Sr|12th IIT-JEE MAIN|NEET|PHYSICS:VOL-III

 -particle  The observation (a) indicates that most of the

Rutherford’s Scattering
portion of the atom is hollow inside.
Experiment:  Because  - particle is positively charged, from the
-particles -particles observations (b), (c) and (d) atom also have
detector

positive charge and the whole positive charge of
the atom must be concentrated in small space which
Radio Lead cavity is at the centre of the atom is called nucleus. The
active Lead Plate
source with slit
Thin Gold Foil
remaining part of the atom and electrons are
revolving around the nucleus in circular objects of
Experimental Observations: all possible radii. The positive charge present in
a) Most of the - particles were found to pass through the nuclei of different metals is different . Higher
the gold- foil without being deviated from their the positive charge in the nucleus, larger will be the
paths. angle of scattering of - particle.
b) Some - particles were found to be deflected
through small angles   90 . Distance of Closest Approach :
c) Few - particles were found to be scattered at  An - particle which moves straight towards the
fairly large angles from their initial path   90 nucleus in head on direction reaches the nucleus
i.e, it moves close to a distance r0 as shown the
figure.

Incident beam of -particles

1
2   As the - particle approaches the nucleus, the
3
4 electrostatic repulsive force due to the nucleus
5 Space
6
 Nucleus increases and kinetic energy of the alpha particle
7 (+Ze) occupied
8 by Electrons goes on converting into the electrostatic potential
9
10  
11 energy. When whole of the kinetic energy is
  converted into electrostatic potential energy, the
- particle cannot further move towards the
nucleus but returns back on its initial path i.e
r0 - particle is scattered through an angle of 180°.
d) A very small number of - particles about 1 in The distance of - particle from the nucleus in this
8000 practically retracted their paths or suffered stage is called as the distance of closest approach
deflections of nearly 180° . and is represented by r0.

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 Sr|12th IIT-JEE MAIN|NEET|PHYSICS:VOL-III
 Let m and v  be the mass and velocity of the the foil.
- particle directed towards the centre of the  Impact Parameter(b):The perpendicular
nucleus. Then kinetic energy of the - particle distance of the initial velocity vector of the
 - particle from centre of the nucleus is called
 1 “impact parameter”.
K    m  v 2
 2  
Ze 2 cot  
Because the positive charge on the nucleus is Ze 2
and that on the  -particle 2e, hence the electrostatic b
1
potential energy of the - particle, when at a 4 0  mv 2
2
distance r0 from the centre of the nucleus, is given Bohr’s model of atom :
1 (2e)(Ze)  Electron can revolve round the nucleus only in
by U  4 . r certain allowed orbits called stationary orbits and
0 0
the Coulomb’s force of attraction between electron
Because at r = r0 kinetic energy of the - particle
and the positively charged nucleus provides
appears as its potential energy, hence, K=U
necessary centripetal force.
1 (2e)(Ze) 1
.  m v2
4 0 r0 2 –e

1 4Ze2
r0  k ( Ze ) e mV 2 +Ze
4 0 m  v 2 
r2 r Nucleus
Alpha partical scattering (additional)
 When a mono energitic beam of  particles is
projected towards a thin metal foil, some of the  Suppose m is the mass of electron, V is the
particles are found to deviate from their original path. velocity and ‘r’ is the radius of the orbit, then in
This phenomenon is called  ray scattering stationary orbits the angular momentum of the
 It is caused by coulomb repulsive force between h
 particles and positive charges in atom. electron is an integral multiple of, where h is
2
The number of  -particles scattered at an angle 
the Planck’s constant. The angular momentum
Q n t z 2 e4 h
is given by N = L  I   mVr  n where n is called principal
(8πε 0 ) r E sin   2
2 2 2 θ 4
 2 quantum number.
where  An electron in a stationary orbit has a definite
Q  Total number of  particles striking the foil amount of energy. It posses kinetic energy
n  number of atoms per unit volume of the foil because of its motion and potential energy on
r  distance of screen from the foil account of the attraction of the nucleus. Each
t  thickness of the foil allowed orbit is therefore associated with a certain
z  Atomic number of the foil atoms quantity of energy called the energy of the orbit,
  angle of scettering which equals the total energy of the electron in it.
E  kinetic energy of  particles In these allowed orbits electrons revolve without
radiating energy .
1
N  t; N  z2 ; N 
 Energy is radiated or absorbed when an electron
sin4 jumps from one stationary orbit to another
2
stationary orbit. This energy is equal to the energy
1 1 difference between these two orbits and emitted
N 2 or N 
E 4 or absorbed as one quantum of radiation of
where  is the velocity of  particles falling on
242 ATOMIC PHYSICS
Sr|12th IIT-JEE MAIN|NEET|PHYSICS:VOL-III
frequency v given by Planck’s equation
 c  Z
hc Vn    . m / s .......... (6)
E2  E1  hv  . This is called Bohr ’ss  137  n

Where ‘c’ is the speed of light in vacuum.
frequency condition.
Conclusion (iii) Time period of electron in the orbit :
 (i) Radius of Bohr’s orbit : When mass of the Angular velocity of electron in nth orbit
nucleus is large compared to revolving electron,
then electron revolves around the nucleus in circular Vn  0 Z 2 8 3 k 2 e 4 m
n   3 where  0  ......(7) is
orbit. According to first postulate rn n h3
k ( Ze)e mV 2  1  the angular velocity of electron in first Bohr’s orbit.
  where k   The time period of rotation of electron in nth orbit
r 2
r  4 0  ......(1)
According to second postulate 2 n3 n3
T  2 .......... (8) i.e T  2 .
h  n 2 0 Z Z
mVr  n where n = 1,2,3,4.............
2 The time period of rotation increases as n increases
nh and is independent on the mass of the electron.
(or) V  .................. (2) (iv) Kinetic Energy of the electron in the orbit
2 mr
After solving the equations, radius of the orbit  The kinetic energy of the electron revolving round
n2h2 the nucleus in nth orbit is given by
r
4 2 kZme 2 1 1  2 ke 2 Z 
2

K n  mV 2  m  . 
h2   n2 2 2  h n
For nth orbit rn  . 
 ............(3)
4 2 ke 2  mZ

