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ATOMIC STRUCTURE
The atom consists of two parts:
The positively charged nucleus containing most
the atom’s mass.
Negatively charged electrons found in the
empty space around the nucleus.
The Rutherford Scattering Experiment
In 1910, Ernest Rutherford (1871 - 1937) and an
associate, Hans Geiger (1882 - 1945) and a student,
Ernest Marsden (1889 - 1970) performed an
experiment in which they bombarded various metal
foils with alpha particles. They observed that most of
the alpha particles passed through the metal foil with
little or no deflection. Some of the alpha particles
actually bounced back towards the alpha source.
An Enlarged View of Rutherford’s Experiment
As a result of Rutherford’s experiment, scientists were
able to conclude two important features of the atom:
1. The atom is mostly empty space.
2. The atom’s positive charge is concentrated in a
central core within the atom.
Some Properties of the Gold Atom
Gold Atom
Gold Nucleus
An electron
Mass
(in grams)
Diameter
(in meters)
3.27 x 10-22 g.
2.9 x 10-10 m.
-3.27 x 10-22 g.
-1. x 10-14 m.
9.02 x 10-28 g.
3.8 x 10-15 m.
What can you conclude about the relative size of the
gold atom, nucleus, and an electron?
If an atom were the size of a football field
(- 100 meters), the nucleus would be about
3 millimeters in diameter (the size of a BB) and an
electron would be roughly half the size of the nucleus.
That is, the atom is mostly empty space.
In other words, when Rutherford fired alpha particles
at the gold foil, it would be like trying to hit a BB in
the middle of a football field with a BB gun from many
miles away, nearly impossible.
Rutherford also observed that only one in twenty
thousand alpha particles were significantly deflected as
they passed through the gold foil. Explain
Rutherford’s observation in terms of the relative size of
the atom and the nucleus.
What comparison can you make about the mass of the
nucleus compared to the mass of the atom?
Almost all of the mass of an atom is contained in the
nucleus.
The Discovery of an Electron
Early scientists knew about charges and in fact,
Benjamin Franklin gave the names positive and
negative to the two different charges. But Michael
Faraday (1791-1867) discovered electrons as “cathode
rays” by applying a high voltage to the ends of a
cathode ray tube. The electrons emitted provide an
image of their path when they strike a fluorescence
zinc sulfide screen.
J.J. Thomson (1856-1940) used a specially designed
cathode ray tube to apply both electric and magnetic
fields simultaneously to the beam of cathode rays. By
balancing the effect of the electric field against that of
the magnetic field he was able to calculate the charge
to mass (e/m) ratio for the particles in the beam.
Thomson’s experiments also demonstrated that
electrons had a negative charge. In addition, he
obtained the same mass to charge ratio with twenty
different metals. These results suggested that electrons
are present in atoms of all elements.
The Millikan Oil Drop Experiment
American Physicist Robert Millikan (1868 - 1953)
performed an experiment in which he sprayed oil
droplets into a chamber from an atomizer. The oil
droplets were allowed to settle slowly towards the
bottom of the chamber. Millikan had two charged
plates on the top and bottom of the chamber which
allowed him to control the rate at which the oil droplets
fell. By varying the voltage on the plates, he was able
to just stop the oil droplets from falling.
As a result of his experiment, Millikan was able to
calculate the following charges on the oil droplets?
-3.2 x 10-19 C
-1.6 x 10-18 C
-4.8 x 10-19 C
-3.2 x 10-18 C
-1.6 x 10-19 C
-9.6 x 10-19 C
-6.4 x 10-19 C
-4.0 x 10-18 C
-8.0 x 10-19 C
-4.8 x 10-18 C
What can you conclude about the charges which
Millikan obtained?
They are all integer multiples of -1.6 x 10-19 C. Since
the charge on an electron is the smallest possible
charge, it is called an “elementary charge.”
SOME IMPORTANT ATOMIC STRUCTURE
TERMS
Atomic number - The atomic number equals the
number of protons in the nucleus. The atomic
number determines the element. Every atom of a
given element has exactly the same number of
protons.
