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

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

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/

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

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

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

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?

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

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

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 )

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,...

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

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)

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/

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

Balmer Series

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