The Nature of the Atom

The Nature of the Atom
-------------------------------------------------------------

Joseph F. Alward, PhD
Department of Physics
University of the Pacific

Neils Bohr
(1885-1962)

Applets used in this eLecture:
Emission and Absorption of a Photon
Laser Animation
(Link to applets before lecture and they'll
be ready when you need them.)

Important Terms and Concepts

nucleus: the extremely dense center of
an atom, discovered by Ernest Rutherford
line spectrum: the particular wavelengths
of light emitted or absorbed by an atom

Bohr model: an electron orbiting a proton
in a circular path (Neils Bohr)

energy levels in hydrogen:
E_n = -13.6 eV / n²

emission of photons: hf = E_i - E_f
Rule of 1234: E = (1234 eV-nm) / l

deBroglie wavelength:l = h / mv

spontaneous emission: electron in a
higher-lying energy orbit makes a transition
to a lower-lying orbit without stimulation

stimulated emission: a photon "tickles"
an electron in a higher-lying energy state;
electrons falls down into lower-lying state,
emitting a photon of exactly the same
frequency as the original photon.

The Plum Pudding Model of the Atom

The atom was once thought to be a solid
ball of positive material of pudding-like
consistency in which were embedded
discrete negatively charged objects
called electrons.
In this (incorrect) model of the atom, the
atom is a ball of uniform density.

Ernest Rutherford was the first to show
that the the atom does not have uniform
density, and that most of the atom's mass
is located at its center.

The Nucleus

Ernest Rutherford won
1908 Nobel Prize for
studies in radioactivity;
discovered nucleus in
1911.

Approaching a Nucleus

Assume the nucleus of a gold atom has a radius
r = 7 x 10^-15 m. Given that the gold nucleus has
79 protons, what must be the speed of a proton
if it is to be fired at the gold nucleus and reach its
"surface" before its direction of travel is reversed
by the repulsion of the nucleus?
-------------------------------------------------------------------

Potential outside nucleus: V = kQ / r

Potential energy of proton: PE = kqQ / r

Conservation of Energy:

E = E₀
kqQ / r = 1/2 mv²
q = e

Q = 79 e

m = 1.667 x 10^-27 kg

Atomic Spectra

Atoms in heated
gases emit and
absorb light of
certain
wavelengths.

Shown at the
left are three
emission spectra
and one
absorption
spectrum.

Line Spectrum of Atomic Hydrogen

In 1885 Johann Balmer discovered an
equation which describes the emission-
absorption spectrum of atomic
hydrogen:

1 / l = 1.097 x 10⁷(1 / 4 - 1 / n²)

where n = 3, 4, 5, 6, ...

Balmer found this by trial and error, and
had no understanding of the physics
underlying his equation.

Balmer's Formula

1 / l = 1.097 x 10⁷ (1 / 4 - 1 / n²)
for n = 3, 4, 5, 6, .....
-------------------------------------------
Consider the case where n = 3:

1 / l = 1.097 x 10⁷ (1 / 4 - 1 / 9)

= 1.524 x 10⁶ m^-1

l = 6.56 x 10^-7 m
= 656 x 10^-9 m
= 656 nm

The ElectronVolt (eV) and the Rule of 1234

The kinetic energy of an electron
accelerated across a potential
difference of one volt is one
electronvolt (eV).

The eV is not a unit of charge,
or a unit of voltage; it is a unit
of energy. The energy E in electronvolts (eV) of a
photon is related to its wavelength l in
nanometers (nm) through the following
relationship:

E = (1234 eV-nm) / l

This equation is important not because
of any essential physics underlying it,
but because is is a time-saver.

E = (1234 eV-nm) / l

What is the energy of a photon
in a beam of light of wavelength
l = 620 nm?
-------------------------------------------

E = 1234 /617
   = 2 eV What is the energy of a photon in
light of frequency f = 3 x 10¹⁴ Hz?
----------------------------------------------
c = l f

l = c / f
  = (3 x 10⁸ m/s) /3 x 10¹⁴ Hz
  = 1 x 10^-6 m
   = 1000 x 10^-9 m
= 1000 nm

E = 1234 / 1000
= 1.23 eV

   l = (1234 eV-nm) / E

What is the wavelength of light whose
photons have energy E = 6.17 eV?
----------------------------------------

l = 1234 /6.17

   = 200 nm What is the frequency of light whose
photons have energy E = 6.17eV?
--------------------------------------

200 nm = 200 x 10^-9 m
c = l f

f = c / l
  = (3 x 10⁸) / 200 x 10^-9
  = 1.5 x 10¹⁵ Hz

Neils Bohr Explains the Balmer Equation

Neils Bohr, a Danish physicist, treated the
hydrogen atom as if it were an electron orbiting
in a circular path about a proton, and assumed
that the angular momentum mvr of the electron
was an integer multiple of h/2p.