2  2 k 2 e 4  mZ 2  mZ 2
For hydrogen atom Z = 1, radius of the first orbit Kn  .   .......(9); K 
h2  n2  n
n2
(n = 1) is given by r1  0.529  10  10 m  0.53 Å
(v) Potential Energy of the electron in the orbit :
This value is called as Bohr’s radius and the orbit is
called Bohr’s orbit. In general, the radius of the k(Ze)e  4  2 kmZe 2 
Un     kZe 2  
nth orbit of a hydrogen like atom is given by rn 2 2
 n h 
 n2 
rn  0.53   Å where n  1, 2, 3,....... (4) 4  2 k 2 e 4  mZ 2 
Z  Un    2  ....... (10)
h2  n 
(ii) Velocity of the Electron in the orbit :
The velocity of an electron in n th orbit (vi)Total energy of the electron in nth orbit
Total energy of the electron in nth orbit
nh
Vn  2 2 k 2 mZ 2 e 4
2 mrn hence En  K n  Un  
n2 h2
2 ke 2  Z   n2 h2 
Vn  .   rn   ..(5) 2  2 k 2 e 4  mZ 2 
h n  4 2 kZme 2  En    2  ........ (11)
h2  n 
i.e the velocity of electron in any orbit is
independent of the mass of electron. The above The expression of total energy for hydrogen like
equation can also be written as atom may be simplified as

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 Sr|12th IIT-JEE MAIN|NEET|PHYSICS:VOL-III
n2 to a lower energy level n1 in stationary atom,
Z2 the difference in energy is radiated as a photon
E n  13.6 eV , n = 1,2,3..... (12)
n2 whose frequency v is given by Planck’s formula.
Where -13.6 eV is the total energy of the electron E n 2 - E n1 = hn
in the ground state of an hydrogen atom.
From the equations (9),(10)&(11) it is clear that 1 1 
PE : K.E : T.E = -2 : 1 : -1 (or) hv  E2  E1  13.6 Z 2  2
 2  eV
.
 n1 n2 
PE KE TE
i.e    13.6 Z 2 
2 1 1  En   n 2 e.V  since i.e., V  1.6  10 19 J
 
 The state n = 1 is called ground state and n > 1
states are called excited states. When electron go  
hence h c  (12.8 1018 ) Z 2  12  12  J
from lower orbit to higher orbit speed and hence   n1 n2 
kinetic energy decrease, but both potential energy
21  1
1 1 
1 (or) wave number v    RZ .  n 2  n 2  m
and total energy increases. E tells us that  1 2 
n2
where R is called for “Rydberg constant”, when
the energy gap between the two successive levels
the nucleus is infinitely massive as compared to the
decreases as the value of n increases. At infinity
revolving electron. In other words the nucleus is
level the total energy of the atom becomes zero.
Energy level diagram of hydrogen atom (Z = 1) considered to be stationary. The numerical value
for normal and excited states as shown the figure. of R is 1.097×107 m-1.
The energy level diagram of hydrogen like atom Emission Spectrum of Hydrogen atom :
with atomic number Z for normal and excited states
Electron in hydrogen atom, can be in excited state
as shown in Figure.
for very small time of the order of 10-8 second.
n= 0eV This is because in the presence of conservative
force system particles always try to occupy stable
n=5 –0.544eV equilibrium position and hence minimum potential
n=4 –0.850eV
nd
2 excited state energy, which is least in ground state. Because of
n=3 –1.511eV
instability, when an electron in excited state makes
1sr excited state
n=2 –3.4eV a transition to lower energy state, a photon is
emitted. Collection of such emitted photon
n=1
ground state
–13.6eV frequencies is called an emission spectrum. This is
as showing in figure.
 The total energy of the electron is negative implies
the atomic electron is bound to the nucleus. To e–
Transition
remove the electron from its orbit beyond the photon
e– +
attraction of the nucleus , energy must be required. N N

 The minimum energy required to remove an

electron from the ground state of an atom is called
its ionization energy and it is 13.6 Z2 eV.
Emission spectrum
 In hydrogen atom the ground state energy of
electron is –13.6 eV, so 13.6 eV is the ionization
energy of the Hydrogen atom.
Emission of radiation:
 When an electron jumps from higher energy level

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1 1 1
n=  R 2  2   max  6568Å
max 2 3 
n=5 bluegreen P fund series
n=4 (I.R. region) 1 1 1 
Bracket series  R 2  2    m in  3636 Å
n=3 L
(I.R. region) min 2  
L L  Paschen series
(I.R. region)  Balmer series lies in the visible region of
n=2 k Balmer series
k (Visible region) electromagnetic spectrum. The wavelength of L
k
n=1 line is 656.8 nm (red). The wavelength of L line
Lyman series
(u.v. region)
is 486 nm (blue green). The wavelength of L  line
The Spectral Series of Hydrogen Atom as shown is 434 nm (violet). The remaining lines of Balmer
in figure, are explained below. series closest to violet light wavelength. The
a) Lyman Series : Lines corresponding to transition speciality of these lines is that in going from one
end to other, the brightness and the separation
from outer energy levels n2  2, 3, 4,.......... to
between them decreases regularly.
first orbit (n1= 1) constitute Lyman series. The  This series is obtained only in emission spectrum.
wave numbers of different lines are given by, Absorption lines corresponding to Balmer series
1 1 1  do not exist, except extremely weakly, because very
v  R 2  2 
 1 n2  few electrons are normally in the state n = 2 and
 Line corresponding to transition from n2 = 2 to n1 only a very few atoms are capable of having an
= 1 is first line; its wavelength is maximum. electron knocked from the state n = 2 to higher
states. Hence photons that correspond to these
1 1 1  1 1 
 R  2  2   1.1107      energies will not be strongly absorbed. In highly
max 1 2  1 4  max  1212 Å
excited hydrogen gas there is possibility for
Similarly transition from n2   to n1  1 gives detecting absorption at Balmer-line wavelengths.
line of minimum wavelength. c) Paschen Series: Lines corresponding to n2 =
4,5,6,.......  to n1 = 3 constitute Paschen series.
1 1 1 
 R  2  2   1.1107   The wave number of different lines are given by
m in 1   min  912 Å

 Lyman series lies in ultraviolet region of electro 1 1

v R 2  2
magnetic spectrum. 3 n2 
 Lyman series is obtained in emission as well as in
 Line corresponding to transition n2 = 4 to
absorption spectrum.
n1 = 3 is first line, having maximum wavelength.
b) Balmer Series: Lines corresponding to
n2  3, 4, 5,........ to n1 = 2 constitute Balmer Line corresponding to transition n2   to n1 = 3
series. The wave numbers of different lines are is last line, having minimum wavelength
1 1 1
1 1
1  R  2  2  
max  18747Å
given by, v   R  2  2   max 3 4 
 2 n2 
1  1 1 1 
 Line corresponding to transition n2=3 to n1=2 is  R  2    1.1  107    0 
 min 3   9 
first line, wavelength corresponding to this transition
is maximum. Line corresponding to transition  min  8202Å
n2   to n1 = 2 is last line; wavelength of last  Paschen series lies in the infrared region of
line is minimum. electromagnetic spectrum.
 This series is obtained only in the emission spectrum.

