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Practice Paper–1
Time: 2 Hours
Maximum Marks: 35
General Instructions:
(i) There are 12 questions in all. All questions are compulsory.
(ii) This question paper has three sections: Section A, Section B and Section C.
(iii) Section A contains three questions of two marks each, Section B contains eight questions of three marks each,
Section C contains one case study-based question of five marks.
(iv) There is no overall choice. However, an internal choice has been provided in one question of two marks and
two questions of three marks. You have to attempt only one of the choices in such questions.
(v) You may use log tables if necessary but use of calculator is not allowed.
SECTION-A
1. Mention the important considerations required while fabricating a p-n junction diode to be used as a Light Emitting
Diode (LED). What should be the order of band gap of an LED if it is required to emit light in the visible range?
Ans. (i) The semiconductors used for fabrication of visible LEDs must have a band gap from 1.8 eV to 3 eV.
(ii) Both p-type and n-type semiconductor are heavily doped.
(iii) It should be encapsulated with a transparent material.
For emitting light in visible range the band gap should be from 1.8 eV to 3 eV.
2. A monochromatic source, emitting light of wavelength, 600 nm, has a power output of 66W. Calculate the number
of photons emitted by this source in 2 minutes. Use h = 6.6 × 10–34 Js.
Ans. Here, l = 600 nm = 6 × 10–7 m; P = 66 W,
N = ?, T = 2 × 60s
Energy of each photon,
E=
hc (6.6 × 10 −34 ) × (3 × 108 )
=
= 3.3 × 10–19 J
λ
6 × 10 −7
Energy emitted by source in time t,
ES = Pt = 66 × 2 × 60 J
\ No. of photons emitted by the source in 2 minutes
N=
ES Pt 66 × 2 × 60
=
=
= 2.4 × 1022 photons.
E
E 3.3 × 10 −19
3. Distinguish between intrinsic and extrinsic semiconductors on the basis of energy band diagram.
Ans.
Intrinsic semiconductor
Extrinsic semiconductor
1. They are pure semiconductors, not doped with any impurity. They are doped with trivalent or pentavalent
2. Number density of free electrons is equal to number density impurity.
of holes. i.e., ne = nh
ne ≠ nh
Electron
CB
} Eg
3.
Hole
VB
7
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SECTION-B
4. Radiations of frequency v1 and v2 are made to fall in turn, on a photosensitive surface. The stopping potentials
required for stopping the most energetic photoelectrons in the two cases are V1 and V2 respectively. Obtain a
formula for determining the threshold frequency in terms of these parameters.
Ans. If v0 is the threshold frequency, then from photoelectric equation, we have
eV1 = hv1 – f0 and eV2 = hv2 – f0
\e(V2 – V1) = h(v2 – v1) or h =
Now e (V2 − V1 )
(v2 − v1 )
eV1 = hv1 – f0 = hv1– hv0
Or v0 = v1 −
eV1
= v1 − eV1
h
= v1 −
 v2 − v1 


 e(V2 − V1 ) 
V1 (v2 − v1 ) ν1V2 − ν1V1 − ν2 V1 + ν1V1
( ν V − ν2 V1 )
=
= 1 2
(V2 − V1 )
(V2 − V1 )
(V2 − V1 )
5. Two material bars A and B of equal area of cross-section, are connected in series to a DC supply. A is made of
usual resistance wire and B of an n-type semiconductor.
(a) In which bar is drift speed of free electrons greater?
(b) If the same constant current continues to flow for a long time, how will the voltage drop across A and B be
affected?
Justify each answer.
Ans. (a) Drift speed in B (n-type semiconductor) is higher
Reason: I = neAvd is same for both
n is much lower in semiconductors.
(b) Voltage drop across A will increase as the resistance of A increases with increase in temperature.
Voltage drop across B will decrease as resistance of B will decrease with increase in temperature.
6. Calculate and compare the energy released by
(a) fusion of 1 kg of hydrogen deep within the sun and
(b) the fission of 1 kg of 235U in a fission reactor.
Ans. (a) In sun 4 hydrogen nuclei combine to form a helium nucleus and 26 MeV of energy is released
4 11H 
→ 42 He + 2e + + 26 MeV
No. of atom in 1g of hydrogen = 6.023 × 1023
No. of atom in 1kg of hydrogen = 6023 × 1023
Energy released by 1 kg of hydrogen
=
26 × 6023 × 1023
 39 × 1026 MeV
4
(b) No. of atom in 235g of
235
U = 6.023 × 1023
No. of atom in 1 kg of
235
U = 6.023 × 1023 ×
1000
235
Energy released in per fission of
235
U = 200 MeV
Energy released in per fission of 1 kg of 235U
8 n
=
Physics-XII
6.023 × 1023 × 1000 × 200
 5.1 × 1026 MeV
235
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7. A convex lens made up of glass of refractive index. 1.5 is dipped in turn in (i) a medium of refractive index 1.65
(ii) a medium of refractive index 1.33 (a) will it behave as a converging or diverging lens in two cases (b) How
will its focal length changes in two media.
