21 keV (c ) Ke/Kp=6.05 10-3

3. Calculate the de-Broglie wavelength of an -particle accelerated through a potential difference of 200Volts. 0.00716 Ao 4. What voltage must be applied to an electron microscope to produce electrons of wavelength 0.40 Ao? 934.76 Volt 5. Can a photon and an electron of the same momentum have the same wavelength? Compare their wavelength if the two have the same energy. e=p p/e =(2mc2/E)-1/2

6. Neutrons are in thermal equilibrium with matter at room temperature 27o C . Determine the average energy and de-Broglie wavelength of neutrons. k=1.3810-23 J/K. 6.21 10-21 J,1.45 Ao 7. An enclosure filled with helium is heated to a temperature of 400 K. A beam of helium atoms emerges out of the enclosure. Calculate de-Broglie wavelength corresponding to He atoms. MHe=6.710-27 kg. 0.628Ao

8. A beam of electrons of kinetic energy 100eV passes through a thin metal foil. On a screen at a distance of 20cm most intense ring observed has a diameter 2.44 cm. calculate the spacing of the related lattice planes in the metal. 10.11Ao 9. Calculate the wavelength associated with a (i) 1MeV electron (ii) 1 MeV proton , and (iii) 1MeV photon. e=0.00875Ao p=0.000287Ao photon=0.0124 Ao

Energy and Momentum for Massive vs. Massless Particles Massive Particles (e.g. electrons) Ek = ½mv2 = p2/2m = h2/2mλ2 p = (2Em)1/2 λ = h/p = h/(2Em)1/2 Massless Particles (e.g. photons) E = pc = hc/λ p = E/c λ = h/p = hc/E Energy of e-s Ek = 25eV p = 2.7x10-24 kg m/s v = 3x106 m/s = c/100 λ = 0.24 nm Typical photons E = 2.5eV p = 1.3x10-27 kg m/s v = c = 3x108 m/s λ = 500 nm deBroglie relationship is universal