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 Sr|12th IIT-JEE MAIN|NEET|PHYSICS:VOL-III
d) Bracket Series: The series corresponds to
transitions from n2 = 5,6,7.......,  to n 1 = 4.
The wave number are given by, n2

1 1 1
v  R 2  2 n1
 4 n2 
 Line corresponding to transition from n2=5 to n1=4
has maximum wavelength and n2   to Note 1: for n 2  4 , and n1 = 1, the number of
n1 = 4 has minimum wavelength. possible lines are 6.
1 1 1 Note 2 : If E is the energy difference between
 R  2  2  
 max max  40477Å
4 5  two given energy states, then due to transition
between these two states wavelength of emitted
1 1 1 
 R  2  2  
 min 4   min  14572Å photon is ( Å ) 
12400
E (eV )
 This series lies in the infrared region of
electromagnetic spectrum. Limitation of Bohr’s model :
e) Pfund Series: This series corresponds to Despite its considerable achievements, the Bohr’s
transitions from n 2  6,7,8,....,  to n1  5 . The model has certain short coming.
1  1 1   It could not interpret the details of optical spectra
wave numbers are given by v    R  
2
5
 n 22  of atoms containing more than one electron.
 Line corresponding to transition from n2 = 6 to n1  It involves the concept of orbit which could not be
= 5 has maximum wavelength and n2   to n1 = checked experimentally
4 has minimum wavelength.  It could be successfully applied only to single-
1 1 1 electron atoms (e.g., H, He+ , Li2+, etc.)
 R 2  2
 max 5 6   Bohr’s model could not explain the binding of atoms
into molecules.
1 1 1 
 max  74563Å   R 2  2   No justification was given for the “principle of
min 5  
quantization of angular momentum”.
 min  22768Å  Bohr’s model could not explain the reason why
 This series lies in infrared region of electromagnetic atoms should combine to form chemical bonds and
spectrum.
Note : In an atom emission transition may start why do the molecules become more stable on such
from any higher energy level and end at any energy combinations.
level below of it. Hence in emission spectrum the  Bohr had assumed that an electron in the atom is
total possible number of emission lines from some located at definite distance from the nucleus and is
excited state n2 to another energy state n1 ( n 2 ) is
revolving with a definite velocity around it. This is
(n 2  n1 )(n 2  n1  1)
against the Heisenberg uncertainty principle. With
2
the advancements in quantum mechanics, it be came
clear that there are no well defined orbits; rather
there are clouds of negative charge.

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Final Maximum
Suppose r be the radius of nth orbit. Then the
S.No Name of State Initial State Formula Series Region
the series
(n1)
(n2) limit wavelength
necessary centripetal force is provided by the
mv 2
1 æ1 1 ÷ ö 4
1. Lyman = Rççç 2 - 2 ÷ 1 UV
n1 = 1 2,3,4,....¥ l =
l çè1 n2 ÷
÷
ø
l =
R
= 911A°
3R above force. Thus,  2ar -------- (i)
r
1 æ1 1 ö÷ 4 36
2. Balmer n1 = 2 3, 4,5....¥
l
= R ççç 2 - 2 ÷
÷
çè2 n2 ø÷
l =
R
l =
5R
Visible
Further, the quantization of angular momentum
1 æ1 1 ö÷ 9 144 nh
3. Paschen n1 = 3 4, 5,6....¥ = R ççç 2 - 2 ÷ l = Near IR
l
÷
çè3 n2 ø÷
l =
R 7R gives, mvr  -------- (ii)
2
1 æ1 1 ö÷ 16 400
4. Brackett n1 = 4 5, 6, 7....¥ = R ççç 2 - 2 ÷ ÷ l = l = Middle IR
l çè 4 n 2 ø÷ R 9R 1/ 4
 n2 h2 
5. Pfund n1 = 5 6, 7 , 8 ....¥ 1 æ1 1 ö÷
= R ççç 2 - 2 ÷ l =
25 l =
9000
Far IR Solving Eqs. (i) and (ii) for r, we get r 
 8am 2 
÷
l èç5 n 2 ø÷ R 11R

EX. 1: The electron in a hydrogen atom makes a EX. 4: Consider a hydrogen-like atom whose
transition n1  n 2 where n1 and n2 are the energy in n th excited state is given
principal quantum numbers of the two states. 13.6Z 2
Assume the Bohr model to be valid. The time by En   when this excited atom makes
n2
period of the electron in the initial state is eight transition from excited state to ground state
times that in the final state. What are the most energetic photons have energy
possible values of n1 and n2 ? E max  52.224eV and least energetic photons
3
T1 n13 n  have energy E min  1.224eV . Find the atomic
Sol. Since, T  n3 ; T  n3 , As T1 = 8T2,  1   8
2 2  n2  number of atom and the state of excitation.
(or) n1 = 2n2. Thus the possible values of n1 and n2 Sol. Maximum energy is liberated for transition En  1
are n1  2, n 2  1,n1  4, n2  2, n1  6, n2  3; and so and minimum energy for En  En 1
on. E1
EX. 2: Find the kinetic energy, potential energy Hence,  E1  52.224eV ... (1)
n2
and total energy in first and second orbit of
hydrogen atom if potential energy in first orbit E1 E1
and n 2  (n  1)2  1.224 eV ...... (2)
is taken to be zero.
Sol. E1 = -13.60eV; K1 = -E1 = 13.60eV Solving above equations simultaneously, we get
U1 = 2E1 = -27.20eV E 2  3.40eV K2 = 3.40eV 13.6Z 2
and U2 = -6.80eV E1  54.4eV and n = 5 Now E1    54.4eV .
12
Now, U1 = 0, i.e., potential energy has been Hence, Z = 2 i.e, gas is helium originally excited to
increased by 27.20eV. So, we will increase U and n = 5 energy state.
E in all energy states by 27.20eV while kinetic EX. 5: A hydrogen-like atom (atomic number Z) is
energy will remain unchanged.
in a higher excited state of quantum number n.
Hence K(eV), U(eV),E(eV)
This excited atom can make a transition to the
First orbit are 13.6, 0, 13.6
first excited state by successively emitting two
in Second orbit 3.40, 20.40, 23.80
EX. 3: A small particle of mass m moves in such a photons of energies 10.20 eV and 17.00 eV
way that the potential energy U = ar2 where a respectively. Alternatively the atom from the
is a constant and r is the distance of the same excited state can make a transition to the
particle from the origin. Assuming Bohr’s second excited state by successively emitting
model of quantization of angular momentum two photons of energies 4.25 eV and 5.95 eV
and circular orbits, find the radius of nth respectively. Determine the values of n and Z
allowed orbit. (ionization energy of hydrogen atom = 13.6 eV)
dU Sol. The electronic transitions in a hydrogen-like atom
Sol. The force at a distance r is, F    2ar
dr from a state n2 to a lower state n1 are given by

