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Chapter 27
Early Quantum theory and models of
the atom
Rutherford Scattering and the Nuclear Atom
Rutherford model of the atom. A small positive nucleus with electrons orbiting around it – mostly empty space.
−
−− −−
−−
−− −−
−− −−
−−
“Plum Pudding” model of the atom. Electrons uniformly distributed in a homogeneous sphere of positive charge.
positive material
electron
Rutherford Scattering and the Nuclear Atom
Rutherford scattering experiment (c. 1911): Scattering α particles (4He atoms) off a thin gold foil.
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 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 orbitRnucleus
=10−10
10−15=105
REarth orbit estimated( ) =Relectron orbitRnucleus
×RSun = 105( ) 7×108( ) = 7×1013m
REarth orbit estimated( )REarth orbit actual( )
=7×1013
1.5×1011= 470 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
Lyman series
Balmer series
Paschen series
…,4,3,2 1111
22 =⎟⎠
⎞⎜⎝
⎛ −= nn
Rλ
…,5,4,3 1211
22 =⎟⎠
⎞⎜⎝
⎛ −= nn
Rλ
…,6,5,4 1311
22 =⎟⎠
⎞⎜⎝
⎛ −= nn
Rλ
ultraviolet visible infrared
R = Rydberg const. =1.097×107m−1
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
In the Bohr model, a photon is emitted when the electron drops from a larger, higher-energy orbit to a smaller, lower energy orbit.
hfEE fi =−
The Bohr Model of the Hydrogen Atom
THE ENERGIES AND RADII OF THE BOHR ORBITS
E =KE +EPE = 12mv
2 −kZe2
rmv2
r= Fcentrip =
kZe2
r2
⇒ 12mv
2 =kZe2
2r
∴E = kZe2
2r−kZe2
r= −
kZe2
2r
The Bohr Model of the Hydrogen Atom
…,3,2,1 2
=== nhnrmvL nnn π
∴rn =h2
4π 2mke2
"
#$
%
&'n2
Z n =1, 2, 3,…
( ) …,3,2,1 m1029.52
11 =×= − nZnrn
Radii for Bohr orbits
Angular momentum is quantized.
We can get rn by combining this with (from the previous slide),
mv2
r= Fcentrip =
kZe2
r2
or, m2vn2 =
mkZe2
rn
Bohr radius
The Bohr Model of the Hydrogen Atom
…,3,2,1 22
2
2
422
=⎟⎟⎠
⎞⎜⎜⎝
⎛−= n
nZ
hemkEn
π
( ) …,3,2,1 J1018.2 2
218 =×−= − n
nZEn
( ) …,3,2,1 eV 6.13 2
2
=−= nnZEn
Bohr energy levels
E = − kZe2
2r
Substituting r = rn
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+.
( ) ( ) eV 12213eV 6.13eV 6.13 2
2
2
2
−=−=−=nZEn
eV 122energy Ionization +=
The Bohr Model of the Hydrogen Atom
THE LINE SPECTRA OF THE HYDROGEN ATOM
⇒1λ=
2π 2mk2e4
h3c"
#$
%
&'Z 2 1
nf2 −
1ni
2
"
#$$
%
&''
ni,nf =1, 2, 3,… ni > nf
2π 2mk2e4
h3c"
#$
%
&'=1.097×107m−1 = R Rydberg const.
For transitions from a higher energy state, ni, to a lower one, nf, the photon energy emitted is given by
E = hcλ= Ei −Ef = −
2π 2mk2e4
h2"
#$
%
&'Z 2
1ni2 −
1nf2
"
#$$
%
&''
Agrees with experiment!
Example: Find the wavelength of the second Balmer line in hydrogen, i.e. the ni = 4 to nf = 2 transition.
1λ= (1.097×107m−1) 1
nf2 −
1ni2
#
$%%
&
'((
= (1.097×107m−1) 122−142
#
$%
&
'(= 2.057×106m−1
⇒ 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 − E1hcλ= KE + E1 =11.0+13.6 = 24.6 eV = 3.94×10−18 J
λ =hc
3.94×10−18=6.63×10−34( ) 3.00×108( )
3.94×10−18
= 5.05×10−8m = 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.
2π r = nλ n =1, 2, 3,…
where, λ =hp=hmv
⇒ 2πr = n hmv
⇒ mvr = n h2π
Fit an integer number of standing de Broglie waves around the circumference of an orbital,
Same as Bohr’s assumption!