Ans. (i) refractive index of medium = m1 = 1.65, m2 = 1.5,
R1 = R, R2 = – R
µ
 1
1 
 1.5
  2  − .15  2 
1
−
− 1   =
=  2 − 1 
=


  R  1.65  R 
R
R
µ
1
.
65
f
 1
 1
2
It will behave as a diverging lens as f is –ve.
If m1 = 1.33, then
1  1.5
 2 .17 2
=
− 1 =
f  1.33  R 1.33 R
f is + ve so converging lens.
(b) In air
1
2 1
= (1.5 − 1) = or f = R
fa
R R
In medium of refractive index 1.65,
165 R
= − 5.5 R
f= −
15 2
In medium of refractive index 1.33,
f=
133 R 7.8
=
R = 3.9R
17 2
2
f is maximum in medium of r-I 1.65.
8. Draw a schematic diagram of a reflecting telescope (Cassegrain). Write two important advantages that a reflecting
telescope has over a refracting telescope.
Ans. Objective is concave mirror of large focal length and large
Objective
aperture having a hole around its pole. The light coming
Secondary Mirror
from a far off object after reflection from the objective
is made to fall on a secondary mirror (convex)placed in
A
Parallel rays
between its pole and focus. The light reflected from the
from a distant
B
convex mirror goes into the eye piece, which acts as a
object
Eye piece
simple microscope.
Advantage of reflecting telescope over a refracting
telescope:
(i) Reflecting telescope is free from chromatic aberration.
(ii) Image is free from spherical aberration.
(iii) Image formed is brighter and resolving power is high.
9. How is a wavefront defined? Using Huygen’s construction draw a figure showing the propagation of a plane wave
reflecting at the interface of the two media. Show that the angle of incidence is equal to angle of reflection.
Ans. Consider a plane wavefront AB reflected at the interface of the two media. When the wavefront strikes the reflected
surface at point B the secondary wavelets travel back into the some medium.
During the time the disturbance from A reach the point C, the secondary wavelets from B have spread over a
hemisphere of radius BD = vt. Where v in the speed of the wave and t is the time taken from the wave to reach
from A to C.
Incident wavefront
Reflected wavefront
Also
AC = vt
N′
N
The tangent plane CD is the new reflected wavefront.
Y
D
A
Let i and r be the angle of incidence and angle of reflection
respectively.
P
r
i
∠ABC = ∠i, ∠BCD = ∠r
X
i
B
r
C
Practice Paper–1
Y
n 9
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In DABC and DDCB, ∠BAC = ∠CDB = 90°
BC = BC common
AC = BD = vt
So
DABC ≅ DDCB
So by cpct
∠ABC = ∠BCD or ∠i = ∠r
10. A monochromatic radiation of wavelength 975 Å excites the hydrogen atom from its ground state to a higher
state. How many different spectral lines are possible in the resulting spectrum? Which transition corresponds to
the longest wavelength amongst them?
Ans. Energy corresponding to the given wavelength:
E (in eV) =

λ(in A)
En – E1 = 12. 71
The excited state:
12400
−13.6
n2
= 12.71 eV
1
+ 13.6 = 12.71
∴
n = 3.9 ≈ 4
n(n − 1)
=6
Total no. of spectral lines emitted :
2
Longest wavelength will correspond to the transition
n = 4 to n = 3
11. Identify the following electromagnetic radiation as per the wavelength given below:
(a) 10– 4 nm
(b) 10–3 m
(c) 1 nm.
Write one application of each.
Ans. (a) 10– 4 nm → gamma radiation
Application : Radio therapy or to initiate nuclear reactions.
(b) 10–3 m → microwaves
Application : in RADAR for aircraft navigation, microwave cooking (oven).
(c) 1 nm → X-rays
Application : In medical science for detection of fractures in bones.
SECTION-C
12. Radio waves are produced by the accelerated motion of charges in conducting wires. Microwaves are produced by
special vacuum tubes. Infrared waves are produced by hot bodies and molecules also known as heat waves. UV
rays are produced by special lamps and very hot bodies like sun.
10 n
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Based on the above facts, answer the following questions:
(i) Solar radiation is:
(a) transverse em wave
(b) longitudinal em wave
(c) both (a) and (b)
(d) none of these
(ii) The cause of greenhouse effect is
(a) infrared rays
(b) ultraviolet rays
(c) X-rays
(d) radio waves
(iii) Biological importance of ozone layer is
(a) it stops UV rays
(b) it reflects radio waves
(c) its layer reduces greenhouse effect
(d) none of these
(iv) Ozone is found in
(a) stratosphere
(b) ionosphere
(c) mesosphere
(d) troposphere
(v) Earth’s atmosphere is richest in
(a) ultraviolet rays
(b) infrared rays
(c) X-rays
(d) microwaves
Ans. (i) (a) transverse em wave
(ii) (a) infrared rays
(iii) (a) it stops UV rays
(iv) (a) stratosphere
(v) (b) infrared rays
Practice Paper–1
n 11
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Practice Paper – 2
Time: 2 Hours
Maximum Marks: 35
General Instructions: Same as Practice Paper-1
SECTION-A
1. A semiconductor has equal electron and hole concentration of 2 × 108 m–3, concentration increases to 4 × 1010 m–3.
(i) What type of semiconductor is obtained on doping? (ii) Calculate the new electron hole concentration of the
semiconductor. (iii) How does the energy gap vary with doping?
