17 pius augustine structure of atom

Dr. Pius Augustine, S H College, Kochi

Sphere of +vely charged
matter of size 10-10m
and electrons are
embeded. (plum
pudding) (Lord Kelvin
and J.J Thomson)
Thomson model of atom
J. J. Thomson
1856 –1940

Plum pudding model
Raisin cake model

Could not explain
atomic spectra.
Results of gold foil
experiments could
not be explained
Failure: Thomson model of atom

Rutherford Scattering Experiment
Earnest Rutherford
1871 -1937

α emitted by radioactive material
(Po) kept inside a thick lead
box.
A lead screen with a fine hole
select fine beam of alpha.
Rutherford’s α particle scattering

Collimated beam is allowed to fall
on gold foil.
Scattered α particles are collected
at angles varying from 0 to 180o
by fluorescent screen.
Rutherford’s α particle scattering

Observations
i. Most of α particles passed through
the gold foil with small deviation.
ii. Few (1/8000) suffered large angles
even greater than 90o.
iii. Few scattered even backward
(180o).

Rutherford atom model
i. Atom has a small, +vely charged
nucleus.
All +ve charges and most of the mass of the
atom are concentrated in the nucleus.
Electrons have no place inside nucleus.
Dimensions of nucleus and electrons are very
small compared to the overall size of atom

Rutherford atom model
ii. For stability, electrons must
revolve around the nucleus in
closed orbits
(electrostatic force will be used
for centripetal force)
Dynamic planetary model

Theory of alpha particle scattering
Assumption:
i. Alpha and nucleus are point
charges.
ii. Scattering is due to coulombian F
iii.Nucleus is massive compared to α
iv. α do not penetrate nucleus.
v. E prop to 1/r2, path of alpha
particles is hyperbola with nucleus
at the outer focus. Dr. Pius Augustine, S H College, Kochi

Impact parameter(b):
Minimum distance to which the
alpha particle would approach the
nucleus if there were no forces
between them.
For head on collision, b = 0.

Distance of closest approach:
When there is head on collision,
distance from the nucleus at which
alpha particle is completely
stopped is distance of closest
approach (vary depend on K.E of
approaching alpha)Dr. Pius Augustine, S H College, Kochi

At the stopping point, K.E = P.E
½ mv2 = 2e Ze
4πεo D
D = 2e Ze = 2 e Ze
4πεo ½ mv2 4πεo K
D - distance of closest approach

Relation b/w b and D
Impact parameter b = Ze2 cotθ/2
4πεo K
D = 2 Ze2
4πεo K
b = D cot θ/2
2

Scattering angle
Angle between asymptotic
direction of approach of the
alpha particle and the
asymptotic direction in
which it recedes.Dr. Pius Augustine, S H College, Kochi

Plot a graph showing the variation of impact
parameter with scattering angle.

Keeping other factors fixed, it is found
experimentally that for a small thickness t, the
number of α-particles scattered at moderate
angles is proportional to t. What clue does this
linear dependence on t provide.
Scattering is mostly due to single
collision and this increases linearly
with the number of target atoms.
Scattering is proportional to the
thickness of the foil. Dr. Pius Augustine, S H College, Kochi

Q. In Rutherford’s scattering
experiment, most of the alpha
particles go unscattered while,
some of them are scattered
through large angles. What can
be concluded from this?

Draw backs of Rutherford model
i. Accelerated charge emit
electromagnetic waves.
Energy decreases and will spiral in.
Questions stability of atoms.
But majority of atoms are stable ??

Draw backs of Rutherford model
2. Classical em theory, frequency of
emitted radiation is proportional to ω.
As electron spiral in, angular velocity
increases, result in a continuous
spectrum with all possible wavelength .
But atoms like H, emit only fixed line
spectra wavelength. ??

