Atomic spectra and atomic
structure.
8.3 Atomic Physics
Atomic Spectra
Bohr Model
Atomic structure
Geiger and Marsden
Ernest Rutherford 1911
The scattering of alpha particles (He2+) nuclei from a thin
gold foil. The back scattering of a few alpha particles
showed that the nucleus is a small compact object.
The spectra of atoms provide information about
the energies of the electron in the atom.
Sharp peaks at discrete wavelengths indicate that
only specified energies are allowed in the atom.
For the Hydrogen atom the Bohr theory explains
the energies in a simple manner based on a
quantization of angular momentum.
The quantization is explained by the de Broglie
theory in terms of standing waves for the
electron.
Planetary model of the atom
Alpha
particle
Scattering from a small compact nucleus
Atomic spectra
Emission
gas of atom A
Absorption
A
e-
high voltage
spectrometer
Emission
gas of atoms
A
light source
(white light)
hf
Excitation
Atomic Spectra
spectrometer
light minus
absorbed wavelengths
e- (fast) + A -> e- (slow) + A*
A* -> A + hf
1
Atomic Spectra
Balmer series for Hydrogen
ultraviolet
visible
Emission
Absorption
(dark lines)
Lowest λ
Highest λ
A series of peaks closer together (continuum) at low λ
Discrete spectral lines are observed.
Rydberg Constant
Disagreement with classical theory
The Balmer series could be analyzed mathematically in
terms of an empirical equation.
Classical physics predicts that the energy of the electron
can have any value- not discrete values observe.
The classical theory could not explain why the electron
did not fall into the nucleus. Like a satellite falling into the
earth.
1
⎛ 1 1⎞
= RH ⎜ 2 − 2 ⎟
λ
⎝2 n ⎠
Rydberg Constant RH = 1.0973732x107 m-1
n = 3,4, 5 ..........
Integers larger than 2.
Bohr Theory
Angular momentum of a tennis ball
1. Electrons move in circular orbits.
2. Only specified atomic energy levels are allowed.
3. Energy is emitted when electron go from one energy
level to another.
4. The orbital angular momentum of the electron is
“quantized” in units of h/2π = = (called h bar)
r= 0.5 m
m = 0.1 kg
v= 2 m/s
What is n?
is quantized.
r
L =mvr = (0.1kg )(2m / s )(0.5m ) = 0.1
L = mvr = n=
r
n=1, 2, 3 .......
L = nh
v
m
v
m
kgm 2
= 0.1J ⋅ s
s
h
6.6 x10−34 J ⋅ s
=
= 1.05 x10−34 J ⋅ s
2π
2π
L
0.1J ⋅ s
L=
= 1033 h
1.0 x10−34 J ⋅ s / h
h=
n
n=1033 n is so large that quantization is not apparent
2
Classical dynamics
For central force
Angular momentum of an electron
r
m=9.1x10-31kg
r =0.1 x10-9 m
v=?
F
v
m
L is much smaller.
Quantization is apparent
F = ma =
L
mv 2 =
1 2 3 4 5
n ->
Bohr theory for hydrogen atom
Classical energies
k qq
k e
1
mv 2 − e 1 2 = − e
2
r
2r
k e e2
r
Results from Bohr theory
2
rn =
n2= 2
m ek e e 2
any value of r is allowed
n=1, 2, 3, ....... integers
Quantum condition
Bohr restricted the values of r
mvr = n=
v and r are not
independent
variables
Only specific values of r are allowed
E=KE+PE
E=
mv 2 k e e2
= 2
r
r
n=3
n=2
n=1
radius increases as n2
For n=1
angular momentum is quantized
n= 1, 2, 3, ........ integers
nh
momentum increases as 1/r
mv =
2πr
(1)2 (1.05 x10−34 Js )
= 5.3 x10 −11 m
(9.1x10−31kg )(8.9 x109 NmC −2 )(1.6 x10 −19 C )2
Size of the Hydrogen atom in the ground state
0.053 nm
Excited state energy levels
Agreement with Rydberg equation
r1 =
Energy levels are quantized
(proportional to 1/n2)
En = −
1 ∆E mek e2 e 4 ⎛ 1
1 ⎞
=
=
− 2 ⎟
⎜
λ hc
4πc= 3 ⎝ n2final ninitial
⎠
mek e2 e 4 ⎛ 1 ⎞
13.6
= − 2 eV
2= 2 ⎜⎝ n2 ⎟⎠
n
mek e2 e 4
4πc= 3
9.109x10 −31(8.987x109 )2 (1.602x10 −19 )4
R=
= 1.099x107 m−1
4π(2.997x108 )(1.054x10−34 )3
R=
Emission energies
⎛ 1
1 ⎞
∆E = Einitial − Efinal = 13.6 ⎜ 2 − 2 ⎟
⎝ nfinal ninitial ⎠
∆Emax
hfmax =13.6 eV
In excellent agreement with the experimental
value 1.097x107 m-1
Predicts spectral lines in the ultraviolet (Lyman series)
and infrared (Paschen series), maximum energies, continuum.