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Sol. (i) Let n1 be initial state of electron. Then
 1
2 1
E  13.6Z  2  2  . For the transition from 13.6
 n1 n 2  E1   eV Here E1 = - 0.85 eV, therefore
n12
a higher state n to the first excited state n1 = 2,
the total energy released is (10.2 + 17.0) eV or 13.6
0.85   or n1 = 4
27.2eV. Thus E = 27.2 eV,, n12
n1 = 2 and n2 = n. (ii) Let n2 be the final excitation state of the
2 1 1 electron. Since excitation energy is always
We have 27.2  13.6Z  4  2  ...... (1)
 n  measured with respect to the ground state, therefore
For the eventual transition to the second excited
state n1 = 3, the total energy released is (4.25 +
 1 
5.95) eV or 10.2eV. E  13.6 1  2  here E  10.2eV, therefore,
 n 2 
2 1 1
Thus 10.2  13.6Z  9  2  .....(2)
 n   1
10.2 = 13.6 1  n 2  or n2 = 2 Thus, the electron
27.2 9n 2  36  2 
Dividing the Eq. (1) by Eq. (2) we get  .
10.2 4n 2  36 jumps from n1 = 4 to n2 = 2.
Solving we get n2 = 36 or n = 6 (iii) The wavelength of the photon emitted for a
Substituting n = 6 in any one of the above equations, transition between n1 = 4 to n2 = 2, is given by
we obtain Z2 = 9 (or) Z = 3, Thus n=6 and Z=3.
EX. 6: A doubly ionized lithium atom is hydrogen 1 1 1 1 7 1 1
 R   2  2  (or)   1.09  10  2  2  =4860Å .
like with atomic number Z = 3. Find the   n 2 n1  2 4 
wavelength of the radiation required to excite EX. 8: A hydrogen atom initially in the ground level
the electron in Li2+ from the first to the third absorbs a photon, which excites it to the n = 4
Bohr orbit. Given the ionization energy of level. Determine the wavelength and
hydrogen atom as 13.6 eV. frequency of photon. To find the wavelength
Sol. The energy of nth orbit of a hydrogen-like atom is and frequency of photon use the relation of
13.6Z 2 energy of electron in hydrogen atom is
given as E n   Thus for Li2+ atom, as Z =
n2 13.6
3, the electron energies of the first and third Bohr En=  eV .
n2
orbits are For n = 1,E1 = –122.4eV, for n = 3, E3
Sol. For ground state n1 = 1 to n2 = 4.
= –13.6eV. Thus the energy required to transfer an
Energy absorbed by photon, E = E2  E1
electron from E1 level to E3 level is, E  E 3  E1
 13.6  (122.4)  108.8eV . Therefore, the  1 1 
= 13.6  2  2  1.6 1019 J
radiation needed to cause this transition should have  n1 n 2 
photons of this energy. hv = 108.8 eV. The
1 1  19
wavelength of this radiation is or  
hc
 = 13.6   2  1.6 10
108.8eV 1 4 
114.25 Å
 15  19
EX. 7: A hydrogen atom in a state of binding energy = 13.6 1.6 10   = 20.4  10-19.
0.85 eV makes a transition to a state of  16 
excitation energy of 10.2 eV. or E = h = 20.4  10-19
(i) What is the initial state of hydrogen atom? 20.4 1019
20.4 1019
(ii)What is the final state of hydrogen atom ? Frequency  = =
h 6.631034
(iii) What is the wavelength of the photon
emitted ? = 3.076  1015 = 3.1  1015 Hz.

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For n = 2, radius rn = n2r1
c 3 108
Wavelength of photon λ = = 15 =
r2 = 22 .r1 = 4  0.53  10-10 m and
 3.076 10
v1
9.74  10-8 m. Thus, the wavelength is 9.7  10-8 m velocity vn=
n
and frequency is 3.1  1015 Hz.
v1 2.19 106
 v2 = =
EX. 9: (a) Using the Bohr’s model calculate the 2 2
speed of the electron in a hydrogen atom in
2  3.14  4  0.53 10 10  2
the n = 1, 2 and 3 levels. Time period T2 =
2.19 106
(b) Calculate the orbital period in each of these = 1.216  10-15 s.
levels. For n = 3, radius r3 = 32
Sol. (a) Speed of the electron in Bohr’s nth orbit r1 = 9 r1 = 9  0.53  10-10 m
c 2πKe 2 v1 2.19 106
 = α where, α = and velocity v3 = = m/s
v ch 3 3
c 2πr3
α = 0.0073  v=  0.0073 Time period T3 = v
n 3

c
For n = 1,  0.0073
v1 = 2  3.14  9  0.53 10-10  3
1 = = 4.1  10-15 s.
2.19 106
= 3  108  0.0073 = 2.19  106 m/s
EX. 10: The radius of the innermost electron orbit
c
For n = 2 v2 =  0.0073 of a hydrogen atom is 5.3  10-11 m. What are
2
the radii of the n = 2 and n = 3 orbits?
3 108  0.0073 Sol. Given, the radius of the innermost electron orbit of
= = 1.095  106 m/s
2 a hydrogen r1 = 5.3  10-11 m.
As we know that rn = n2r1
c
For n = 3 v3 =  0.0073 For n = 2, radius r2 = 22
3
r1 = 4  5.3  10-11 = 2.12  10-10 m.
3 108  0.0073 For n = 3, radius r3 = 32
= = 7.3  105 m/s.
3 r1 = 9  5.3  10-11 = 4.77  10-10 m.
2πr EX. 11: A 12.5 eV electron beam is used to bombard
(b) Orbital period of electron is given by T = gaseous hydrogen at room temperature. What
v
series of wavelength will be emitted?
2
 n 2   h  4 0 Sol. Energy of electron beam E=12.5 eV
Radius of nth orbit rn     
 m   2  e = 12.5  1.6  10-10 J
2

Planck’s constant h=6.63  10-34 J-s

(1) 2  (6.63 10 34 ) 2 Velocity of light c=3  108 m/s
 r1 = 4  9.87  (9 10)9  9 1031  6.6 10 19 )
34
hc 6.62 10  310
8

