Section 1A
Section 1A
Structure of the Atom
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Page 1
Structure of the Atom
Atoms and Sub-atomic Particles
One of the oldest ideas in science is that matter can be divided and further divided until the
smallest possible particles of matter are obtained. This idea was put forward by the Greek
philosopher, Dermocritus, in 400 B.C. He called the particles atoms (Greek: atomos, indivisible).
In 1808, the British chemist John Dalton, formulated his Atomic Theory. He postulated that all
matter consists of atoms, minute particles which cannot be created, destroyed or split. He
theorized that all the atoms of an element are identical in every aspect, for example, their size
and mass. These particles were indivisible and remained unchanged during a chemical
reaction.
The idea of indivisible atom was shown to be wrong by the works of a number of scientists in
the late nineteenth century.
The electron was the first sub-atomic particle to be identified. In 1897, Sir Joseph John
Thomson discovered the electron through his work on cathode rays. He also determined the
charge-to-mass (e/m) ratio of electron.
In 1909, Milikan, through his famous oil-drop experiment, determined the charge e on an
electron. Combined with Thomson's value for e/m ratio, it was then possible to calculate the
mass
m of an electron. Nowadays, the accepted values for these are
e
me
=
=
1.602 x 10
9.110 x 10
-19
-28
C
g
The Nucleus - Rutherford's Scattering Experiment
In 1909, Lord Rutherford and his colleagues, Geiger and Marsden, performed the goldfoil
scattering experiment.
The Scattering Experiment
Fast-moving alpha particles were directed to hit a piece of gold foil which was only a few
hundred atoms thick.
Result:
1.
Most of the alpha particles passed through without changing direction.
2.
A few were deflected through large angles.
3.
Some even rebounded back.
to vacuum pump
thin gold foil
movable fluorescent screen
lead shield
moveable microscope
alpha
particle
source
deflected beam of
alpha particles
rebounded beam of alpha particles
Section 1A
Structure of the Atom
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Page 2
Explanation:
Rutherford proposed that an atom consisted of a central, minute nucleus where the
mass and positive charges of the whole atom were concentrated; the electrons occupied the
remaining space in the atom, revolving around the nucleus.
alpha particles
gold atoms
Summary
i]
ii]
iii]
of
Rutherford's
hypothesis:
In the centre of the atom is a very small dense region called the nucleus.
The electrons move around the nucleus at great speed. The number of electrons equals the
number of protons.
Most of the atom is empty space.
The neutron was predicted by Rutherford in 1920 to account for the difference between atomic
mass and atomic number. It was finally detected experimentally in 1932 by Sir James
Chadwick.
Therefore, an atom was found to consist of three fundamental sub-atomic particles, these are
electrons which are negatively
charged, protons which are positively
charged and
neutrons which are uncharged.
Particle
Relative mass
Charge
Proton
1
+1
Neutron
1
0
1
1836
Electron
-1
Radioactivity
A number of elements have atoms which are unstable and split up to form smaller atoms. This
process is called radioactive decay.
Three types of radiation are given off by radioactive substances:
i)
-particles:
These are helium nuclei (He2+) in nature, i.e. positively charged. They have the lowest
penetrating power.
ii)
-particles:
These are electrons, i.e. negatively charged. Their penetrating power is higher than that
of -particles.
iii)
-rays
These are uncharged rays, similar to
able to pass through 0.1 m of metal.
X-rays. They have high penetrating power, being
Equations for nuclear reactions
A
The symbol
is used to represent the nuclide X with atomic number Z and mass
ZX
number A. The symbols used to represent the various types of radiation or particles are listed
below:
Particles /
Radiation
alpha particle
Symbols
4
2
He or 
beta particle
0
-1
proton
1
1
e or 
p or 11 H
neutron
1
0
gamma radiation

n

when
Mass no. (A)
Atomic no.(Z)
Mass no. (A)
Atomic no.(Z)
Mass no. (A)
Atomic no.(Z)
Mass no. (A)
Atomic no.(Z)
Mass no. (A)
Atomic no.(Z)
emitted
____ by ____ .
____ by ____ .
_______________ .
____ by ____ .
____ by ____ .
____ by ____ .
____ by ____ .
______________ .
______________ .
______________ .
Section 1A
Example:
Structure of the Atom
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Page 3
State what type of emission occurs at each of the following stages in the following
decay chain:
228
88
( a)
( b)
(c )
Ra 