Nucleons - particles in the nucleus (protons plus
neutrons)
Nuclear charge - The charge on the nucleus (the
nuclear charge equals the number of protons)
Mass number - The mass number (an integer) is the
sum of the neutrons plus protons. The mass
number also equals the number of nucleons.
Isotopes - Isotopes are atoms of the same element
(same atomic number) with a different number of
neutrons.
Nuclide - A particular type of atom having a
characteristic nucleus.
Ion - An ion is a charged particle.
IMPORTANT SUBATOMIC PARTICLES
Particle
Symbol
Charge
Mass
(Atomic mass
units, amu)
How Many?
Proton
p
+1
1.007276
Atomic Number
Neutron
n
0
1.008665
Mass Number Atomic Number
Electron
e
-1
0.0005486
(
the mass Number of Protons
of a proton)
Give the number of subatomic particles in each of the
following atoms:
This particle is called carbon - 14.
Protons:
_______
Neutrons:
_______
Electrons:
_______
Protons:
_______
Neutrons:
_______
Electrons:
_______
Nuclear charge: ______
Number of nucleons: ______
Mass number: ______
What is the name of this particle? ______________
Uranium - 238
Determine the number of particles based on the symbol
below:
Protons:
_______
Neutrons:
_______
Electrons:
_______
Number of nucleons: _____
Nuclear charge: ______
Name this particle.
Give each of the following quantities in an atom of
boron - 11?
Protons: _____
Electrons: _____
Neutrons: _____
Nuclear charge: _____
Mass number: _____
Give the symbol for the nucleus of boron -11.
A certain particle has 11 protons, 13 neutrons, and 10
electrons.
1. What is the name of the element?
2. What is the mass number of this particle?
3. How many nucleons are found in this particle?
4. What is the overall charge on this particle?
5. Is this particle an atom or an ion?
6. Give the symbol for the nucleus of this particle.
Answer each of the following questions based on the
following symbol:
1. What is the mass number of this particle?
2. How many protons, neutrons, and electrons are
found in this particle?
3. What is the nuclear charge on this element?
Answer each of the following questions based on the
following symbol:
1. What is the mass number of this particle?
2. How many protons, neutrons, and electrons are
found in this particle?
3. What is the nuclear charge on this element?
Isotopes of Hydrogen
Hydrogen
Deuterium
(Heavy Hydrogen)
Tritium
Give the number of protons, neutrons, and electrons for
each of the isotopes of hydrogen listed above.
Which of the following nuclides are isotopes?
(a)
(b)
(c)
(d)
(e)
(f)
Properties of Heavy Water and Ordinary Water
Ordinary Water
1
H2O
Heavy Water
2
H2O (D2O)
Molecular mass
18 grams/mol
20 grams/mol
Density @25°C
0.997 g/cc
1.105 g/cc
Temperature at
max. density
4.0 °C
11.6 °C
Melting point
0.00 °C
3.802 °C
Boiling point
100.0 °C
101.42 °C
Heat of fusion
1436 cal/mol
1510 cal/mol
10,484 cal/mol
10,743 cal/mol
Dielectric
constant
81.5
80.7
Refractive index
1.333
1.328
Surface tension
72.75 dynes/cm
72.8 dynes/cm
13.10 mpoise
16.85 mpoise
Property
Heat of
vaporization
Viscosity @25°C
Units of Atomic Mass
An atomic mass unit (amu or u) is defined as
exactly
the mass of a carbon-12 atom.
or the mass of a carbon-12 atom is exactly
12.000 . . . amu.
The atomic masses of all the other elements are
determined by comparing their masses with
carbon-12.
For example, the mass of an average hydrogen
atom is 8.400% the mass of a carbon-12 atom.
Thus, the mass of an average hydrogen atom is
0.08400 x 12 amu = 1.008 amu.
Likewise, an average atom of magnesium has a
mass which is 2.0254 times the mass of a carbon12 atom. Therefore, the mass of an average
magnesium atom is 2.0254 x 12 amu = 24.305 amu.
