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Published byGlenn Lyman Modified over 4 years ago

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Hydrogen Atom Coulomb force “confines” electron to region near proton => standing waves of certain energy + -

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Electron in n=2 level makes a transition to lower level by emitting a photon of frequency f=E/h = (E 2 -E 1 )/h =c/

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Transitions Electron in ground state E 1 can move to the excited state E 3 if it absorbs energy equal to E 3 -E 1 absorb a photon hf= E 3 -E 1 electron will not stay in the excited state for long => emits a photon or a sequence of photons hf= E 3 - E 1 hf = E 3 - E 2 = hc/ hf = E 2 - E 1 = hc/ Emission spectrum 1 2 3 photon

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Line spectra from H, He, Ba, Hg Continuous visible spectrum

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Hydrogen Atom Coulomb force “confines” electron to region near proton => standing waves of certain energy + -

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Atoms In 1913 Neils Bohr proposed a model of hydrogen based on a particle in an orbit electron with charge -e in a circular orbit about a nucleus of charge +Ze Coulomb attraction provides centripetal force mv 2 /r= kZe 2 /r 2 potential energy is U= kq 1 q 2 /r = -kZe 2 /r kinetic energy K=(1/2)mv 2 =(1/2)kZe 2 /r hence U= -2K (same for gravity!) total E= K +U = -K = -(1/2)kZe 2 /r e/m theory states that an accelerating charge radiates energy! Should spiral into the nucleus! Why are atoms stable?

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Bohr’s postulates Bohr postulated that only certain orbits were stable and that an atom only radiated energy when it made a transition from one level to another the energy of a photon emitted was hf = E i - E f since the energies of the orbits are related to their radii, f= (1/2)(kZe 2 /h)(1/r 2 - 1/r 1 ) experimentally observed photon frequencies satisfied f=c/ = cR(1/n 2 2 - 1/n 1 2 ) where n 1 and n 2 are integers Rydberg-Ritz formula do the allowed values of r n 2 ? If we think of the allowed orbits as standing waves then we need 2 r= n for constructive interference

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Stable orbits 2 r= n for constructive interference but de Broglie says =h/p hence pr= nh/2 but L=rp for circular orbits hence L= mvr =n ħ n=1,2,3,… angular momentum is quantized! Bohr Atom

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Bohr Theory How do we find the allowed radii? Coulomb force = kZe 2 /r 2 = mv 2 /r => v 2 =kZe 2 /mr but Bohr says mvr= n ħ => v 2 = n 2 ħ 2 /m 2 r 2 solve for r: r= n 2 (ħ 2 /mkZe 2 ) = n 2 a 0 /Z where a 0 is a radius corresponding to n=1 and Z=1(Hydrogen) a 0 = ħ 2 /mke 2 = 0.0529 nm (called the Bohr radius) hence only certain orbits are allowed => only certain energies energy differences = (1/2)kZe 2 (1/r 2 - 1/r 1 ) = (1/2)(kZ 2 e 2 /a 0 )(1/n 2 2 - 1/n 1 2 )

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Bohr Atom compare with Rydberg-Ritz formula for observed wavelengths in Hydrogen 1/ = R(1/n 2 2 - 1/n 1 2 ) where R is Rydberg constant frequency of photons f=c/ = c R(1/n 2 2 - 1/n 1 2 )= (E 2 - E 1 )/h using Z=1, R=mk 2 e 4 /4 cħ 3 =1.096776 x 10 7 m -1 in agreement with experiment! Energy levels can be determined from allowed radii E=-(1/2)kZe 2 /r = -(mk 2 e 4 /2ħ 2 )(Z 2 /n 2 ) = -E 0 Z 2 /n 2 E 0 is the lowest energy for hydrogen = 13.6 eV hence hydrogen atom(Z=1) has energies E n = -13.6eV/n 2 n=1,2,...

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E n = -13.6eV/n 2 n=1,2,3,... E 1 = -13.6eV

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Hydrogen Atom E n = -13.6eV/n 2 n=1,2,3,… ground state has E 1 = -13.6eV ionization energy is 0- E 1 = 13.6eV => energy needed to remove electron excited state: n=2 E 2 = -(13.6/4)eV electron must absorb a photon of energy hf= E 2 - E 1 =hc/ = (3/4)(13.6eV)

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Electron in n=2 level makes a transition to lower level by emitting a photon of frequency f=E/h = (E 2 -E 1 )/h =c/

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E n -E 1 = 13.6eV ( 1 -1/n 2 ) = hf = hc/ max =hc/(13.6eV)(.75) ~ 122 nm E n -E 2 = 13.6eV ( 1/4 -1/n 2 ) = hf = hc/ max =hc/(13.6eV)(5/36) ~ 658 nm => 4 lines visible

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Balmer Series

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