Hydrogen Atom

+

U

 1 4  0

q q

1 2

r

• Coulomb force “confines” electron to region near proton => standing waves of certain energy

E pot

 1 4  0

r



e

)  0

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 3 2 1 hf= E 3 - E 1 hf = E 3 - E 2 = hc/  hf = E 2 - E 1 = hc/  Emission spectrum photon

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

Hydrogen Atom

+

U



k q q

1 2

r

• Coulomb force “confines” electron to region near proton => standing waves of certain energy

U



k r



e

)  0

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 • 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!

mv 2 /r= kZe 2 /r 2 • 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 2  r= n  for constructive interference then we need

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: where a 0 r= n 2 ( ħ 2 /mkZe 2 ) = n 2 a 0 /Z 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,...

U



k r



e

)  0 E n = -13.6eV/n 2 n=1,2,3,...

E 1 = -13.6eV

Hydrogen Atom

• E n = -13.6eV/n 2 • ground state has E 1 n=1,2,3,… = -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 2 = 13.6eV ( 1/4 -1/n 2 ) = hf = hc/   max =hc/(13.6eV)(5/36) ~ 658 nm => 4 lines visible E n -E 1 = 13.6eV ( 1 -1/n 2 ) = hf = hc/   max =hc/(13.6eV)(.75) ~ 122 nm

Balmer Series