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Atoms
Physics
Montwood High School
R. Casao
More than 100 different elements have been
discovered.
• Each element consists of atoms that contain:
•
•
•
•
a number of protons Z,
an equal number of electrons ,
and a number of neutrons N.
The number of protons Z is called the atomic
number.
• The lightest atom, hydrogen (H), has Z = 1;
helium (He) has Z = 2; lithium (Li) has Z = 3;
and so forth.
• Nearly all the mass of the atom is concentrated
in a tiny nucleus, which contains the protons
and neutrons.
•
The nuclear radius is about 1 to 10 fm (1 fm =
10-5 m).
• The distance between the nucleus and the
electrons is 0.1 nm; this distance determines
the size of the atom.
• The chemical and physical properties of an
element are determined by the number and
arrangement of the electrons in the atom.
• Each proton has a positive charge +e so the
nucleus has a total positive charge +Z·e.
• The electrons are negatively charged (-e), so
they are attracted to the nucleus and repelled
by each other.
•
Since electrons and protons have equal but
opposite charges and there are an equal
number of electrons and protons in the atom,
atoms are electrically neutral.
• Atoms that lose or gain one or more electrons
are electrically charged and are called ions.
•
The Bohr Model
Developed by Niels Bohr to explain the spectra
emitted by hydrogen atoms.
• By the beginning of the 1900’s, data had been
collected on the emission of light by atoms in a
gas when they are excited by an electric
discharge.
•
Viewed through a spectroscope with a narrowslit aperture, the light appears as a discrete set
of lines of different colors and wavelengths.
• The spacing and intensities of the lines are
characteristic of the element.
• The wavelengths of these spectral lines could be
accurately determined, and much effort went
into finding regularities in the spectra.
• The line spectra for sunlight, hydrogen, and
mercury:
•
•
Johann Balmer found that wavelengths of the
lines in the visible spectrum of hydrogen can be
represented by the formula
2
m
  3.64.6 nm  2
, where m  3, 4,5,
m 4
Balmer suggested that this might be a special
case of a more general expression that would be
applicable to the spectra of other elements.
• Such an expression was found by Rydberg and
Ritz, known as the Rydberg-Ritz formula, gives
the reciprocal wavelength as:
•
 1
1 
 R  2  2 

 n2 n1 
1
•
Rydberg-Ritz formula:
 1
1 
 R  2  2 

 n2 n1 
1
where:
1.
2.
3.
•
n1 and n2 are integers with n1 > n2
R is the Rydberg constant, which is the same for
all spectral series of the same element and varies
only slightly in a regular way from element to
element.
For hydrogen, RH = 1.096776 x 107/m.
Rydberg-Ritz formula gives the wavelengths
for all the lines in the spectra of hydrogen as
well as alkali metals like lithium and sodium.
•
According to Bohr’s model, the electron of the
hydrogen atom moves under the influence of
the Coulomb attraction to the positive nucleus
according to classical mechanics, which
predicts circular or elliptical orbits with the
force center at one focus.
Energy in a Circular Orbit
For an electron (-e) moving in a
circular orbit of radius r about
a positive charge Z·e such as
the nucleus of a hydrogen
atom (Z = 1).
• The total energy of the electron
can be related to the radius of
the orbit.
•
•
The potential energy of the electron of charge –e
at a distance r from a positive charge Z·e is:
k  q1  q2 k   Z  e    e  k  Z  e2
U


r
r
r
•
where k is the Coulomb constant.
Kinetic energy K can be expressed as a function
of r by using Newton’s second law ΣF = m·a.
• Setting the Coulomb attractive force equal to
the mass times the centripetal acceleration
gives:
•
k Z e
mv

2
r
r
2
2
•
Then:
1
1 k  Z e
2
K  mv  
2
2
r
2
The kinetic energy varies inversely with r like
the potential energy.
• The magnitude of the potential energy is twice
that of the kinetic energy; U = -2·K.
• This is a general result in 1/r2 force fields and
also holds for circular orbits in a gravitational
field.
• The total energy is the sum of the kinetic energy
and the potential energy:
•
1 k Z e k Z e
1 k Z e
E  K U  

 
2
r
r
2
r
2
2
2
Mechanical stability is achieved because the
Coulomb attractive force provides the
centripetal force necessary for the electron to
remain in orbit.
• Classical electromagnetic theory says that such
an atom would be unstable electrically because
the electron must accelerate when moving in a
circle and therefore radiate electromagnetic
energy of frequency equal to its motion.
• According to classical theory, such an atom
would quickly collapse, the electron spiraling
into the nucleus as it radiates away its energy.
• Bohr solved the difficulty of the collapsing atom
by postulating that only certain orbits, called
stationary states, are allowed, and in these
orbits the electron does not radiate.
•
An atom radiates only when the electron makes
a transition from one allowed orbit (stationary
orbit) to another.
• Bohr’s first postulate: nonradiating orbits
The electron in the hydrogen atom can move only
in certain nonradiating, circular orbits called
stationary orbits.
• Bohr’s second postulate relates the frequency of
radiation to the energies of the stationary
states.
• If Ei and Ef are the initial and final energies of
the atom, the frequency of the emitted radiation
during a transition is given by:
•
f 
Ei  E f
h
•
Bohr’s second postulate: photon frequency from energy
conservation
f 
•
•
Ei  E f
h
where h is Planck’s constant
This postulate is equivalent to the assumption of
conservation of energy with the emission of a photon of
energy h·f
f 
•
Ei  E f
h
1 k  Z e
 
