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Explore de Broglie's wave-particle theory, Davisson-Germer Experiment, significance of the Hydrogen Atom, and early models of the atom. Understand hydrogen's quantum nature and energy levels through absorption and emission spectra calculations. Dive into electron orbits, energies, and wavelengths in the hydrogen atom.
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Wave Properties of Particles • In 1924, Louis de Broglie postulated that because photons have wave and particle characteristics, perhaps all forms of matter have both properties • Furthermore, the frequency and wavelength of matter waves can be determined
de Broglie Wavelength and Frequency • The de Broglie wavelength of a particle is • The frequency of matter waves is
The Davisson-Germer Experiment • They scattered low-energy electrons from a nickel target • They followed this with extensive diffraction measurements from various materials • The wavelength of the electrons calculated from the diffraction data agreed with the expected de Broglie wavelength • This confirmed the wave nature of electrons • Other experimenters have confirmed the wave nature of other particles
Importance of Hydrogen Atom • Hydrogen is the simplest atom • The quantum numbers used to characterize the allowed states of hydrogen can also be used to describe (approximately) the allowed states of more complex atoms • This enables us to understand the periodic table
More Reasons the Hydrogen Atom is so Important • The hydrogen atom is an ideal system for performing precise comparisons of theory and experiment • Also for improving our understanding of atomic structure • Much of what we know about the hydrogen atom can be extended to other single-electron ions • For example, He+ and Li2+
Early Models of the Atom • J.J. Thomson’s model of the atom • A volume of positive charge • Electrons embedded throughout the volume • A change from Newton’s model of the atom as a tiny, hard, indestructible sphere
Early Models of the Atom, 2 • Rutherford • Planetary model • Based on results of thin foil experiments • Positive charge is concentrated in the center of the atom, called the nucleus • Electrons orbit the nucleus like planets orbit the sun
Difficulties with the Rutherford Model • Atoms emit certain discrete characteristic frequencies of electromagnetic radiation • The Rutherford model is unable to explain this phenomena • Rutherford’s electrons are undergoing a centripetal acceleration and so should radiate electromagnetic waves of the same frequency • The radius should steadily decrease as this radiation is given off • The electron should eventually spiral into the nucleus • It doesn’t
+ 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.
de Broglie Waves in the Hydrogen Atom • In this example, three complete wavelengths are contained in the circumference of the orbit • In general, the circumference must equal some integer number of wavelengths • 2 r = n λ n = 1, 2, …
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
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
-1.5 eV n = 3 -3.4 eV n = 2 + -13.6 eV n = 1 Etot - n = ∞ 0 - - - Ionized state of Hydrogen: proton
-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 -
E = hc/l in eV (Rydberg const.) E(nm) or l < 0 Absorb photons of a given l E (nm) or l > 0 Emit photons of a given l
-1.5 eV n = 3 -13.6 eV -3.4 eV n = 2 n = 1 Etot E = hc/l n = ∞ 0 E (12) = 10.2 eV = 10.2 x (1.6 x 10-19 J/eV) = 1.63 x 10-18 J l = 121 nm Ultraviolet range -
Ex. 29.6 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)
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.
ABSORPTION SPECTRUM ABSORPTION SPECTRUM Hydrogen Spectrum
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