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Hydrogen Atom
• Coulomb force “confines” electron to region near proton => standing waves of certain energy
+-
Uq q
r
1
4 0
1 2
Ee e
rpot
1
40
0( )( )
Electron in n=2 level makes a transition to lowerlevel by emitting a photon of frequency f=E/h = (E2-E1)/h =c/
Transitions• Electron in ground state E1 can move to the excited
state E3 if it absorbs energy equal to E3-E1
• absorb a photon hf= E3-E1
• electron will not stay in the excited state for long => emits a photon or a sequence of photons
hf= E3 - E1
hf = E3 - E2 = hc/
hf = E2 - E1 = hc/
Emission spectrum
1
2
3
photon
Line spectra fromH, He, Ba, Hg
Continuous visiblespectrum
Hydrogen Atom
• Coulomb force “confines” electron to region near proton => standing waves of certain energy
+- 1 2q q
U kr
( )( )0
e eU k
r
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 mv2/r= kZe2/r2
• potential energy is U= kq1q2/r = -kZe2/r
• kinetic energy K=(1/2)mv2=(1/2)kZe2/r
• hence U= -2K (same for gravity!)
• total E= K +U = -K = -(1/2)kZe2/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 = Ei - Ef
• since the energies of the orbits are related to their radii,f= (1/2)(kZe2/h)(1/r2 - 1/r1)
• experimentally observed photon frequencies satisfiedf=c/ = cR(1/n2
2 - 1/n12) where n1 and n2 are integers
Rydberg-Ritz formula
• do the allowed values of r n2 ?
• If we think of the allowed orbits as standing waves then we need2r= n for constructive interference
Stable orbits
• 2r= 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!
BohrAtom
Bohr Theory• How do we find the allowed radii?
• Coulomb force = kZe2/r2 = mv2/r => v2=kZe2/mr
• but Bohr says mvr= n ħ => v2= n2 ħ2/m2r2
• solve for r: r= n2 (ħ2/mkZe2) = n2 a0/Zwhere a0 is a radius corresponding to n=1 and
Z=1(Hydrogen)
• a0 = ħ2/mke2 = 0.0529 nm (called the Bohr radius)
• hence only certain orbits are allowed => only certain energies
• energy differences = (1/2)kZe2(1/r2 - 1/r1)= (1/2)(kZ2e2/a0)(1/n2
2 - 1/n12)
Bohr Atom• compare with Rydberg-Ritz formula for observed wavelengths in
Hydrogen 1/ = R(1/n22 - 1/n1
2) where R is Rydberg constant
• frequency of photons f=c/= c R(1/n22 - 1/n1
2)= (E2 - E1)/h
• using Z=1, R=mk2e4/4cħ3 =1.096776 x 107 m-1 in agreement with experiment!
• Energy levels can be determined from allowed radii
• E=-(1/2)kZe2/r = -(mk2e4/2ħ2)(Z2/n2) = -E0 Z2/n2
• E0 is the lowest energy for hydrogen = 13.6 eV
• hence hydrogen atom(Z=1) has energies En = -13.6eV/n2 n=1,2,...
( )( )0
e eU k
r
En = -13.6eV/n2 n=1,2,3,...
E1 = -13.6eV
Hydrogen Atom
• En = -13.6eV/n2 n=1,2,3,…
• ground state has E1 = -13.6eV
• ionization energy is 0- E1 = 13.6eV=> energy needed to remove electron
• excited state: n=2 E2 = -(13.6/4)eV
• electron must absorb a photon of energy hf= E2 - E1 =hc/ = (3/4)(13.6eV)
Electron in n=2 level makes a transition to lowerlevel by emitting a photon of frequency f=E/h = (E2-E1)/h =c/
En-E1 = 13.6eV ( 1 -1/n2) = hf = hc/
max=hc/(13.6eV)(.75) ~ 122 nm
En-E2 = 13.6eV ( 1/4 -1/n2) = hf = hc/max=hc/(13.6eV)(5/36) ~ 658 nm => 4 lines visible
Balmer Series