A bit of Physics Humour – Magnify your mind as you magnify it to read.
• A BIT OF PHYSICS HUMOUR
• Sir Ezrnest Rutherford, President of the Royal Academy, and recipient of the Nobel Prize in Physics, narrated the following story:
• "Some time ago I received a call from a colleague. He was about to give a student zero for his answer to a physics question, while the student claimed a perfect score. The instructor and the student agreed to an
impartial arbiter, and I was selected.
• I read the examination question: "Show how it is possible to determine the height of a tall building with the aid of a barometer."
• The student had answered: "Take the barometer to the top of the building,attach a long rope to it, lower it to the street, and then bring it up, measuring the length of the rope. The length of the rope is the height of the
building."
• The student really had a strong case for full credit since he had really answered the question completely and correctly! On the other hand, if full credit were given, it could well contribute to a high grade in his physics
course and certify competence in physics, but the answer did not confirm this. I suggested that the student have another try. I gave the student six minutes to answer the question with the warning that the answer should
show some knowledge of physics.
• At the end of five minutes, he hadn't written anything. I asked if he wished to give up, but he said he had many answers to this problem; he was just thinking of the best one. I excused myself for interrupting him and
asked him to please go on. In the next minute, he dashed off his answer, which read: "Take the barometer to the top of the building and lean over the edge of the roof. Drop the barometer, timing its fall with a stopwatch.
• Then, using the formula x=0.5*a*t^2, calculate the height of the building."
• At this point, I asked my colleague if he would give up. He conceded, and gave the student almost full credit. While leaving my colleague's office, I recalled that the student had said that he had other answers to the
problem, so I asked him what they were.
• "Well," said the student, "there are many ways of getting the height of a tall building with the aid of a barometer. For example, you could take the barometer out on a sunny day and measure the height of the barometer,
the length of its shadow, and the length of the shadow of the building, and by the use of simple proportion, determine the height of the building."
• "Fine," I said, "and others?"
• "Yes," said the student, "there is a very basic measurement method you will like. In this method, you take the barometer and begin to walk up the stairs. As you climb the stairs, you mark off the length of the barometer
along the wall. You then count the number of marks, and his will give you the height of the building in barometer units."
• "A very direct method."
• "Of course. If you want a more sophisticated method, you can tie the barometer to the end of a string, swing it as a pendulum, and determine the value of g [gravity] at the street level and at the top of the building.
• From the difference between the two values of g, the height of the building, in principle, can be calculated."
• "On this same tack, you could take the barometer to the top of the building, attach a long rope to it, lower it to just above the street, and then swing it as a pendulum. You could then calculate the height of the building by
the period of the precession".
• "Finally," he concluded, "there are many other ways of solving the problem."
• "Probably the best," he said, "is to take the barometer to the basement and knock on the superintendent's door. When the superintendent answers, you speak to him as follows: 'Mr. Superintendent, here is a fine
barometer. If you will tell me the height of the building, I will give you this barometer."
• At this point, I asked the student if he really did not know the conventional answer to this question. He admitted that he did, but said that he was fed up with high school and college instructors trying to teach him how to
think.
• The name of the student was.
Neils Bohr
The Nobel Prize in Physics 1922
•

Niels Bohr 1885 - 1962 Bohr Atom Model

Postulates
1.Stationary orbits
2.Quantum condition
3.Non radiating orbits
4.Frequency condition

Bohr atom Model
Combination of Rutherford
model and Plank’s
quantum theory of
radiation.

Bohr atom Model
2. Electrons revolve around nucleus only in
permissible orbits (stationary orbits) for
which angular momentum of electron is
an integral (n)multiple h/2π
(quantisation condition)
Electrons will not radiate energy when
they are in stationary orbits.
n – principal quantum number.Dr. Pius Augustine, S H College, Kochi

Bohr atom Model
3. Atom radiates energy only when an
electron jumps from a stationary
orbit of higher Einitial energy to one of
lower Efinal energy.
Frequency emitted = (Ei – Ef )
h
Energy corresponds to n=1, ground state

Bohr atom Model
i. Radii of stationary orbits ?
ii. Total energy of electron in the
orbit?

