Ch 15: Electric force and energy quantization determine atomic structure.

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Oct 18, 2013 (3 years and 9 months ago)

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Ch 15: Electric force and energy
quantization determine atomic structure.

Based on experimental observations scientists propose
mental models of atom. Atom is not observed directly.

In 400 B.C. Democritus theorized that matter consisted of
tiny particles
called atoms.

Dalton Model


Billiard ball




To explain the ratio of the elements
combining to make a compound.

CO
2

is 1:2 , NaCl is 1:1



No direct or experimental evidence is provided to
support the model.



Atom is indivisible.


15.1 The Discovery of the el
ectron

Cathode Ray led Thomson to discover the
electron and to change Dalton’s Model to
accommodate the electron in the atom.




Thomson discovered the electron in his
cathode ray
tube
, since
cathode rays are deflected by electric and
magnetic fields.


1.


Thomson used his cathode ray tube to measure
q/m

ratio for electr
ons and protons. He did not know the
exact mass (m) or charge (q) for ele&proton.




Thomson’s experiments showed that
q/m

for an electron
is roughly 10
11

C/kg. This ratio is over a 1000 times
larger than the
q/m

for a hydrogen ion [ex
plain]


Ex1: A charged object is moving at a speed of 3.5 x 10
4

km/s perpendicular to a magnetic field with a
magnitude of 0.60 T. If the radius of deflection is 0.33
mm, find charge
-
to
-
mass ratio and name the charged
object.




Thomson used perpendicular
cr
ossed
electric and
magnetic fields to determine the speed of the cathode
rays.

For electrons
passing undeflected through the fields,
forces acting on electrons are balanced:



Ex2:
A beam of charged particles moves undeflected
perpen
dicular to electric and magnetic fields. The
magnetic field has a magnitude of 2.0

10

3

T. The
electric field is produced by two parallel plates 3.0 cm
apart with a potential difference of 3.0

10
2

V. Find the
speed of the particles.








An atom is a
positive sphere of electricity in which
electrons are embedded (atom is not the smallest part of
matter and it is divisible).
Thomson

knew:



An atom had an equal number of protons and electrons
(neutral).



The proto
ns had more mass than electrons but same
ch
arge.



The size of the atom was10
-
10

m.



J.J. found that regardless of the type of cathode
material used, he got the same q/m, therefor C.R.
particles are contained in all types of matter.



䍒⁰C牴楣汥r⁡牥⁢畩汤i湧⁢潣歳o潦浡瑴敲m



Dalton’s billiard bal
l model is inadequate.

Thomson’s model was known as the plum
-
pudding model
or the raisin bun model.






Mass of electron and proton were found after Millikan
found the elementary charge (1.60 x 10
-
19
C) and that is
next topic.

Tompson’s Cathode
-
Rays work
sheet Name: ………….

Ex1: A charged object is moving at a speed of 3.5 x 10
4

km/s perpendicular to a magnetic field with a magnitude of
0.60 T. If the radius of deflection is 0.33 mm, find charge
-
to
-
mass ratio and name the charged object.





Ex2:
A beam of charged particles moves undeflected perpendicular to electric and magnetic fields. The magnetic
field has a magnitude of 2.0

10

3

T. The electric field is produced by two parallel plates 3.0 cm apart with a
potential difference of 3.0

10
2

V. Find the speed of the particles.




1.

What was the outcome of Thomson's experiment with cathode rays?

2.

Name the four parts of a cathode ray tube.


3.

Draw a sketch of the plates that produce an electric field.


4.

What happens to the cathode ray in the pre
sence of an electric field?

5.

Which way is the ray deflected?

6.

What happens to the cathode ray in the presence of a magnetic field?

7.

Which way is the ray deflected?

8.

What happens to the cathode ray when both the electric field and the

magnetic fields are presen
t?

9.

Explain

why the cathode ray

is deflected upward in the presence of an

electric field.

10.

Describe the behavior of particles in the presence of a magnetic field?

11.

How did Thomson solve the problem of deflections?

12.

What were his conclusions?

13.

What was he able t
o calculate?

14.

What fact did he find even though he used different sources of cathode rays?

15.

What was the mass to charge ratio?

16.

What was the name given to the new particle?


15.2
Quantization of Charge


Millikan experiment

Millikan found the charge on the ele
ctron and showed that
it was a fundamental unit of electrical charge. All charges
are multiple of 1.6 x 10
-
19

C.


Millikan

s Oil
-
drop Experiment












X
-
Rays to
ionize drops






Robert Millikan measured the charge on the electron.



A fine spray of oil falls through a hole

into a chamber
where the drops can be observed.



The plates at the top and bottom of the chamber are
charged (the top plate is positive).



X
-
rays are shot onto the oil
drops which cause

the
drops to be negatively charged.



