CLUSTER LEVEL WORK SHOP
SUBJECT PHYSICS
QUESTION BANK
Dual Nature of Matter and radiation
Question Question and answers
No.
1.
The wavelength of electromagnetic radiation is doubled. What will happen to the energy
ofphoton?
Ans : It becomes half. ( E = hc/λ )
2.
Ultraviolet light is incident on two photosensitive materials having work function ϕ1 &ϕ2 (ϕ1 >
ϕ2). In which of the case will K.E. of emitted electrons be grater? Why?
Ans : Material having work function ϕ2 has greater Kinetic energy. ( K = hv – ϕ0 )
3.
An increase in the intensity of incident light does not change the maximum velocity of the emitted
photo electrons. Why?
Ans: Kinetic energy is independent of intensity of incident light. Hence no change in velocity.
4.
The stopping potential in an experiment on photoelectric effect is 1.6 volt.what is the maximum
kinetic energy of the photoelectrons emitted?.
5
6
7
Ans: K = eV0 = 1.6 eV
Name the experiment which verified the wave nature of particles.
Ans :Devison and Germer experiment.
What is the stopping potential applied to a photocell if the maximum kinetic energy of a
photoelectron is 𝟓𝒆𝑽?
Ans: K.E.=eV0
5eV= eV0
V0 =5 volt(negative)
8.
Define threshold wavelength.
Ans: maximum wavelength of the incident radiation above which no photoelectric emission is
possible.
Show graphically, the variation of de-Broglie wavelength (𝝀)with the momentum of an electron.
9.
2 Marks Questions
An 𝛼-particle and a proton are accelerated from rest by the same potential. Find the ratio of their
de-Broglie wavelengths.
Ans = Since de Broglie wavelength 𝝀 ∝
10.
𝟏
𝒎
√ 𝒒
𝒐𝒓
𝝀𝛼
𝝀𝑝
𝒎 𝒒
= √𝒎 𝒑 𝒒𝒑 =
𝛼 𝛼
𝟏
√𝟖
The two lines A and B shown in the graph plot the de-Broglie wavelength λ as function of 1/ √V
(V is the accelerating potential) for two particles having the same charge. Which of the two
represents the particle of heavier mass?
Ans: 𝑆𝑖𝑛𝑐𝑒
𝝀=
𝟏
𝟏
( 𝒔𝒍𝒐𝒑𝒆
√𝟐𝒎 𝒒 √𝑽
𝟏
=
) 𝒕𝒉𝒆𝒓𝒆𝒇𝒐𝒓𝒆 𝒉𝒊𝒈𝒉𝒆𝒓 𝒕𝒉𝒆 𝒔𝒍𝒐𝒑𝒆 𝒔𝒎𝒂𝒍𝒍𝒆𝒓 𝒕𝒉𝒆 𝒎𝒂𝒔𝒔. 𝑴𝒂𝒔𝒔 𝑩 𝒊𝒔 𝒉𝒂𝒗𝒊𝒆𝒓.
√𝟐𝒎 𝒒
11
Deduce de Broglie wavelength of electron accelerated by potential of 𝑽 volt.
Expression for de Broglie wavelength associated with Accelerated Electron:
The de Broglie wavelength associated with electron of momentum 𝑝 is given by
ℎ
ℎ
𝜆= =
… . (𝑖)
𝑝 𝑚𝑣
Where 𝑚 is the mass and 𝑣 is velocity of electron. If 𝐸𝑘 is the kinetic energy of electron, then
1
1
𝑝 2
𝑝2
𝑝
𝐸𝑘 = 𝑚𝑣 2 = 𝑚 ( ) =
(𝑠𝑖𝑛𝑐𝑒 𝑝 = 𝑚𝑣 ⟹ 𝑣 = )
2
2
𝑚
2𝑚
𝑚
⟹ 𝑝 = √2𝑚𝐸𝑘
ℎ
∴ Equation (ii) gives
𝜆=
… (𝑖𝑖)
√2𝑚𝐸𝑘
If 𝑉 volt is accelerating potential of electron, then kinetic energy,
𝐸𝑘 = 𝑒𝑉
ℎ
∴ Equation (ii) gives
𝜆 = 2𝑚𝑒𝑉
√
12
3 Marks Questions
If the frequency of incident radiation on a photocell is doubled for the same intensity, what
changes will you observe in (i)K.E.of photoelectron emitted (ii)photoelectric current and
(iii)stopping potential. Justify your answer in each case.
Ans: . (i)K.E. of emitted photoelectron becomes more than double.
(ii)P.E. current remains unchanged.
(iii)Stopping potential becomes more than double.
… (𝑖𝑖𝑖)
13
The work function for a certain metal is 4.2 eV. Will this metal give photoelectric emission for
incident radiation ofwavelength 330 nm?
Ans: No
, Work function of the metal,
Charge on an electron, e = 1.6 × 10−19 C ,Planck’s constant, h = 6.626 × 10−34Js
Wavelength of the incident radiation, λ = 330 nm = 330 × 10−9 m
Speed of light, c = 3 × 108 m/s ,The energy of the incident photon is given as:
It can be observed that the energy of the incident radiation is less than the work function of the
metal. Hence, nophotoelectric emission will take place.
14
Monochromatic light of wavelength 632.8 nm is produced by a helium-neon laser. The power
emitted is 9.42 mW.
(a) Find the energy and momentum of each photon in the light beam,
(b) How many photons per second, on the average, arrive at a target irradiated by this beam?
(Assume the beamto have uniform cross-section which is less than the target area), and
(c) How fast does a hydrogen atom have to travel in order to have the same momentum as that of
the photon?
Ans: Wavelength of the monochromatic light, λ = 632.8 nm = 632.8 × 10−9 m
Power emitted by the laser, P = 9.42 mW = 9.42 × 10−3 W
Planck’s constant, h = 6.626 × 10−34Js,Speed of light, c = 3 × 108 m/s
Mass of a hydrogen atom, m = 1.66 × 10−27 kg
(a)The energy of each photon is given as:
The momentum of each photon is given as:
(b)Number of photons arriving per second, at a target irradiated by the beam = n
Assume that the beam has a uniform cross-section that is less than the target area.
Hence, the equation for power can be written as:
(c)Momentum of the hydrogen atom is the same as the momentum of the photon,
Momentum is given as:
Where, v = Speed of the hydrogen atom
15
Write three basic properties of photons which are used to obtain Einstein’s photoelectric equation.
Use this equation to draw a plot of maximum kinetic energy of the electron emitted versus the
frequency of the incident radiation.
Ans: Propreties : each photon has fixed energy and momentum, photons not deflected by electric
field and magnetic field, in photon electron collision the total energy and momentum is
conserved.
16
Write Einstein’s photoelectric equation. State and explain clearly the three salient features
observed in photoelectric effect on the basis of this equation.
Refer NCERT text book page no 394
17.
Describe Davisson and Germer experiment to establish the wave nature of electrons. Draw a
labeled diagram of apparatus used.
Refer NCERT Text book page no 403.
ATOM AND NUCLEI
1 MARK QUESTIONS
Q.1 What is the distance of closest approach when a 5Mev proton approaches a gold nucleus (Z=79).
(1)
Ans.
r0=
1
Ze 2
= 2.3 * 10-14m.
4  F2
Q.2 Which has greater ionizing power: alpha or beta particle?
(1)
Ans. Alpha particle.
Q.3In Bohr’s theory of model of a Hydrogen atom, name the physical quantity which equals to an integral
multiple of h/2 π ?
(1)
Ans: Angular momentum
Q.4 What is the relation between ‘n’ & radius ‘r’ of the orbit of electron in a Hydrogen atom according to
Bohr’s theory?
(1)
2
Ans: r α n
Q.5 What is Bohr’s quantization condition?
Ans: The angular momentum of an electron in a circular orbit must be an integral multiple of h/2 π ,
where h is Plank’s constant
(1)
Q.6 What is the relation between the radius of the atom & the mass number?
(1)
Q.7 What is the ratio of the nuclear densities of two nuclei having mass numbers in the ratio 1:4?
Ans: 1:1
(1)
Q.8 Is a  particle different from an electron?
(1)
Ans: There is no difference between beta particle and electron.
Q9. Draw graph between no. of nuclei un-decayed with time for a radioactive substance (1)
Ans
Q.10 Among the alpha, beta & gamma radiations, which are the one affected by a magnetic field? (1)
Ans: alpha & beta
Q.11 Why do α particles have high ionizing power?
(1)
Ans: because of their large mass & large nuclear cross section
Q.12 Write the relationship between the half life & the average life of a radioactive substance.
(1)
Ans: T =1.44t1/2
Q.13. Among alpha, beta and gamma radiations, which get affected by electric field?
Ans. Alpha and beta radiations are charged, so they are affected by electric field.
.
Q.14 State the condition for controlled chain reaction to occur in a nuclear reactor.
Ans. In nuclear fission, two or three neutrons are released per fission. If on the average one neutron
causes further fission, the chain reaction is said to be controlled.
Q.15 Why is the heavy water used as a moderator in a nuclear reactor?
Ans. The basic principle of mechanics is that momentum transfer is maximum when the mass of
colliding particle and target particle are equal. Heavy water has negligible absorption cross-section for
neutrons and its mass is small, so heavy water molecules do not absorb fast neutron, but simply slow
them.
2 Marks questions
Q.1. For an electron in the second orbit of hydrogen, what is the moment of linear momentum as per the
Bohr’s model?
(2)
Ans: L=2(h/2 π) =h/ π (moment of linear momentum is angular momentum)
Q.2 The ground state energy of hydrogen atom is -13.6eV.What is the K.E & P.E of the electron in this
state?
(2)
Ans: K.E=-E=13.6 eV, P.E=-2K.E=-27.2 eV
Q.3 What is the shortest wavelength present in the Paschen series of hydrogen spectrum?
(2)
Ans: n1=3, n2=infinity, λ=9/R=8204Ǻ
Q.4 Calculate the frequency of the photon which can excite an electron to -3.4 eV from -13.6 eV.
(2)
15
Ans: 2.5x10 Hz
Q.5 The wavelength of the first member of Balmer series in the hydrogen spectrum is 6563Å.Calculate
the wavelength of the first member of Lyman series in the same spectrum.
Ans: 1215.4Å
(2)
Q.6 The ground state energy of hydrogen atom is -13.6eV.What is the K.E & P.E of the electron in this
state?
(2)
Ans: K.E=-E=13.6 eV, P.E=-2K.E=-27.2 eV
Q.7 Find the ratio of maximum wavelength of Lyman series in hydrogen spectrum to the maximum
wavelength in Paschen Series?
(2)
Q.8 How many electrons, protons & neutrons are there in an element of atomic number (Z) 11& mass
number (A) 24?
(2)
Hint:
ne = np =11, nn = (A – Z) = 24 -11 = 13
Q.9 Select the pairs of isotopes & isotones from the following:
(2)
i. 13C6
ii.14N7
iii.30P15iv.31P15
Ans: isotopes-iii &iv ,isotones-i& ii
Q.10 By what factor must the mass number change for the nuclear radius to become twice?
(2)
1
Ans:
3
2 or 2 3 timeA
Q.11 What is the nuclear force? Mention any two important properties of it.
Q.12 Obtain the binding energy of the nuclei 56Fe26 &209Bi83in MeV from the following
data: mH=1.007825amu,mn=1.008665amu, m(56Fe26)=55.934939amu,
m(209 Bi 83)=208.980388amu, 1amu=931.5MeV
(2)
Q.13 Binding Energy of 8O16 &17C35 one 127.35 Mev and 289.3 Mev respectively. Which of the two nuclei
is more stable stability & BE/N?
Q14 A radioactive sample having N nuclei has activity R. Write an expression for its half life in terms of
R & N.
(2)
Ans: R=Nλ, t1/2=0.693/λ =0.693N/R
Q.15 Tritium has a half life of 12.5 years against beta decay. What fraction of a sample of pure tritium
will remain un- decayed after 25 years?
(2)
Ans: N0/4
Q.16 What percentage of a given mass of a radioactive substance will be left un-decayed after 5 half-life
periods?
(2)
Ans: N/N0 =1/2n =1/32 =3.125%
Q.17 A radioactive nucleus ‘A’ decays as given below:
β
γ
A
A1
A2
If the mass number & atomic number of A1 are 180 & 73 respectively, find the mass number & atomic
number
of A & A2
(2)
Ans: A—180 & 72, A2—176 & 71
Q.18 Two nuclei P & Q have equal no: of atoms at t=0.Their half lives are 3 & 9 hours respectively.
Compare the rates of disintegration after 18 hours from the start.
(2)
Ans: 3:16
Q.19. A radio nuclide sample has N0 nuclei at t=0.Its number of undecayed nuclei get reduced to N0/e
at t=t.What does the term t stand for? Write the term of t,the time interval `T`in which half of the original
number of nuclei of this radionuclide would have got decayed.
Ans. It is the mean life time of radio nuclide T is the half life period of radio nuclide,the relation t=1.44T
.i.e. mean life period =1.44*half life period.
Q.20 Distinguish between isotopes & isobars.Give one examples for each of the species.
Ans. Isotopes: The nuclides having same atomic number Z but different atomic masses(A) are called
isotopes.
Isobars:The nuclides having same atomic masses(A) but different atomic number(Z) are called isobars.
3Marks Questions
Q.1.Define half life of a radioactive sample.Which of the following radiations: 𝛂-rays,β-rays,&γ-rays
1)are similar to x-rays
2)are easily absorbed by matter
3)travel with greatest speed
4)are similar in nature to cathode rays?
Ans: Half life: It is a radioactive sample is defined as the time in which the mass of sample is left one half
of the original mass.
1)γ-rays are similar to x-rays.
2) α-rays are easily absorbed by matter.
3) γ-rays are travel with greatest speed .
4)β-rays are similar in nature to cathode rays.
238
Q.2. Calculate the energy released if U
–nucleus emits an α-particle.
Yes, the decay is spontaneous (since Q is positive).
Q.3. Calculate the binding energy per nucleon
nucleus.
Binding energy per nucleus=341.87MeV/40 nucleon. =8.55MeV/nucleon.
Q.4. Draw the graph showing the variation of binding energy per nucleon with mass numbers. Give the
reason for decrease of binding energy per nucleon.
Ans.
The graph of binding energy per nucleon versus mass number(A) is shown in figure .The decrease of
binding energy per nucleon for nuclei with high mass number is due to increased coulomb repulsion
between protons inside the nucleus.
Graph between binding energy per nucleons & mass number shows that
(1) Some nuclei with mass number A < 20 have large binding energy per nucleon than their neighbour
nuclei. For example 2 He 4 , 4 Be 8 , 6 C 12 , 8 O 16 and 10 Ne 20 . These nuclei are more stable than their neighbours.
(2) Due to high value of B.E per nucleons 2He4 , 6C12 and 8O16 are stable and shows peak value. Binding
energy of 3Li7 is greater than that of 2He4 but the value of binding energy per nucleus is lesser. Also a
nucleus having even A and even Z is more stable while an odd nucleus is least stable.
(3) Binding energy per nucleon is maximum for iron ( 26 Fe 56 ) . It's value is 8.8 MeV per nucleon.
(4) For nuclei having A > 56, binding energy per nucleon gradually decreases for uranium (A = 238), the
value of binding energy per nucleon drops to 7.5 MeV.
Q.5. The trajectories, traced by different α-particle , n Geiger-Marsden experiment were observed as
shown in figure.