Radii of stationary orbits
Electrostatic force of attraction = Ze *e
b/w nucleus and electron 4πεo r2.
Centrifugal force = mv2/r
Combining the above two, r = Ze2
4πεo mv2

Radii of stationary orbits
Bohr quantum condition mvr = nh/2π
v = nh / 2πrm
Sub v in r,
Radius of nth orbit rn = n2h2εo
πZe2m
rn α n2 or rn = n2 r1
Radii is in the ratio 1: 4 : 9 : 16 ……
For H – atom , Z = 1 r1(H-atom) = 0.53 Å

n2h2
4π2m2r2v2 =

Velocity of electron
mv2/r = Ze *e
4πεo r2.
mvr = nh/2π
Velocity of electron
V = (1/n) e2/2ε0h
V1 = e2/2ε0h = c/137 m/s
V2= V1/2, V3= V1/3

Calculate the velocity of electron in
the second orbit of H-atom (ε0 = 8.84
x 10-12 F/m)
V = (1/n) e2/2ε0h
=1.1 x 106 m/s

State whether T2-r3 law is valid for
Bohr’s orbits, where T-is time period
and r-radius of the orbit?
Law of mechanics are valid in
stationary orbits.

An electron of H-atom makes f
revolutions per second when it is in
the first Bohr’s orbit. How many
revolutions does it make in the
second Bohr’s orbit?
T2 = 1/f2 r = n2r0
Using T2-r3 law
f2
2/f1
2 = r1
3/r2
3 = (n1
2/n2
2)3

Centripetal force
Bohr’s quantum condition
Bohr’s frequency condition.

Centripetal force

Total energy of electron in the orbit = K.E + P.E
It is sum of Kinetic + Potential
Potential energy = - Ze *e
4πεo r
Kinetic energy = ½ mv2 = Ze *e
r = Ze2 8πεo r
4πεo mv2
Total energy = - Ze4 = - me4Z2
rn = n2h2εo 8πεo r 8εo
2n2h2
πZe2m

The energy of an electron in an excited
hydrogen atom is -3.4 eV. Calculate the
angular momentum of the electron
according to Bohr theory.
En = -13.6/n2 eV
-3.4 = -13.6/n2 n = 2
L = nh/2π = 2.1 x 10-34 Js

K.E
P.E
T.E

For H-atom, Z =1
Energy of electron in the first orbit
E1 = -13.58 eV
E2 = -(13.58/4) eV = -3.39 eV
E3 = -(13.58/9) eV = -1.51 eV
So on….
En = E1/n2

En α -1/n2
ie. as the value of n increases En
increases, (decrease in –ve
value or energy increases)
Outer orbits have greater energies
than the inner orbits

Bohr’s interpretation for H spectrum
If electron jump from ni to nf (ni > nf )
Frequency of radiation emitted,
ν = (Eni - Enf ) / h
= me4
8εo
2h3
{1/nf
2 – 1/ni
2}

Bohr’s interpretation for H spectrum
Wave number(ν) is reciprocal of
wavelength = 1/λ = ν / c
ν = (Eni - Enf ) / hc
= me4
8εo
2 ch3
= R
R – Rydberg constant = 1.091 x 107m-1
{1/nf
2 – 1/ni
2}
{1/nf
2 – 1/ni
2}

1 eV = 1.602 x 10-19 J
R = 1.091 x 107m-1

Spectral line of H atom
i. Lyman series : nf = 1, ni = 2,3,4,5,6,7….
ii. Balmer series : nf = 2, ni = 3,4,5,6,7 ….
iii. Paschen series nf = 3, ni = 4,5,6,7 ….
iv. Brackett series nf = 4, ni = 5,6,7….
v. Pfund series nf = 5, ni = 6,7…
vi. Humphries seris nf = 6, ni = 7,8.

