Chapter 27
Early Quantum
theory and models of
the atom
Rutherford Scattering and the Nuclear Atom
positive
material
electron
−−
−−
−−
−−
−−
−−
−
−−
−−
“Plum Pudding” model of the atom.
Electrons uniformly distributed in a
homogeneous sphere of positive
charge.
Rutherford model of the atom.
A small positive nucleus with electrons
orbiting around it – mostly empty space.
Rutherford Scattering and the Nuclear Atom
If the “Plum Pudding” model were
correct, the massive α (4He)
particles would not deflect much.
But, large deflections of the
α  particles were seen, showing
that a massive and small nucleus
containing most of the mass of the
atom existed at the center of the
atom, confirming Rutherford’s
model.
Rutherford scattering experiment (c. 1911):
Scattering α particles (4He atoms) off a thin
gold foil.
Rutherford Scattering and the Nuclear Atom
Conceptual Example: Atoms are Mostly Empty Space
In the planetary model of the atom, the nucleus (radius = 10-15 m)
is analogous to the Sun (radius = 7 x 108 m). Electrons orbit
(radius = 10-10 m) the nucleus like the Earth orbits (radius = 1.5 x 1011 m)
the Sun. If the dimensions of the solar system had the same proportions
as those of the atom, would the Earth be closer to or farther away from
the Sun than it actually is?
Relectron orbit
Rnucleus
10 −10
= −15 = 10 5
10
REarth orbit ( estimated ) =
REarth orbit ( estimated )
REarth orbit ( actual )
Relectron orbit
Rnucleus
× RSun = (10 5 ) ( 7 ×108 ) = 7 ×1013 m
7 ×1013
=
= 470
11
1.5 ×10
times larger than the actual Earth’s
orbit à farther than Pluto’s orbit.
à  An atom has a greater fraction of
empty space than the Solar System!
Line Spectra
The individual wavelengths emitted by two gases and the
continuous spectrum of the sun.
Line Spectra
The Line Spectrum of Hydrogen
1
⎛ 1 1 ⎞
= R⎜ 2 − 2 ⎟
λ
⎝ 1 n ⎠
Lyman series
Balmer series
1
⎛ 1 1 ⎞
= R⎜ 2 − 2 ⎟
λ
⎝ 2 n ⎠
Paschen series
1
⎛ 1 1 ⎞
= R⎜ 2 − 2 ⎟
λ
⎝ 3 n ⎠
n = 2, 3, 4, …
n = 3, 4, 5, …
n = 4, 5, 6, …
R = Rydberg const. = 1.097 ×10 7 m −1
ultraviolet
visible
infrared
Bohr model of the atom
(Niels Bohr, c. 1913)
The main goal of the Bohr model was to explain the experimental
line spectra.
Main assumptions of the Bohr model:
1) Electrons exist in quantized circular orbits around the nucleus in an
atom. These orbits are called stationary orbitals or stationary states.
2) Electrons in these orbitals do not radiate E&M waves (so they do not
decay with the electrons spiraling into the nucleus).
3) Each orbital has a specific angular momentum (and thus energy)
associated with it.
The Bohr Model of the Hydrogen Atom
Ei − E f = hf
In the Bohr model, a photon is emitted when the electron drops
from a larger, higher-energy orbit to a smaller, lower energy
orbit.
The Bohr Model of the Hydrogen Atom
THE ENERGIES AND RADII OF THE BOHR ORBITS
2
kZe
E = KE + EPE = 12 mv 2 −
r
mv 2
kZe 2
= Fcentrip = 2
r
r
2
kZe
2
1
⇒ 2 mv =
2r
kZe 2 kZe 2
kZe 2
∴E =
−
=−
2r
r
2r
The Bohr Model of the Hydrogen Atom
Angular momentum is quantized.
Ln = mvn rn = n
h
2π
n = 1, 2, 3, …
We can get rn by combining this
with (from the previous slide),
mv 2
kZe 2
= Fcentrip = 2
r
r
2
mkZe
or, m 2 vn2 =
rn
Radii for Bohr orbits
" h2
% n2
∴rn = $ 2
2'
# 4π mke & Z
2
n
rn = 5.29 ×10 −11 m
Z
(
Bohr radius
)
n = 1, 2, 3, …
n = 1, 2, 3, …
The Bohr Model of the Hydrogen Atom
kZe 2
E =−
2r
Substituting r = rn
⎛ 2π 2 mk 2e 4 ⎞ Z 2
⎟⎟ 2
En = −⎜⎜
2
h
⎝
⎠ n
n = 1, 2, 3, …
Bohr energy levels
2
Z
En = − 2.18 ×10 −18 J 2
n
(
)
Z2
En = −(13.6 eV ) 2
n
n = 1, 2, 3, …
n = 1, 2, 3, …
The Bohr Model of the Hydrogen Atom
ENERGY LEVEL DIAGRAMS
The Bohr Model of the Hydrogen Atom
Example: The Ionization Energy of Li2+
Li2+ is a lithium atom (Z=3) with only one electron. Obtain the ionization
energy of Li2+.
Z2
32
En = −(13.6 eV ) 2 = −(13.6 eV ) 2 = −122 eV
n
1
Ionization energy = +122 eV
The Bohr Model of the Hydrogen Atom
THE LINE SPECTRA OF THE HYDROGEN ATOM
For transitions from a higher energy state, ni,
to a lower one, nf, the photon energy emitted
is given by
" 2π 2 mk 2 e 4 % 2 " 1 1 %
hc
E=
= Ei − E f = − $
' Z $$ 2 − 2 ''
2
λ
h
#
& # ni n f &
⇒
1 " 2π 2 mk 2 e 4 % 2 " 1 1 %
=$
' Z $$ 2 − 2 ''
3
λ # h c & # n f ni &
ni , n f = 1, 2, 3, …
ni > n f
" 2π 2 mk 2 e 4 %
7
−1
=
1.097
×10
m
= R Rydberg const.
$
'
3
# hc &
Agrees with experiment!
Example: Find the wavelength of the second Balmer line in hydrogen,
i.e. the ni = 4 to nf = 2 transition.
#1 1&
1
7
−1
= (1.097 ×10 m ) %% 2 − 2 ((
λ
$ n f ni '
#1 1&
7
−1
= (1.097 ×10 m ) % 2 − 2 ( = 2.057 ×10 6 m −1
$2 4 '
⇒ λ = 486 nm blue
Example: Ionization of a hydrogen atom.
What wavelength photon is needed to ionize a hydrogen atom in the
ground state and give the ejected electron 11.0 eV of kinetic energy?
KE = hf − E1
hc
= KE + E1 = 11.0 +13.6 = 24.6 eV = 3.94 ×10 −18 J
λ
−34
8
6.63×10
3.00
×10
(
)
(
)
hc
λ=
=
−18
3.94 ×10
3.94 ×10 −18
= 5.05 ×10 −8 m = 50.5 nm ultraviolet
De Broglie’s Explanation of Bohr’s Assumption About Angular Momentum
De Broglie suggested standing particle waves as an explanation
for Bohr’s angular momentum assumption.
Fit an integer number of standing
de Broglie waves around the
circumference of an orbital,
2 π r = nλ
n = 1, 2, 3, …
h h
where, λ = =
p mv
h
h
⇒ 2π r = n
⇒ mvr = n
mv
2π
Same as Bohr’s
assumption!