r0
What names are given to symbols ‘b’ and ’θ’ shown here.Nucleus
0
What can we say about the
value of b for (1)θ=0 ,(2)θ=π radians.
-particle
Ans.
a) The symbol ’b’ represents
(energyimpact
E)
parameter & ‘θ’ represents the scattering.
0
b)(1) when θ=0 , the impact parameter will be maximum & represent the atomic size.
(2)When θ=π radians, the impact parameter ‘b’ will be minimum & represent the nuclear size.
Q.6.
group the following six nuclides into three pairs of (1)isotones (2)isotopes(3)isobars.
Q.7. How does the size of nucleus depend on its mass number ?Hence explain
why the density of nuclear matter is independent of the size of the nucleus.
Q.8 Half lives of two substances A and B are 20 min and 40 min respectively. Initially the sample had
equal no of
nuclei. Find the ratio of the remaining no: of nuclei of A and B after 80 min.
Ans: 1:4
(3)
235
Q.9 If 200MeV energy is released in the fission of single nucleus of 92U, how much fission must occur to
produce a
power of 1 kW.
(3)
Q.10 State the basic postulates of Bohr’s atomic model & derive an expression for the energy of an
electron in any
orbit of hydrogen atom.
Q.11 Derive an expression for the radius of stationary orbit. Prove that the various stationary orbits are
not
equally spaced.
Q.12 Derive mathematical expressions for: (i) kinetic energy, & (ii) potential energy of an electron
revolving in an
orbit of radius ‘r’; how does the potential energy change with increase in principal quantum number
(n) for
the electron and why?
Q.13 Define the decay constant for a radioactive sample. Which of the following radiations α, β, & λ rays
are: (i)
Similar to X-rays? (ii) easily absorbed by matter? & (iii) similar in nature to cathode rays?
Q.14The energy levels of an atom are as shown below. a) Which of them will result in the transition of a
photon of
wavelength 275 nm? b) Which transition corresponds to the emission of radiation maximum
wavelength?
0eV
A
-2eV
B
C
-4.5eV
D
-10eV
Ans:
E=hc/λ=4.5eV, transition B
Eα1/λ, transition A
Q.15 The trajectories, traced by different α-particle, n Geiger-Marsden experiment were
observed as shown in figure.
a)What names are given to symbols ‘b’ and ’θ’ shown here.
0
b)What can we say about the value of b for (1)θ=0 ,(2)θ=π radians
Ans.
A) The symbol ’b’ represents impact parameter & ‘θ’represents the scattering
0
B) when θ=0 , the impact parameter will be maximum & represent the atomic size.
C)When θ=π radians, the impact parameter ‘b’ will be maximum & represent the
nuclear size

r0
Nucleus
4 Mark Questions:(VBQ)
-particle
Q.1. Albert Einstein, the great
physicist is known for his pioneering work in the field of special theory of
relativity, mass energy equivalence and explanation of photoelectric effect phenomenon.
(energy E )
Based on mass-energy equivalence equation E=mc2 given by Einstein, atom bombs were designed and
used in japan causing large scale destruction. Einstein was shocked and thereafter started working
actively with United Nations Organisation.
In your opinion, what applications did Einstein want for scientific concepts evolved/ formulated?
What human values are reflected about Einstein’s contribution to UNO?
State Einstein’s Equation of photoelectric effect.
Use the equation to describe the change in KE of photoelectrons with increase in frequency of incident
radiations
Ans.
(a) Advocated peaceful application for welfare of man
(b) Concern for others; Love for peaceful co-existence.
(c) KE(K) = hv – W
(d) As frequency increases, the maximum kinetic energy of photoelectrons emitted also increases
5 MARKS QUESTIONS
Q.1 State postulates of Bohr’s theory. Derive expression for radius and energy of hydrogen like atom by
Bohr’s
Theory.
Ans:
Bohr atomic model: - in 1931 Neil Bohr proposed a model, which is theoretically and experimentally
correct. For this Bohr awarded a Nobel Prize in 1922.bohr atomic model is application of quantum theory.
Fundamental postulates of Bohr atom model are following.
Stationary orbit: - electron revolve round the nucleus in only specific circular orbits. Since electrons move
in specific circular orbits so it does not radiate any energy i.e its energy remains constant.
Stability: -electron revolves round the nucleus then for stability necessary centripetal force for the circular
motion is provided by electrostatic attraction between electron and nucleus. Thus these two forces are
equal.
Quantisation condition: - electron can move only in those orbits for which an angular momentum is an
integer whole multiple of h/2 . Angular momentum L = m v r = n h/2 
Where h is a planks constant, m=mass of electron,
V = velocity of electron, r = radius of permitted orbit n =1,2,3---------Transition: - when an element
jumps from higher to lower energy orbit, the energy is radiated when electron jumps from lower to higher
energy orbit then energy is absorb from outside. The difference of energies of the orbits is converted in
electromagnetic radiation.
5) Frequency condition: - if energy of initial orbit is Ei and final orbit is Ef then frequency of radiation
absorbed or light emitted is Ef – Ei =h this is called Bohr’s frequency condition.
Bohr's Orbits (for Hydrogen and H2-like Atoms): Consider an electron of charge ‘e’ and mass ‘m’
revolving around nucleus with speed v in a circular orbit of radius ‘r’.
(1) Radius of orbit : From first postulates of Bohr atom model, for an electron around a stationary nucleus
1 ( Ze)e mv 2

the electrostatics force of attraction provides the necessary centripetal force i.e.
….
4 0 r 2
r
(i)
nh
nh
From second postulates of Bohr atom model m v r 
….(ii)
 v
2
2 m r
From equation (i) and (ii) radius of nth orbit
r
2 2
2 2
2
2
n h 0
n h
n
n
m, – e
rn 

 0.53
Å (iii)
 rn 
2
2
2
Z
Z
4 k Z m e
 mZ e
(2) Speed of electron : From the above relations (ii) & (iii), speed of electron in nth orbit can be calculated
as
2 k Z e 2
Z e2
 c Z
6 Z
vn 


m / sec
where (c = speed of light 3 
.  2.2  10
nh
2  0 n h  137  n
n
108 m/s)
1 1 ( Ze)e
1 ( Ze)e
(3) Total energy : From eq (i)
.
 mv 2 so kinetic energy K 
2 40 r
40 r
1 ( Ze)(e)
40
r
Total energy (E) is the sum of potential energy and kinetic energy i.e.
1 (Ze)(e) 1 1 (Ze)e
E  K U 

40
r
2 40 r
Electrostatic potential energy, U 
E
 m e4  z 2
1 1 Ze 2
n 2 h 2 0
 2 2 . 2 (iv)
E


From eq (iii) rn 
.
Hence
 8 h  n
2 40 rn
 m z e2
 0 
En  13.6
OR
Z2
eV
n2
U
m e4
& R  2 3 = Rydberg's constant = 1.09  107 per m.
2
8 0 c h
(4) Transition of Electron:When an electron makes transition from higher energy level having energy
E2(n2) to a lower energy level having energy E1 (n1) then a photon of frequency  is emitted
(a) Energy of emitted radiation : From eq (iv)
Note: Total energy E   K 

