ATOMS

CLASS - XII
In this unit, we shall discuss the Rutherford's a-ray Scatterin g
models of atoms in some detail. The first Experiment :
contribution. in this regard come from The experimental setup used by
Dalton, who proposed that matter is Rutherford and his collaborators, Geiger
made of atoms, which are indivisible J.J.
ES
and Marsden is shown in figure.
Thomson proposed a structure for the
A SS
atom, which is moolified by Rutherford L
C m o row
and Later by Niels Bohr. N
IO for t om
Thomson Model of Atom :
A T you
E V
L repa i ng
According to the Thomson
E p
r
model,
every atem consist of a positively chargeol
sphare of radius of the order of 10–10m
in which entire mass and positive charge
of the actom are uniformly distributed. S is a piece of radioactive source

Inside this sphare, the electrons are (B3Bi214) contained in a lead cavity. The

embeded like seeds in a watermelon or –particles emitted by the source are

like plums in a pudding. The number of collimated into a narrow beam with the

electron is such that their negative help of a lead slit (Collimator).

charge is equal to the positive charge of The Collimated beam is allowed to

atom. Thus the atom is electrically fall on a thin gold foil of thickness of the
neutral. order of 2.1×10 7 m. The -particles

E S
Sthrough
scattered in different directions are
Limitations of Thomson Atom model :
A S
(i) It could not explain the origin of
C L
observed
ro w a rotatable detector
moof a zine sulphide screen and
spectral series of hydrogen and other
IOou for N to
consisting
m
T
a microscope.
atoms, observed
VA in g y
experimentally.
E E
L ar p
The -particles produce bright
flashes on the Zn S screen. These are
(ii) I t could not pre observed in the microscope and counted
explain the large
at different angles from the direction of
angle scattering of
incidence of the beam. The angle  of
–particles from
deviation of an -particle from its original
thin metal foils, as
direction is called its scattering angle .
observed by Rutherford.
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Observations : An -particle carries two units of

A graph is plotted between the positive charge and has mass of a helium

scattering angle  and the number of atom. Charge on gold nucleus = ze,

–particles N, scattered at  for a large where Z of gold is 79. As gold nucleus is

number of -particles. We find that - about 50 times heavier than an

-particle, we assume that it would
remain stable in the scattering process.

E S
Therefore, the trajectory of
S can
-particle
S
LA molaw
be computed using
w
ro of motion and coulomb
C
Newton's
N forcetoofmrepulsion between -particle and
T IOou gold
for nucleus i.e.
VA in g y
E LE ar p
1
F=
 ze   2e 
pre 4 0 r2
Where r is the distance of
-particle from the centre of the nucleus.
(i) Most of the –particles pass straight
The magnitude and direction of the force
through the gold foil. It means they do
on an -particle changes continuously
not suffer any collision with gold atoms.
as it approaches the nucleus first and
(ii) Only about 0.14% of incident then moves away from it.
-particles scatter by more than 1°.

(iii) About one -particle in every 8000

–particles deflects by more than 90°.

S
Explanation :

An a particle is over 7000 more

SS E
massive than an electron and in this LA
C tending o
As shown
r w in figure an –particle
mo to collide with the nucleus,
experiment, –particle is travelling at a
high speed, therefore, very strong forceIO
N r t om (1)

T fo down due to repulsive force of the

slow
A ing yo nucleus, finally stops and is then
u
V
ELEpreparto repelled
alone could have deflected them through
large angles. This led Rutherford back. This -partide, therefore
postulate that the entire positive charge retrace its path, scattering through 180°.
of the atom must be concentrated in a The -particles 2 and 2' tending
tiny central core of the atom. This tiny to hit the nucleus at its periphery,
central core of each atom was called experience strong repulsive force and get
atomic nucleus. scattered through large angles (>90°).
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 The -particles 3 and 31, which Potential Energy of -particle at
pass at a distance from the nucleus di st a nce r0 from t he nucleus
experience small repulsive forces and get = Potential × charge
scattered through small angles. ze
P.E. =  2c
4 0 r0
The -particles which pass at large
distances from the nucleus go almost Kinetic Energy of -particle of
undeviated. mass in moving with velocity v is

