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Atoms
Atomic Constituents
The work of J.J. Thomson, E. Rutherford and others toward the end of the 19th century and into
the beginning of the 20th century established that atoms consist of a positively charged nucleus
and negatively charged electrons. In a neutral atom, the positive charge on the nucleus exactly
balances the total negative charge of the electrons. All experiments up to the present time
indicate that the electron is an elementary particle, i.e., it has no internal structure. However, the
nucleus was found to have structure. Rutherford discovered that the positive charge on the
nucleus is carried by protons. Each proton has a charge of +1e, where e is the magnitude of the
electron’s charge. The nucleus also contains neutrons, which carry no charge. Modern
elementary particle theory and recent experiments indicate that protons and neutrons are not
elementary, but are made up of quarks. These quarks carry charges that are a fraction of the
electron's charge ( ± 13 e or ± 23 e ). However, free fractional charges have never been observed,
because quarks are always confined within other particles (such as protons and neutrons). Free
single quarks do not exist. Thus, all charge in nature is found to be integer multiples of e.
Protons and neutrons are collectively called nucleons. The mass of a neutron is only slightly
greater than that of a proton. Both are approximately 2000 times as massive as the electron. The
number of protons in the nucleus is called the atomic number (or proton number) Z.1 The number
of neutrons (neutron number) is given the symbol N. The total number of nucleons is the mass
number (or nucleon number) A. Evidently, A = Z + N. The label mass number is appropriate
because the electrons contribute very little mass to the atom. To good approximation, the mass
of the electrons can be neglected.
Symbols for the various atoms are usually written in the form ZA X , where X is the one- or twoletter symbol for the atom. For example, 12
6 C denotes a carbon atom with a mass number of 12.
(The text omits the atomic number from the symbol). Each atom has a different atomic number
Z. This is because the characteristics of a neutral atom are determined by the number of
electrons in the atom. Atoms of the same element (same Z) may have different neutron numbers
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and therefore different mass numbers. Such atoms are called isotopes. Examples: 12
6 C, 6 C,
14
6
C ; 32 He , 42 He .
Atomic Mass Units and the Mole
An atomic mass unit is defined in the following way: one atom of the isotope
12
6 C is defined to have a mass of exactly 12 atomic mass units (12 u). Using this definition we
find that the atomic masses of the other elements are not whole numbers. This is mainly because
of the binding energy associated with nuclei. Since nuclei represent bound states of nucleons,
the mass of a nucleus is less than the total mass of the individual nucleons by an amount
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In a neutral atom Z = number of electrons.
1
Δm = binding energy/ c2. Note that 1 u = 1.66 × 10-27 kg.
One mole (of any substance) contains the same number of objects as there are atoms in exactly
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12 g of the isotope 12
6 C . This number is Avogadro’s number NA = 6.022 × 10 .
Note: NA × (mass of 12
6 C atom) = 12 g. Thus, NA × (12 u) = 12 g, i.e.,
-3
1 u = (1 g)/ NA = (10 kg)/ NA = = 1.661 × 10-27 kg, as given above.
Atomic and Nuclear Sizes
The size of an atom is dictated by the size of the “orbit” of the outermost electrons. Roughly
speaking, this size is of the order of 1 angstrom unit, i.e., 10-10 m. The symbol for the angstrom

unit is A . The typical size of a nucleus on the other hand is of the order of 10-15 m or 1 fermi.
One fermi is equal to 1 femtometer (fm).
Probing Atomic Structure
Discovery of the Electron
With his cathode ray tube2 apparatus, J. J. Thomson was able to measure the charge to mass ratio
of the electron. You will do a similar experiment in the Modern Physics Laboratory course.
Important elements of the apparatus include: (1) cathode – emits of electrons; (2) anode –
accelerates and collimates electrons; (3) deflector plates – a voltage is applied between these
plates to deflect electrons; (4) low-pressure gas – for viewing electrons' path (due to ionization
and light emission)3 or fluorescent screen for seeing position at which the electron beam impacts
the screen.
Consider plates of length L and separation d. If a potential difference V is applied to the plates,
then the electric field E is given by E = V/d. If the electrons have a horizontal velocity u, then
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3
Cathode rays are beams of what we now know as electrons.
The apparatus has to be evacuated anyway, to permit the electrons to move freely.
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the time taken to travel the length of the plates is given by t = L/u. After emerging from the
region between the plates, the y-component of the velocity is given by
⎛ eE ⎞ ⎛ L ⎞ eEL
u y = ayt = ⎜ ⎟ ⎜ ⎟ =
.
(4.1)
⎝ m ⎠ ⎝ u ⎠ mu
Note that the electrons are non-relativistic. Also, the acceleration of an electron due to the
gravity is negligible compared to that due to the electric field. Now,
u
u
eEL
(4.2)
tan θ = y = y =
,
u x u mu 2
where θ is the deflection angle.
Note that in this experiment, the deflection is usually very small; therefore, the approximation
tan θ ≈ θ (where θ is in radians) is a very good one. From Eq. (4.2), we can solve for e/m:
e u2
=
tan θ .
(4.3)
m EL
Thus, e/m could be measured if θ , E, L, and u are known. To find u, Thompson added a