 228

 228

 224
89 Ac 
90Th 
88Ra
(a)
(b)
(c)
.
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1.
Q
P
-
+
R
S
The above diagram shows how alpha and beta particles and gamma rays, emitted from
a radioactive source, S, behave in an electric field.
a)
Use the information on the diagram to identify the type of emission present at
Q, and R.
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P,
b)
By what other means could a similar pattern of deflection of the three types of
radiation be caused.
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Uses of Radioactivity
i)
Leak detection:
A radioactive source is introduced into systems like storage tanks and buried pipelines,
and a detector is used to locate the position of leakage.
ii)
Radiotherapy:
-rays is used in cancer treatment.
iii)
Carbon-14 dating:
By comparing the concentration of carbon-14 in the archaeological(考古學的) specimen
and similar materials at present time, the age of the specimen can be estimated.
iv)
Nuclear power:
Nuclear reactions result in the release of a vast quantity of energy, which is used to
generate electricity.
v)
As tracers:
Radioactive isotopes can be used as tracers to study metabolism(新陳代謝) in living
organisms.
.
2. a)
b)
234
An isotope of uranium is represented by the symbol
92 U . What is the
significance of the two numbers ?
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Identify A, B, C, D and E
234
230
i)

A +
92 U
90Th
ii)
239
92
U 
235
92
iii)
U 
B
+
239
93
C
+
in the following equations:
Np
235
92
iv)
238
92
U +
2
1
H

239
92
v)
235
92
U +
1
0
n

95
42
U
U + D
Mo +
139
57
La + 2
1
0
n + 7E
Section 1A
Structure of the Atom
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Page 4
The Mass of an Atom
Atomic Number
This is the number of protons in the nucleus of an atom. It is given the symbol
represents:
i]
the number of protons in the nucleus;
ii] the number of electrons in a neutral atoms; and
iii] ordinal number in the periodic table.
Number
This is the number of protons and neutrons in a nucleus. It is given the symbol
This is always a whole number (c.f. atomic mass).
Z . It
Mass
A .
Nuclides
This is any nuclear species of given mass number (A) and atomic number (Z), for
example:
16
12
; 6C .
8O
Isotopes
These are atoms of an elements which differ only in the number of neutrons they
contain. They thus have the same atomic number but different mass number.
12
14
Examples:
and 6 C
6C
Isotopes Abundance
Most elements consist of mixtures of isotopes. The abundance of each in the mixture is
called its isotopic abundance. For example, silicon occurs in naturally occurring compounds as
29
30
92.28% 28 Si ;
4.67%
Si and 3.05%
Si.
_____________________________________________________________________________________
.
3. The table below shows the mass number and number of neutrons in the nucleus, for
four atoms, W, X, Y and Z.
Mass number
Neutrons in nucleus
W
36
18
X
39
20
Y
40
21
Z
40
22
a)
Write down the atomic numbers of the four atoms.
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__________________________________________________________________
b)
Which of the four atoms are isotopes of the same element ?
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Relative Isotopic Mass
This is the ratio of the mass of a particular atom to
12 C.
nuclide
That is ,
Relative isotopic mass=
1
12
of the mass of the atom of the
mass of a partic ular atom
12
1
C
12  mass of n uclide
Relative Atomic Mass ( Ar )
This is the weighted average of the relative isotopic masses of an element. It is
calculated by multiplying the relative isotopic mass of each isotope by its fractional abundance
and adding all these values together.
35
37
Example 1: Chlorine consists of 2 natural isotopes,
Cl and
Cl, with percentage
abundance of 75.4% and 24.6% respectively. Calculate the atomic mass of
chlorine.
Section 1A
.
4. Neon in the air contains
mass of neon.
(20.2)
90% of
20
Ne
and
10%
Structure of the Atom
of
22
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Page 5
Ne. Calculate the atomic
Determination of Atomic Mass by Mass Spectrometer
negatively charged
accelerating grids
electromagnet
vaporization
chamber
ionization
chamber
to
vacuum
pump
detector
There are FIVE main operations performed by the mass spectrometer:
1]
The sample of the material to be analysed, which may be an element or a compound, is
injected and heated to vapourized in the vaporization chamber. The vaporized sample
then passes into the ionization chamber.
2]
Electrons are knocked out from the atoms of the element by bombardment with highenergy electrons produced by the hot-filament (thermionic emission). This cause
ionization of the atomic sample and result in the production of positively charged ions
which are mainly singly charged.
X (g) + e 
X + (g) + 2 e fast
slow
3]
These positively charged ions are then acted upon by a 'velocity selector' (an electric field
and magnetic field applied perpendicular to the pathway of the ions) which select those
ions with the same velocity (though may differ in mass) to enter the circular chamber.
4]
They were then deflected along a circular path by another known magnetic field applied
perpendicular to the direction of movement of the ions. The lighter the positive ions, the
greater is the deflection. Hence, at a given electric and magnetic field strength, only ions of
a particular mass/charge ratio can hit the ion detector.
5]
By varying the accelerating electric field or the deflecting magnetic field, ions of any
mass/charge ratio can be brought to the ion detector. A mass spectrum can then be traced
out by a recorder. Also, the relative abundance of each isotopes can be found from the
relative magnitude of the signal (current) produced in the detector.
In the mass spectrum of an element, the peaks can give information about various isotopes of
the element. Whereas in the mass spectrum of a compound, the peak with the highest m/e
ratio will most likely correspond to the ‘molecular ion’, i.e. the molecule which has lost only a
single electron. In most mass spectra, the values of the m/e ratio can be converted to the
relative masses of the particles if the charges on the ions are taken to be one.
Therefore from the measurement of the mass spectrometer, two data about the sample element
can be obtained:
[i]
the isotopic masses of each isotope ; and
[ii]
the relative abundance of each isotopes in the sample.
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Example:
Section 1A Structure of the Atom
The following diagram is the mass spectrum of the element lead.
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Page 6
de te ctor
r e adi ng
204
206
208
(i)
Give a full interpretation of this trace.
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__________________________________________________________________
__________________________________________________________________
(ii)
Calculate the atomic mass of lead (correct to 4 sig. figures ).
.
5. The mass spectrum of neon consists of three lines corresponding to mass/charge ratios
of
20,
21
and
22
with relative intensities of
0.910;
0.0026;
0.088
respectively. Explain the significance of these data and, hence, calculate the relative
atomic mass of neon.
6. The isotopic composition of the gas radon was investigated using a mass spectrometer,
part of which is shown in Fig. 4 below.
magnetic field
perpendicular to
plane of
diagram
Y
X
detector
a)
Radon has two isotopes,
i)
222
86
R n and
220
86
Rn.
Write the formulae of the two singly-charged ions that would form in the
instrument.
ii)
State which ion will follow the path marked
X on the diagram.
__________________________________________________________________
-1
b)
Mention two adjustments that could be made to the instrument to bring the ions
from Y on to the detector.
__________________________________________________________________
__________________________________________________________________
c)
If Rn
ions were to form in the instrument, would you expect them to be
deflected less than or more than the ions at X and Y ?
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2+
Give an account of the use of a mass spectrometer for determining the relative
masses of particles.
(6marks)
(902A3(a))
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Section 1A Structure of the Atom
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2 Name the particles
charge and mass.
229
i)