In the absence of voltage, the forc
e on the drops is
determined by their mass

only, weight, F
g
.



When a voltage is applied, negatively charged drops
will slow down, stop or begin moving upwards. The
behavior of the drop is determined by the applied
voltage

and the charge on the oil drop and
the weight
of the oil drop.



Millikan used these measurements to determine that
the
charges on the drops were multiples of 1.6 x 10
-
19

C.

[Diagrams showing: suspended, constant speed,
accelerating up &down]
[Copy from board]

[Note for the same mass F
g

vector

has same lenghth and

points

downwards]

Ex
-

Latex sphere with a mass of 5.00


10

16

kg is
suspended in an electric field between plates directed
upwards that are 9.00

10
-
3

m apart. To keep the sphere
suspended, the potential difference between the plate
s is
55 V. How many elementary charges has the oil drop?


Ex


A positively charged particle with a mass of 4.5 x 10
-
14

kg is accelerating upwards at 1.5 m/s
2

under the
influence of an electric field between two horizontal plates
separated by 45 mm. If the

potential difference across the
plates is 110 V, what is the charge on the particle? (make
FBD diagram and lines of electric fields)



The Discovery of the Nucleus

The Rutherford Model
-

Alpha,

2+
,

scattering
experiment

In 1911,
Ernest Rutherford

directed
+2
-
particles at a thin
gold foil. They placed a zinc
sulphide screen near enough to
the foil to detect any alpha
particles that got through.


The zinc sulphide would give
off a flash of light whenever struck by an
-
particle.


They expected

that the
-
particles would go right
through the foil with hardly any deflection because in the
Thomson model, the

positive and negative electric charges
inside an atom were assumed to be uniformly distributed
through a solid atom. Consequently the positively charged
-
particles would only encounter weak
net
electric forces
and so pass through
the thin foil with only slight
deflections (less than a degree).



The

Thomson mode
l



Rutherford model




-
p
articles undeflected


-
particles scattered

Results of the
alpha
-
scattering

experiment were quite
different. Rutherford made three observations:

1.Most alpha particles were undeflected.

2. Some were scattered (deflected)

3.Some wer
e reflected.



Rutherford’
s explanation of these observations:

1.

atom

must be
largely empty space surro
unding a tiny
nucleus
,

2.

atom

positive charge and nearly all its mass are
concentrated

in
nucleus
,

3.

electrons revolve around

the nucleus like planets revolve
around the sun
.
This model is known as
the planetary
model

of the atom.

[Thomson model failed and
modified]

http://micro.magnet.fsu.edu/electromag/jav
a/rutherford/

The fact that an occasional alpha particle can be stopped
by a gold nucleus and returned along its initial path
provides with a way of estimating
an upper limit for the
size of the nucleus using conservation of energy:

Ek=Ep=Fe x d

Drawback
s

1.

E
m
is
si
o
n

and
absorption

Spectral lines cannot be
explained by Rutherford model.




Helium emission spectrum Absorption spectrum

Accordi
ng to Maxwell, an accelerating charge gives off
energy.
Then the orbiting
electron, with centripetal
acceleration, should continuously
radiate energy and spiral into the
nucleus, which it does not do.


The Bohr Model of the Atom

In 1913 Bohr presented a mo
del of the atom that begins
with Rutherford's picture of an atom as a nucleus
surrounded by electrons moving in circular orbits.

Bohr's work was primarily with the hydrogen atom.

Spectroscopy

A diffraction grating can spread light out into a spectrum
with
colours according to their wavelengths

continuous
spectrum
-

contains all
wavelengths



Hot dense, solid,
liquid, or gas,
material


emission line
spectrum
bright
selected lines at
distinct
wavelengths




Hot low dense
gas

absorption line
spectrum
dark
selected
wavelengths



Cold low dense
gas




Lines are ordered from short to high wavelength.



Each element has its unique spectrum lines.

Kirchhoff

used spectra to identi
fy unknown elements.

http://jersey.uoregon.edu/vlab/elements/Elements.html



No one had, before Bohr, explained why this pattern
occurs. The reason why elements produce spectral lines
was

still not explained.


Balmer Series



The hydrogen atom bright line spectra were a major key
to understanding atomic structure. In this spectrum there
are four lines in the visible region. J
ohann Balmer,
a
high school teacher
, searched for a mathematical
relationship and found, by trial and error, that he could
calculate the wavelength of each of the four lines
in the
hydrogen spectrum

using the formula:


n
f

= 2

R
H
=Rydb
erg's constant=1.10x10
7
/m

n
i
=3,4,......(only for the Balmer series).