magnetic field B at right angles to the electric field and the path of the electrons. The magnetic
field was adjusted so that the electron beam was no longer deflected. In this case,

  
(4.4)
F = q ( E + u × B) = 0.
Because the electric field, the magnetic field, and the velocity are all perpendicular to one
another, Eq. (4.4) reduces to
E
u= .
(4.5)
B
Substituting in Eq. (4.3) gives
e
E
V tan θ
=
tan θ =
.
(4.6)
2
m LB
dLB 2
(Show picture of Thomson's data.) Thomson's result was low by a factor of approximately 2
compared to the currently accepted value because he did not explicitly account for other sources
of magnetic field in his apparatus. (Note that the derivation of Eq. (4.6) given above is
essentially the solution to Prob. 3.40 in the textbook.)
Millikan Oil Drop Experiment
Knowing e/m, one could find m if e were known. Robert Millikan devised his famous oil drop
experiment in order to obtain the value of e.
[For a brief video clip, see: http://www.youtube.com/watch?v=XMfYHag7Liw]
Tiny droplets of oil (about 5 microns in diameter) were sprayed in the region between two
conducting plates, and were viewed by a telescope.

E=0

Fdrag

mg


Felec = qE

Fdrag

mg
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Without an electric field, the droplets fall under the influence of gravity alone. A drag force
opposes their motion and eventually the drop reaches terminal velocity when
mg = Fdrag = buT 0 ,
(4.7)
where uT0 is the terminal velocity with no electric field and b is the drag constant given by
b = 6πη R
(4.8)
for a sphere. In the expression for the drag constant, η is the viscosity of air and R is the radius
of the droplet. Now, the mass m of the drop is given by
m = ρV = 43 π R 3 ρ ,
(4.9)
where ρ is the density of the oil. Substituting Eqs. (4.8) and (4.9) into Eq. (4.7), one finds that
9uT 0η
(4.10)
.
2g ρ
Hence, the radius of the drop can be calculated and from that, the mass can be obtained. (Typical
size of a drop: ~3 × 10-6 m or 3 µ m .)
R=
In the process of spraying, a droplet may acquire a net charge. Also, x-rays can be used to
bombard the region containing the droplets, thereby producing charged droplets (positive or
negative). If a downward electric field is applied, a negatively charged droplet will rise if
Felec > mg . The drag force now points downward. When terminal velocity is reached, we have
(4.11)
qE = mg + buTE ,
where uTE is the terminal velocity with the electric field in place. But we can eliminate b using
Eq. (4.7):
⎛ mg ⎞
qE = mg + ⎜
⎟ uTE ,
⎝ uT 0 ⎠
i.e.,
⎛ u ⎞
mg ⎜1 + T 0 ⎟
⎝ uTE ⎠ .
q=
(4.12)
E
Thus, by measuring uT 0 and uTE , q can be calculated. Millikan found that the charge on a droplet
was always an integer multiple of 1.6 × 1019 C. From this he deduced the quantization of electric
charge.
Show picture of Millikan’s data (p. 14 Rohlf).
Mention Begeman's role and history.
The Rutherford Nuclear Atom
Play: Mech. Univ. video “Atoms” Ch. 15.
After the discovery and characterization of the electron, Thompson proposed a model of the
atom in which electrons were dispersed like raisins in a uniform distribution of positive charge
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(“plum pudding" model). In order to test this model the following experiment was carried out by
Geiger and Marsden.
Incident
Alpha particles (He-4 nuclei) resulting from radioactive decay
alpha
particles
were used to bombard thin (10-6 m) gold foil, and the angles at
θ
which the alpha particles were scattered were measured. It
would be expected that the scattering (or deflection) angles
would be very small (less than a few degrees) because the
Foil
electrons were too small to affect the alpha particles
significantly. Also, the positive charge is distributed over a