E1S
SS 2 mv
We can show that the number of
 K.E. = 2
of –particles scattered per unit area N()
at scattering angle  varies inversely as LA o row
C Asmat the distance of closest
N
IO fo r t om :

sin4   .
AT you approach
2
E
L aV i ng K.E. = P. E.
N  
1

sin4  
E r
pre
p 1
mv 2 
ze  2e
2 2 4 0 r0

Distance of closest Approach : ze  2e

r0 
When -particle is directed or 1 
4 0  mv2 
2 
towards the nucleus, the kinetic energy
of –particle goes on decreasing and in Impact Parameter (b) :
turn electrical potential energy goes on It is defined as the perpendicular
increasing due to coulomb's repulsive distance of the initial velocity of -particle
force between nucleus and -particle. from the central line of the nucleus, when

At a certain distance r0 from the the particle is far away from the nucleus.

nucleus, K.E. of -particle reduces to

ES
zero. The particle stops and it cannot go
A SS
closer to the nucleus. It is repelled by
L
C mo ro w

paith, turning through 180°. ThisIO

the nucleus and therefore, it retrace its
N r t om
T fo When the impact parameter is
Aofing yo large, an -particle will deviate through
u
V
ELEprepar
distance r is known as the distance
0

clasest approach. a much smaller angle. However, when

impact parameter is small force
Electrical potential at distance r 0
experienced is large and the -particle
due to nucleus
will scatter through a large angle.
1 Ze
V  Rutherford calculated analytically
4 0 r0
the relation between the impact
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parameter band scattering angle 8, FC = F C
which is given by -
mv 2 e.e e2
  mv 2

2  r 4 0 r 2 4 0 r
1 ze cot 2  1 2
b
4 0 K.E. K.E.  2 mv 
1
K.E. of the electron in the orbit = mv²
2
Rutherford's Atom Model :

(1) Every atom consist of a tiny central e2

Hence K.E. =
core, called the atonic nudeus in which
potential S
E S 8 0 r

the entire positive charge and almost

A S energy of electron in orbit :
L row
N C tomU m= e  e  e
entire mass of the atom are concentrated. o 2

(2)
IO or 4   r 4  r
T you  Total energy of electron
The size of the nucleus is of the
f
A
0 0

order of 10 m which is very small

V as
–15

E
EL repa
compared to the size of the atom r
whichi ng
hydrogen atom -
in

is of the order of 10 m.
–10
p E = K.E. + U
(3) The atomic nucleus is a number
e2 e2
of electrons. As atom is electrically E= 
8 0 r 4 0 r
neutral, the total negative charge of
electrons surrounding the nucleus is e2
or E
equal to total positive charge on the 8 0 r

nucleus. Hence the total energy of electron

(4) These electrons revolve around the in orbit of hydrogen atom is negative.
nucleus in various circular orbits. The Hence the electron is bound to the
centripetal force required by electron for nucleus i.e. the electron is not free to
revolution is provided by the electrostatic
E S
leave the orbit around the nucleus.

S ofwRutherford Atom Model :

force of attraction between the electrons
S
A moro to the
Limitation
(1) L According
and the nucleus.
C
N classical
tom EM theory,
Energy of the electron in orbit :
T IO u the
for revolving electrons
V A ing yo must radiate energy in
Let F = Centripetal force required

E Eprepar
to keep a revolving electron inLorbit
C

the form of EM waves.

FC = electrostatic force of attraction
As revolving electron
between the revolving electron and the
loses energy continuously, it must spiral
nucleus.
inwards and finally fall into the nucleus,
then for a dynamically stable orbit but as matter is stable, we cannot expect
in a hydrogen atom (z=1). the atoms to collapse.
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(ii) As the revolving electrons spiral Thus the angular momentum of a
inwards, their angular velocities and orbiting etechon is quantised.
hence their frequencies of revolution
nh
would change continuously. Therefore, mvr = n = 1, 2, 3, ..
2
frequency of EM waves emitted must
Here n is called principle
change continuously.
quantum, number.
Therefore, atoms should emit
* The electron, while revolving in
continuous spectrum but we observe only
ES
SS
such orbits, shall not lose energy ie. its

A moro
line spectrum.

Bohr Model of Hydrogen Atom : C L

energy would w
stay constant.

