ATOMIC STRUCTURE CLASS 11 JEE BEST TEACHER NOTES

JEE-Physics

ATOMIC STRUCTURE
VARIOUS MODEL S FOR STRUCTURE OF ATOM
• Dalton's Theory
Every material is composed of minute particles known as atom. Atom is indivisible i.e. it cannot be subdivided.
It can neither be created nor be destroyed. All atoms of same element are identical physically as well as
chemically, whereas atoms of different elements are different in properties. The atoms of different elements are
made up of hydrogen atoms. (The radius of the heaviest atom is about 10 times that of hydrogen atom and its
mass is about 250 times that of hydrogen). The atom is stable and electrically neutral.

• Thomson's Atom Model
The atom as a whole is electrically neutral because the positive charge present on the atom (sphere) is equal to
the negative charge of electrons present in the sphere. Atom is a positively charged sphere of radius 10–10 m
in which electrons are embedded in between. The positive charge and the whole mass of the atom is uniformly
distributed throughout the sphere.

electron


positively
charged
matter


• Shortcomings of Thomson's model
(i) The spectrum of atoms cannot be explained with the help of this model
(ii) Scattering of –particles cannot be explained with the help of this model

RUTHERFORD ATOM MODEL

• Rutherford experiments on scattering of  – particles by thin gold foil
The experimental arrangement is shown in figure. –particles are emitted by some radioactive material (polonium),
kept inside a thick lead box. A very fine beam of –particles passes through a small hole in the lead screen. This
well collimated beam is then allowed to fall on a thin gold foil. While passing through the gold foil, –particles
are scattered through different angles. A zinc sulphide screen was placed out the other side of the gold foil. This
screen was movable, so as to receive the –particles, scattered from the gold foil at angles varying from 0 to
180°. When an –particle strikes the screen, it produces a flash of light and it is observed by the microscope. It
was found that :
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vacuum ZnS
lead screen
mi




lead box gold foil
screen
cr o




–7
10 m
s co




N() cosec 
4
pe




2
beam of
N ( )




-particle  most -pass
through
source of
-particle

about 1 in 8000 is some are deviated 90° 180°
repelled back through large angle  



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• Most of the  – particles went straight through the gold foil and produced flashes on the screen as if there
were nothing inside gold foil. Thus the atom is hollow.
• Few particles collided with the atoms of the foil which have scattered or deflected through considerable large
angles. Few particles even turned back towards source itself.
• The entire positive charge and almost whole mass of the atom is concentrated in small centre called a nucleus.
• The electrons could not deflected the path of a  – particles i.e. electrons are very light.
• Electrons revolve round the nucleus in circular orbits. So, Rutherford 1911, proposed a new type of model
of the atom. According to this model, the positive charge of the atom, instead of being uniformly distributed
throughout a sphere of atomic dimension is concentrated in a very small volume (Less than 10 –13m is
diameter) at it centre. This central core, now called nucleus, is surrounded by clouds of electron makes.
The entire atom electrically neutral. According to Rutherford scattering formula, the number of  – particle

N 0 nt(2Ze 2 )2 1
scattered at an angle  by a target are given by N  2 2 2 2

16(4  0 ) r (mv 0 ) sin 4 
2

Where N0 = number of  – particles that strike the unit area of the scatter
n = Number of target atom per m3
t = Thickness of target
Ze = Charge on the target nucleus
2e = Charge on  – particle
r = Distance of the screen from target
v0 = Velocity of  – particles at nearer distance of approach the size of a
nucleus or the distance of nearer approach is given by

1 (2Ze) 2 1 (2Ze) 2
r0    where EK = K.E. of particle
4 0  1 2 4  0 E K
 2 mv 0 
 




r0
us
e le c

cle
nu




-particle
tro n




+
 + nucleus
Ze
b target
nucleus
area = b2


Bohr's Atomic Model
In 1913 Neils Bohr, a Danish Physicist, introduced a revolutionary concept i.e., the quantum concept to explain the NODE6\E\Data\2014\Kota\JEE-Advanced\SMP\Phy\Unit No-12\Modern Physics\Eng\01_Modern Physics.p65

stability of an atom. He made a simple but bold statement that "The old classical laws which are applicable to bigger
bodies cannot be directly applied to the sub–atomic particles such as electrons or protons.