Balmer did not know why his formula worked. Balmer
predicted other lines in the UV and infra
-
red region.

http://www.avogadr
o.co.uk/light/bohr/spectra.htm

Ex
-

Find the wavelength and frequency of the spectral
line n
i
=3 in the Balmer series

1/

=R
H
(1/nf
2
-
1/ni
2
)


=1.10x10
7
(1/4
-
1/9)


=6.55x10
-
7
m

v=f


3.00x108=fx6.545454x10
-
7


f=4.58x10
14

Hz


Bohr used Balmer idea and Plan
ck’s quantum ideas to
establish a new atomic model.

Bohr made the following assumptions:
In hydrogen atom

1.

there can be only certain values
of the total energy (electron's
kinetic energy +potential energy).
Quantized energy levels
.


2.

These allowed energy l
evels
correspond to different orbits for
the electron as it moves around
the nucleus.

3.

Larger orbits have larger energies.

*
-

4.

When moving in an allowed orbit, called it
stationary
state
, the electron is exempt from the classical laws of
electromagnetism an
d does not radiate energy as it
moves along its orbital path.

REM: In other cases when electrons are accelerated they emit
(Radio or microwave) EMR as Maxwell stated.

5.

The energy of the electron in an atom is quantized
(used Plank idea).

6.

If an electron go
es from a higher to a lower energy
level, energy is given off

(emission spectrum)
. If an
electron goes from a lower level to a higher level,
energy is absorbed

(absorption spectrum)
.

http://www.colorado.edu/physics/2000/quantumzone/bohr.html


Consequences of the Bohr Model

1. Radius of the hydrogen orbit

Bohr’s model of the hydrogen a
tom states that electrons
can orbit the nucleus only at specific locations given by
the formula:

r
n
=r
1
n
2


r
n
=radius of the n
th

orbit (any orbit).

r
1
=radius of the first orbit=5.29x10
-
11
m
(see formula sheet)


r
1

is the radius of the lowest possible energy
level or
ground state
of the hydrogen atom.

n=1,2,3,....

Complete the chart:

Level
(n)

Radius of orbit
(m)

1

5.29x10
-
11

2


3


4


5




Level
(n)

Radius of orbit
(m)

1


5.29x10
-
11

2


2.12x10
-
10

3


4.76x10
-
10

4


8.46
x10
-
10

5


1.32x10
-
9


2.Energy levels for the allowed orbits of
hydrogen




E
n
=Energy of n
th

level

E
1
=Energy of first level=
-
13.6 eV

n=1,2,3,....

Orbit
(n)

Energy of
orbit (eV)

Frequency

Hz

E=hf

Wavelength

(m)

c=λf

1

-
13.6 eV



2




3




4




5

























The energy of the energy levels
is always
negative (energy
levels like a well under ground
level). This is because the
energy at ∞ is 0 and the closer
to the nucleus the less the
Orbit
(n)

Energy of
orbit (eV)

1


-
13.6

2


-
3.40

3


-
1.51

4


-
0.850

5


-
0.544


††

†††
〮〠0†


n=1


n=2


n=4


n=3


n=5


n=6

-
13.6eV

-
0.85eV

-
3.4eV

-
0.54eV

-
1.5 eV

-
0.38eV

energy. At ∞ is the greatest possible energy.



The closer the electron to the nucleus the more energy
needed to free the

electron from the atom (
ionization
energy
).



an electron making a transition from lower to a higher
energy level results in a
dark
-
line spectrum



an electron making a transition from a higher to

a lower energy level results in a
bright
-
line spectrum



The ene
rgy in an electron transition between energy
levels can be calculated by:

E = E
f



E
i

E = energy of transition (absorbed or emitted),
J or eV

E
f

= energy of the final energy level, J or eV

E
i

= energy of the initial energy level, J or eV



We can calculate t
he wavelength and frequency of the
absorbed or emitted photon from the transition:

E
photon

= E
f



E
i

= hf =hc/λ



The energy of the photon is always positive.



An electron excited to higher energy levels does not
have to return to ground state in a single ju
mp. They do
it in few intermediate steps. At each step, a photon is
emitted. Each photon emitted has a smaller energy.

Example

An electron in hydrogen

atom, shown, undergoes a transition


from the 6
th

energy level to the 2nd
energy level. What is the fr
equency and

n=1


n=2


n=4


n=3


n=5


n=6

-
13.6eV

-
0.85eV

-
3.4eV

-
0.54eV

-
1.5 eV

-
0.38eV

wavelength of light emitted? And what type?


E
2

=
-
3.40 eV

E
6

=
-
0.378 eV


E
photon
=E
f
-
E
i
= E
6



E
2

=
-
3.40
-
(
-
0.377777)=


=3.022223eV x (1.6 x 10
-
19
)


= 4.83555 x 10
-
19

J

=hf

f=7.29344 x 10
14
Hz

v=fx


3.00x10
8
=7.29344 x 10
14



=4.11x10
-
7
m

Type: Blue light

Example

An electron undergoes a transition


from the 2nd level to 4th level in a
hydrogen atom. What is the

frequency and wavelength of the
radiation absorbed?