IOou energy
for
N (3) tomThe emission/absorption of
There are three basic postulates
T occurs only when an electron
A ing y jumps from one of its specified non-
of this model -
V
LE ar
(1) E epcore
Every atom consist of a central
pr
radiating orbit to another. The difference
in the total energy of electron in the two
called nucleus, in which entire positive
charge and almost entire mass of the orbits is absorbed when the electron
atom are concentrated. A suitable jumps from an inner to an outer orbit
number of electrons revolve around the and emitted when electron jumps from
nucleus in circular orbits. The centripetal outer to the inner orbit.
force required for revolution is provided h = E2 – E1
by the electrostatic force of attraction
Where  is the frequency of
between the electron and the nucleus.
radiation emitted on jumping from outer
Centripetal force = Electrostatic to inner orbit of energy E 2 and E 1
force of attraction respectively.

ES
SS
mv 2 1  ze   e  Radius of Bohr's stationary Orbits :

r 4 0 r2
C LA mo rowfor stationary orbits -
We know that

mv 2 Kze2  K  14 0
IOou N tomfor
nh
 v 
nh

T
  2  Z 1 mvr =
2mr
Acaning y By putting the value of  in mv
r r 2

L V
E par
2
kze2
E
(2) According to Bohr, electron 
r r2
revolve only in certain discrete
p r e non
m n2h2 Kze2
radiating orbits, called stationary orbits,    2
r 4 2 m 2 r 2 r
for which total angular momention of the
n2 h 2
revolving electron is an integral multiple r for hydrogen atom z=1
42mke2
of h/2, where h is plank's constant.
It shows that – r  n2

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 Hence the radius of stationary Total Energy of electron in Bohi's
orbits are in the ratio 12 : 22 : 32 : and so Stationary Orbit :
on i.e. 1 : 4 : 9 : ...... clearly the stationary Kinetic energy of election revolving
orbits are not equally spaced. in a stationary orbit is –
Velocity of electron in Bohr's K.E. of electron =
Stationary Orbit :
1 kze2  mv2 kze2 
mv 2
kze 2
mv² =  by  2 
r r 
S
As we know that –  2 2 2r 
r r

SS E
Potential energy of electron = Potential
A
2
kze
or r
mv 2
–––– (1)
L
C m row
× charge
o
nh N
IO fo r t oofmelectron = kze   e   kze 2

T
and also r –––– (2) P.E.
2mv
V A o u r r
g y Total energy of electron in the orbit –

kze 2
nh E E
L repar
by equation (1) and (2) we get
2kze
-
i n
2 E = K.E. + P.E.
mv 2

2mv
or v  p nh 1 kze2 kze2 kze 2
E  
2ke2 2 r r 2r
for hydrogen atom, z=1 v
nh n2h2
by putting r  we get-
1 42mkze2
As v  , hence the orbital velocity
n 22mk 2z 2e4
E
of electron in outer orbits is smaller as n2 h 2
compared to its value in the inner orbits. by substituting the standard values, we
get –
Frequency of electron in Bohr's
13.6 1
Stationary Orbit- E ev E  ev
h2 n2
It is the number of revolutions

ES
Hence the total energy of electron

SS
completed per second by the electron in
in a stationery orbit is negative, which

CLA m
a stationary orbit, around the nucleus. w
It is represented by . orothat the electron is bound
mean means

N
IO fo
to the
r t om
nucleus and is not free to leave it.
As = rw = r(2)
A T you *
V
When n=1 then this state of lowest
2kze
E g
2 2

2r nh.2 r nhrL

v kze
   in
E prep r energy of the atom is called ground state.
a
The energy of this state is E = –13-6ev. 1
kze2 ke2 * Therefore, the minimum energy
   z=1 for hydrogen
nhr nhr
required to free the electron from the
The frequency of electron in subsequent
ground state of hydrogen atom is 13-6ev.
1 This is called ionisation energy of
stationary orbits is smaller as   .
n
hydrogen atom.
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* As n increases, the value of regative 1 22mk 2e4 z 2 1 1
Or   2  2
energy decreases ie energy is  ch3  n1 n2 
progressively larger in the outer orbits.
1  
Now  v , where v is the wave
Origin of Spectral Lines : 

At room temperature, most of the number of radiation emitted i.e. number

hydrogen atoms are in grourd state. of complete wave in unit length. and

When a hydrogen atom receives energy 22mk 2e 4

ES R (Rydberg Constant)

SS
by processes such as electron collisions Ch3
or heat, the atom may require sufficient Or
LA
C m row
R = 1.097×10
o
7
m–1
energy to raise the electron to higher
N
IO Hence
r t om  1 1 
T you
energy state i.e. from n=1 to n=2, 3, 4,....
fo v R   For hydrogen Z=1
A
2
 n1 n22 
The atom is then said to be in an excited
E
L aV i ng
state.
E r
pre
p Above equation is called Rydberg
formula for the spectrum of hydrogen atom.