Lyman series : nf = 1, ni = 2,3,…
Lies in the UV region
ν1 = ¾ R
ν2 = 8/9 R
ν3 = 15/16 R
Note : wave number increases,
wavelength decreases.
Higher transitions shift low
wavelength side.

Wave number = 1/λ = ν / c
R – Rydberg constant = 1.0 973 7316 x 107 m-1

Kossel diagram

How many spectral lines are emitted by
atomic hydrogen excited to the nth
energy level?
The number of transition from states higher than
n=1 to the state n=1 is (n-1)
The number of transition from states higher than
n=2 to the state n=2 is (n-2)
And so on.
Total number is
N= (n-1) + (n-2) + (n-3) + …..+ [n-(n-1)]
= ½ n(n-1)
([n-(n-1)]) - only one transition

A transition in H-atom is taking place
from energy level n = 3 to n = 2. Find the
wavelength of the photon emitted. Will
the photon be visible? To which spectral
series will this photon belong? R = 1.097
x 107 m-1.

Is there any connection between the
frequency of revolution of electron about
the nucleus in the H-atom and the
frequency of its radiation?
There is no connection between the two.
When n is very large (~10,000), the
frequency of radiation in the transition
from n to (n-1) is aproximately equal to
frequency of revolution.

Given the ionization energy of H-atom is
13.6 eV, what is the ionization energy of
He+?
E = -13.6 Z2/n2.
= -54.4 eV
(Z=2 for He and n=1)

Q. Given the ionization energy of H-atom
is 13.6 eV, what is the ionization energy
of doubly ionized Li?
Q. The wavelength of the first line of
Lyman series for H-atom is identical to
that of the second line of Balmer series
for some hydrogen like ion ‘x’. Calculate
energies of the first four levels, of ‘x’.

Can H-atom absorb a photon of energy
greater than 13.6 eV. Explain.
Yes.
Atom will be ionized (using 13.6 eV) and
excess energy will go as the kinetic
energy of the electron. The process is
called photoionization.

Obtain the units of wavenumber and
Rydberg constant.
Wavelength of some of the lines emitted
by H-atoms are given below. Pick up the
lines which belong to Lyman series.
6563 Å, 1216 Å, 9546 Å,
4861 Å, 1026 Å

Absorption spectrum of Hydrogen
If white light is passed through atomic hydrogen,
only those photons having energy hν equal to
the energy difference between two direct states
will be absorbed.
Outcoming light will have dark lines in the
continuous background -absorption spectrum
The wavelength corresponds to darklines is
same as that of the wavelenth emitted by HDr. Pius Augustine, S H College, Kochi

Absorption spectrum of Hydrogen at
ordinary temperature contains only
Lyman series. Comment.
At ordinary temperatuers, large fraction of
atoms will have electrons only in n=1.
In solar spectrum Balmer series is also seen –
in the outer layer of sun, H-atoms with
electrons in the n=2 are available.

The energy required to ionize H-atom is
21.7 x 10-19 J. What will be the
wavelength of the radiation emitted if a
proton captures a stationary electron to
form H-atom in the ground state.
Hint:
When a proton and electron combine the
energy released will be
21.7 x 10-19 J = hν = hc/λ

Compute the shortest and longest
wavelength of the Balmer series.
Given R = 1.097 x 107 m-1
Hint:
1/λ = ν (read it new bar)= R (1/22 – 1/ni
2)
λmin - when energy hν is maximum – ie. ni is
infinity 1/ni = 0
λmax - when energy hν is lowest – ie. ni = 3

The wavelength of the first memberof
Balmer series of hydrogen atom ha a
wavelength of 6565 Å. Compute the
wavelength of the second member.
Hint:
1/λ = R (1/22 – 1/ni
2)
1/λ1 = R (1/22 – 1/32) = 5R/4x9
1/λ2 = R (1/22 – 1/42) = 12R/4x16
Take ratio and eliminate R. Then solve for λ2.