= 0.53  10 -10
m. Using the relation E= =
λ 12.5 1.6 10 19
2πr1 = 0.993  10-7 m = 993  10-10 m = 993 A0
For n = 1 T1 = v
1 This wavelength falls in the range of Lyman series
(912 A0 to 1216A0)
2  3.14  0.53 10 10 thus, we conclude that Lyman series of wavelength
= = 1.52  10-16 s
2.19 106
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 Sr|12th IIT-JEE MAIN|NEET|PHYSICS:VOL-III
993 A0 is emitted. Mr mr
EX. 12:In accordance with the Bohr’s model, find r1  , r2 
M m M m
the quantum number that characterises the
vi) For both the eletron and the nucleus the necessary
earth’s revolution around the sun in an orbit
centripetal force to revolve in circular orbits is
of radius 1.5  1011 m with orbital speed 3  104
provided by the electrostaic force between them.
m/s. (Mass of earth = 6.0  1024 kg).
A. Given, radius of orbit r = 1.5  1011 m 1 Ze 2 Mr 1 Ze 2
i.e., mr1   
2 2
; m
Orbital speed v=3  104m/s; 4 0 r 2 M m 4 0 r 2
Mass of earth M=6  1024kg
nh 2πvrm Ze 2
i.e.,  r 3 2  ----- (1)
Angulalr momentum, mvr = or n = 4 0
2π h
[where, n is the quantum number of the orbit] Mm
where   called reduced mass
2  3.14  3 10 1.5 10  6 10
4 11 24
M m
= vii) From Bohr’s theory of quantization of angular
6.63 10 34
moentum, total angular momentum of the system
= 2.57  1074 or n = 2.6  1074 . Thus, the
quantum number is 2.6  1074 which is too large. nh nh
L  I1  I 2  ; mr1   Mr2  
2 2

The electron would jump from n=1 to n=3 2 2

13.6 M 2r 2 Mm2 r 2 nh
E3 = =  1.5 eV.. m   
32 ( M  m) 2
( M  m) 2
2
So, they belong to Lyman series.
mMr 2 nh nh
EFFECT OF FINITE MASS OF NUCLEUS ( M  m)  ie  r  
2
--- (2)
( M  m) 2
2 2
ON BOHR’S MODEL OF AN ATOM
i) In the atomic spectra of hydrogen and hydrogen viii) A system of this type is equalent a single particle of
like atoms a very small deviation with Bohr’s model mass  revolving around the position of the heavier
results particle(nucleus) in an orbit of radius r.
ii) This is in the assuption that the nucleus is infinitely  0 n2 h 2
massive when compared to mass of eletron so that From (1) and (2) r 
 z  e2
it remains stationary during the rotation of eletron
around it ix) Radius of orbit of such a particle in a quantum state
iii) Infact the nucleus is not infinitely massive and hence  0n2h2 n2
both the nucleus and eletron revolve around their n is rn   r 
 z  e2 n
z
centre of mass with same angular velocity 
 Ze 2
x) Potential energy of the system PE  and
4 0 r
cm
M
r2 r1
m kinetic energy of the system
iv)
1 1 1
Electron KE  I1 2  I 2 2   I1  I 2   2
Nucleus 2 2 2
v) let m be the mass of eletron, M be the mass of the
1
nucleus , Z be its atomic number and r be the 
2
 mr12  Mr22   2
separation between them. If r1 , r2 are distances of
centre of mass from electron and nucleus 1 M 2r 2 m2 r 2  2
 m  M 2 

respectively then r1  r2  r and 2   M  m 2  M  m  

250 ATOMIC PHYSICS

Sr|12th IIT-JEE MAIN|NEET|PHYSICS:VOL-III
deformation phase)is less than the energy required
1 mMr 2 2
 M  m   
1 mM  2 2
 r  to excite the eletron to next higher energy state,
2  M  m 2
2 M m eletron can’t be excited and the loss of KE of the
system during deformation phase again converts
1 2 2 1 Ze2  Ze2  into KE of the system and totally there will be no
ie KE   r     r 
3 2
 
2 2 4 0 r  40  loss of KE of of the system and hence the collision
 Total energy of the system (or equivalent is elastic
particle of mass  ) E  PE  KE iii) If loss in KE of the system during deformation
phase is more than or equal to the energy required
Ze 2 Ze 2  Z  e 2 Z 2 e4 to excite the electron to next higher state, excitation
E   2 2  2 2 2
8 0 r 8 0 n h  0 8 0 n h of the electron may take place and hence kinetic
ie Energy of nth quantum state energy of the system may not be conserved hence
the collision may be inelastic or even perfectly
-Z2 e4 mZ 2 e4 me4  Z 2   inelastic.
En =    
8 02 n2 h2 8 02 n2 h2  m 8 02 h2  n2 m  iv) If loss in KE is sufficient even ionization may take
place.Even though the possible loss in KE is greater
 z2    Z2  M than or equal to excitation energy of electron,
En  13.6 2   eV  13.6 2   eV
n  m  n  M m excitation may not take place necessarily and hence
collision may be elastic.
13.6  Z 2  v) Consider a particle of mass m moving with velocity
 eV
m  n 2  u which strikes a stationary hydrogen like atom of
1 mass M which is in ground state.
M
vi) Loss in KE will be maximum in perfectly inelastic
The formulae for rn and En can be obtained simply collision. In this case if V is common velocity after
by replacing m by  in the formulae for stationary collision, from conservation of linear momentum
nucleus, If  is wave length of photon emitted due mu
mu+0=(M+m)V  V 
to transition from a quantum state n2 to a quantum M m
 Maximum possible loss in KE is
hc me 4 z 2 1  1 1  1 1
    2 K  mu 2   M  m  V 2
state n1 , then  8 0 h  m   n1 n2 
2 2 2

1  
2 2
 M  1  Mm  2
i.e  K   u
2M m
1 me 4 z 2 1  1 1 
     vii) If E is minimum excitation energy (ex: n=1 to
 8 0 h c 
2 3
m   n12 n22 
1   n=2 in ground state) and if K  E eletron can’t
 M 
be excited hence there will be no loss of KE of the
 1 system hence the collision is elastic.
1 R0 Z 2 1 
  2 2 viii) If K  E the electron may get excited and the
  m   n1 n2  where R is Rydberg’ss
1   0 collision may be perfectly in elastic.
 M 
ix) If K  E the electron may get excited to higher
constant when the nucleus is stationary energy states or even removed from the atom and
EXCITATION BY COLLISION may have some kinetic energy. In this case the
i) When an atom is bombarded by particles like collision may be inelastic or may be perfectly
electron, proton, neutron,   particle etc, the loss inelastic as there is loss in KE of the system, or
in KE of the system during collision may be used even elastic if excitation dose not take place.
in excitation of the atom x) If KEmin is the minimum KE that should be
ii) If loss in KE of the system during collision (during
ATOMIC PHYSICS 251
 Sr|12th IIT-JEE MAIN|NEET|PHYSICS:VOL-III
pocessed by the colliding particle to excite the  m . ( continuous spectrum ) For a given accelerating
electron, K  E for excitation potential,  m is called cut off wave length.
1 mM 2 1 M m c) Properties of continuous x - rays spectra are
u  E  mu 2  E   independent of nature of target metal and they
2 M m 2  M 
depend only on accelerating potential.
M m  m
 KEmin  E    E  1   12
 M   M 10 50KV
8 40KV
In the same way we can calculate KEmin in other
6