COMPARISON OF ATOMIC MASS SCALES
Based on
Oxygen =
exactly 16
Based on
Carbon-12 =
exactly 12
Oxygen-16
15.99560 amu
15.99491 amu
Oxygen
16 amu exactly
15.9999 amu
Carbon-12
12.00052 amu
12 amu exactly
Carbon
12.011 amu
12.010 amu
Silver
107.873 amu
107.868 amu
Determining the Mass of Atoms
The mass of atoms is determined by use of a mass
spectrograph. The sample of matter in ionized in a
vacuum chamber. The resulting positive ions are then
accelerated by means of a negatively charged screen.
Most of the ions pass through the screen, though a slit
to focus the ion beam, and then into a magnetic field.
By varying the accelerating voltage, the ions can be
made to strike the detector. Velocities up to 150,000
miles/sec can be obtained with a voltage of 400-4000
volts.
Atomic Mass of an Element
The atomic mass of any element is the weighed
average of its naturally occurring isotopes.
To calculate the atomic mass of the element, a
weighed average, multiply the mass of each isotope
by its percent abundance divided by 100, and find
the sum of these values.
Average atomic mass =
Find the atomic mass of boron if there are two
naturally occurring isotopes, 10B and 11B. 19.6% of the
atoms have a mass of 10.01292 amu and 80.4% of the
atoms have a mass of 11.00931 amu.
A sample of element X contains 90. percent 35X atoms,
8.0 percent 37X atoms, and 2.0 percent 38X atoms. The
average isotopic mass is closest to
(1) 32
(3) 37
(2) 35
(4) 38
Element X has two isotopes. If 72.0% of the element
has an isotopic mass of 84.9 atomic mass units, and
28.0% of the element has an isotopic mass of 87.0
atomic mass units, the average atomic mass of element
X is numerically equal to
(1)
(2)
(3)
(4)
If 75.0% of the isotopes of an element have a mass of
35.0 amu and 25.0% of the isotopes have a mass of
37.0 amu, what is the atomic mass of the element?
(1) 35.0 amu
(3) 36.0 amu
(2) 35.5 amu
(4) 37.0 amu
Given the following table containing the isotopic
abundance of magnesium, calculate the atomic mass
of an average atom of magnesium.
Percent
Abundance
Isotopic mass
24
78.70%
23.985 amu
25
10.13%
24.985 amu
26
11.17%
25.983 amu
Isotope
Mg
Mg
Mg
There are two naturally occurring isotopes of
copper, copper-63 and copper-65. Calculate the
atomic mass for an average atom of copper if
69.17% of the atoms have a mass of 62.9396
amu and 30.83% of the atoms have a mass of
64.9278 amu.
A naturally occurring sample of an element
contains 20.% of the atoms with a mass of
120 amu and 80.% of the atoms with a mass of
110 amu. Calculate the average atomic mass of
the element.
A sample of a naturally element was found to
contain 60.0% of its atoms have an isotopic
mass of 70.0 amu and 40.0% of its atoms have
an isotopic mass of 65.0 amu. Calculate the
average atomic mass of the element.
Converting Atomic Mass Units to Grams
The SI unit for the amount of substance is the mole
(mol), which is the amount of substance that
contains as many elementary particles (atoms,
molecules, ions, etc.) as there are atoms in exactly
12 grams of the carbon-12 isotope. The term mole is
the name of a number of items similar to a pair (2
items), a dozen (12 items), and a gross (144 items).
The currently accepted value for one mole is
1 mole = 6.022045 x 1023 particles
This number is called Avogadro’s number, N0.
Likewise, the mass of 6.022 x 1023 atoms of the
carbon-12 isotope is 12.00 grams.
Similarly, the mass of one Avogadro’s number of
any particle in grams is equal to its mass in atomic
mass units (amu) x 6.022 x 1023.
For example, the mass of 6.022 x 1023 atoms of an
average oxygen atom is 15.99 grams.