Ans. ni = 2 × 108 m–3
On doping nh = 4 × 1010 m–3
(i) As on doping the number density of holes have increased so it is a p-type semiconductor.
(ii) ni2 = nenh
n 2 (2 × 108 ) 2
ne = i =
= 106 m −3
10
nh
×
4
10
(iii) Energy gap decreases with doping due to the creation of acceptor energy level.
2. The energy levels of an atom are shown in figure. Which transition corresponds to emission of radiation of (i)
maximum wavelength (ii) minimum wavelength?
0 eV
– 2 eV
– 4.5 eV
A
B
C
D
– 10 eV
OR
Figure shows variation of stopping potential (V0) with frequency (v) for two photosensitive material M1 and M2.
(i) Why is the slope same for both lines?
(ii) For which material will the emitted electrons have greater kinetic energy for the incident radiations of the same
frequency? Justify your answer.
M1
M2
V0
v
Ans. For transition A, Energy of emitted photon = 0 – (– 2) = 2 eV
For transition B, Energy of emitted photon = 0 – ( – 4.5) = 4.5 eV
For transition C, Energy of emitted photon = – 2 – (– 4.5) = 2.5 eV
For transition D, Energy of emitted photon = – 2 – (– 10) = 8 eV
hc 1
∝
λ λ
Therefore, for maximum wavelength, E should be minimum. Transition A for which energy difference E is minimum
corresponds to emission of radiation of max. wavelength.
As energy emitted, E =
12
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Similarly, transition D for which energy diff. E is minimum, corresponds to emission of radiation of minimum
wavelength.
(i) From Einstein’s photoelectric equation
eV0 = hv – f0
OR
or V0 = h v − φ0
e
e
h
h
The slope of stopping potential vs frequency graph is a constant equal to , and in same for all metals. So
e
e
the slope is same for both lines.
(ii) As (K.E.)max = hv – hv0
For metal M1, the threshold frequency v0 is lesser. So the (K.E.)max is larger for metal M1 so metal M1 will emit
photoelectrons of higher K.E.
3. In V-I characteristics of a p-n junction diode:
(a) Why is the current under reverse bias almost independent of the applied potential upto a critical voltage?
(b) Why does the reverse current show a sudden increase at critical voltage?
Ans. (a) When a p-n junction is reverse biased, the minority charge carriers i.e., electrons from p side and holes from n
side drift towards the junction. The number density of minority charge carrier remains almost constant upto a
critical voltage. So the current under the reverse bias is independent of applied voltage upto a critical voltage.
(b) When the reverse biased voltage is increased beyond critical voltage, the electric field at the junction becomes
strong enough to pull out electrons from the host atoms on p-side and accelerates them towards n-side. Due to
the movement of these electrons across the junction, the current increases suddenly at critical voltage.
SECTION-B
4. Ground state energy of hydrogen atom is – 13.6 eV
(i) What is the K.E. of an electron in the second excited state?
(ii) What is the P.E. of an electron in the third excited state?
(iii) If the electron jumps to the ground state from the third excited state, calculate the wavelength of the photon
emitted.
E
Ans. E1 = − 13.6 eV, E n = 21
n
(i) For second excited state n = 3
E
− 13.6
= − 1.51 eV
So
E3 = 21 =
9
3
E
− 13.6
E4 = 21 =
= − 0.85 eV
16
4
Kinetic energy in the second excited state = – E3 = 1.51 eV
(ii) P.E. in 3rd excited state = 2E4 = 2 × (– 0.85) = – 1.7 eV
hc
(iii)
E = hv =
= E 4 − E1
λ
hc
6.6 × 10 −34 × 3 × 108
−34
8
=
l=
= 6.6 × 10 × 3 × 10 = 970 Å
−19
E 4 − E1 [ − 0.85 − ( − 13.6)] × 1.6 × 10 −19
12.75 × 1.6 × 10
5. Describe briefly using the necessary circuit diagram, the three basic processes which take place to generate the
emf in a solar cell when light falls on it. Draw the I-V characteristics of a solar cell.
Write two important criteria required for the selection of a material for solar cell fabrication.
Ans. The generation of emf by a solar cell when light falls on it is due to :
(i) Generation of e-h pairs due to light (with hv > Eg) close to the junction.
Practice Paper – 2 n 13
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(ii) Separation: of electrons and holes due to electric field of the depletion region. Electrons are swept to n-side
and holes to p-side.