Intensity
cases where atom is also moving using the 4 30KV
conservation of linear momentum. 2 20KV
X-Rays : o -1 -2 -3 -4 -5 -6 -7 -8 -9 -10
Roentgen discovered the X-rays. 0
Wave length in A
i) Most commonly x-rays are produced by the
deceleration of high energy electrons bombarding
d) hc 12400 0
a hard metal target. min   A
eV V
ii) The target should have
a) high atomic weight
b) high melting point 1
min it is Duane and Hunt’s law
c) high thermal conductivity V
e) Maximum frequency of emitted x - ray photon is
iii) They are electromagnetic waves of very short
wavelength. i.e., order of wavelength 0.1A° to ev
 max 
100A° , order of frequency 1016Hz to 1019 Hz, h
order of energy 124eV to 124keV f) In this spectrum intensity first increases, reaches a
iv) Most of the kinetic energy of electrons is converted maximum value I and then decreases.
into heat and only a fraction is used in producing x- max
g) Every spetrum starts with certian minimum wave
rays (less than 1% x - rays and more than 99% length called limiting wave length or cut off wave
heat).
length min .
v) Intensity of x-rays depends on the number of
electrons striking the target which inturn depends h) With the increase in target potential, min and
on filament current. wavelength corresponding to maximum intensity
vi) Quality of x - rays (hard /soft) depends on P.D
0 shifts towards minimum wavelength side.
applied to x - rays tube.
vii) high frequency x-rays are called hard x-rays i) At a given potential the range of wave length of
viii) low frequency x-rays are called soft x-rays continous x - rays produced is min to  .
ix) Penetrating power of x-rays is a function of j) Efficiency of x - ray tube
potential difference between cathode and target. out put power
x) Interatomic distance in crystals is of the order of the  x 100
input power
wavelength of x-rays hence crystals diffract x-rays.
xi) Production of x-rays is converse of photoelectric input power P = VI. Where V is P.D applied to x -
effect. ray tube I = anode current
ii) Characterstic X-ray spectrum:
X-Ray spectrum
i) Continuous X-ray spectrum:
a) It is produced when high speed electrons are
suddenly stopped by a metal target.
b) It contains all wave lengths abovea minimum wavelength

252 ATOMIC PHYSICS

Sr|12th IIT-JEE MAIN|NEET|PHYSICS:VOL-III
12
10 K K 
8
35KV 

Intensity
6
4
2
min I
o -1 -2 -3 -4 -5 -6 -7 -8 -9 -10
Wave length in A
0
min 0 k k
a) Produced due to transition of electrons from higher 
energy level to lower energy level in target atoms As target potential V is increased
b) Wavelengths of these x-rays depend only on atomic a)  0  min  decreases
number of the target element and independent of b) Wavelength of k remains constant.
target potential.
c) Characteristic x-rays of an element consists of K, c) diffrence between k and min increases
L,M and N series. d) diffrence between k line and  k line remains

d) K-series of lines are obtained when transition takes constant.
place from higher levels to k shell
e) Difference between  k    0 increases.
K L
L Moseley’s Law
K M
M
i) “The square root of frequency ( v ) of the spectral
line of the characteristic x-rays spectrum is directly
+
K proportional to the atomic number(z) of the target
L element.
M
N   Z or  =a(Z-b)
O
e) This spectrum is useful in identifying the elements
by which they are produced. 
v

f) Relation among the energies Ek  Ek   Ek , 

Ek  EL 0 b=1 Z

g) Intensity of x - rays Ik  Ik  Ik ii) The slope(a) of  -Z curve varies from series to
series and also from line to line of a given series.
h) Relation among frequences k , k  and  L is
 1  Z1  1  2  Z1  1 
1 1 1 For K series       Z 1 
 k    k   L     2  Z2 1   2 
 K L
1
K

hc iii) ak  ak   ak

h) E K  E L  h K  
K  iv) The intercept on ‘Z’ axis gives the screening
he constant ‘b’ and it is constant for all spectral lines
E K  E M  h K   in given series but varies with the series.
K 
he b = 1 for k series ( k  , k  , k  )
E L  E M  h L  b = 7.4 for L series
K
v) The wavelength of characteristic X-rays is given
iii) Intensity and wavelength      graph 1 1
1 
by =R(Z-b)  n12 n22 
2

vi) Ratio of k and k  lines from a given target is

ATOMIC PHYSICS 253

 Sr|12th IIT-JEE MAIN|NEET|PHYSICS:VOL-III

k 32 1
   2 Z 1
k  27 Sol.  2   1
vii) Significance :  1  Z2 1
a) The elements must be arranged in the periodic table EX. 16: An X-ray tube produces a continuous
as per their atomic numbers but not on their atomic spectrum of radiation with its short-wavelength
end at 0.45A0. What is the maximum energy
weights.
of a photon in the radiation? (b) From your
b) Helped to discover new elements like masurium
answer to (a), guess what order of accelerating
(43) and illinium (61) etc. voltage (for electrons) is required in such a
c) Decided the positions and atomic numbers of rare tube?
earth metals. Sol. a) min  0.45A0
EX. 13: Electrons with de-Broglie wavelength 
hc
fall on the target in an X-ray tube. The cut-off Emax  h max 
wavelength of emitted X-rays is min

2mc 2 2h 12431
A) 0  B) 0    27624.44eV  27.624 KeV
h mc 0.45
b) The minimum accelerating voltage for electrons
2m 2c 2  3
C) 0  D) 0   27.6keV
h2 is  27.6kV
e
Pe2  h /  
2
hc h2 i.e. of the order of 30 kV
Sol.  KE   
0 e
2me 2me 2m 2 EX. 17: The wavelength of the characteristics X-
ray K line emitted from zinc (Z=30) is
2mC  2
 0  1.415A0. Find the wavelength of the K line
h
EX. 14: In coolidge tube the potential difference of emitted from molybdenum (Z=42).
electron gun is increased from 12.4 KV to 24.8 Sol. According to Moseley’s law, the frequency for K
series is given by
KV. As a result the value of K  C increase
v   Z  1
2

two fold. The wave length of K line is ( C 

c
cut off wave length) or   Z  1
2

A) 1A0 B) 0.5A0 C) 1.5A0 D) 1.25A0 

EX. 15: Wavelength the K X-ray of on element A 1
or  k  Z  1 ...(3.6)
2


is 1 and wavelength of K X-ray element B
Where k is a constant. Let  ' be the wavelength
1 1
is 2 .  is equal to and Z1 & Z2 are atoms of K line emitted from molybdenum, then
2 4
1
 k  Z  1 ...  3.7 
2
number of A and B respectively then
A) 2Z2-Z1=1 B) Z2-2Z1=1 '
Dividing (3.6) and (3.7) we get
Z2 Z1
C) Z  4 D) Z  4 2 2
 Z 1   30  1 
1 2 '      1.415 A  0.708 A
0 0