relatively large region, so the repulsive electrical force between the atom's positive charge and
that of the alpha particle is relatively weak. When Geiger and Marsden performed the
experiment, most of the alpha particles were deflected through small angles as expected.
However, very surprisingly, there were a few large-angle scattering events with the scattering
angle > 90º.
On the basis of these results, Rutherford proposed his nuclear model of the atom, in which the
positive charge of the atom is concentrated in an extremely small region at the center of the
atom. The electrons orbit the nucleus at relatively large distances away. Thus, the atom’s
volume is mostly empty space. This would explain why most alpha particles are not scattered.
However, on the rare occasions when an alpha particle approaches a nucleus closely, very large
electrical forces are experienced by the alpha particle, which is then scattered through a large
angle.
b
θ
b
φ
r
Nucleus
[Show: Figs 6.8 & 6.9 & 6.10, p. 155, Krane's Modern Physics]
The trajectory is in general a hyperbola. The scattering rate for each angle θ can be found using
classical mechanics. In polar coordinates, the equation of the hyperbola is
1 1
zZe 2
(4.13)
= sin φ +
(cos φ − 1),
r b
8πε 0b 2 K
where z is atomic number of the scattered particle, Z is the atomic number of nucleus, and K is
the initial kinetic energy of the incoming particle. Now, for r → ∞ , φ → π − θ . Substituting
these values yields
zZ e 2
1
(4.14)
b=
cot θ .
2 K 4πε 0
2
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For foil with n atoms per unit volume and thickness t, the number of atoms per unit area = nt.
Particles that are incident within an area π b 2 centered on the scattering nucleus will be scattered
at angles greater than θ . The fraction f of projectiles that are scattered at angles ≥ θ is given by
(number of scattering centers/unit area) ×π b 2 . Thus,
(4.15)
f = ntπ b 2 .
We wish to find the number of particles per unit area scattered in an annular (ring-shaped) region
between θ and θ + dθ . The fraction of projectiles entering between b and b+db is given by
df = nt 2π bdb.
(4.16)
By differentiating b [given in Eq. (4.14)] with respect to θ , we find
zZ e 2
(4.17)
db =
− csc 2 12 θ )( 12 dθ )
(
2 K 4πε 0
so that
2
2
⎛ zZ ⎞ ⎛ e ⎞
2 1
1
df = π nt ⎜
⎟ csc 2 θ cot 2 θ dθ .
⎟⎜
2
K
4
πε
⎝
⎠⎝
0 ⎠
(4.18)
The area of the ring shaped region into which the particles are scattered is given by
dA = (2π r sin θ )rdθ ,
(4.19)
where r is the distance of the detector from the foil. The total number of particles scattered per
unit area at θ is given by
N df
N sc (θ ) =
,
(4.20)
dA
where N is the total number of projectiles. Substituting for |df| and dA, we get
2
2
Nnt ⎛ zZ ⎞ ⎛ e 2 ⎞
1
N sc = 2 ⎜
.
⎟
⎟ ⎜
4r ⎝ 2 K ⎠ ⎝ 4πε 0 ⎠ sin 4 12 θ
(4.21)
This is the Rutherford scattering formula. This formula was found to be consistent with many
experiments. [Show data and fit (Fig 6.16 pg.159 Krane)]
Problem 3.48, Textbook
b
umin
v
b
rmin
Nucleus
Assume particles re non-relativistic. Energy conservation gives
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1
2
2
mu 2 = 12 mumin
+
1 zZe 2
,
4πε 0 rmin
------- (i)
where u is the speed of the projectile when it is far from the nucleus. Now, angular momentum
is also conserved in this interaction because the force is a central force; therefore, it exerts no
torque. Thus,
mub = mumin rmin ,
i.e.,
ub
umin =
. ------- (ii)
rmin
Thus,
1
2
⎛ b 2u 2 ⎞
1 zZe 2
mu 2 = 12 m ⎜ 2 ⎟ +
. -------- (iii)
⎝ rmin ⎠ 4πε 0 rmin
If b = 0,
1
2
mu 2 = K =
1 zZe 2
,
4πε 0 d
where d = rmin for b = 0. Thus,
e 2 zZ
d=
,
4πε 0 K
which is the answer obtained in the text.
[Also, do Prob. 3.47. From Gauss' Law, E =
1 Qr
4πε 0 R3
inside a uniformly charged sphere.]
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