IL
p
n
Depletion region
(iii) Collection : Electrons reaching n-side are collected by front contact and holes reaching p-side are collected
by the back contact.
Thus p-side becomes +ve and n-side –ve giving rise to photo-voltage
When an external load is connected, photo-current IL flows through the load.
V - I characteristics of solar cell is drawn in 4th quadrant because a solar cell does not draw current but
supplies it to the load.
Semiconductors with band gap close to 1.5 eV are ideal materials for solar cell fabrication. They are made with
semiconductor like Si (Eg = 1.1 eV), Ga As (Eg = 1.43 eV), Cd Te (Eg = 1.45 eV), Cu In Se2 (Eg = 1.049 eV) etc.
Important criteria for selection of material:
(i) Band gap (~ 1.0 to 1.8 eV)
(ii) High optical absorption (~ 104 cm–1)
(iii) Electrical conductivity
(iv) Availability of raw material
(v) Cost. (Any two)
6. Explain the concept of nuclear forces. Discuss their characteristic properties. Which properties distinguish them
from electrostatic forces?
Ans. Nuclear force : Inside the nucleus large number of protons are held along with the neutrons in a very small space.
The Coulombian force of repulsion between protons (due to like charges) is nullified by a strong attractive force
called nuclear force.
So Nuclear forces are the strong forces of attraction which hold together the nucleus in a small space called nucleus,
inspite of strong electrostatic force of repulsion between protons.
Characteristic/properties of nuclear forces:
Refer theory topic no. 1 under the heading Nuclear force.
14 n
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Comparison between electrostatic and nuclear force:
Electrostatic
1. These are weak forces.
2. Depending upon the charge it could be attractive or repulsive
and magnitude depends upon the amount of charge.
3. It is a long range force.
4. It is a central force.
5. It obeys inverse square law.
Nuclear Force
1. These are the strongest attractive forces
2. It is independent of charge.
3. It is a short range force.
4. It is a non central force.
5. It does not obey inverse square law.
7. (a) State two important features of Einstein’s photoelectric equation.
(b) Radiation of frequency 1015 Hz is incident on two photosensitive surface P and Q. There is no photoemission
from surface P. Photoemission occurs from surface Q but photoelectrons have zero kinetic energy. Explain
these observations and find the value of work function for surface Q.
Ans. (a) According to Einstein’s photoelectric equation (KE)max = hn – hn0
(i) If frequency of incident photon n is less than the threshold freq. n0, no emission of photo-electrons is possible
because for n < n0
(KE)max is negative, which is not possible
(ii) With increase in frequency of incident radiation there will be increase in (KE) of photoelectrons but no change
in photo-electric current.
(b) n = 1015 Hz
As there in no photoemission from surface P, so threshold freq. of P is more than 1015 Hz
For metal Q, (KE) = 0
So
hn = f0 + KE
\
hn = f0
Work function of
Q = hn = 6.6 × 10–34 × 1015 = 6.6 × 10–19 J
=
6.6 × 10 −19
= 4.13 eV
1.6 × 10 −19
8. A compound microscope uses an objective lens of focal length 4 cm and eyepiece lens of focal length 10 cm. An
object is placed at 6 cm from the objective lens. Calculate the magnifying power of the compound microscope.
Also calculate the length of the microscope.
OR
An optical instrument uses eye-lens of power 12.5 D and objective lens of power 50 D and has a tube length of
20 cm. Name the optical instrument and calculate its magnifying power, if it forms the final image at infinity.
Ans.
fo = 4 cm, fe = 10 cm, uo = – 6 cm, m = ?
m= −
vo 
D
1+ 

fe 
| u0 | 
To find vo using lens formula,
1
1
1
1 1 1
1
1 1 3−2 1
=
=
− , = −
; = − =
vo
f0
vo u0 4 vo − 6
4 6
12
12
So vo = 12 cm
12 
25 
35
m = − 1 +  = − 2 ×
=−7
10
6  10 
Length of the microscope = ue + vo
1
1
1
1 1
1
1
1
1 −2 − 5 − 7
=
−
For eye piece,
= − ,
; = −
− =
=
10
−
25
u
ue
ve ue
e
fe
25 10
50
50
− 50
= − 7.14 cm
7
Distance between lenses = | vo | + | ue | = 12 + 7.14 = 19.14 cm
ue =
Practice Paper – 2 n 15
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OR
The optical instrument is compound microscope.
Given, Pe = 12.5 D, L = 20 cm, Po = 50 D
1
1
100
100
We know f =
\ fe =
m=
cm = 8 cm; f o =
cm = 2 cm
P
12.5
12.5
50
The magnifying power of compound microscope
L D 20 25
×
=
×
= 31.25
m =
(∵ D = least distance of distinct vision = 25 cm)
2
8
fo fe
9. In Young’s double slit experiment, explain with reason in each case, how the interference pattern changes, when
(i) Width of the slits is doubled
(ii) Separation between the slits is increased, and
(iii) Screen is moved away from the plane slits.
