Vacuum tube

1. l, only for shorter than certain wavelength Current 0 3I 2I I V - Fixed wavelength Varying intensity Vacuum tube V Vo Maximum electron energy

2. Current V • Why the maximum electron energy • depend on the wavelength (frequency)? • Why strong beams of long wavelength • cannot knock out electrons? l1 < l2 < l3 l1 Classical EM theory (wave theory) tells us that the max electron energy should depend on only the intensity of the light. l2 l3 Huge water wave, large amplitude sound wave In 1900, Max Planck postulated that electromagnetic energy is emitted in discrete packets, or quanta. The energy of these quanta is proportional to the frequency of the radiation. E = hf = hc/l h = 6.6 x 10-34 Js (Planck’s constant)

3. Work function: depends on material ~ eV In explaining the photoelectric effect, Einstein picked up the idea of Planck, and proposed in 1905 that light was not only emitted by bundles of energy E = hf, but it was also absorbed in such bundles. Light Quanta = Photons Kinetic Energy of an electron = Ephoton - Ethreshold Ephoton = hc/l > Ethreshold light quantum: photon -

4. In this case, light can be considered as a massless particle, photon with energy solely determined by its wavelength (frequency). There is no conflict with Einstein’s relativistic mechanics: massless particle can have speed of light! E = hf = hc/l h = 6.6 x 10-34 Js (Planck’s constant) l = 580 nm photon (yellow light) carries E = (6.6 x 10-34 Js)(3 x 108 m/s)/(580 x 10-9 m) = 3.4 x 10-19 J

5. How do we understand the intensity of light? number of photons with same wavelength or frequency Intensity of light (EM radiation) ≈ Wave behaves like a particle. Does a particle behaves like wave?

6. Relativistic case particle quantity wave quantity It would seem that the basic idea of the quantum theory is the impossibility of imagining an isolated quantity of energy without associating with it a certain frequency de Broglie in 1923 as a graduate student massless light quantum (photon) E = hf = hc/l for mo = 0 l = h/p E = pc = hc/l

7. wave particle De Broglie wavelength l = h/p De Broglie proposed that all particles (electrons) should have wavelength associated with their momentum in exactly the same manner. DUALITY

8. Q. What is the de Broglie wavelength of an electron that has a kinetic energy of 100 eV? After an electron is accelerated in 100 V potential difference, its kinetic energy is 100 eV. eV unit has to be converted into SI unit, Joule. 1 eV = 1.6 x 10-19 J Ek = (1/2)mov2 = 1.6 x 10-17 J v2 = 2Ek/mo = 2(1.6 x 10-17 J)/(9.1 x 10-31 kg) = 3.52 x 1013 m2/s2 v = 5.93 x 106 m/s low speed: no need to use relativistic l = h/p = h/mov = (6.6 x 10-34 Js)/(9.1 x 10-31 kg x 5.93 x 106 m/s) = 1.23 x 10-10 m = 0.123 nm

11. C.J. Davisson and L.H Germer scattered low energy electrons off The Ni crystal and observed a peculiar pattern of scattered electrons. Diffraction pattern from the regular Crystalline structure of order of 0.1 nm size.

13. ABSORPTION SPECTRUM Emission SPECTRUM Hydrogen Spectrum

14. Louis de Broglie (1892 – 1987) Niels Bohr (1885 – 1962) Bohr’s Hydrogen Atom and the Electron as a Wave In 1913, Rutherford’s atom received a quantitative description from Niels Bohr who explained experimentally observed discrete nature of atomic spectrum of Hydrogen. In spite of its immediate success in providing theoretical account of the spectrum and other nature of Hydrogen atom, a complete understanding of Bohr’s atom came only after de Broglie’s conjecture (1923) that electrons should display wave properties. + -

15. + Bohr’s Hydrogen Atom (1 proton and 1 electron) centripetal force = Coulomb force - - - electron as a wave (de Broglie) 2pr = l, 2l, 3l, … l = h/mv This is the necessary condition for electron to maintain an orbit.

16. n n 2pr = l, 2l, 3l, … l = h/mv 2prn = nh/mv rn = (5.3 x 10-11)n2 (m); vn = 2 x106/n (m/s) 2 x 0.053  0.1 nm is the size of Hydrogen atom

17. Now, let’s think about the total energy of the electron in the nth orbit. (1/2)mvn2 -ke2/rn Etot = Potential Energy + Kinetic Energy

18. -1.5 eV n = 3 -3.4 eV n = 2 + -13.6 eV n = 1 Etot - n = ∞ 0 - - - Ionized state of Hydrogen: proton

19. -1.5 eV n = 3 -13.6 eV -3.4 eV n = 2 n = 1 Electron Energy diagram of Hydrogen atom Etot n = ∞ 0 Electron has to absorb 12.1 eV energy for this Electron has to absorb 10.2 eV energy for this lowest energy: ground state - -

20. E = hc/l in eV (Rydberg const.) E(nm) or l < 0 Absorb photons of a given l E (nm) or l > 0 Emit photons of a given l

21. -1.5 eV n = 3 -13.6 eV -3.4 eV n = 2 n = 1 Etot E = hc/l n = ∞ 0 E (12) = 10.2 eV = 10.2 x (1.6 x 10-19 J/eV) = 1.63 x 10-18 J  l = 121 nm Ultraviolet range -

22. Q What is the longest wavelength em radiation that can ionize unexcited hydrogen atom? smallest energy ground state (n = 1) n = 1  m = ∞ 1/l = (1.097 x 107) x (0 – 1) = - 1.097 x 107 (m-1) l = -9.12 x 10-8 (m) = -91.2 (nm) (- means absorption)

24. Q. What is the shortest wavelength for the Balmer series? Largest energy difference in Balmer series From n = ∞ to m = 2 (Balmer) l = (R x (1/4 – 0))-1 = 365 nm The Balmer series are in the visible range.