 Z ' 1   42  1 
EX. 18: If the short series limit of the Balmer series
for hydrogen is 3644A0, find the atomic
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number of the element which give X-ray
1 2  1 
wavelengths down to 1A0. Identify the element   Z  1  2  2   3 R  Z  12
K 1 2  4
Sol. If short series limit of the Balmer series is
corresponding to transition n   to n  2 which 4
or  Z  1  3R
2
is given by
K
1  1 1  R
 R 2  2   4
 2   4   1599.25
3  1.097  10    0.76  1010 
7

44
 A0 
1
or R   or (Z-1)2=1600
 3644
or Z-1=40
The shortest wavelength corresponds to n   to
or Z=41
n=1. Therefore c is given as EX. 21: The K-absorption edge of an unknown
1 2 1 1  element is 0.171A0
 R  Z  1  2  2  or a) Identify the element
c 1  
b) Find the average wavelengths of the
1 1 3644
 Z  1 K , K  & K  lines.
2
    911
c R 1A0  4 A0
  4
 1
c) If a 100 eV electron strike the target of
3644
this element, what is the minimum wavelength
or Z-1 = 30.2 or Z=31.2  31
of the X-ray emitted?
Thus the atomic number of the element is 31 which
Sol. From Moseley’s law, the wavelength of k series of
is gallium.
X-rays is given by taking   1 in modified in
EX. 19: A material whose K absorption edge is 0.2A0
rydberg’s formula given as
is irradiated by X-rays of wavelength 0.15A0.
Find the maximum energy of the 1 2 1 
 R  Z  1  1  2  for K lines where,
photoelectrons that are emitted from the K   n 
shell. n=2,3,4,...
Sol. The binding energy for k shell in eV is a) For K-absorption edge, we put n   , in above
hc 12431 expression gives
Ek   eV  62.155 KeV
k 0.2 1
 Z  1 
The energy of the incident photon in eV is R
hc 12431
E   82.873KeV 1
 or Z   1  74
0.15
Therefore, the maximum energy of the
 0.17110 1.097 10 
10 7

photoelectrons emitted from the K shell is The element is Tungsten

Emax  E  Ek  82.873  62.155 KeV 1 2  1
= 20.718 KeV b) For K line :   R  74  1 1  22 
K  
EX. 20: The wavelength of K X-rays produced 0
by a X-ray tube is 0.76A0. What is the atomic K  0.228 A
number of the anode material of the tube?
1 2  1
Sol. K X-rays are produced when an electron makes For K  line :   R  74  1 1  32 
K
a transition from n=2 to n=1 to fill a vacancy in K-
shell. The wavelength of X-ray lines is given by 0
K   0.192 A

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 Sr|12th IIT-JEE MAIN|NEET|PHYSICS:VOL-III
now experiences viscous force and buoyancy
1 2  1
For K  line :   R  74  1 1  2  directed upwards and weight downwards. The
K  4  drop soon moves with terminal velocity Vg
0 downwards. Vg can be measured by measuring
K  0.182 A distance travelled by the drop and time taken for
c) The shortest wavelength corresponding to an it.
electron with kinetic energy 100 eV is given by  Charge on the drop can be calculated using
1/ 2
hc 12431 0 6ph (Vg + Ve )æ ö÷
0
c   A  124.31 A çç 9hVg ÷
E 100 the formula q = çç ÷
E çè (
2 d - d 1
) ø÷÷
g
Determination of charge of an electron :
where h = Coefficient of viscosity of air
Millikan’s oil drop method.[Additional] d = density of oil; d1 = density of air
 Millikan’s method to find charge of an oil drop is E = Intensity of electric field between the plates
based on the measurement of terminal velocity of g = acceleration due to gravity
the charged oil drop under the action of gravity and Ve = terminal velocity of the oil drop in electric field
under the combined action of gravity and electric Vg=terminal velocity of the oil drop in gravitational field
field.  After large number of observations Millikan found
that the value of ‘q’ is an integral multiple of a
common value ‘e’ such that q = ne. Where n is an
integer and e is the charge of an electron. It is found
that charge of electron, e = 1.6x10-19 C.
 When the electric field is applied between the plates
w2 A and B, suppose a negatively charged drop is
moving upwards with constant velocity Ve.
upward forces are electric force (Fe) and Buoyancy
(Fb) while downward forces are viscous force (Fv)
 Electric field can be produced between the plates and weight (Fg).
A and B. Small droplets of non volatile oil (clock Net force on the drop is zero.
Fe
oil or apiezon oil) are sprayed using an atomizier
above the hole in plate A. Some of the droplets
acquire charge due to friction. The chamber is Fb
illuminated by sending light horizontally through it. ve
The drops can be seen by using a telescope placed \ Fe + Fb = Fg + Fv
Fg
perpendicular to the light beam. A drop looks like Fv
a bright star moving upwards or downwards.
 An electric field is produced between the plates A 4p 3 1 4p 3
and B by connecting upper plate to the positive qE + R d g= R dg + 6ph RVe ® (1)
3 3
terminal and lower plate to the negative terminal of
R - Radius of the drop
a battery. The field is adjusted so that an oil drop
If electric field is switched off, suppose the drop
slowly moves upwards. The drop experiences
falls downwards with terminal velocity Vg .
electric force and buoyancy in the upward direction
upward forces are Buoyancy and viscous force
where as weight and viscous force in the downward
while downward force is weight.
direction.
Net force on the drop is zero.
 The uniform velocity attained by the drop Ve is
measured by measuring the distance travelled and
time taken for it. Now electric field is switched off
and the drop is allowed to fall freely. The oil drop

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Sr|12th IIT-JEE MAIN|NEET|PHYSICS:VOL-III

Fv Fe

Fb Fb
vg v=0
\ Fb + Fv = Fg \ Fe + Fb = Fg
Fg
Fg
4p 3 1 4p 3 4p 4p
R d g + 6ph RVg = R dg ® (2) qE + R 3d 1g = R 3dg
3 3 3 3
subtracting (2) from (1) Case-3: In Millikan’s oil drop method, if a
qE - 6 p h RV g = 6 p h RVe positively charged drop is moving upwards with
constant velocity V in the presence of electric field
6 p h R (Ve + V g )
q= ® (3) directed upwards then, upward forces are
E buoyancy (Fb) and electric force (Fe) while
4p 3 downward forces are weight (Fg) and viscous force
from equation (2) R (d - d 1 )g = 6ph RVg (Fv). Net force on the drop is zero.
3
9 hV g 9hVg Fe
\ R2 = R= ® (4)
2 (d - d 1 )g ; 2 (d - d 1 )g Fb
v
1/ 2
\ Fe + Fb = Fg + Fv Fg
6ph (Vg + Ve )æ
çç 9hVg
ö
÷
÷
÷ ® (5)
\ q= çç
çè 2 (d - d )g ÷
1
E ÷
ø Fv
Case-1 : In Millikan’s oil drop method, if the drop
4p 3 1 4p 3
is falling under gravity in the absence of electric field qE + Rd g = R dg + 6ph RV
then, upward forces are Buoyancy (Fb) and viscous 3 3
force (Fv) while downward force is weight (Fg). Case-4: In Millikan’s oil drop method if a positively
Net force on the drop is zero. charged drop is moving downwards with constant
velocity V in the presence of electric field directed
Fv
upwards then, upward forces are Buoyancy (Fb),
Fb electric force (Fe) and viscous force (Fv) while
downward force is weight (Fg) .
\ Fb + Fv = Fg v
Net force on the drop is zero.