L

v
 m e4
 8 2 h 2
 0
E  E2  E1   
 z2
 m e4
.



 n2
 8 2 h 2
 2
 0
 z2
.
 n2
 1
E 
m Z 2 e4  1
1 
  2
2 2  2
8  0 h  n1 n2 
OR
 1
1 
E  13.6Z 2  2  2 
 n1 n2 
(b) Frequency of emitted radiation; E  h  h v 
m Z 2 e4  1
1 
 2
2 2  2
8 0 h  n1
n2 
The frequency
E2
mZ e  1
1 
 2  E1
2 3  2
8 0 h  n1
n2 
(c) Wave number/wavelength: If c be the velocity of light and  its wavelength, then v = c/
 1
1
me 4  1
1 
1 
c
me 4  1
1 
or

 2  = R 2  2 
v 



2 3 
2
2 3
2
2
 8 0 h  n1 n2 
 8 0 h c  n1
n 2 
n 2 
 n1
of the emitted radiations can be found from the following relation v 
where R =
me 4
2 4
, R is known as Rydberg’s constant and its value is 1.097  10 7 m 1 .
8 0 h c
Wave number ν is the number of waves in unit length. It is reciprocal of wavelength is given by
 1
1
1 
This equation is the general expression for the wave number of radiation
ν   R 2  2 

n 2 
 n1
2
3
emitted by the electron when it jumps from higher orbit n2 to lower orbit n1.
Q.2. Explain the spectral lines of hydrogen specra.
It has been shown that the energy of the outer orbit is greater than the energy of the inner ones. When the
hydrogen atom is subjected to eternal energy, the electron jumps from lower energy state to a higher
energy state i.e. the hydrogen atom is excited. The excited state is not stable hence the electron returns to
its ground state in about 10-8 seconds. The excess of energy is now radiated in the form of radiations of
different wavelengths. The different wavelengths constitute spectral series, which are characteristics of
atom emitting them. The wavelength of different members of the series can be found from the following
relation. v =
 1
1
 R 2  2


n2
 n1
1




This relation explains the complete spectrum of
hydrogen.
(a)
Lyman series - The series consists of all wavelengths which are emitted when electron jumps
from an outer orbit to the first orbit i.e., the electronic jumps to K orbit gives rise to Lyman series.
Here n1 = 1 and n2 = 2, 3, 4, …, 
The wavelengths of different members of Lyman series are :
(i)
First member - In this case n1 = 1 and n2 = 2, it is called line of Lyman series, hence
1  3R
4
1
 R 2  2  
or  
or
3R

2  4
1
1
(ii)
4
 1216 10 10 m = 1216 Å
7
3 1.097 10
Second member - In this case n1 = 1 and n2 = 3, it is called line of Lyman series, hence
1  8R
1
 R 2  2 
or

3  9
1
1


9
or
8R

9
 1026  10 10 m = 1026
7
8  1.097  10
Å
(iii)
Similarly, the wavelengths of other members can be calculated.
Limiting member - In this case n1 = 1 and n2 = , hence
E2 – E1 = h
1 1
 R 2    R


1
1
or

1
R
or

1
 912  10 10 m  912 Å
7
1.097  10
This series lies in ultraviolet region.
(b)
Balmer series - This series consists of all wavelengths which are emitted when an electron jumps
from an outer orbit to the second orbit i.e., the electron jumps to L orbits give rise to Balmer sereis.
Here n1 = 2 and n2 = 3, 4, 5, …, .
The wavelengths of different members of Balmer series are :
(i)
First member - In this case n1 = 2 and n2 = 3, it is called line of Balmer series, hence
1  5R
1
 R 2  2  

3  36
2
1
Å.
(ii)
or

36
or
5R

36
 6563  10 10 m = 6563
7
5  1.097  10
Second member - In this case n1 = 2 and n2 = 4, it is called line of Balmer series hence
1  3R
1
 R 2  2  
or

4  16
2
1

16
or
3R

16
 4861  10 10 m = 4861
7
3  1.097  10
Å.
(iii)
Similarly the wavelengths of other members can be calculated.
Limiting case - In this case n1 = 2 and n2 = , hence
1 R
1
 R 2   

 4
2
1
or

4
= 3646 Å.
R
This series lies in visible and near ultraviolet region.
(c)
Paschen series - This series consists of all wavelengths which are emitted when an electron jumps
from an outer orbit to the third orbit i.e., the electronic jumps to M orbit give rise to Paschen series.
Here n1 = 3 and n2 = 4, 5, 6, …, . The different wavelengths of this series can be obtained from the
formula
1
1 
 R 2  2 

n 2 
 3
1
where n2 = 4, 5, 6, …, .
For the first member, the wavelength is 18750 Å. This series lies in infra-red region.
(d)
Brackett series - This series consists of all wavelengths which are emitted when an electron jumps
from an outer orbit to the fourth orbit i.e., the electronic jumps to N orbit give rise to Brackett series.
Here n1 = 4 and n2 = 5, 6, 7, …, . The different wavelengths of this series can be obtained from the
formula
1
1 
 R 2  2 

n 2 
 4
1
where n2 = 5, 6, 7, …, .
This series lies in infra-red
region.
(e)
Pfund series - The series consists of all wavelengths which are emitted when an electron jumps
from an outer orbit to the fifth orbit i.e., the electronic jumps to O orbit give rise to Pfund series.
Here n1 = 5 and n2 = 6, 7, 8, …, . The different wavelengths of this series can be obtained from the
formula
1
1 
 R 2  2 

n 2 
 5
1
spectrum.
S.No.
1.
2.
3.
4.
5.
where n2 = 6, 7, 8, …, .
Series observed
Lyman series
Balmer series
Paschen series
Brackett series
Pfund series
This series lies in the infra-red region of the
Value of n1
1
2
3
4
5
Value of n2
2, 3, 4, …, 
3, 4, 5, …, 
4, 5, 6, …, 
5, 6, 7, …, 
6, 7, 8, …, 
Question Bank (Atoms and Nuclei)
1 Mark Questions:
Q.1.
Among alpha, beta and gamma radiations, which get affected by electric field?
Ans. Alpha and beta radiations are charged, so they are affected by electric field.
125
Q.2.
Ans.
What is the nuclear radius of
27
Fe if that of Al is 3.6 fermi? ]
Position in the spectrum
Ultraviolet
Visible
Infra-red
Infra-red
Infra-red
Q.3.
Assuming the nuclei to be spherical in shape, how does the surface area of a nucleus of mass
number A1 compare with that of a nucleus of mass number A2?
Q.4.
A nucleus of mass number A, has a mass defect Δm. Give the formula, for the binding energy per
nucleon, of this nucleus.
Q.5.
Write a typical nuclear reaction in which a large amount of energy is released in the process of
nuclear fission.
Ans. Nuclear fission reaction is
Q.6.
process.
Give the mass number and atomic number of elements on the right hand side of the decay
Q.7.
State the condition for controlled chain reaction to occur in a nuclear reactor.
Ans. In nuclear fission, two or three neutrons are released per fission. If on the average one neutron
causes further fission, the chain reaction is said to be controlled.
Q.8. Why is the heavy water used as a moderator in a nuclear reactor?
Ans. The basic principle of mechanics is that momentum transfer is maximum when the mass of
colliding particle and target particle are equal. Heavy water has negligible absorption cross-section for
neutrons and its mass is small, so heavy water molecules do not absorb fast neutron, but simply slow
them.
Q.9. Which observation led to the conclusion in the α-particle scattering exp. That atom has vast empty
space?
Ans. A larger number of alpha particles went through undeflected.
Q.10. What changes takes place in the nucleus when a γ – rays is emitted?
Ans. The nucleus looses energy, but remains same isotope it was.
Q.11. Why is the ionization power of ά – particle of greater than γ – rays?
Ans. Owing to greater mass and charge, it is able to knock out/pull out electrons which colliding with
atoms and molecules in its path.
Q.12. Why neutrons are considered as ideal particle for nuclear reactions?
Ans. They are neutral in nature and get absorbed by nucleus, thus distributing the neutron proton ratio.
Q.13. What conclusions were drawn from the observation in which few alpha-particle were seen
rebounding from gold foil?
Ans. The entire positive charge and the mass were concentrated at one place inside the atom, called the
nucleus.
Q.14. Two nuclei have mass numbers in the ratio 1:8.What is the ratio of their nuclear radii?
Ans. Since R= R0A1/3 ,therefore R1/R2=A11/3 /A21/3=(1/8)1/3 =1/2
R1: R2=1:2
Q.15. Which have greater ionizing power ;α-particles or β-particles?
Ans. α-particles have more ionizing power than β-particles.
2 Mark Questions:
Q.3.
A radio nuclide sample has N0 nuclei at t=0.Its number of undecayed nuclei get reduced to N0/e
at t=t.What does the term t stand for? Write the term of t,the time interval `T`in which half of the original
number of nuclei of this radionuclide would have got decayed.
Ans. It is the mean life time of radio nuclide T is the half life period of radio nuclide,the relation t=1.44T
.i.e. mean life period =1.44*half life period.
Q.4.
Distinguish between isotopes & isobars.Give one examples for each of the species.
Ans. Isotopes: The nuclides having same atomic number Z but different atomic masses(A) are called
isotopes.
Isobars:The nuclides having same atomic masses(A) but different atomic number(Z) are called isobars.
3 Mark Questions:
Q.1.Define half life of a radioactive sample.Which of the following radiations: 𝛂-rays,β-rays,&γ-rays
1)are similar to x-rays
2)are easily absorbed by matter
3)travel with greatest speed
4)are similar in nature to cathode rays?
Ans. Half life:It is aradioactive sample is defined as the time in which the mass of sample is left one half
of the original mass.
1)γ-rays are similar to x-rays.
2) α-rays are easily absorbed by matter.
3) γ-rays are travel with greatest speed .
4)β-rays are similar in nature to cathode rays.
238
Q.2. Calculate the energy released if U
–nucleus emits an α-particle.
Yes, the decay is spontaneous(since Q is positive).
Q.3. Calculate the binding energy per nucleon
nucleus .
Binding energy per nucleus=341.87MeV/40 nucleon. =8.55MeV/nucleon.
Q.4. Draw the graph showing the variation of binding energy per nucleon with mass numbers. Give the
reason for decrease of binding energy per nucleon.
Ans.
The graph of binding energy per nucleon versus mass number(A) is shown in figure .The decrease of
binding energy per nucleon for nuclei with high mass number is due to increased coulomb repulsion
between protons inside the nucleus.
Graph between binding energy per nucleons & mass number shows that
(1) Some nuclei with mass number A < 20 have large binding energy per nucleon than their neighbour
nuclei. For example 2 He 4 , 4 Be 8 , 6 C 12 , 8 O 16 and 10 Ne 20 . These nuclei are more stable than their neighbours.
(2) Due to high value of B.E per nucleons 2He4 , 6C12 and 8O16 are stable and shows peak value. Binding
energy of 3Li7 is greater than that of 2He4 but the value of binding energy per nucleus is lesser. Also a
nucleus having even A and even Z is more stable while an odd nucleus is least stable.
(3) Binding energy per nucleon is maximum for iron ( 26 Fe 56 ) . It's value is 8.8 MeV per nucleon.
(4) For nuclei having A > 56, binding energy per nucleon gradually decreases for uranium (A = 238), the
value of binding energy per nucleon drops to 7.5 MeV.
Q.5. The trajectories, traced by different α-particle , n Geiger-Marsden experiment were observed as
shown in figure.