Dλ
Ans. The fringe width β =
d
(i) When the width of the slit is doubled; the intensity of interfering wave becomes four times, intensity of maxima
becomes 16 times i.e., fringes become brighter.
1
(ii) β ∝ , when separation between the slits is increased the fringe width decreases i.e., fringes comes closer.
d
(iii) b ∝ D, when screen is moved away from the plane of the slits, the fringe width increases i.e., fringes become
farther away.
10. The following table gives data about the single slit diffraction experiment:
Wavelength of light
Half angular width of the principal maxima
l
q
pl
qq
Find the ratio of the width of the slits used in the two cases. Would the ratio of the half angular width of the first
secondary maxima, in the two cases, be also equal to q?
Ans. Let the slit width be a1 and a2 in the two cases
a
pλ a1
q
λ
pλ
qθ
×
or 1 =
and qθ =
q=
;
=
a
λ
a
p
a1
a2
θ
2
2
3 pλ
Half angular width of second case
2 a2
a1
q
= 3 λ = p = p× =q
a
p
Half angular width of firsst case
2
2 a1
11. Why are infra-red radiations referred to as heat waves? Name the radiations which are next to these radiations in
the electromagnetic spectrum having (i) shorter wavelength (ii) longer wavelength.
OR
Name the em waves in the wavelength range 10 nm to 10–3 nm. How are these waves generated? Write their two
applications.
Ans. Infrared rays are produced by hot bodies or by vibrations of molecules or atoms.
IR rays are referred to as heat waves because they get easily absorbed by water molecules in most materials. Due
to the increased thermal agitation their temperature increases and hence get heated. They are used in muscular
pain relief.
(i) EM waves having short wavelength than infrared waves are visible, UV, X-rays and g-rays.
(ii) EM waves having longer wavelength than infrared waves are microwaves, short radio waves, television and
FM radio.
OR
–3
Em waves in the wavelength range 10 nm to 10 nm are X-ray. X-rays are generating by bombarding a metal
target with high frequency electrons.
16 n
Physics-XII
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Uses:
(i) Diagnosis of bone fractures
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(ii) Treatment of some forms of cancer
SECTION-C
12. When the light from a monochromatic source is incident on a single monochromatic slit, it gets diffracted and a
pattern of alternate bright and dark fringes is obtained on screen, called diffraction pattern of single slit. In diffraction
pattern of single slit, it is found that.
(i) Central bright fringe is of maximum intensity and the intensity of any secondary bright fringe decreases with
increase in its order.
(ii) Central bright fringe is twice as wide as any other secondary bright or dark fringe.
Incident
light
Slit
Viewing screen
Based on the above facts, answer the following questions.
(i) A single slit of width 0.1 mm is illuminated by a parallel beam of light of wavelength 6000 Å and diffraction
bands are observed on a screen 0.5 m from the slit. The distance of the third dark band from the central bright
band is
(a) 3 mm
(b) 1.5 mm
(c) 9 mm
(d) 4.5 mm
(ii) In Fraunhofer diffraction pattern, slit width is 0.2 mm and screen is at 2 m away from the lens. If wavelength
of light used is 5000 Å then the distance between the first minimum on either side the central maximum is
(a) 10–1 m
(b) 10–2 m
(c) 2 × 10–2 m
(d) 2 × 10–1 m
(iii) Light of wavelength 600 mm is incident normally on a slit of width 0.2 mm. The angular width of central
maxima in the diffraction pattern is (measured from minimum to minimum)
(a) 6 × 10–3 rad
(b) 4 × 10–3 rad
(c) 2.4 × 10–3 rad
(d) 4.5 × 10–3 rad
(iv) A diffraction pattern is obtained by using a beam of red light. What will happen if the red light is replaced by
the blue light?
(a) bands disappear
(b) bands become broader and farther apart
(c) no change will take place
(d) diffraction bands become narrower and crowded together
(v) To observe diffraction the size of obstacle
λ
(a) should be
where l is wavelength
(b) should be of the order of wavelength
2
(c) has no relation to wavelength
(d) should be much larger than the wavelength
Ans. (i) (c) 9 mm
(ii) (b) 10–2 m
(iii) (a) 6 × 10–3 rad
(iv) (d) diffraction bands become narrower and crowded together
(v) (b) should be of the order of wavelength
Practice Paper – 2 n 17
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Practice Paper – 3
Time: 2 Hours
Maximum Marks: 35
General Instructions: Same as Practice Paper-1
SECTION-A
1. Explain how the width of depletion layer in a P-n junction diode changes when the junction is (i) forward biased
(ii) reverse biased.
Ans. (i) In forward bias, positive terminal of the battery is connected to p side and negative terminal to n-side of the
p-n junction. So the holes from p side which are majority carrier are pushed towards the junction and electron
from n-side are pushed towards the junction. This results in reduction of the width of depletion layer.
