Download The Atom Hypothesis

The Atom Hypothesis
As long ago as 60 BC, Lucretius reasoned that all matter must
be composed some atoms – smallest indivisible units that
have a given character in chemical and physical properties.
With the advent of chemistry and the discovery of ever-more
elements, in 1869 Mendeleev realized that these elements
could be ordered in a periodic table according to
increasing mass and recurring chemical properties. This
recursion was a clue that all atoms are made of a few
more-fundamental building blocks. The building blocks
are arranged in different combinations to make the various
elements.
Mendeleev’s Periodic Table in 1869
Reihen
Gruppe
I
Gruppe
II
Gruppe
III
Gruppe
IV
Gruppe
V
Gruppe
VI
Gruppe
VII
1
H=1
2
Li=7
Be=9,4
B=11
C=12
N=14
O=16
F=19
3
Na=23
Mg=24
Al=27,3 Si=28
P=31
S=32
Cl=35,5
4
K=39
Ca=40
44?
Ti=48
V=51
Cr=52
Mn=55
5
Cu=63
Zn=65
68?
72?
As=75
Se=78
Br=80
6
Rb=85
Sr=87
Yt=88
Zr=90
Nb=94
Mo=96
100?
7
Ag=108 Cd=112
In=113
Sn=118
Sb=122
Te=125
I=127
8
Cs=133
Di=138
Ce=140
10
Er=178
La=180
Ta=182
W=184
11
Au=199 Hg=200 Tl=204
Pb=207
Bi=208
Ba=137
9
Th=231
U=240
Mendeleev’s Original Periodic Table, 1869
``I saw in a dream a table where all the
elements fell into place as required.
Awakening, I immediately wrote it down on a
piece of paper. Only in one place did a
correction later seem necessary."
Mendeleev’s Postulates
1.The elements, if arranged according to their atomic weights, exhibit an apparent periodicity of
properties.
2.Elements which are similar as regards their chemical properties have atomic weights which
are either of nearly the same value (eg. Pt, Ir, Os) or which increase regularly (eg. K, Ru, Cs).
3.The arrangement of the elements, or of groups of elements in the order of their atomic weights,
corresponds to their so-called valencies, as well as, to some extent, to their distinctive chemical
properties; as is apparent among other series in that of Li, Be, Ba, C, N, O, and Sn.
4.The elements which are the most widely diffused have small atomic weights.
5.The magnitude of the atomic weight determines the character of the element, just as the
magnitude of the molecule determines the character of a compound body.
6.We must expect the discovery of many as yet unknown elements-for example, elements
analogous to aluminum and silicon- whose atomic weight would be between 65 and 75.
7.The atomic weight of an element may sometimes be amended by a knowledge of those of its
contiguous elements. Thus the atomic weight of tellurium must lie between 123 and 126, and
cannot be 128.
8.Certain characteristic properties of elements can be foretold from their atomic weights. (18)
The Periodic Table Today
= predicted by Mendeleev, discovered later
The Constituents of the Atom
From stoichiometry of chemical reactions, it was known that elements
combine only in fixed integer ratios their atoms must form bonds to one
another, and the character of the bonds must be an inate property of each
element.
The size of an atom (~1 Å) was obtained from electrochemistry. The
electron was discovered, from radioactivity ( rays) and from emission in
cathodes ray tubes. Its charge e and mass m were measured. In
electrochemical reactions, the number of electronic charges needed to react
a molecular weight of a substance was measured Avogadro’s number.
But the electron mass m was tiny compared to that of an atom.
Conclusion: there must be positively charged particles (nucleons) within
each atom, in equal number to the electrons so that the atom is neutral. The
positive particles must be much heavier than the electron. The  particle of
radioactivity is the positive nucleus of a helium atom.
What is the structure of the
nucleons and electrons in an atom?
Two possible pictures emerged:
•The nuclear atom:
positive nucleus surrounded
by “solar system” of electrons
•The “plum pudding” atom:
uniform distribution of
nucleons and electrons
Which picture is right?
Rutherford scattering
In 1911, Ernest Rutherford set out to probe the charge
distribution within the atom by scattering  particles from a
thin foil of gold.
Plum pudding: charge distribution is everywhere locally
balanced – little force, little scattering.
Nuclear atom: when  happens to make a near-head-on
collision, it can penetrate within electron cloud – strong force,
large-angle scattering.
There were troubles with either picture!
Nuclear atom: What keeps the + charged nucleons bound in a
small nucleus, when their Coulomb repulsion pushes them apart??
Plum pudding: How can the atoms of a substance retain their
identity when the atoms are pushed together in a solid? Why don’t
they just coalesce into a larger atom of another substance??
Rutherford scattering
Rutherford set out to measure the distribution of scattering angle
for  particles, produced by radioactive decay from Polonium,
scattered as they pass through a thin gold foil.
Rutherford’s students observed the scattered ’s on a phosphor
screen (PbS film on glass, similar to the screen of a television):

distance of closest approach
kinetic energy of incident : T = ½ mv2 ~ 4 MeV
Potential energy when  (z=2) is a distance r from
nucleus:
2Ze 2
U
40 r
1
In a head-on collision,  would come to a stop at a
distance d where T = U:
d min
1 2Ze 2 (9 x109 )  2  79  (1.6 x1019 )
14



5
.
7
x
10
m  57 F
6
40 U
4 x10
Let’s calculate the scattering angle
for an impact parameter d~2dmin:
We will simplify the analysis by using the impulse approximation.
Let the  pass by the atom on a straight line path (no scattering).
Calculate the impulse (change in momentum) from the Coulomb
force acting along the trajectory. Then apply the impulse to obtain
the scattering angle.



Analyze the impulse delivered by the Coulomb force
separately for the directions parallel and perpendicular to
the incident direction.
By symmetry, the parallel impulse delivered before the 
reaches the atom exactly cancels the parallel impulse
delivered after it has past the atom: p‖= 0.
For the perpendicular impulse, calculate the perpendicular
component of the Coulomb force, acting at a distance x
along the path of the :

d

F  F 
2
2
 x d
2 Ze 2
F
,
2
2
40 x  d
1






dx
p   F dt   F
v


 1
2 Ze 2 
d



p   F dx   
2
2 
2
2
4

v
x

d
0
 x  d

 



2 Ze 2
du
p 
40 vd  (1  u 2 ) 3 / 2
u  tan  ,

du  sec 2  ,
 /2

 cosd  sin 
 / 2
4 Ze 2
p 
40 vd
 /2
 / 2
1  u 2  sec 2 
2

dx


We can now calculate the scattering angle:
tan θ 
p
p 
d

 min  0.5
p
2U(d min ) / v
d
  40
The  can only scatter at large angle if the positive
charges in the atom are concentrated on a small distance
scale dmin.
Rutherford’s confirmation of the nuclear model of the
atom raised another, deeper question: what holds the
positive nucleons within such a small nucleus?
Answer: the strong interaction – an immensely strong,
short-range interaction that acts only within the size of
the nucleus.