90Th
ii)
225
88
X
and
X
+
Y
225
88
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Page 7
in the following nuclear reactions, and give their
X
Ra
is
225

Y + 89 Ac
Y is
Compare the action of a magnetic field upon the paths of
Ra
X
and
Y.
(4 marks)
(93IA1(a))
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Answer
1.
to Exercise
(902A3a)
The modern method of
the mass spectrometer.
determining
relative
atomic
B
4
sample
the
sample
charged
ions( 21
element)
are
ions
travel along
-
magnetic
into
uses
@
@
1
2
1
2
marks
marks
is
bombarded
by
electrons( 21 )
to
form
positively
).
ions
focused
2.
(an
particles
magnetic field
- the
B.
- the
the
parts
explain
electron beam
-
of
ion detector
electric field
A
masses
accelerated( 21 )
field
the
until
is
ion -
by
the
electric
deflected( 21 )
adjusted( 21 )
detector( 21
so
by
that
field( 21 )
the
ions
between
magnetic
of
a
plates
A
and
mass
are
field( 21 ).
particular
).
- knowing the magnetic field strength , accelerating voltage
of the path , the atomic mass can be calculated.(1)
and
-
( m / e )
the mass spectrometer
ionized species.(1)
responds
to
the mass - to - charge
the
radius
of
the
(931A1a)
(i)
(ii)
X
Y
is
4
2
is
0
1
He / helium nucleus /  - particles
e / fast moving electron /  - particles
The path of  and  - particles are deflected /
but they are deflected in opposite directions (1).
or
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x 
x particle
x or X
x
x
x
 - particle
or Y
bent
(1)
direction (1)
(1)
(1)
bent (1) by
a
magnetic
field