(ii) in reverse bias, n-side is connected with positive terminal of the battery and p-side to negative terminal of the
battery. The minority carriers i.e., holes from n-side are pushed towards the junction whereas electrons which
are minority carriers in p-side are pushed towards the junction. The majority carrier from p as well as n-side
move away from the junction. This increases the width of depletion layer.
Forward Biased:
A′
A
n
p
B
B′
A
A′
AA′ → Original depletion layer
BB′ → Depletion layer after forward biasing
Reverse Biased
n
p
B
B′
AA′ → Original depletion layer
BB′ → Depletion layer after reverse biasing
2. You are given two nuclei 3X7 and 3Y4. Explain giving reasons, as to which one of the two nuclei is likely to be
more stable?
OR
56
Calculate the binding energy per nucleon of 26Fe nucleus. Given that mass of 26Fe56 = 55.934939 u, mass of proton =
1.007825 u, mass of neutron = 1.008665 u and 1u = 931 MeV.
Ans. In case of 3X7,
No. of neutron 7 − 3 4
=
= = 1.33
3
3
No. of proton
In case of 3Y4 ,
No. of neutron 4 − 3 1
=
= = 0.33
3
3
No. of proton
For stability, this ratio has to be close to one. Obviously, nucleus 3X7 is more stable than the nucleus 3Y4.
OR
56
For 26Fe ,
no. of protons = 26 = Z
No. of neutrons = A – Z = 56 –26 = 30
Dm = [Zmp + (A – Z) mn – MN]
18
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\ Mass defect = 26 mp + 30 mn – MFe
= 26 × 1.007825 + 30 × 1.008665 – 55.934939
= 26.20345 + 30.25995 – 55.934939
= 0.528461 u.
931
B.E./nucleon = 0.528461 ×
= 8.79 MeV.
56
3. What do the acronyms LASER and LED stand for? Name the factor which determines (i) frequency and
(ii) intensity of light emitted by LED.
Ans. LASER stands for Light Amplification by Stimulated Emission of Radiation.
LED stands for Light Emitting Diode
(i) The frequency of light emitted by LED is determined by the band gap of the semiconductor used in LED.
(ii) Intensity of light emitted by LED depends on the doping level of the semiconductor and forward bias voltage.
SECTION-B
4. (a) In photoelectric effect, do all the electrons that absorb a photon come out as photoelectrons irrespective of their
location? Explain.
(b) A source of light, of frequency greater than the threshold frequency, is placed at a distance ‘d’ from the cathode
of a photocell. The stopping potential is found to be V. If the distance of the light source is reduced to d/n
(where n > 1), explain the changes that are likely to be observed in the (i) photoelectric current and (ii) stopping
potential.
Ans. (a) No, it is not necessary that if the energy supplied to an electron is more than the work function, it will come
out.
The electron after receiving energy, may lose energy to the metal due to collisions with the atoms of the metal.
Therefore, most electrons get scattered into the metal. Only a few electrons near the surface may come out of
the surface of the metal for whom the incident energy is greater than the work function of the metal.
(b) on reducing the distance, intensity increases.
Photoelectric current increases with the increase in intensity. Stopping potential is independent of intensity,
and therefore remains unchanged.
5. (a) Draw the necessary energy band diagrams to distinguish between conductors, semiconductors and insulators.
(b) How does the change in temperature affect the behaviour of these materials? Explain briefly.
Ans. (a)
Insulator
Eg > 3eV
Semiconductor
}
Energy
gap (Eg)
Filled
valence
band
{
}
Eg < 3eV
Overlapped conduction and valence band
(b) With increase in temperature, resistivity of a conductor increases because of increase in no. of collisions. So
relaxation time decreases which increases resistivity.
With increase in temperature, resistivity of a semiconductor decreases. Because of increase in temperature
more number of electrons gain energy to jump to conduction band and are available for conducting current.
In case of insulator for moderate change of temperature, there is no change in behaviour.
Practice Paper – 3 n 19
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6. Binding energy per nucleon versus mass number curve is as shown.
A
A1
A2
A3
Z S, Z1 W, Z2 X and Z3 Y are four nuclei indicated on the curve.
Based on the graph:
(a) Arrange X, W and S in the increasing order of stability.
(b) Write the relation between the relevant A and Z values for the
following nuclear reaction.
S → X + W
(c) Explain why binding energy for heavy nuclei is low.
Ans. (a) S, W, X
(b) Z = Z1 + Z2
A = A1 + A2
(c) Reason for low binding energy:
In heavier nuclei, the Coulombian repulsive effects can increase considerably and can match/ offset the attractive
effects of the nuclear forces. This can result in such nuclei being unstable.
7. In the figure given below, three light rays red (R), green (G) and blue (B) are incident on an isosceles right-angled
prism abc at face ab. Explain with reason, which ray of light will be transmitted through the face ac. The refractive
index of the prism for red, green, blue light are 1.39, 1.44, 1.47 respectively.
Trace the path of rays after passing through face ab.
a
B
G
R
45°
b
1
sin ic
so m for critical angle of 45° is
1
= 2 = 1.414
m =
sin 45°
Angle of incidence for all colours is 45°.