* By abore formula it is clear that

wavelength/frequencies / ware numbers
of radiations emitted by the excited
From there excitted states, the hydrogen atom are not continuous. They
electron can fall back to a state of lower have specific values depending upon the
energy and emitting a photon of values of n1 and n2.
particular energy (= difference in * Frequency of radiation emitted
energies of the two states). when the atom makes transition from

Let E1 and E2 are the total energy the higher energy state (n2) to the lower

of election in the inner and outer orbit

ES
energy state (n1) -

SSc
LA mo r
respectively. 1 1
v w
o v  RC  2  2 
When an electron jumps from an C
N tom
 n1 n2 

T I
outer to an inner orbit, the energy ofO u Bohr's
for Explanation of spectral series of
radiation emited is given by -
V A ing yo Hydrogen Atom :
h = E – E 2 ELEprepar
1 When an atomic gas or vapour at

1 low pressure is excited usually by

22mk 2z 2e2 1
hn   2  2
h2 passing an electric current though it, the
 n1 n2 
gas/vapour emits radiations of cortain
hc 2 mk z e  1
2 2 2 4
1 specific wavelengths only. This kind of
  2  2
 h 2
 n1 n2  spectrum is called "line emission

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spectrum" and it consists of a few bright These values of v lie in the "ultra
lines on a dark background. violet region" of the spectrum.

2. Balmer Series :

According to Bohr, Balmer series

is obtained when an electron jumps to
the second orbit (n1=2) from any outer
orbit (n2 = 3, 4, 5,...)

ES
SS
When the white light is passed Wave number of these spectral

LA mor
through the same gas/vapour, we
ow
lines were calculated as –
observe a bright background crossed by C
N tom  1
a few dark lines signifying the missing
wavelengths or the wavelengths thatA
O
TI you f o r v R 
2
1

n  (n = 3, 4, 5....)
2 2 2

V are 2

E L
absorbed by the gas. They formE a "Line
p a r i ng This set of spectral lines lie in the
absorption spectrum" pr e visible part of the spectrum.
It was found that missing
wavelengths are the same as the
wavelength present in the emission
spectrum of the gas/vapour.

Bohr gave his theory of hydrogen

atom about spectral series which had
observed experimentally by various
scientists. Bohr offered a theoretical
explanation of these spectral series as

S
3. Paschen Series :
E
follows :

1. Lyman Series :
A mo SS
According to Bohr, paschan series

C L ow an electron jumps to
is obtainedrwhen
Bohr postulates that Lyman series

I N the and
tom orbit (n =3) from any outer orbit
O u (nfor= 4, 5, 6, ...). 1

T
is obtained when an electron jumps to

V A ing yo
the first orbit (n =1) from any outer orbit
2

(n = 2, 3, 4....)
2
1

E E
L r e pa r
Bohr calculated the wave number
of spectral lines of Paschen series from
p
Wave numbers of spectral lines of the relation :
lyman series were calculated as -
1 1
v R 2  2 (n2 = 4, 5, 6....)
1 1 3 n2 
v R 2  2 (n2 = 2, 3, 4....)
1 n2  The value of lie in the infrared region of
the spectrum.
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4. Brackett Series : Putting n = 1, 2, 3, ..... we get the

According to Bohr, Brackett series energies of electrons in various

is obtained when an electron jumps to stationary orbits as?

the 4th orbit (n1 = 4) from any outer orbit 13.6 13.6
E1    13.6 ev , E 2   2  3.4 ev ,
(n2 = 5, 6, 7...) 12 2

13.6 13.6
1 E3    1.51ev , E 4   2  .85 ev ,
1 32
4
Wave number v  R  2  2  (n2=5, 6, 7 ...)
4 n2 
13.6
ES 13.6

SS
E5   2
 0.54 ev , E6   2  0.37ev ,
Brackett series was discovered in 5 6

the infrared region of the spectrum.