3
Example
de Broglie wavelength explanation
of Bohr theory
Find the wavelength of the transition from
n=3 to n=2.
⎛ 1
1
1 ⎞
1⎞
⎛ 1
= R ⎜ 2 − 2 ⎟ = 1.097x107 ⎜ 2 − 2 ⎟
λ
3 ⎠
⎝2
⎝ n final ninitial ⎠
⎛ 1 1⎞
= 1.097x107 ⎜ − ⎟ = 1.52x106 m−1
⎝4 9⎠
λ = 6.56x10−7 m = 656nm
red line in Balmer series
Bohr theory
Shows that the energy levels in the hydrogen atom are
quantized.
Correctly predicts the energies of the hydrogen atom (and
hydrogen like atoms.)
The Bohr theory is incorrect in that it does not obey the
uncertainty principle. It shows electrons in well defined
orbits.
Quantum mechanical theories are used to calculate the
energies of electrons in atoms. (i.e. Shrödinger equation)
Characteristic X-rays are due to
emission from heavy atoms
excited by electrons
mvr = n
h
2π
quantization of angular momentum
⎛ h ⎞
2πr = n ⎜
circumference = nλ
⎟ = nλ
⎝ mv ⎠
Quantization of angular momentum is equivalent to forming
circular standing waves.
Extension of the Bohr Theory
Bohr theory can only be used to predict energies of
Hydrogen-like atoms. (i.e. atoms with only one electron)
This includes H, He+, Li2+ ....
For example He+ ( singly ionized helium has 1 electron
and a nucleus with a charge of Z = +2)
For this case the energy for each state is multiplied by
Z2 =4
m k 2 z2e4 ⎛ 1 ⎞
En = − e e 2 ⎜ 2 ⎟
2=
⎝n ⎠
⎛ 1⎞
⎛ 1⎞
2 1
En = −13.6(Z ) 2 = −13.6(22 ) ⎜ 2 ⎟ = −54.4 ⎜ 2 ⎟ eV
n
⎝n ⎠
⎝n ⎠
for He+
Characteristic x-rays
The wavelength of characteristic
x-ray peaks due to emission from
high energy states of heavy atoms
(high Z).
Mo
fast emetal
atom
Upper
shell
K shell
dislodge
K shell e-
hf
Kα
4
X-ray emission
Calculate the wavelength for Kα x-ray emission of Mo
(Z=+42) The electron in the L shell must be in a l=1 state
Bohr model with Shielding
Effective Z = 42-3
L shell 2p
2s
hf
K shell 1s
Kα
Good agreement
Calculated
value
71 pm
Mo
Effective Z =42-1
2⎛ 1 ⎞
⎛ 1⎞
E(Lshell) = −13.6 ( Z − 3 ) ⎜ 2 ⎟ = −13.6(42 − 3)2 ⎜ ⎟ = −5.17x103 eV
⎝2 ⎠
⎝4⎠
⎛1⎞
EKshell = −13.6(Z − 1)2 ⎜ 2 ⎟ = −2.28x104 eV
⎝1 ⎠
71 pm
∆E = 2.28x104 − 5.17x103 = 1.76x104 eV
∆E = hf =
hc
λ
λ=
−34
8
hc
(6.63x10 J)(3.0x10 m / s)
=
= 7.1x10−11m
∆E (1.6x10 −19 J / eV)(1.76x10 4 eV)
5