The ray will be allowed to pass through ac if i < ic.
Red colour:
m = 1.39,
1
, i > 45°, i.e., ic > i, or i < ic
sin ic =
1.39 c
Ans. As
c
m=
a
45°
B
G
R
Then it will be refracted.
Green colour : m = 1.44,
1
45°
R
b
c
sin ic =
, i < 45°, i.e., ic < i, or i > ic
1.44 c
Then total internal reflection will take place.
B G
Blue colour : m = 1.47,
1
, i < 45°, i.e., ic < i, or i > ic
sin ic =
1.47 c
Then total internal reflection will take place.
8. (a) A giant refracting telescope at an observatory has an objective lens of focal length 15 m. If an eye piece of
focal length 1.0 cm is used, what is the angular magnification of the telescope?
20 n
Physics-XII
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(b) If this telescope is used to view the moon, what is the diameter of the image of the moon formed by the objective
lens. The diameter of the moon is 3.48 × 106 m and radius of lunar orbit is 3.8 × 108 m.
OR
(a) Draw the labelled diagram for the formation of image by a compound microscope.
(b) Define the magnifying power of compound microscope.
(c) Explain why both the objective and the eyepiece of a compound microscope must have short focal length.
Ans. (a) fo = 15 m, fe = 1 cm = 10–2 m
Angular magnification
f o − 15
=
m = −
= – 1500
f e 10 −2
–ve sign shows that image is inverted.
(b) Let D be the diameter of the moon in meters. r is radius of the
lunar orbit. Then Objective
d is the diameter of the image of the moon.
3.48 × 106 d
D d
=
⇒
=
a=
r
fo
3.8 × 108 15
(a)
d=
Moon
fo
α
r
α
Image
3.48 × 106 × 15
= .1373 m = 13.73 cm
3.8 × 108
OR
B
A
D
E
Eye
vo
uo
B
A
2Fo A Fo
A
O
F
o
Fe
Objective
lens
O
B
Eyepiece
D
(b) The magnifying power of a compound microscope is defined as the ratio of angle subtended at the eye by the
final virtual image to the angle subtended at the eye by the object when both are at the teast distance of distinct
vision from the eye
Practice Paper – 3 n 21
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(c) (i) If the focal lengths are less than their magnifying power will be more.
(ii) To avoid any aberration in refraction due to larger bend on passing through the eye piece.
9. (a) If one of two identical slits producing interference in Young’s experiment is covered with glass, so that the
light intensity passing through it is reduced to 50%, find the ratio of the maximum and minimum intensity of
the fringe in the interference pattern.
(b) What kind of fringes do you expect to observe if white light is used instead of monochromatic light?
Ans. (a) Let the maximum intensity be Imax and minimum intensity Imin
Imax µ (a1 + a2)2
Imin µ (a1 – a2)2
I0 µ a12
or a1 µ
I0
2
or a2 µ
a22 µ

 I 0 +
I0 

2
2
I max
=
2
I min 
I0 
 I 0 −

2
I max ( 2 + 1) 2
=
or I
( 2 − 1) 2 or
min
\
I0
I0
2
I max (3 + 2 2 )
=
I min (3 − 2 2 )
(b) Time varying fringes patterns in which relative positions of fringes changes with time because of the presence
of various wavelengths in white light.
The interferences patterns due to different constituent colours of white light overlap (incoherently) coloured
fringes of different width are obtained.
10. Given the ground state energy E0 = – 13.6 eV and Bohr radius a0 = 0.53 Å. Find out how de-Broglie wavelength
associated with the electron orbiting in the ground state would change when it jumps into the first excited state.
Ans. As de-Broglie wavelength
h
h
l= =
p
2mE
For ground state n = 1
For First excited state n = 2
13.6
En = − 2 eV ; E1 = – 13.6 eV
n
E2 1
− 13.6
=
E2 =
eV , So
E1 4
4
λ1
E2 1
=
=
⇒ l2 =2l1
λ2
E1 2
11. In a plane electromagnetic wave, the electric field oscillates sinusoidally at a frequency of 2.0 × 1010 Hz and
amplitude 48 Vm–1.
(a) What is the wavelength of a wave?
(b) What is the amplitude of the oscillating magnetic field?
(c) Show that the average energy density of the E field equals the average energy density of the B-field
[c = 3 × 108 ms–1]
OR
Name the parts of electromagnetic spectrum which is:
(a) Suitable for radar system used in air craft navigation
(b) Used to treat muscular strain.
(c) Used as a diagnostic tool in medicine.
Write in brief, how these waves can be produced.
22 n
Physics-XII
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Ans. Given : v = 2 × 1010 Hz, E0 = 48 Vm–1
E0
c 3 × 108
48
= 1.5 × 10 −2 m
=
= 1.6 × 10 −7 T
(a) l = =
(b) B0 =
v 2 × 1010
c 3 × 108
(c) To show that the energy density of electric field = average energy density of the magnetic field.