r0
What names are given to symbols ‘b’ and ’θ’ shown here.Nucleus
0
What can we say about the
value of b for (1)θ=0 ,(2)θ=π radians.
-particle
Ans.
a) The symbol ’b’ represents
(energyimpact
E)
parameter & ‘θ’ represents the scattering.
0
b)(1) when θ=0 , the impact parameter will be maximum & represent the atomic size.
(2)When θ=π radians, the impact parameter ‘b’ will be minimum & represent the nuclear size.
Q.6.
group the following six nuclides into three pairs of (1)isotones (2)isotopes(3)isobars.
Q.7. How does the size of nucleus depend on its mass number ?Hence explain
why the density of nuclear matter is independent of the size of the nucleus.
4 Mark Questions:(VBQ)
Q.1. Albert Einstein, the great physicist is known for his pioneering work in the field of special theory of
relativity, mass energy equivalence and explanation of photoelectric effect phenomenon.
Based on mass-energy equivalence equation E=mc2 given by Einstein, atom bombs were designed and
used in japan causing large scale destruction. Einstein was shocked and thereafter started working
actively with United Nations Organisation.
In your opinion, what applications did Einstein want for scientific concepts evolved/ formulated?
What human values are reflected about Einstein’s contribution to UNO?
State Einstein’s Equation of photoelectric effect.
Use the equation to describe the change in KE of photoelectrons with increase in frequency of incident
radiations
Ans.
(a) Advocated peaceful application for welfare of man
(b) Concern for others; Love for peaceful co-existence.
(c) KE(K) = hv – W
(d) As frequency increases, the maximum kinetic energy of photoelectrons
CLUSTER LEVEL WORK SHOP
SUBJECT PHYSICS
QUESTION BANK
Dual Nature of Matter and radiation
emitted also increases.
Question
No.
1.
Question and answers
2.
Ultraviolet light is incident on two photosensitive materials having work function ϕ1 &ϕ2 (ϕ1
> ϕ2). In which of the case will K.E. of emitted electrons be grater? Why?
The wavelength of electromagnetic radiation is doubled. What will happen to the energy
ofphoton?
Ans : It becomes half. ( E = hc/λ )
Ans : Material having work function ϕ2 has greater Kinetic energy. ( K = hv – ϕ0 )
3.
An increase in the intensity of incident light does not change the maximum velocity of the
emitted photo electrons. Why?
Ans: Kinetic energy is independent of intensity of incident light. Hence no change in velocity.
4.
The stopping potential in an experiment on photoelectric effect is 1.6 volt.what is the
maximum kinetic energy of the photoelectrons emitted?.
5
6
7
Ans: K = eV0 = 1.6 eV
Name the experiment which verified the wave nature of particles.
Ans :Devison and Germer experiment.
What is the stopping potential applied to a photocell if the maximum kinetic energy of a
photoelectron is 𝟓𝒆𝑽?
Ans: K.E.=eV0
5eV= eV0
V0 =5 volt(negative)
Define threshold wavelength.
Ans: maximum wavelength of the incident radiation above which no photoelectric emission is
possible.
8.
9.
Show graphically, the variation of de-Broglie wavelength (𝝀)with the momentum of an
electron.
2 Marks Questions
An 𝛼-particle and a proton are accelerated from rest by the same potential. Find the ratio of
their de-Broglie wavelengths.
Ans = Since de Broglie wavelength 𝝀 ∝
10.
𝟏
√𝒎 𝒒
𝒐𝒓
𝝀𝛼
𝝀𝑝
𝒎𝒑 𝒒𝒑
= √𝒎
𝛼 𝒒𝛼
=
𝟏
√𝟖
The two lines A and B shown in the graph plot the de-Broglie wavelength λ as function of 1/
√V (V is the accelerating potential) for two particles having the same charge. Which of the
two represents the particle of heavier mass?
Ans: 𝑆𝑖𝑛𝑐𝑒
𝝀=
𝟏
𝟏
( 𝒔𝒍𝒐𝒑𝒆
√𝟐𝒎 𝒒 √𝑽
𝟏
=
) 𝒕𝒉𝒆𝒓𝒆𝒇𝒐𝒓𝒆 𝒉𝒊𝒈𝒉𝒆𝒓 𝒕𝒉𝒆 𝒔𝒍𝒐𝒑𝒆 𝒔𝒎𝒂𝒍𝒍𝒆𝒓 𝒕𝒉𝒆 𝒎𝒂𝒔𝒔. 𝑴𝒂𝒔𝒔 𝑩 𝒊𝒔 𝒉𝒂𝒗𝒊𝒆𝒓.
√𝟐𝒎 𝒒
11
De
Deduce de Broglie wavelength of electron accelerated by potential of 𝑽 volt.
Expression for de Broglie wavelength associated with Accelerated Electron:
The de Broglie wavelength associated with electron of momentum 𝑝 is given by
ℎ
ℎ
𝜆= =
… . (𝑖)
𝑝 𝑚𝑣
Where 𝑚 is the mass and 𝑣 is velocity of electron. If 𝐸𝑘 is the kinetic energy of electron, then
1
1
𝑝 2
𝑝2
𝑝
𝐸𝑘 = 𝑚𝑣 2 = 𝑚 ( ) =
(𝑠𝑖𝑛𝑐𝑒 𝑝 = 𝑚𝑣 ⟹ 𝑣 = )
2
2
𝑚
2𝑚
𝑚
⟹ 𝑝 = √2𝑚𝐸𝑘
∴ Equation (ii) gives
𝜆=
ℎ
√2𝑚𝐸𝑘
… (𝑖𝑖)
If 𝑉 volt is accelerating potential of electron, then kinetic energy,
𝐸𝑘 = 𝑒𝑉
∴ Equation (ii) gives
𝜆=
ℎ
… (𝑖𝑖𝑖)
2𝑚𝑒𝑉
√
12
3 Marks Questions
If the frequency of incident radiation on a photocell is doubled for the same intensity, what
changes will you observe in (i)K.E.of photoelectron emitted (ii)photoelectric current and
(iii)stopping potential. Justify your answer in each case.
Ans: . (i)K.E. of emitted photoelectron becomes more than double.
(ii)P.E. current remains unchanged.
(iii)Stopping potential becomes more than double.
13
The work function for a certain metal is 4.2 eV. Will this metal give photoelectric emission for
incident radiation ofwavelength 330 nm?
Ans: No
, Work function of the metal,
Charge on an electron, e = 1.6 × 10−19 C ,Planck’s constant, h = 6.626 × 10−34Js
Wavelength of the incident radiation, λ = 330 nm = 330 × 10−9 m
Speed of light, c = 3 × 108 m/s ,The energy of the incident photon is given as:
It can be observed that the energy of the incident radiation is less than the work function of the
metal. Hence, nophotoelectric emission will take place.
14
Monochromatic light of wavelength 632.8 nm is produced by a helium-neon laser. The power
emitted is 9.42 mW.
(a) Find the energy and momentum of each photon in the light beam,
(b) How many photons per second, on the average, arrive at a target irradiated by this beam?
(Assume the beamto have uniform cross-section which is less than the target area), and
(c) How fast does a hydrogen atom have to travel in order to have the same momentum as that
of the photon?
Ans: Wavelength of the monochromatic light, λ = 632.8 nm = 632.8 × 10−9 m
Power emitted by the laser, P = 9.42 mW = 9.42 × 10−3 W
Planck’s constant, h = 6.626 × 10−34Js,Speed of light, c = 3 × 108 m/s
Mass of a hydrogen atom, m = 1.66 × 10−27 kg
(a)The energy of each photon is given as:
The momentum of each photon is given as:
(b)Number of photons arriving per second, at a target irradiated by the beam = n
Assume that the beam has a uniform cross-section that is less than the target area.
Hence, the equation for power can be written as:
(c)Momentum of the hydrogen atom is the same as the momentum of the photon,
Momentum is given as:
Where, v = Speed of the hydrogen atom
15
Write three basic properties of photons which are used to obtain Einstein’s photoelectric
equation. Use this equation to draw a plot of maximum kinetic energy of the electron emitted
versus the frequency of the incident radiation.
Ans: Propreties : each photon has fixed energy and momentum, photons not deflected by
electric field and magnetic field, in photon electron collision the total energy and momentum is
conserved.
Write Einstein’s photoelectric equation. State and explain clearly the three salient features
observed in photoelectric effect on the basis of this equation.
16
Refer NCERT text book page no 394
17.
Describe Davisson and Germer experiment to establish the wave nature of electrons. Draw a
labeled diagram of apparatus used.
Refer NCERT Text book page no 403.
ATOM AND NUCLEI
1 MARK QUESTIONS
Q.1 What is the distance of closest approach when a 5Mev proton approaches a gold nucleus (Z=79).
(1)
Ans.
r0=
1
Ze 2
= 2.3 * 10-14m.
4  F2
Q.2 Which has greater ionizing power: alpha or beta particle?
(1)
Ans. Alpha particle.
Q.3In Bohr’s theory of model of a Hydrogen atom, name the physical quantity which equals to an integral
multiple of
h/2 π ?
(1)
Ans: Angular momentum
Q.4 What is the relation between ‘n’ & radius ‘r’ of the orbit of electron in a Hydrogen atom according to
Bohr’s
theory?
(1)
2
Ans: r α n
Q.5 What is Bohr’s quantization condition?
Ans: The angular momentum of an electron in a circular orbit must be an integral multiple of h/2 π ,
where h is
Plank’s constant
(1)
Q.6 What is the relation between the radius of the atom & the mass number?
(1)
Q.7 What is the ratio of the nuclear densities of two nuclei having mass numbers in the ratio 1:4?
Ans: 1:1
(1)
Q.8 Is a  particle different from an electron?
(1)
Ans: There is no difference between beta particle and electron.
Q9. Draw graph between no. of nuclei un-decayed with time for a radioactive substance (1)
Ans
Q.10 Among the alpha, beta & gamma radiations, which are the one affected by a magnetic field? (1)
Ans: alpha & beta
Q.11 Why do α particles have high ionizing power?
(1)
Ans: because of their large mass & large nuclear cross section
Q.12 Write the relationship between the half life & the average life of a radioactive substance.
(1)
Ans: T =1.44t1/2
Q.13. Among alpha, beta and gamma radiations, which get affected by electric field?
Ans. Alpha and beta radiations are charged, so they are affected by electric field.
.
Q.14 State the condition for controlled chain reaction to occur in a nuclear reactor.
Ans. In nuclear fission, two or three neutrons are released per fission. If on the average one neutron
causes further fission, the chain reaction is said to be controlled.
Q.15 Why is the heavy water used as a moderator in a nuclear reactor?
Ans. The basic principle of mechanics is that momentum transfer is maximum when the mass of
colliding particle and target particle are equal. Heavy water has negligible absorption cross-section for
neutrons and its mass is small, so heavy water molecules do not absorb fast neutron, but simply slow
them.
2 Marks questions
Q.1. For an electron in the second orbit of hydrogen, what is the moment of linear momentum as per the
Bohr’s
model?
(2)
Ans: L=2(h/2 π) =h/ π (moment of linear momentum is angular momentum)
Q.2 The ground state energy of hydrogen atom is -13.6eV.What is the K.E & P.E of the electron in this
state?
(2)
Ans: K.E=-E=13.6 eV, P.E=-2K.E=-27.2 eV
Q.3 What is the shortest wavelength present in the Paschen series of hydrogen spectrum?
(2)
Ans: n1=3, n2=infinity, λ=9/R=8204Ǻ
Q.4 Calculate the frequency of the photon which can excite an electron to -3.4 eV from -13.6 eV.
(2)
Ans: 2.5x1015Hz
Q.5 The wavelength of the first member of Balmer series in the hydrogen spectrum is 6563Å.Calculate
the
wavelength of the first member of Lyman series in the same spectrum.
Ans: 1215.4Å
(2)
Q.6 The ground state energy of hydrogen atom is -13.6eV.What is the K.E & P.E of the electron in this
state?
(2)
Ans: K.E=-E=13.6 eV, P.E=-2K.E=-27.2 eV
Q.7 Find the ratio of maximum wavelength of Lyman series in hydrogen spectrum to the maximum
wavelength in
Paschen Series?
(2)
Q.8 How many electrons, protons & neutrons are there in an element of atomic number (Z) 11& mass
number (A) 24?
(2)
Hint:
ne = np =11, nn = (A – Z) = 24 -11 = 13
Q.9 Select the pairs of isotopes & isotones from the following:
i. 13C6
ii.14N7
iii.30P15iv.31P15
Ans: isotopes-iii &iv ,isotones-i& ii
Q.10 By what factor must the mass number change for the nuclear radius to become twice?
Ans:
3
(2)
(2)
1
3
2 or 2 timeA
Q.11 What is the nuclear force? Mention any two important properties of it.
(2)
Q.12 Obtain the binding energy of the nuclei 56Fe26 &209Bi83in MeV from the following
data: mH=1.007825amu,mn=1.008665amu, m(56Fe26)=55.934939amu,
m(209 Bi 83)=208.980388amu, 1amu=931.5MeV
Q.13 Binding Energy of 8O16 &17C35 one 127.35 Mev and 289.3 Mev respectively. Which of the two nuclei
is more
stable stability & BE/N?
Q14 A radioactive sample having N nuclei has activity R. Write an expression for its half life in terms of
R & N.
(2)
Ans: R=Nλ, t1/2=0.693/λ =0.693N/R
Q.15 Tritium has a half life of 12.5 years against beta decay. What fraction of a sample of pure tritium
will remain
un- decayed after 25 years?
(2)
Ans: N0/4
Q.16 What percentage of a given mass of a radioactive substance will be left un-decayed after 5 half-life
periods?
(2)
Ans: N/N0 =1/2n =1/32 =3.125%
Q.17 A radioactive nucleus ‘A’ decays as given below:
β
γ
A
A1
A2
If the mass number & atomic number of A1 are 180 & 73 respectively, find the mass number & atomic
number
of A & A2
(2)
Ans: A—180 & 72, A2—176 & 71
Q.18 Two nuclei P & Q have equal no: of atoms at t=0.Their half lives are 3 & 9 hours respectively.
Compare the rates
of disintegration after 18 hours from the start.
(2)
Ans: 3:16
Q.19. A radio nuclide sample has N0 nuclei at t=0.Its number of undecayed nuclei get reduced to N0/e
at t=t.What does the term t stand for? Write the term of t,the time interval `T`in which half of the original
number of nuclei of this radionuclide would have got decayed.
Ans. It is the mean life time of radio nuclide T is the half life period of radio nuclide,the relation t=1.44T
.i.e. mean life period =1.44*half life period.
Q.20 Distinguish between isotopes & isobars.Give one examples for each of the species.
Ans. Isotopes: The nuclides having same atomic number Z but different atomic masses(A) are called
isotopes.
Isobars:The nuclides having same atomic masses(A) but different atomic number(Z) are called isobars.
3Marks Questions
Q.1.Define half life of a radioactive sample.Which of the following radiations: 𝛂-rays,β-rays,&γ-rays
1)are similar to x-rays
2)are easily absorbed by matter
3)travel with greatest speed
4)are similar in nature to cathode rays?
Ans: Half life: It is a radioactive sample is defined as the time in which the mass of sample is left one half
of
the original mass.
1)γ-rays are similar to x-rays.
2) α-rays are easily absorbed by matter.
3) γ-rays are travel with greatest speed .
4)β-rays are similar in nature to cathode rays.
238
Q.2. Calculate the energy released if U
–nucleus emits an α-particle.
Yes, the decay is spontaneous (since Q is positive).
Q.3. Calculate the binding energy per nucleon
nucleus.
Binding energy per nucleus=341.87MeV/40 nucleon. =8.55MeV/nucleon.
Q.4. Draw the graph showing the variation of binding energy per nucleon with mass numbers. Give the
reason for decrease of binding energy per nucleon.
Ans.
The graph of binding energy per nucleon versus mass number(A) is shown in figure .The decrease of
binding energy per nucleon for nuclei with high mass number is due to increased coulomb repulsion
between protons inside the nucleus.
Graph between binding energy per nucleons & mass number shows that
(1) Some nuclei with mass number A < 20 have large binding energy per nucleon than their neighbour
nuclei. For example 2 He 4 , 4 Be 8 , 6 C 12 , 8 O 16 and 10 Ne 20 . These nuclei are more stable than their neighbours.
(2) Due to high value of B.E per nucleons 2He4 , 6C12 and 8O16 are stable and shows peak value. Binding
energy of 3Li7 is greater than that of 2He4 but the value of binding energy per nucleus is lesser. Also a
nucleus having even A and even Z is more stable while an odd nucleus is least stable.
(3) Binding energy per nucleon is maximum for iron ( 26 Fe 56 ) . It's value is 8.8 MeV per nucleon.
(4) For nuclei having A > 56, binding energy per nucleon gradually decreases for uranium (A = 238), the
value of binding energy per nucleon drops to 7.5 MeV.
Q.5. The trajectories, traced by different α-particle , n Geiger-Marsden experiment were observed as
shown in figure.