C L A m o row
Clearly as n increases E becomes n

5. Pfund Series : N
IO for t om less negative until at n=, E =0. n

A T you The energy level diagram is shown

L V
According to Bohr Pfund series is
E i ng in figure for hydrogen atom. The
E preorbit
obtained when an electron jumps
5 orbit (n =5) from any outer
th
1
to
p a r
the
heighest energy state corresponds to
n= and has energy E=0 ev. This is the
(n2 = 6, 7, 8,...).
energy of the atom, when the electron
1 1 is removed (r=) from the nucleus and
Wave number v  R  2  2  n2 = 6, 7, 8....
5 n2  the electron is at rest.
Pfund series was discoverd in the
infrared region of the spectrum.

Energy Level Diagram :

A diagram which represents the

total energies of electron in different
stationary orbits of an atom is called the
ES
energy level diagram of that atom.
A SS
In this diagram, total energies of L
C mo ro w
electron in various stationary orbits are N r t om
IO *foAs in Increases energies of the excited
A T you state Come closer togather.
V
represented by the parallel horizontal
lines drawn according to someE ng
EL prepa i
suitable
r De Broglie's Explanation of Bohr's
energy scale.
second postulate of Quantization :
Total energy of electron in nth orbit
of hydrogen atom is – The second postulate of Bohr
model says that angular momentun of
13.6
E   2 ev electron orbiting around the nucleus is
n
quantized.
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 nh nh nh
i.e. mvr= where n = 1, 2, 3...  2r = mvr =
2 mv 2

Louis de Broglie explained this  h 

Angular momentum mvr = n  
puzzle. According to de Broglie, the  2 
election in its circular orbit, as proposed
Hence angular moment of
by Bohr, must be seen as a particle wave.
electron revolving in nth orbit must be
We know that when a string fixed an integral multiple of h 2 . Which is
at ture ends is plucked, a large number
ES
SS
the quantum condition proposed by

LA mor
of wavelengths are excited. But only
ow postulate.
Bohr in second
those waves which have nodes at the two
C
N Limitation
tom of Bohr's Theory :
ends from the standing waves and
T IOou (1)for This theory is applicable to only
survive.
V A ing y simplest atom like hydrogen, with Z=1.
E LEstanding
It means that in a string,
e p ar
pr
waves form when total distance travelled The theory falls in case of atoms of other

by a wave down the string and back is elements for which Z>1.

any integer multiple of the wavelengths. (2) The theory does not explain why

Waves with other wavelengths orbits of electrons are taken as circular,

interfere with them selves upon reflection while elliptical orbits are also possible.

and their amplitudes vanish quickly. (3) Bohr's theory does not say
anything about the relative intensities
of spectral lines.

(4) Bohr's theory does not take into

account the wave properties of electrons.
Hence, according to de Broglie, a
ES
stationary orbit is that which contains
A SS
an integral number of de Broglie waves
L
C mo ro w
associated with the revolving election.
N
IO for t om
For an electron revolving in nth
AT you
circular orbit of radius r,
E
L aV i ng
E re
For the permissible orbit, 2r =pn
r p
h
According to de Broglie  
mv

Where V is speed of electron

revolving in nth orbit

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Questions (b) Which of these lines

Q.1 Find the ratio of energies of photons corresponds to the absorption of

producad due to transition of an radiation of maximum wavelength.

election of hydrogen atom from its (a) Q.6 (a) The energy levels of an atom
Second permitted energy level to the are shown cort in fig. which of them
first leve. (b) The heighest permitted will result in the emission of a
energy level to the first permitted level. photon of wavelength 275 mm?

Q.2 What is the ratio of radius of the orbits (b)

E S Which transition corresponds
S w of radiation of maximum
S
CLAwavelength?
corresponding to first excited state to emission
ro
mo
N
and ground state in a hydrogen atom?
om
T you
Q.3 The radius of innermost electron orbit IO for t

V A
E
of a hydrogen atom is 5.3×10–11 m.
ng
E L repa
What is the radius of orbit in r
thei
second excited state? p
Q.4 Calculate the wavelength of H line
(n2=3) in Balmer series of hydrogen
atom, given Rydberg Constant of
R= 1.097×107 m–1

Q.5 Photons, with a continuous range

of frequencies are made to pass
through a simple of hydrogen. The
transitions shown here, indicate
three of the spectral absorption
lines in the continuous spectrum.
ES
A SS
L
C mo ro w
N
IO for t om
AT you
E
L aV i ng
E r
pre
p

(a) Identify the spectral series of

the hydrogen emission spectrum, to
which each of these three lines
corresponds.
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