2
h
2
1 1
  
 r2 r1 
where r1 and r2 are the radii of the initial and final orbits.
•
To obtain the frequencies implied by the RydbergRitz formula, the radii of stable orbits must be
proportional to the squares of integers.
 1
1 
f   c  R  2  2 

 n2 n1 
c
•
Bohr looked for a quantum condition for the radii
of the stable orbits that would yield this result
and found he could obtain it if he postulated that
the angular momentum of the electron in a stable
orbit equals an integer times ħ (h bar; Planck’s
constant divided by 2·π).
•
Bohr’s third postulate: quantized angular
momentum
nh
mvr 
 n
2 
where n = 1, 2, …
ħ = h = 1.055 x 10-34 J·s = 6.582 x 10-16 eV·s
2 
• From the Coulomb attractive force equal to the
mass times the centripetal acceleration
equation, we can solve for v:
2
2
2
2
k Z e
mv

2
r
r
k  Z e
v 
mr
•
Combining the angular momentum equation
and the velocity equation:
k  Z e
v 
mr
n
v
mr
nh
mvr 
 n
2 
n
v
mr
k  Z e
 n 

 
mr
 mr 
2
2
n 
2
 k Z e
mr
2
2
2
2
n 
k  Z e

2
2
m r
mr
2
2
n 
r
2
mk  Z e
2
2
2
•
Solving for the radius of the Bohr atom:
2
ao
rn 
n 
2
mk Z e
Z
2
•
2
where ao is called the Bohr radius.
ao 
2
mk e
2
 0.0529 nm
Energy Levels
•
The total energy of the electron in the hydrogen
atom is related to the radius of the circular
orbit by:
2
1 k  Z e
E 
2
r
•
Substituting the quantized values of r given by:
ao
rn 
Z
2
•
We get:
1 k  Z  e2
1 k  Z  e2
1 k  Z  e2
En   
  2
 
2
r
2 n  ao
2 n2  2
2
mk  Z e
Z
m  k 2  e4 Z 2
2 Eo
En  
 2  Z  2
2
2
n
n
•
where n = 1, 2, …
mk e
1 k e
Eo 
 
 13.6 eV
2
2
2 ao
2
4
2
The energies En with Z = 1 are the quantized
allowed energies for the hydrogen atom.
• Transitions between these allowed energies
result in the emission or absorption of a photon
whose frequency is given by
•
f 
and whose wavelength is
•
Ei  E f
h
c
hc
 
f Ei  E f
It is convenient to have the value of h·c in
electron-volt·nanometers: h·c = 1240 eV·nm
Since the energies are
quantized, the frequencies and
wavelengths of the radiation
emitted by the hydrogen atom
are quantized in agreement
with the observed line
spectrum.
• The figure shows the energylevel diagram for hydrogen.
• The energy of the hydrogen
atom in the ground state is
E1 = 13.6 eV.
• As n approaches infinity the
energy approaches zero, the
highest energy state.
•
The process of removing an
electron from an atom is called
ionization, and the energy
required to remove the electron
is the ionization energy.
• The ionization energy of the
hydrogen atom, which is the
binding energy of the atom, is
13.6 eV.
• Transitions from a higher to a
lower state are indicated in the
figure.
• Transitions from n2 = 2 to and
n1 = 3, 4, 5, … are known as the
Balmer series.
•
Transitions from n2 = 3 to and
n1 = 4, 5, 6, … are known as
the Paschen series.
• Lyman found the series
corresponding to n2 = 1.
• Brackett found the series
corresponding to n2 = 4.
• Pfund found the series
corresponding to n2 = 5.
• The Balmer series lies in the
visible portion of the
electromagnetic spectrum.
•
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•
•
•
When electrons are excited, they
absorb energy and move to a higher
energy level (upward gold arrow).
When they emit light, they move to a
lower energy level.
Blue light (blue arrow) is produced
when electrons move from level 5 to
level 2.
Green light (green arrow) is produced
when electrons move from level 4 to
level 2.
Red light (red arrow) is produced when
electrons move from level 3 to level 2.
Problem example: determine
the energy and wavelength of
the line with the longest
wavelength in the Lyman series.
• The Lyman series corresponds
to transitions ending at the
ground-state energy Ef = E1 =
-13.6 eV.
• Since λ varies inversely with
energy, the transition with the
longest wavelength is the
transition with the lowest
energy, which is that from the
first excited state n = 2 to the
ground state n = 1.
•
•
The energy of the photon is the difference in the
energies of the initial and final atomic state:
E  Ei  E f  E2  E1
13.6 eV 13.6 eV
E

2
2
2
1
E  3.40 eV  13.6 eV
E  10.2 eV
•
The wavelength of the photon is:
hc
1240 eV  nm


 121.6 nm
E2  E1
10.2 eV