Chapter 27 Early Quantum theory and models of the atom - Physics

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 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.


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 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


…,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


…,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


ENERGY LEVEL DIAGRAMS


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 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!

Chapter 27 Early Quantum theory and models of the atom - Physics

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