E=E
f



E
i

=E2
-
E4

=
-
0.850
-
(
-
3.40)

f=6.15x10
14

Hz

c=f


3.00x108=6.15x1014



=4.88x10
-
7

m

States of the atom

1.

When an electron has the smallest allowable amount of
energy, it is in the lowest energy level called the
ground
state(stationary).

The atom is stable.

2.

If a
n electron absorbs energy, it can make transition to a
higher energy level called an
excited state
. Atomic
electrons remain in excited levels only a fraction of a
second before returning to the ground state and emitting
energy (neon light is an exited neon

gas).

3.

The
ionization state

level is the state at which the
electron is totally removed from the atom to E=0.
This
is equivalent to work function in photoelectric effect

http://www.avogadro.
co.uk/light/bohr/spectra.htm


Exact match


which photon will be absorbed?

http://www.stmary.ws/physics/home/animations3/modernPhysics/bohr_transitions.html


The Northern Lights and the Emission Line
Spectrum of Oxygen

Alberta skies often display aurora
borealis, or northern lights. At
high altitudes above the surface
of Earth, high
-
energy electrons,
trapped by Earth’s magnetic field,
interact with oxygen
and nitrogen
at
oms. During these interactions,
the electrons in these atoms are
excited and move i
nto higher
energy
levels. Eventually, the
excited electrons

return to their
ground states.
In doing so, they
emit light that for
ms the
auror
a
borealis.



Phys
ics 30 Practice Examples:


1.

Find the energy needed to ionize a hydrogen atom
whose electron is in n=4.






(E=E
f
-
E
i
=0
-
(
-
0.850)=

0.850 eV


2.

A hydrogen atom absorbs a photon of wavelength


434.1 nm.

(a)How much energy did the atom absorb?

(b)What were the ini
tial and final states of the hydrogen
electron?



(a)E=hc/(=6.63x10
-
34
3.00x108/434.1x10
-
9

=2.86eV

(b) Using the energy level

diagram n=2 to n=5



3.

An electron dropped from n=4 to n=1. How many
spectral lines could be produced ?



6 lines


4.

If the wavelength

of emitted spectrum is 486 nm,

what are the corresponding stationary states?




E=hc/
λ(=6.63x10
-
34 x3.00x108/486x10
-
9

=4.09259x10
-
19

J

=2.55 eV. The two levels that have a difference of 2.55 eV
are the second and the fourth.



5.

Find the longest wavelength photons that will be
absorbed by a hydrogen atom in its ground state.

Longest wavele
ngth=smallest energy absorbed=n=1to
n=2.

(=1.22x10
-
7m

n=1to n=3 (=1.02x10
-
7 m

n=1to n=4 (=9.75x10
-
8m


n=1


n=2


n=4


n=3


n=5


n=6

-
13.6eV

-
0.85eV

-
3.4eV

-
0.54eV

-
1.5 eV

-
0.38eV


Physics 30 Practice Examples: Name………………..


1.

An electron in a hydrogen atom drops from the 4
th

to
2
nd

energy level. Is this an absorption or
emission photon? Find the frequency and wavelength of the photon. (Make a diagram to show the
transition)









2.

Find the energy needed to ionize a hydrogen atom whose electron is in n=4.






3.

A hydrogen atom abs
orbs a photon of wavelength 434.1 nm.

a)

How much energy did the atom absorb?




b)

What were the initial and final states of the hydrogen electron?




4.

An electron dropped from n=4 to n=1. How many possible spectral lines could be produced? (make
diagram)






5.

Find the longest wavelength photons that will be absorbed by a hydrogen atom in its ground state.



5. An electron undergoes a transition from the 2nd level to 4th level in a
hydrogen atom. What is the frequency and wavelength of the radiation
absorbed?





Successes of the Bohr model

1.

It explained the properties of elements.

2.

It explained the hydrogen spectra.


Drawbacks of the Bohr model

1.

It could not explain the spectra of elements containing
many electrons.

2.

It could not explain why only certain orbits
were
allowed (see 5).

3.

It does not explain why each spectral line split into
several lines when put in a magnetic field or electric
field.

4.

It does not explain the relative intensities of spectral
lines.

5.

The wave nature of the electron is not used in this
mo
del.


The quantum mechanical model

In about 1920 Heisenberg and
Schroedinger developed a quantum
mechanical model to describe the nature
of the atom.



It is a mathematical model.



The atom cannot be visualized.



There are no exact levels. The position of th
e
electron is given in terms of probabilities. These
probabilities are called orbital (clouds).





END OF CHAPTER 15