Bohr incorporated the following new ideas now regarded as postulates of Bohr's theory.
1. The centripetal force required for an encircling electron is provided by the electrostatic attraction between the

1  Ze  e mv 2
nucleus and the electron i.e.  ...(i) v
4 0 r 2 r
0 = Absolute permittivity of free space = 8.85 × 10–12 C2 N–1 m–2 r Electron

m = Mass of electron +
v = Velocity (linear) of electron Nucleus
r = Radius of the orbit in which electron is revolving. +Ze
Z = Atomic number of hydrogen like atom.

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2. Electrons can revolve only in those orbits in which angular momentum of electron about nucleus is an integral

h nh
multiple of . i.e., mvr = ...(ii)
2 2
n = Principal quantum number of the orbit in which electron is revolving.


3. Electrons in an atom can revolve only in discrete circular orbits called stationary energy levels (shells). An
electron in a shell is characterised by a definite energy, angular momentum and orbit number. While in any of
these orbits, an electron does not radiate energy although it is accelerated.


4. Electrons in outer orbits have greater energy than those in inner orbits. The orbiting electron emits energy when
it jumps from an outer orbit (higher energy states) to an inner orbit (lower energy states) and also absorbs energy
when it jumps from an inner orbit to an outer orbit. En – Em = hn,m E3
E2
where, En = Outer energy state
E1
Em = Inner energy state
+
n,m = Frequency of radiation


5. The energy absorbed or released is always in the form of electromagnetic radiations. Nucleus




MATHEM ATICAL ANALYSIS OF BOHR'S THEORY


1  Ze  e mv 2 nh
From above equation (i) and (ii) i.e.,  and mvr = ...(ii)
4 0 r 2 r 2

We get the following results.


1. Velocity of electron in nth orbit : By putting the value of mvr in equation (i) from (ii) we get

1  nh  Z  e2  Z
Ze 2     v  v     v 0 ....(iii)
4  0  2  n  2 0 h  n


19 2

Where, v0 =
1.6  10  = 2.189 × 106 ms–1 =
c
= 2.2 × 106 m/s
2  8.85  10 12
 6.625  10 34 137
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where c = 3 × 108 m/s = speed of light in vacuum




2. Radius of the nth orbit : From equation (iii), putting the value of v in equation (ii), we get

Z e2  nh n 2  0 h 2  n 2
m  r   r=   r0 ...(iv)
 n 2 0 h  2 Z  me 2  Z

2

8.85  10 12  6.625  10 34 
where r0 = 2 = 0.529 × 10–10 m  0.53 Å

3.14  9.11  10 31  1.6  10 19 

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3. Total energy of electron in n th orbit :

1 Ze 2 1  Ze  e 
From equation (i) KE  mv 2  and PE   2K.E.  |PE| = 2 KE
2 8  0 r 4 0 r

 Ze 2
Total energy of the system E = KE + PE = –2KE + KE = – KE =
8  0 r


Z2  me 4  Z 2
By putting the value of r from the equation (iv), we get E    8 2 h 2   n 2  E 0 ...(v)
n2  0 

4

where E 0 
 
 9.11  10 3 1.6  10 19   13.6 eV
2 34 2

8  8.85  10 12   6.625  10 
2 r
4. Time period of revolution of electron in nth orbit : T 
v

n 3  4 20 h 3  n 3
By putting the values of r and v, from (iii) and (iv) T    T0
Z 2  me 4  Z 2

2 34 3

4  8.85  10 12   6.625  10 
where, T0  4 = 1.51 × 10–16 second
9.11  10 31  1.6  10  19




5. Frequency of revolution in nth orbit :

4
1 Z2 me 4 Z2
f   3  2 3  3  f0 where, f0 
9.11  10 31  1.6  10 19   = 6.6 × 1015 Hz
2 3
T n 4 0 h n 4  8.85  10 12 6.625  10 34   

6. Wavelengt h of photon

me 4 1 1  2 1 1  2 hc 1 me 4  1 1 
E  E n 2  E n1   2  2  Z  13.6  2  2  Z  E      2 3  2  2  Z2
8 20 h 2 n
 1 n 2  n
 1 n 2 
  8  0 h c n
 1 n 2 

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 1 1  2
 R   2  2  Z where,  is called wave number..
n1 n2 

R   R H = Rydberg constant

4

=

9.11  10 31  1.6  10 19  = 1.097 × 107 m–1 = 1.097 × 10–3 Å–1 (for stationary nucleus).
2 34 3

8  8.85  10 12  6.625  10   3  10 

If nucleus is not stationary (i.e., mass of nucleus is not much greater than the mass of the revolving particle like
R
electron), then R = where, m = mass of revolving particle and M = mass of nucleus
1m/M



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