1
Energy density of E. field = ue = ε 0 E 2 ...(i)
2
1 2
B ...(ii)
u
=
Energy density of M-field = B
2µ 0
2
1
1  1  2
E
1
1
2
2
2
Now,
ue = ε o E = ε 0 (cB) as = c = ε 0 
 B = 2µ × B
2  µ0 ε0 
2
2
B
0
Thus,

ue = uB  as c =



µ0 ε0 
1
OR
(a) Microwaves
Production: Klystron/magnetron
(b) Infrared radiation
Production: Hot bodies/vibrations of atoms and molecules
(c) X-rays
Production: Bombarding high energy electrons on metal target.
SECTION-C
12. A prism is a portion of a transparent medium bounded by two plane faces inclined to each other at a suitable angle.
A ray of light suffers two refractions on passing through a prism and hence deviates through a certain angle from its
original path. The angle of deviation of a prism is, d = (m – 1) A, through which a ray deviates on passing through
a thin prism of small refracting angle A
sin (A + δ m ) / 2
If m is refractive index of the material of the prism, then prism formula is, m =
sin A/ 2
Based on the above facts, answer the following questions.
(i) The angle of deviation is minimum for
(a) red colour
(b) yellow colour
(c) violet colour
(d) blue colour
(ii) When white light moves through vacuum
(a) all colours have same speed
(b) different colours have different speeds
(c) violet has more speed than red
(d) red has more speed than violet
(iii) The deviation through a prism is maximum when angle of incidence is
(a) 45°
(b) 70°
(c) 90°
(d) 60°
(iv) What is the deviation produced by a prism of angle 6°? (Refractive index of the material of the prism is 1.644)
(a) 3.864°
(b) 4.595°
(c) 7.259°
(d) 1.252°
(v) A ray of light falling at an angle of 50° is refracted through a prism and suffers minimum deviation. If the angle
of prism is 60°, then the angle of minimum deviation is
(a) 45°
(b) 75°
(c) 50°
(d) 40°
Ans. (i) (a) (ii) (a)
(iii) (c)
(iv) (a) (v) (d)
Practice Paper – 3 n 23
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Practice Paper – 4
Time: 2 Hours
Maximum Marks: 35
General Instructions: Same as Practice Paper-1
SECTION-A
1. How does the energy gap in an intrinsic semiconductor vary, when doped with (a) a pentavalent impurity and
(b) a trivalent impurity? Draw their energy band diagrams.
2. In the ground state of hydrogen atom, its Bohr radius is 5.3 × 10–11 m. The atom is excited such that the radius
becomes 21.2 × 10–11 m. Find the value of principal quantum number and total energy of the atom in excited state.
OR
The electron in hydrogen atom is initially in the third excited state. What is the maximum number of spectral lines
which can be emitted, when it finally moves to the ground state?
3. Mention the important considerations required while fabricating a p-n junction diode to be used as a Light Emitting
Diode (LED). What should be the order of band gap of an LED if it is required to emit light in the visible range?
SECTION-B
4. In a plot of photoelectric current versus anode potential, how does:
(i) The saturation current vary with anode potential for incident radiations of different frequency but same intensity?
(ii) The stopping potential vary for incident radiations of different intensities but same frequency?
(iii) Photoelectric current vary for different intensities but same frequency of incident radiations? Justify your
answer is each case.
5. Distinguish between n-type and p-type of semiconductors on the basis of energy band diagrams. Compare their
conductivities at absolute zero temperature and at room temperature.
6. Using Rutherford’s model of an atom, derive the expression for the total energy of the electron in hydrogen atom
what is the significance of total negative energy possessed by the electrons.
7. (a) An equiconvex lens with radii of curvature of magnitude r each, is put over a liquid layer poured on top of a
plane mirror. A small needle, with its tip on the principal axis of the lens, is moved
along the axis until its inverted real image coincides with the needle itself. The
distance of the needle from the lens is measured to be ‘a’. On removing the liquid
layer and repeating the experiment the distance is found to be ‘b’.
Liquid
Given that two values of distances measured represent the focal length values in the
two cases, obtain a formula for the refractive index of the liquid.
(b) If r = 10 cm, a = 15 cm, b = 10 cm, find the refractive index of the liquid.
8. Draw ray diagram to explain the working of a refracting astronomical telescope when the final image is formed at
near point.
Define its magnifying power and obtain expression for it.
OR
(a) Draw a labelled ray diagram showing the formation of a final image by a compound microscope at least distinct
vision.
(b) The total magnification produced by a compound microscope is 20. The magnification produced by the eye
piece is 5. The microscope is focussed on a certain object. The distance between the objective and the eye piece
is observed to be 14 cm. If least distance of distinct vision is 25 cm, calculate the focal length of the objective
and the eye piece.
9. A plane wavefront, of width x, is incident on an air-water interface and the corresponding refracted wavefront has
a width z as shown. Express the refractive index of air with respect to water, in terms of the dimension shown.
24