r0
What names are given to symbols ‘b’ and ’θ’ shown here.Nucleus
0
What can we say about the
value of b for (1)θ=0 ,(2)θ=π radians.
-particle
Ans.
a) The symbol ’b’ represents
(energyimpact
E)
parameter & ‘θ’ represents the scattering.
0
b)(1) when θ=0 , the impact parameter will be maximum & represent the atomic size.
(2)When θ=π radians, the impact parameter ‘b’ will be minimum & represent the nuclear size.
Q.6.
group the following six nuclides into three pairs of (1)isotones (2)isotopes(3)isobars.
Q.7. How does the size of nucleus depend on its mass number ?Hence explain
why the density of nuclear matter is independent of the size of the nucleus.
Q.8 Half lives of two substances A and B are 20 min and 40 min respectively. Initially the sample had
equal no of
nuclei. Find the ratio of the remaining no: of nuclei of A and B after 80 min.
Ans: 1:4
(3)
235
Q.9 If 200MeV energy is released in the fission of single nucleus of 92U, how much fission must occur to
produce a
power of 1 kW.
(3)
Q.10 State the basic postulates of Bohr’s atomic model & derive an expression for the energy of an
electron in any
orbit of hydrogen atom.
Q.11 Derive an expression for the radius of stationary orbit. Prove that the various stationary orbits are
not
equally spaced.
Q.12 Derive mathematical expressions for: (i) kinetic energy, & (ii) potential energy of an electron
revolving in an
orbit of radius ‘r’; how does the potential energy change with increase in principal quantum number
(n) for
the electron and why?
Q.13 Define the decay constant for a radioactive sample. Which of the following radiations α, β, & λ rays
are: (i)
Similar to X-rays? (ii) easily absorbed by matter? & (iii) similar in nature to cathode rays?
Q.14The energy levels of an atom are as shown below. a) Which of them will result in the transition of a
photon of
wavelength 275 nm? b) Which transition corresponds to the emission of radiation maximum
wavelength?
0eV
A
-2eV
B
C
-4.5eV
D
-10eV
Ans:
E=hc/λ=4.5eV, transition B
Eα1/λ, transition A
Q.15 The trajectories, traced by different α-particle, n Geiger-Marsden experiment were
observed as shown in figure.
a)What names are given to symbols ‘b’ and ’θ’ shown here.
0
b)What can we say about the value of b for (1)θ=0 ,(2)θ=π radians
Ans.
A) The symbol ’b’ represents impact parameter & ‘θ’represents the scattering
0
B) when θ=0 , the impact parameter will be maximum & represent the atomic size.
C)When θ=π radians, the impact parameter ‘b’ will be maximum & represent the
nuclear size

r0
Nucleus
4 Mark Questions:(VBQ)
-particle
Q.1. Albert Einstein, the great
physicist is known for his pioneering work in the field of special theory of
relativity, mass energy equivalence and explanation of photoelectric effect phenomenon.
(energy E )
Based on mass-energy equivalence equation E=mc2 given by Einstein, atom bombs were designed and
used in japan causing large scale destruction. Einstein was shocked and thereafter started working
actively with United Nations Organisation.
In your opinion, what applications did Einstein want for scientific concepts evolved/ formulated?
What human values are reflected about Einstein’s contribution to UNO?
State Einstein’s Equation of photoelectric effect.
Use the equation to describe the change in KE of photoelectrons with increase in frequency of incident
radiations
Ans.
(a) Advocated peaceful application for welfare of man
(b) Concern for others; Love for peaceful co-existence.
(c) KE(K) = hv – W
(d) As frequency increases, the maximum kinetic energy of photoelectrons emitted also increases
5 MARKS QUESTIONS
Q.1 State postulates of Bohr’s theory. Derive expression for radius and energy of hydrogen like atom by
Bohr’s
Theory.
Ans:
Bohr atomic model: - in 1931 Neil Bohr proposed a model, which is theoretically and experimentally
correct. For this Bohr awarded a Nobel Prize in 1922.bohr atomic model is application of quantum theory.
Fundamental postulates of Bohr atom model are following.
Stationary orbit: - electron revolve round the nucleus in only specific circular orbits. Since electrons move
in specific circular orbits so it does not radiate any energy i.e its energy remains constant.
Stability: -electron revolves round the nucleus then for stability necessary centripetal force for the circular
motion is provided by electrostatic attraction between electron and nucleus. Thus these two forces are
equal.
Quantisation condition: - electron can move only in those orbits for which an angular momentum is an
integer whole multiple of h/2 . Angular momentum L = m v r = n h/2 
Where h is a planks constant, m=mass of electron,
V = velocity of electron, r = radius of permitted orbit n =1,2,3---------Transition: - when an element
jumps from higher to lower energy orbit, the energy is radiated when electron jumps from lower to higher
energy orbit then energy is absorb from outside. The difference of energies of the orbits is converted in
electromagnetic radiation.
5) Frequency condition: - if energy of initial orbit is Ei and final orbit is Ef then frequency of radiation
absorbed or light emitted is Ef – Ei =h this is called Bohr’s frequency condition.
Bohr's Orbits (for Hydrogen and H2-like Atoms): Consider an electron of charge ‘e’ and mass ‘m’
revolving around nucleus with speed v in a circular orbit of radius ‘r’.
(1) Radius of orbit : From first postulates of Bohr atom model, for an electron around a stationary nucleus
1 ( Ze)e mv 2

the electrostatics force of attraction provides the necessary centripetal force i.e.
….
4 0 r 2
r
(i)
nh
nh
From second postulates of Bohr atom model m v r 
….(ii)
 v
2
2 m r
From equation (i) and (ii) radius of nth orbit
r
2 2
2 2
2
2
n h 0
n h
n
n
m, – e
rn 