Fg
4p 3 1 4p 3
R d g + 6p h RV = R dg \ Fe + Fb + Fv = Fg
3 3
Case-2: In Millikan’s oil drop method if a positively
charged drop is at rest in the presence of electric
field directed upwards then, upward forces are
Buoyancy (Fb) and electric force (Fe) while 4p 3 1 4p 3
qE + R d g + 6ph RV = R dg
downward force is weight (Fg). 3 3
Net force on the drop is zero. Case-5 : A charged oil drop having charge q1
is falling freely under gravity in the absence
of electric field with a velocity ‘V1’. It is held
stationary in an electric field. As it acquires a
charge it moves up with a velocity V2. Then

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 Sr|12th IIT-JEE MAIN|NEET|PHYSICS:VOL-III
i) The new charge on the drop is q2 = (k + 1)q1
ii) Additional charge aquired is D q = kq1

Where k = V2
V1
In the absence of electric field, drop is falling
downwords with velocity V1.

Fg = Fb + Fv FV cosq = qE
Fg - Fb = Fv
FV sin q = Fg = W
Fg - Fb = 6ph RV1 - (1)
W cot q
\ q=
E
In the presence of electric field when the drop is at EX. 22: In Millikan’s oil drop experiment an oil
rest. drop of radius r and charge Q is held in
Fe equilibrium between the plates of a charged
parallel plate capacitor when the potential
Fg = Fb + Fe Fb
difference is V. To keep another drop of same
v=0
Fg - Fb = q1E - (2) oil whose radius is 2r and carrying charge 2Q
in equilibrium between the plates, find the new
Fg potential difference required.
In the electric field, after acquiring final charge q2 QE

drop is moving upwards with velocity V2.Then

Fe V=0
Fg + Fv = Fb + Fe Sol. Since drop is at rest QE = mg
Fg - Fb = q2E- 6phRV2 but,V2 = kV1 Fb mg

Fg - Fb = q2E- k6phRV1 - (3) v2 3

V 4 3 r3 2 V 2 r  Q
1
Fv Q  r g ; V  Q ; V  r  Q
d 3 1  1 2
F
q1E = q2 E - kq1 E g

V2 1
So,the final charge on the drop is q2 = (k + 1)q1  8 ;  V2  4V
V 2
Additional charge aquired is D q = kq1 EX. 23: A charged oil drop of charge q is falling
Case-6 : In the Millikan’s oil drop experiment under gravity with terminal velocity v in the
the oil drop is subjected to a horizontal electric absence of electric field. A electric field can
field E and the drop moves with a constant keep the oil drop stationary. If the drop
velocity making an angle q to the horizontal.
acquires an additional charge, it moves up with
If the weight of the drop is W, neglecting
velocity 3v in that field. Find the new charge
W cot q on the drop.
buoyancy, the charge on the drop is
E Sol. In the absence of electric field
6RV

V
mg  6 RV  (1)

mg

258 ATOMIC PHYSICS

Sr|12th IIT-JEE MAIN|NEET|PHYSICS:VOL-III
when electric field is applied n 4
qE EX. 26: A charged oil drop is of charge q is falling
freely under gravity in the absence of electric
field with a velocity ‘v’. It is held stationary in
V=0
mg  qE  (2) an electric field, as it acquires a charge it
moves up with a velocity ‘3v. Now the charge
mg on the drop is
If q1 = new charge and drop is moving up then Sol mg  6 r , Eq = mg
qE
1
Eq1  mg  6 r (3 ) , Eq1 = Eq + 3Eq
Eq1 = 4Eq  q1 = 4q
3V EX. 27: A charged oil drop falls with a terminal
q1 E  mg  3  6 RV  (3) mg
velocity V in the absence of electric field. An
6R3V electric field E keeps the oil drop stationary
in it. When the drop acquires a charge ‘q’ it
q1 E  4 mg  4 qE  q1  4 q moves up with same velocity. Find the initial
charge on the drop.
EX. 24: In Millikan’s method of determining the Sol mg  6 r ,Eq = mg
charge of an electron, the terminal velocities E(Q+q) = mg + 6 r ,E( Q+q) = 2EQ
of oil drop in the presence and in the absence Q + q = 2Q  q = Q
of an electric field are xcm/s upwards and y
cm/s downwards respectively. Find the ratio EX. 28: Two oil drops in Millikan’s experiment are
of electric force to gravitational force on the falling with terminal velocities in the ratio 1:4.
oil drop. (Neglect Buoyancy) The ratio of their de-Broglie wave length is
Sol. In Gravitational field, weight = viscous force Sol T  r 2 r  VT
W  6 ry........(1)
In electric field, r1 V 1 m1 1
 1  
Electric force = weight + viscous force r2 V2 2 m2 8
Eq  W  6 rx ...........(2) 1 m2  2
  . 8 4
Substitute (1) in (2) then Eq  6 r ( y  x ) ...........(3)  . = 32 : 1
2 m1 1 1 1
Electric force x y

Gravitational force y

EX. 25: In a Millikan’s experiment an oil drop of Alpha Ray Scattering

radius 1.5 x 10-6 m and density 890 kg/m3 is 1. ‘Coulomb’s law correctly describes the electric
held stationary between two condenser plates force is that (pick the wrong statement)
1.2 cm apart and kept at a p.d of 2.3 kV. If 1) binds the electrons and neutrons in the nucleus
upthrust due to air is ignored, then the number of an atom.
of excess electrons carries by the drop will be 2) binds electrons to nucleus
V 4 4 d 3) binds atoms together to form molecules
Sol Eq = mg; .ne   r 3 . g ; n    r 3  g
d 3 3 Ve 4) binds atoms and molecules to from solids
2. Is the probability of backward scattering
4   3.375 1018  890  9.8 1.2 102 (i.e.scattering of  - particals at angles greater
n 
3 2.3 103 1.6 1019 than 900) predicted by Thomson’s model much

ATOMIC PHYSICS 259