Hydrogen Atom
1
q1q2
U
4 0 r
+
• Coulomb force “confines” electron to region near
proton => standing waves of certain energy
E pot
(e)( e)

0
4 0
r
1
Electron in n=2 level makes a transition to lower
level 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
3
hf= E3 - E1
2
hf = E3 - E2 = hc/
1
hf = E2 - E1 = hc/
Emission spectrum
photon
Continuous visible
spectrum
Line spectra from
H, He, Ba, Hg
Hydrogen Atom
+
q1q2
U k
r
• Coulomb force “confines” electron to region near
proton => standing waves of certain energy
(e)( e)
U k
0
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 satisfied
f=c/ = cR(1/n22 - 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 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 = 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/Z
where 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/n22 - 1/n12)
Bohr Atom
• compare with Rydberg-Ritz formula for observed wavelengths in
Hydrogen
1/ = R(1/n22 - 1/n12) where R is Rydberg constant
• frequency of photons f=c/= c R(1/n22 - 1/n12)= (E2 - E1)/h
• using Z=1, R=mk2e4/4cħ3 =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,...
(e)( e)
U k
0
r
E1 = -13.6eV
En = -13.6eV/n2
n=1,2,3,...
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 lower
level by emitting a photon of frequency
f=E/h = (E2-E1)/h =c/
En-E2 = 13.6eV ( 1/4 -1/n2)
= hf = hc/
max=hc/(13.6eV)(5/36)
~ 658 nm
=> 4 lines visible
En-E1 = 13.6eV ( 1 -1/n2)
= hf = hc/
max=hc/(13.6eV)(.75)
~ 122 nm
Balmer Series