 0.53
Å (iii)
 rn 
2
2
2
Z
Z
4 k Z m e
 mZ e
(2) Speed of electron : From the above relations (ii) & (iii), speed of electron in nth orbit can be calculated
as
2 k Z e 2
Z e2
 c Z
6 Z
vn 


m / sec
where (c = speed of light 3 
.  2.2  10
nh
2  0 n h  137  n
n
108 m/s)
1 1 ( Ze)e
1 ( Ze)e
(3) Total energy : From eq (i)
.
 mv 2 so kinetic energy K 
2 40 r
40 r
1 ( Ze)(e)
40
r
Total energy (E) is the sum of potential energy and kinetic energy i.e.
1 (Ze)(e) 1 1 (Ze)e
E  K U 

40
r
2 40 r
Electrostatic potential energy, U 
E
 m e4  z 2
1 1 Ze 2
n 2 h 2 0
 2 2 . 2 (iv)
E


From eq (iii) rn 
.
Hence
 8 h  n
2 40 rn
 m z e2
 0 
En  13.6
OR
Z2
eV
n2
U
m e4
& R  2 3 = Rydberg's constant = 1.09  107 per m.
2
8 0 c h
(4) Transition of Electron:When an electron makes transition from higher energy level having energy
E2(n2) to a lower energy level having energy E1 (n1) then a photon of frequency  is emitted
(a) Energy of emitted radiation : From eq (iv)
Note: Total energy E   K 

L

v
 m e4
 8 2 h 2
 0
E  E2  E1   
 z2
 m e4
.



 n2
 8 2 h 2
 2
 0
 z2
.
 n2
 1
E 
m Z 2 e4  1
1 
  2
2 2  2
8  0 h  n1 n2 
OR
 1
1 
E  13.6Z 2  2  2 
 n1 n2 
(b) Frequency of emitted radiation; E  h  h v 
m Z 2 e4  1
1 
 2
2 2  2
8 0 h  n1
n2 
The frequency
E2
mZ e  1
1 
 2  E1
2 3  2
8 0 h  n1
n2 
(c) Wave number/wavelength: If c be the velocity of light and  its wavelength, then v = c/
 1
1
me 4  1
1 
1 
c
me 4  1
1 
or

 2  = R 2  2 
v 



2 3 
2
2 3
2
2
 8 0 h  n1 n2 
 8 0 h c  n1
n 2 
n 2 
 n1
of the emitted radiations can be found from the following relation v 
where R =
me 4
2 4
, R is known as Rydberg’s constant and its value is 1.097  10 7 m 1 .
8 0 h c
Wave number ν is the number of waves in unit length. It is reciprocal of wavelength is given by
 1
1
1 
This equation is the general expression for the wave number of radiation
ν   R 2  2 

n 2 
 n1
2
3
emitted by the electron when it jumps from higher orbit n2 to lower orbit n1.
Q.2. Explain the spectral lines of hydrogen specra.
It has been shown that the energy of the outer orbit is greater than the energy of the inner ones. When the
hydrogen atom is subjected to eternal energy, the electron jumps from lower energy state to a higher
energy state i.e. the hydrogen atom is excited. The excited state is not stable hence the electron returns to
its ground state in about 10-8 seconds. The excess of energy is now radiated in the form of radiations of
different wavelengths. The different wavelengths constitute spectral series, which are characteristics of
atom emitting them. The wavelength of different members of the series can be found from the following
relation. v =
 1
1
 R 2  2


n2
 n1
1




This relation explains the complete spectrum of
hydrogen.
(a)
Lyman series - The series consists of all wavelengths which are emitted when electron jumps
from an outer orbit to the first orbit i.e., the electronic jumps to K orbit gives rise to Lyman series.
Here n1 = 1 and n2 = 2, 3, 4, …, 
The wavelengths of different members of Lyman series are :
(i)
First member - In this case n1 = 1 and n2 = 2, it is called line of Lyman series, hence
1  3R
4
1
 R 2  2  
or  
or
3R

2  4
1
1
(ii)
4
 1216 10 10 m = 1216 Å
7
3 1.097 10
Second member - In this case n1 = 1 and n2 = 3, it is called line of Lyman series, hence
1  8R
1
 R 2  2 
or

3  9
1
1


9
or
8R

9
 1026  10 10 m = 1026
7
8  1.097  10
Å
(iii)
Similarly, the wavelengths of other members can be calculated.
Limiting member - In this case n1 = 1 and n2 = , hence
E2 – E1 = h
1 1
 R 2    R


1
1
or

1
R
or

1
 912  10 10 m  912 Å
7
1.097  10
This series lies in ultraviolet region.
(b)
Balmer series - This series consists of all wavelengths which are emitted when an electron jumps
from an outer orbit to the second orbit i.e., the electron jumps to L orbits give rise to Balmer sereis.
Here n1 = 2 and n2 = 3, 4, 5, …, .
The wavelengths of different members of Balmer series are :
(i)
First member - In this case n1 = 2 and n2 = 3, it is called line of Balmer series, hence
1  5R
1
 R 2  2  

3  36
2
1
Å.
(ii)
or

36
or
5R

36
 6563  10 10 m = 6563
7
5  1.097  10
Second member - In this case n1 = 2 and n2 = 4, it is called line of Balmer series hence
1  3R
1
 R 2  2  
or

4  16
2
1

16
or
3R

16
 4861  10 10 m = 4861
7
3  1.097  10
Å.
(iii)
Similarly the wavelengths of other members can be calculated.
Limiting case - In this case n1 = 2 and n2 = , hence
1 R
1
 R 2   

 4
2
1
or

4
= 3646 Å.
R
This series lies in visible and near ultraviolet region.
(c)
Paschen series - This series consists of all wavelengths which are emitted when an electron jumps
from an outer orbit to the third orbit i.e., the electronic jumps to M orbit give rise to Paschen series.
Here n1 = 3 and n2 = 4, 5, 6, …, . The different wavelengths of this series can be obtained from the
formula
1
1 
 R 2  2 

n 2 
 3
1
where n2 = 4, 5, 6, …, .
For the first member, the wavelength is 18750 Å. This series lies in infra-red region.
(d)
Brackett series - This series consists of all wavelengths which are emitted when an electron jumps
from an outer orbit to the fourth orbit i.e., the electronic jumps to N orbit give rise to Brackett series.
Here n1 = 4 and n2 = 5, 6, 7, …, . The different wavelengths of this series can be obtained from the
formula
1
1 
 R 2  2 

n 2 
 4
1
where n2 = 5, 6, 7, …, .
This series lies in infra-red
region.
(e)
Pfund series - The series consists of all wavelengths which are emitted when an electron jumps
from an outer orbit to the fifth orbit i.e., the electronic jumps to O orbit give rise to Pfund series.
Here n1 = 5 and n2 = 6, 7, 8, …, . The different wavelengths of this series can be obtained from the
formula
1
1 
 R 2  2 

n 2 
 5
1
spectrum.
S.No.
1.
2.
3.
4.
5.
where n2 = 6, 7, 8, …, .
Series observed
Lyman series
Balmer series
Paschen series
Brackett series
Pfund series
This series lies in the infra-red region of the
Value of n1
1
2
3
4
5
Value of n2
2, 3, 4, …, 
3, 4, 5, …, 
4, 5, 6, …, 
5, 6, 7, …, 
6, 7, 8, …, 
Question Bank (Atoms and Nuclei)
1 Mark Questions:
Q.1.
Among alpha, beta and gamma radiations, which get affected by electric field?
Ans. Alpha and beta radiations are charged, so they are affected by electric field.
125
Q.2.
Ans.
What is the nuclear radius of
27
Fe if that of Al is 3.6 fermi? ]
Position in the spectrum
Ultraviolet
Visible
Infra-red
Infra-red
Infra-red
Q.3.
Assuming the nuclei to be spherical in shape, how does the surface area of a nucleus of mass
number A1 compare with that of a nucleus of mass number A2?
Q.4.
A nucleus of mass number A, has a mass defect Δm. Give the formula, for the binding energy per
nucleon, of this nucleus.
Q.5.
Write a typical nuclear reaction in which a large amount of energy is released in the process of
nuclear fission.
Ans. Nuclear fission reaction is
Q.6.
process.
Give the mass number and atomic number of elements on the right hand side of the decay
Q.7.
State the condition for controlled chain reaction to occur in a nuclear reactor.
Ans. In nuclear fission, two or three neutrons are released per fission. If on the average one neutron
causes further fission, the chain reaction is said to be controlled.
Q.8. Why is the heavy water used as a moderator in a nuclear reactor?
Ans. The basic principle of mechanics is that momentum transfer is maximum when the mass of
colliding particle and target particle are equal. Heavy water has negligible absorption cross-section for
neutrons and its mass is small, so heavy water molecules do not absorb fast neutron, but simply slow
them.
Q.9. Which observation led to the conclusion in the α-particle scattering exp. That atom has vast empty
space?
Ans. A larger number of alpha particles went through undeflected.
Q.10. What changes takes place in the nucleus when a γ – rays is emitted?
Ans. The nucleus looses energy, but remains same isotope it was.
Q.11. Why is the ionization power of ά – particle of greater than γ – rays?
Ans. Owing to greater mass and charge, it is able to knock out/pull out electrons which colliding with
atoms and molecules in its path.
Q.12. Why neutrons are considered as ideal particle for nuclear reactions?
Ans. They are neutral in nature and get absorbed by nucleus, thus distributing the neutron proton ratio.
Q.13. What conclusions were drawn from the observation in which few alpha-particle were seen
rebounding from gold foil?
Ans. The entire positive charge and the mass were concentrated at one place inside the atom, called the
nucleus.
Q.14. Two nuclei have mass numbers in the ratio 1:8.What is the ratio of their nuclear radii?
Ans. Since R= R0A1/3 ,therefore R1/R2=A11/3 /A21/3=(1/8)1/3 =1/2
R1: R2=1:2
Q.15. Which have greater ionizing power ;α-particles or β-particles?
Ans. α-particles have more ionizing power than β-particles.
2 Mark Questions:
Q.3.
A radio nuclide sample has N0 nuclei at t=0.Its number of undecayed nuclei get reduced to N0/e
at t=t.What does the term t stand for? Write the term of t,the time interval `T`in which half of the original
number of nuclei of this radionuclide would have got decayed.
Ans. It is the mean life time of radio nuclide T is the half life period of radio nuclide,the relation t=1.44T
.i.e. mean life period =1.44*half life period.
Q.4.
Distinguish between isotopes & isobars.Give one examples for each of the species.
Ans. Isotopes: The nuclides having same atomic number Z but different atomic masses(A) are called
isotopes.
Isobars:The nuclides having same atomic masses(A) but different atomic number(Z) are called isobars.
3 Mark Questions:
Q.1.Define half life of a radioactive sample.Which of the following radiations: 𝛂-rays,β-rays,&γ-rays
1)are similar to x-rays
2)are easily absorbed by matter
3)travel with greatest speed
4)are similar in nature to cathode rays?
Ans. Half life:It is aradioactive sample is defined as the time in which the mass of sample is left one half
of the original mass.
1)γ-rays are similar to x-rays.
2) α-rays are easily absorbed by matter.
3) γ-rays are travel with greatest speed .
4)β-rays are similar in nature to cathode rays.
238
Q.2. Calculate the energy released if U
–nucleus emits an α-particle.
Yes, the decay is spontaneous(since Q is positive).
Q.3. Calculate the binding energy per nucleon
nucleus .
Binding energy per nucleus=341.87MeV/40 nucleon. =8.55MeV/nucleon.
Q.4. Draw the graph showing the variation of binding energy per nucleon with mass numbers. Give the
reason for decrease of binding energy per nucleon.
Ans.
The graph of binding energy per nucleon versus mass number(A) is shown in figure .The decrease of
binding energy per nucleon for nuclei with high mass number is due to increased coulomb repulsion
between protons inside the nucleus.
Graph between binding energy per nucleons & mass number shows that
(1) Some nuclei with mass number A < 20 have large binding energy per nucleon than their neighbour
nuclei. For example 2 He 4 , 4 Be 8 , 6 C 12 , 8 O 16 and 10 Ne 20 . These nuclei are more stable than their neighbours.
(2) Due to high value of B.E per nucleons 2He4 , 6C12 and 8O16 are stable and shows peak value. Binding
energy of 3Li7 is greater than that of 2He4 but the value of binding energy per nucleus is lesser. Also a
nucleus having even A and even Z is more stable while an odd nucleus is least stable.
(3) Binding energy per nucleon is maximum for iron ( 26 Fe 56 ) . It's value is 8.8 MeV per nucleon.
(4) For nuclei having A > 56, binding energy per nucleon gradually decreases for uranium (A = 238), the
value of binding energy per nucleon drops to 7.5 MeV.
Q.5. The trajectories, traced by different α-particle , n Geiger-Marsden experiment were observed as
shown in figure.