Deriving Bohr's Energy Equation

E = 1/2 mv² - ke²/r (1)
ma = F (2)
F = ke² /r²   (3)
a = v² /r   (4)
mvr = nh/2p    (5)
E = (1240 eV-nm) /l (6)

E_n = -13.6 eV /n²   (7)

mv²/r = ke²/r² (8)
(1/2) mv²= (1/2) ke²/r    (9)

Substitute (9) into (1):
E = -(1/2) ke²/r (10)

Square (5), divide by 2mr²:
1/2 mv² = n²h²/4mp²r² (11)
Substitute (11) into (9):
n²h²/4mp²r² = (1/2) ke²/r   (12)

Solve (12) for r:
r = n²h²/4mke²p² (13)

Substitute (13) into (10): (14)

E = -(2mk²e⁴p²/h²⁾ / n^{2 (15)}

Use the values of m, k, e, and h
given in the text, and Eqn. (6)
to obtain Eqn. (7).

Energy Levels in Hydrogen

E_n = -13.6 eV /n²
-----------------------
E₁ = - 13.6 eV
E₂ = - 3.40 eV
E₃ = - 1.51 eV
E₄ = - 0.85 eV
E₅ = - 0.54 eV
E₆ = - 0.38 eV

Hydrogen Atom Orbital Radii

From Eqn. (13):
r = n^{2 (}h²/ 4mke²p²)
r_n = (0.529 A) n²

a₀ = 0.529 A

r_n = n² a₀

Hydrogen Energy Level Problem

What is the total energy of an electron that is in
an orbit with a radius of 4.761 x 10^-10 m ?
---------------------------------------------------------------
r = (0.529 x 10^-10) n²

Solve for n.

Energy Transitions in Atoms

Energy of photon = Energy lost by electron
hf = E_i - E_f

Animation of Emission and Absorption of a Photon

Calculating Wavelengths of Emitted Light .

hf = E_i - E_f
E₃ ---> E_2:
E_i = - 1.51 eV
E_f = - 3.40 eV
------------------------------
hf = - 1.51 - (-3.40)
= 1.89 eV
------------------------------
l = (1234 eV-nm) / E

= 1234 / 1.89
= 652 nm
( Click here to compare to
Balmer values given
earlier.)

Other Energy Transitions

The final state in the energy
transitions is n = 3 for the
Paschen series, n = 2 for
the Balmer series, and
n = 1 for the Lyman series.

Recalling that the range of
visible wavelengths is
approximately 300-700 nm,
one can see that only
transitions ending at n = 2
emit light in the visible range.

Louis deBroglie Explains Bohr's Angular Momentum Postulate

Prince Louis deBroglie
(1892-1987)
French aristocrat

l = h / mv
An integral multiple of wavelengths
must fit in the length 2pr, otherwise
destructive interference would
occur.

2pr = nl
= n (h / mv)

Rearranging:
mvr = n (h / 2p)

The Periodic Table

Atomic number is the number
of protons in the nucleus, which
is the same as the number of
electrons in the neutral atom.

Atomic mass is the number of
neutrons and protons in the
nucleus. Shown in the figure
is the average number.

X-Rays

Filament wire is heated by filament
battery; electrons boiled out of the
|wire are accelerated to very high
speeds by the electric field between
the filament and the metal target.

When the electrons slam into the
metal they experience a very large
deceleration which causes the
electrons to emit very energetic
electromagnetic waves.

X-Rays in Computer-Assisted Tomography (CAT)

Variation in intensity
of detected X-rays depends on density
and type of material through which the
rays pass. Triangulation allows the computer to determine position of various structures. tomos, Greek: slice)

CAT Scan Math

CAT Scan Images

Two-dimension CAT scan image. Three-dimension CAT scan image.

Spontaneous Emission

An electron moves from a higher-lying
energy state to an unoccupied energy
state of lower energy. In order that
energy be conserved, a photon is
emitted.

hf = E_i - E_f

This type of transition is called
spontaneous because it happens
without any external stimulus.

Stimulated Emission

The oscillating electric field of the
incident photon causes the electron
to oscillate and to emit a photon
of the same frequency. The result
is two photons in phase (coherent)
with the same frequency.
If there are other electrons in
similar energy states, these two
photons will double themselves,
and so on.

Population of Energy Levels

Normally, a population of electrons
in a gas of atoms seek the lowest
available energy states.

When there are more electrons
in higher-lying energy states than
in lower-lying ones, the population
is said to be inverted

Laser Animation

Helium-Neon Laser

Lasers in Eye Surgery

Compact Disk Reader

Appendix

Measured Hydrogen Line Spectrum [back]