Ionization and ionization potential
The process of electron removal from
an atom is called ionization.
Amount of energy required for
ionization is called ionization potential.
Ionization potential of H is 13.6 eV
Ionization potential of Li is 5.4 eV

Ionization potential/energy vs Z graph

Very large energy is required for He, Ne, Ar etc.
Very small energy is required for Li, Na, K etc.
This indicates all the electrons donot occupy the same
orbit, but take positions in so-called shells around the
nucleus.
Result was supported by what is seen in periodic table
i. Atomic radius decrease from left to right
ii. Atomic radius increase from top to bottom
Ionization potential/energy - Shells

Frank-Hertz
Experiment
Experimental
proof for
discrete
energy levels
in atoms.

Apparatus description
Glass tube – containing vapour of element
under investigation.
Thermionic emitter at one end-filament.
P.D is applied – accelerate the emitted
electrons.
Wire mesh known as grid G.
Franck and Hertz Experiment

Apparatus description
Plate P - small negative potential w. r. to G.
+ve potential on G accelerates electrons.
-ve potential V0 on the plate retards
electrons
Only electrons with energy above a
threshold would reach the plate

Working
As V is increased IP (plate current) increases.
Below certain critical value, electrons undergo
elastic collisions with atoms. (elastic
collision means atom does not take energy)
Beyond a certain value of the p.d (V), electrons
will pass the grid and will have energy sufficient
to reach the plate.

Working
But from graph – after reaching a maximum
current, it drops.
This happens when electron colliding with one of
the atom gives up some or all of its KE in
exciting the atom to higher level. (Inelastic)
So electrons will not reach the plate and IP drops

Working
With further increase in V, more electrons will
reach plate and IP increase.
If electrons have sufficient energy it may cause
more than one inelastic collisions.

Analysis of experimental result
• Gas used was Hg vapours.
• When the accelerating voltage reached 4.9
volts, there was a sudden drop in current.
• This was due to the transfer of energy of the
electrons to the atoms of the vapour – inelastic
collision – energy of electron is transferred to
atoms and it gets excited.

• After which current again increases as the
electrons start reaching the collecting
plate.
• When the voltage is 9.8 (twice 4.9 volts)
again energy of electron is just sufficient
to make second atom excited.

Peaks were observed at voltages equal
to intergral mutiple of 4.9 V.
These excited electrons will come down
and release emission spectra.

Frank and Hertz analyzed the emission
spectra of Hg vapours and found that
photons of energy 253.6 nm was
emitted which is matching with 4.9 eV.
Established the presence of stationary
orbits proposed by Bohr.

Bohr’s Correspondence Principle
Quantum physics give the same result
as classical physics in the limit of
large quantum numbers

Correspondence Principle
It has been noticed that the result of quantum
physics are quite different from those of
classical physics.
However, under extreme situations, when the
electron orbit is very large (say 1cm), the
results of quantum physics are expected to
match with the result of classical physics.

Correspondence Principle
For H atom, a radius of 1cm mean quantum number
n = 10,000.
Classical aproach – frequency of the radiation
ν = 1/T = v/2πr T = 2πr/v
V- orbital velocity and r – radius.
Use radius and velociy of nth orbit electron
Frequency ν = (me4/8ε0
2h3)(2/n3)
If frequency is calculated using Bohr’s theory
ν = ∆E/h, it also gives the same result
ν = (me4/8ε0
2h3)(2/n3)

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Dr. Pius Augustine, SH College, Kochi

93
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Dr. Pius Augustine, Dept of Physics, Sacred Heart College, Thevara
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Dr. Pius Augustine, Asst. Professor, Sacred Heart College, Thevara, Kochi.

17 pius augustine structure of atom

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