r0
What names are given to symbols ‘b’ and ’θ’ shown here.Nucleus
0
What can we say about the
value of b for (1)θ=0 ,(2)θ=π radians.
-particle
Ans.
a) The symbol ’b’ represents
(energyimpact
E)
parameter & ‘θ’ represents the scattering.
0
b)(1) when θ=0 , the impact parameter will be maximum & represent the atomic size.
(2)When θ=π radians, the impact parameter ‘b’ will be minimum & represent the nuclear size.
Q.6.
group the following six nuclides into three pairs of (1)isotones (2)isotopes(3)isobars.
Q.7. How does the size of nucleus depend on its mass number ?Hence explain
why the density of nuclear matter is independent of the size of the nucleus.
4 Mark Questions:(VBQ)
Q.1. Albert Einstein, the great physicist is known for his pioneering work in the field of special theory of
relativity, mass energy equivalence and explanation of photoelectric effect phenomenon.
Based on mass-energy equivalence equation E=mc2 given by Einstein, atom bombs were designed and
used in japan causing large scale destruction. Einstein was shocked and thereafter started working
actively with United Nations Organisation.
In your opinion, what applications did Einstein want for scientific concepts evolved/ formulated?
What human values are reflected about Einstein’s contribution to UNO?
State Einstein’s Equation of photoelectric effect.
Use the equation to describe the change in KE of photoelectrons with increase in frequency of incident
radiations
Ans.
(a) Advocated peaceful application for welfare of man
(b) Concern for others; Love for peaceful co-existence.
(c) KE(K) = hv – W
(d) As frequency increases, the maximum kinetic energy of photoelectrons
emitted also increases.
1 Marks Questions
1)
What is Huygens’s principle?
a) Every point on a given wave front may be considered as the source of secondary
wavelet which spread out with the speed of light in that medium
b) The new wave front is the forward envelop of the secondary wavelets at the
instant.
2)
State Principle of Superposition of wave?
When two or more wave simultaneously passes through the same medium, each
wave acts on every particle of the medium as if the other waves are not present. The
resultant displacement of any particle is the vector addition of the displacements
due to individual wave. This is known as the principle of superposition.
3)
What is coherent source? Why two independent sources are coherent?
4)
Two sources are said to be coherent if they emit light waves of the same wave length
and start with same phase or have a constant phase difference.
Two independent monochromatic sources emit wave of same wavelength. But the
wave are not in phase so they are not coherent this is because atom cannot emit
light wave in same phase and this source are said to be incoherent source
Why diffraction in light wave is less pronounced?
As the wavelength of light is very small, compared to that of sound wave and even
tiny obstacles have large size, compared to the wavelength of light waves, diffraction
effects of light are very small. So the light wave is less pronounced.
5)
What type of wave shows the property of polarisation.?
Only transverse wave shows the property of polarisation.
6)
What is the main condition to produce interference of light?
The sources of light must be coherent.
7)
What are coherent sources of light?
Sources which emit continuously light waves of the same frequency or wavelength with a
zero or constant phase difference between them are called coherent sources.
8)
Two slits in Young’s double slit experiment have widths in the ratio 81:1. What is the
ratio of the amplitudes of light waves from them?
𝐴1
𝐴2
9)
9
How much is the distance in terms of fringe width β between central bright and fourth
dark fringe?
X4 
10)
𝑊1
= √𝑊2 = 1
(2  4  1) D 7
 
2d
2
Define resolving power of a telescope.
It is the reciprocal of the smallest angular separation between two point objects whose
images can be just resolved by it.
2 Marks Questions
1)
Distinguish between constructive and destructive interference?
CONSTRUCTIVE
1. At points where the crest of one
wave meets the crest of other wave
or trough of one wave meets the
trough of other wave.
2. The waves are in phase.
3. The displacement is maximum and
these points appear bright.
4. These points are marked by
crosses(X).
DESTRUCTIVE
1. At point which the crest of one wave
meets the trough of other wave.
2. The waves are in opposite phase.
3. The displacement is minimum and
these points appear dark.
4. These points are marked by circle (O).
2)
What are condition for sustained interference (any two)?
a) The two sources are said to be coherent.
b) Two sources should be very narrow.
c) The source should lie very close to each other to form distinct and broader
fringes.
3)
A parallel beam of light of 500 nm falls on a narrow slit and the resulting diffraction
pattern is observed on a screen 1 m away. It is observed that the first minimum is at
a distance of 2.5 mm from the centre of the screen. Calculate the width of the slit.
The distance of the nth minimum from the center of the screen is,
Where, D = distance of slit from screen, λ = wavelength of the light, a = width of the
slit
For first minimum, n=1
Thus,
4)
State the conditions, which must be satisfied for two light sources to be coherent.
(i) The two sources of light must be obtained from a single source by some method.
(ii) The two sources must give monochromatic light of same frequency.
(iii) The path difference between the waves from the two sources must be zero or
constant.
5)
Draw a graph showing the variation of intensity against the position x on the screen in
Young’s double slit experiment.
6)
Draw a graph showing the variation of intensity with diffraction angle  in a single slit
diffraction experiment.
If we plot a graph between the intensities of maxima and minima against the diffraction
angle  , we get a graph which has a broad central maximum in the direction (  0 ) of
incident light. On either side, it has secondary maxima of decreasing intensity, as shown
below
7)
Two slits in young’s double slit experiment are illuminated by two different lamps
emitting light of the same wavelength. Will you observe the interference pattern? Justify
your answer.
.
No, the light waves emitted by two different lamps cannot be coherent. So the conditions
of maxima and minima of intensity will change rapidly on the screen, producing uniform
illumination.
8)
In Young’s double slit experiment if the distance between two slits is halved and
distance between the slits and the screen is doubled, then what will be the effect on fringe
width ?
D
Original fringe width,  
d
  2D 4  D
 '

 4
New fringe width,
d /2
d
i.e., fringe width increases four times.
9)
If the path difference produced due to interference of light coming out of two slits for
yellow colour of light at a point on the screen be 3 / 2 , what will be the colour of the
fringe at the point? Give reason also.
The given path difference satisfies the condition for the minimum of intensity. Hence a
dark fringe will be formed at the given point. If white light is used, all components of
white light except the yellow one would be present at this point.
10)
Two identical coherent waves, each of intensity I0, are producing an interference pattern.
Write the value of the resultant intensity at a point of (i) constructive interference and (ii)
destructive interference.
(i) Resultant intensity at a point of constructive interference,
I  I 0  I 0  2 I 0 I 0 cos 0  4I 0
.
(ii) Resultant intensity at a point of destructive interference,
I  I 0  I 0  2 I 0 I 0 cos180  2 I 0  2 I 0  0
11)
In Young’s double slit experiment, two disturbances arriving at a point P have a phase
difference of  / 3 . What is the intensity at this point expressed as a fraction of maximum
intensity I0?
The resultant intensity at any point of an interference pattern is given by
I  I 0 cos 2

2
Where I 0 is the maximum value of I .  is the phase difference between two waves.
Given    / 3 , therefore,
2
 3
3
  I0
I  I 0 cos
 I 0 

6
4
 2 
2

12)
Radio waves diffract pronouncedly around the buildings, while light waves, which are
also electromagnetic waves, do not. Why?
For diffraction to take place, the wavelength should be of the order of the size of the
obstacle. The radio waves have wavelengths of the order of the size of the building and
other obstacles coming in their way and hence they easily get diffracted. Since
wavelength of the light waves is very small, they are not diffracted by buildings.
13)
One of the two slits in Yung’s double slit experiment is so painted that it transmits half
the intensity of the other. What is the effect on interference fringes?
Let I 0 be the intensity of light from each slit. When the slit is not painted,
I
 I
I max 
I min
0
0
 I0
 I0
  4I
 0
2
0
2
When one of the slits is painted, it transmits half of the original intensity.

I max
2
2
2
2

I 
1 

  I 0  0   I 0 1 
  2.914 I 0


2 
2



I 
1 

I min   I 0  0   I 0 1 
  0.086 I


2 
2


Hence on painting one of the two slits, the contrast between the bright and dark fringes
decreases.
3 Marks Questions
1)
Sketch the geometrical shape of the wave front:
(i)
Emerging from a point source of light.
(ii)
Emerging from a linear source of light like a rectangular slit.
(iii) Corresponding to a beam of light coming from a far away source.
(i) Spherical wave front - In the case of waves travelling in all directions from a point
source, the wave fronts are spherical in shape.
(ii) Cylindrical wave front - When the source of light is linear in shape, the wave front is
cylindrical in shape.
(ii) Plane wave front - As a spherical or cylindrical wave front advances, its curvature
decreases progressively. So a small portion of such a wave front at a large distance from
the source will be a plane wave front.
2).
Sketch the wave fronts corresponding to:
(i)
Parallel rays.
(ii)
Converging rays.
(iii) Diverging rays.
(i)
(ii)
(iii)
3)
State Huygens ’ Principle. What are the assumptions on which this principle is based?
Huygens’ Principle - According to this principle, each point on a wave front is a source
of secondary waves, which add up to give a wave front at any later time.
This principle is based on the following assumptions:
1. Each point on a wave front acts as a fresh source of new disturbance, called secondary
waves or wavelets.
2. The secondary wavelets spread out in all directions with the speed of light in the given
medium.
3. The new wave front at any later time is given by the forward envelope of the secondary
wavelets at that time.
4).
What is a sustained interference? State the necessary conditions for obtaining a sustained
interference of light.
It is the interference pattern in which the positions of maxima and minima of intensity on
the observation screen do not change with time. It is also called permanent interference.
Conditions for obtaining sustained interference pattern:
(i)
The two sources should continuously emit waves of same frequency or wavelength.
(ii)
The two sources of light should be coherent.
(iii) The amplitudes of the interfering waves should be equal.
(iv)
The two sources should be narrow.
(v)
The interfering waves must travel nearly along the same direction.
(vi)
The sources should be monochromatic.
5)
Define resolving power of a compound microscope. Write its expression. On what factors
does it depend?
The resolving power of a microscope is the reciprocal of the smallest distance between
two point objects at which they can be just seen separately in the microscope.
1 2 sin 
R.P. of a microscope  
d

Clearly, the resolving power of a microscope depends on:
(i) The wavelength ( ) of the light used.
(ii) Half the angle ( ) of the cone of light from point object.
(iii) The refractive index (  ) of the medium between the object and the objective of the
microscope.
6)
Define resolving power of a telescope. Write its expression. On what factors does it
depend?
The resolving power of a telescope is the reciprocal of the smallest angular separation
between two distant objects whose image can be just seen separately by it.
1
D
R.P. of a telescope 

d 1.22
Clearly, the resolving power of a telescope depends on:
(i) the diameter (D) of the telescope objective and
(ii) the wavelength ( ) of the light used.
7)
Describe an experiment to show that light waves are transverse in nature.
When ordinary or unpolarised light is passed through a tourmaline crystal P, called
polarizer, its intensity is cut down to half. We get plane polarized light. This is possible
only if light is a transverse wave. To further confirm this, a similar crystal A, called
analyser, is placed in the path of the beam. When the axes of the two crystals are parallel,
maximum intensity is transmitted by A. If crystal A is rotated in its own plane, the
intensity of light gradually decreases until it becomes zero when the axes of the two
crystals are perpendicular to each other. This confirms the transverse nature of light
waves.
8).
State and prove Brewster’s law of polarization.
When ordinary light is incident on the surface of a transparent medium, the reflected light
is partially plane polarized. For a particular angle of incidence, the reflected light is found
to be completely polarized with its vibrations perpendicular to the plane of incidence.
This angle of incidence at which a beam of unpolarised light falling on a transparent
surface is reflected as a beam of completely plane polarized light is called polarizing or
Brewster angle.
It is denoted by i p .
At the polarizing angle, the reflected and transmitted rays are perpendicular to each other,
as shown in figure. Suppose i p is the polarizing angle of incidence and rp is the
corresponding angle of refraction. Then
i p  rp  90
or
rp  90  i p
From Snell’s law, the refractive index of the transparent medium is
sin i p
sin i p
sin i p



or
  tan i p
sin rp sin 90  i p  cos i p
This relation is known as Brewster law. This law states that the tangent of the polarizing
angle of incidence of a transparent medium is equal to its refractive index.
9)
Explain with reason, how the resolving power of an astronomical telescope will change
when (i) frequency of the incident light on the objective lens is increased, (ii) focal length
of the objective lens is increased, and (iii) aperture of the objective lens is halved.
D
D

R.P. of an astronomical telescope 
1.22  1.22 c
(i)
(ii)
(iii)
When frequency of the incident light is increased, the resolving power of the
telescope increases.
When the focal length of the objective lens is increased, the resolving power does not
change.
When aperture of the objective lens is halved, the resolving power of the telescope is
also halved.
5 Marks Questions
1)
Using Huygens’ principle, show that for a parallel beam incident on a reflecting surface,
the angle of reflection is equal to the angle of incidence.
As shown in figure, consider a plane wave front AB incident on the plane reflecting
surface XY, both the wave front and the reflecting surface being perpendicular to the
plane of paper.
First the wave front touches the reflecting surface at B and then at the successive points
towards C. In accordance with Huygens’ principle, from each point on BC, secondary
wavelets start growing with the speed c. During the time, the disturbance from A reaches
the point C, the secondary wavelets from B must have spread over a hemisphere of radius
BD=AC=ct, where t is the time taken by the disturbance to travel from A to C. The
tangent plane CD drawn from the point C over this hemisphere of radius ct will be the
new reflected wave front.
Let angles of incidence and reflection be i and r respectively. In ABC and DCB , we
have
BAC  CDB
BC  BC
[ Each is 90]
[Common]
Each is equal to vt
AC  BD
 ABC  DCB
Hence
or
ABC  DCB
ir
i.e., the angle of incidence is equal to the angle of reflection. This proves the first law of
reflection.
Further, since the incident ray SB, the normal BN and the reflected ray BD are
respectively perpendicular to the incident wave front AB, the reflecting surface XY and
the reflected wave front CD (all of which are perpendicular to the plane of the paper),
therefore, they all lie in the place of the paper, i.e., in the same plane. This proves the
second law of reflection.
2)
State Huygens’ principle and use it to construct refracted wave front for refraction of a
plane wave front at a plane refracting surface. Hence derive Snell’s law.
Consider a plane wave front AB incident on a plane surface XY, separating two media 1
and 2, as shown in figure. let v1 and v2 be the velocities of light in the two media, with
v2  v1.
The wave front first strikes at point A and then at the successive points towards C.
According to Huygens’ Principle, from each point on AC, the secondary wavelets start
growing in the second medium with speed v2 . Let the disturbance take time t to travel
from B to C, then BC= v2 t. During the time, the disturbance from B reaches the point C,
the secondary wavelets from point A must have spread over a hemisphere of radius AD=
v2 t in the second medium. The tangent plane CD drawn from point C over this
hemisphere of radius v2 t will be the new refracted wave front.
Let the angles of incidence and refraction be i and r respectively.
From right ABC , we have
BC
sin BAC  sin i 
AC
From right ADC , we have
AD
sin DCA  sin r 
AC
sin i BC v1t



sin r AD v 2 t
or
sin i v1 1

 2
sin r v 2
This proves snell’s law refraction. The constant 1  2 is called the refracting index of the
second medium with respect to first medium.
Further, since the incident ray SA, the normal AN and the refracted ray AD are
respectively perpendicular to with incident wave front AB, the dividing surface XY and
the refracted wave front CD(all perpendicular to the plane of the paper),therefore, they all
lie in the plane of the paper, i.e., in the same plane. This proves second law of refraction.
3) Illustrate with the help of suitable diagram, the action of the following on a plane wave front:
(i) a prism,
(ii) a convex lens, and
(iii) a concave mirror.
(i) Behavior of a prism - Since the speed of light in glass in smaller than that in air,
therefore the lower part C of the plane wave front which travels through the greatest
thickness of the glass prism is slowed down the most and the upper part A, which travels
through the minimum thickness of the glass prism, is slowed down the least. This
explains the tilting of a plane wave front after refraction through a glass prism.
(ii) Behavior of a convex lens - The central part B of the plane wave front travels
through the greatest thickness of the lens and is slowed down the most. The marginal
parts A and C of the wave front travel through minimum thickness of the lens and are
slowed down the least. So the emerging wave front is spherical and converges to focus.
(iii) Behavior of a concave mirror - The central part B of the incident wave front has to
travel the greatest distance before getting reflected, compared to the marginal parts A and
C. Therefore, the central portion B’ of the reflected wave front is closer to the mirror than
the marginal portions A’ and C’. Hence the reflected wave front is spherical and
converges to focus.
4)
Derive a mathematical expression for the width of interference fringes obtained in
Young’s double slit experiment with the help of a suitable diagram.
As shown in figure, a narrow slit S is illuminated by monochromatic light of wavelength
. S1 and S 2 are two narrow slits at equal distance from S. The slits S1 and S 2 act as two
coherent sources, separated by a small distance d. Interference fringes are obtained on a
screen placed at distance D from the sources S1 and S 2 .
Consider a point P on the screen at distance x from the centre O. The path difference
between two waves is p  S2 P  S1P
From right angled S 2 BP and S1 AP,
2
2
 2 
d   2 
d 
S 2 P  S1 P  S 2 B  PB  S1 A  PA   D   x      D   x   
2   
2  



2
Or
2

2
2
 
S 2 P  S1 P S 2 P  S1 P   2 xd
2
2
or

S 2 P  S1 P 
2 xd
S 2 P  S1 P
In practice, the point P lies very close to O, therefore S1P  S2 P  D . Hence
p  S 2 P  S1 P 
2 xd
2D
or
p
xd
D
Positions of bright fringes - For constructive interference or bright fringes,
xd
nD
p
 n
or
x
where n  0,1,2,3,..........
D
d
Positions of dark fringes - For destructive interference or dark fringes,
xd

D
p
 (2n  1)
or
x  2n  1
where n  1,2,3,.......
D
2
2d
Fringe width - It is the separation between two successive bright or dark fringes.
nD n  1D
D
Thus fringe width
  x n  x n 1 

or

d
d
d
This expression is independent of n. Hence the fringe width is same irrespective of the
distance from the centre O.
5).
Using Huygens’ principle, explain diffraction of light due to a single slit illuminated by a
mono- chromatic source. Explain the formation of the pattern of the maxima and minima
obtained on the screen. Hence derive expression for the angular and linear widths of
central maxima.
Consider a plane wave front WW of monochromatic light incident normally on a narrow
rectangular slit AB. According to Huygens’ theory, all Parts of the slit AB will become
source of secondary wavelets, which all starts in the same phase. These wavelets spread
out in all directions, thus causing diffraction of light after it emerges through slit AB.
The secondary wavelets diffracted at an angle  are focussed at point P. Then the path
difference between the wavelets from A and B will be
p  BP  AP  BN  AB sin   d sin 
Positions of minima - Let the point P be so located that p   and   1 .
d sin 1  
If we divide the slit into two halves AC and CB, then the path difference between the
wavelets from any two corresponding points of AC and CB will be  / 2 . These wavelets
add up destructively to produce a minimum.
Thus the condition for nth minimum is
d sin  n  n , n  1,2,3,.....
Then
The directions of various minimum are given by
n
 n  sin  n 
d
Positions of secondary maxima - If the point P is so located that p 
Then
d sin  '1 
3

2
3
. When    '1' ,
2
We can divide the slit into three equal parts. The path difference between two
corresponding points of the first two parts will be  / 2. The wavelets from these points
will interfere destructively. However, the wavelets from the third part of the slit will
contribute to same intensity forming a secondary maximum.
Thus the condition for nth secondary maximum is

d sin  ' n  (2n  1) , n  1,2,3,.....
2
The directions of secondary maxima are given by
 ' n  sin  ' n  (2n  1)

2d
Angular width of central maximum - The angular width of the central maximum is the
angular separation between the directions of the first minima on the two sides of the
central maximum.
The directions of first minima on either side of central maximum are given by    / d
This angle is called half angular width of central maximum.
2
 Angular width of centar max imum  2 
d
Linear width of central maximum - If D is the distance of the screen from the single
slit, then the linear width of central maximum will be
 0  D  2 
2 D
d
0 
Arc

2  redius  D 

