(a) 0 these,

(c)t2 ( ) r4 (900) (1988)

ANSWER KEY

() 2 (d) (6) 4. (b) 5 (d) (d) (b) 8. (b)

(b) 14. (c) 15. (d) 16.
9 (d) 10. (b) 11. (a) 12. () 13. (b)
17. (c) 18. (b) 19 (a) 20. (d) 21. (c) 22 (a) 23. (a) 24. (a)
25. (b) 26. (b) 27. () 28. (d) 29 (c) 30. (b) 31 (a) 32 (a)
33. (b) 34 (a) 35. (d) 36. (d) 37. () 38 () 39 (c) 40. (a)
41. (b) 42. (a) 43 (b) 44 () 45. (d) 40. (c) 47. (C) 48 (b)
49. (c) 50. (a) 51. (d) 52. (b) 53. (b) 54. () 55 () 56. (d)
57. (b) 58. (d) 59. (c) 60. () 6l. (d) 62. (a) (b) 0-4. (b)
65. (a) 66. (6) 67. (6) 68. () 69. (6) 70. (6) 71. (d) 72. (a)
73. (c) 74. (d) 75. (a) 76. (a) 77. a)
78 (b) 79 (a) SO. (b)
81. (a) 82. (b) 83. (d) 84. (a) 85. (c) So. (6) $7. (d) SS.
89. (c) 90. (d) 91. (a)
(d)
92. (d) 93. (c) 94. () 95. (a) 96. (b)
97. (a) 98. (c) 99. (b) 100. (b) 101. (c) 102. () 103 (b) 104 (b)
105. (a) 106. (c) 107. (b) 108. (a) 109. (d) 110. (6) 111. (6) 112. (a)
113. (a) 114. (b) 115. (d) 116. (d) 117. (a) 118. ()
SemiconadUcTor E e U D

EXPLANATIONS

to heating, number of electron-

holes are 6. (d): Duce of
so overall resistance
semicondluctors,
In p-1ype hole pairs will increase,
1. (c) and trivalent atoms
are the
carriers diode will change.
the majority and reversed
such as B, Al or Ga.
Due to which forward biasing
dopants
(d): LED bulb will light up if switcns)biasing both are changed.
2. 3 kS2, R,
=

2 kS2,
both A and B is/are open. Hence it 7. (b): Given: V= 3 V, R=
A or B or

gate. B 100
a NAND
represents
1.9 eV Voltage gain of the CE amplifier,
3. (b): Given, energy gap =

Now, for the LED

to operate, electrons need to

cross this energy gap.

Ay--B 10-150
Power gain,Ap =ßxAy= 100 (-150) -15000
=
x
Wavelength of light emitted, is
1242 eV-nm Negative sign represents that output voltage
= 654 nm in opposite phase with the input voltage.
1.9 eV
8. (b) A-
4. (b) A A B

B - Y
A BC C Output (Y)
0 0 1 |0
010 1 0
B
0 0 1|
Y=(A B+A B)

5. (d): Given VBE = 0; VCE = 0

At output, the truth table corresponds to NOR
20 V
gate.
IcRc=4kS2 9. (d): A diode is said to be forward biased if
p-side is at higher potential than n-side of p-n
junction.
R 10. (b): Here, R = 2 kS2 = 2000 2, Vo = 4 V
w
B= 100, Rg = 1 k2 = 1000 2, V, =?
B500k2
E 2000
Voltage gain, A =B=100x =200
Rp 1000

Also, A or V ==-
A 200
Collector current, I = (20-0)
4x10 y100 =20 mV
or
Ic=5 x 10-3 A 5 mA =
11. (a): Diode D, 22
Input voltage, V, VBE + IpRg =
is reverse biased R
or so, it will block
V,=0+IpRg or 20 I x 500 x
10 10 V
the current and R, 2 2
20
= 40 HA diode D is forward
500 x 103 biased, so it will pass
the current.
Current gain, B= C 5 x 1 0 3 = 125
Hence, equivalent circuit becomes
40x10-6
as shown in
the figure.
 Weo N TAIPMI Chapterwise Sotutions y
areve in the i uit ent owing 17. ()
thwaghthe vsistanne W
i4
2S
A

12 V
The potential difference across the resistance R
48N AN is
V SV 05V AV
t 4-8--0then -0-1 By Ohms law,
h-8--othen, -I-0 Thecurrent in the circuit is
AV
AK
10A 010 'A 30 mA
R 100

18, (b) Diode is forward

Outat o the cinruit, Y(A + )C
blas tor positive voltage
i f C 1 and A0, 8 1 V> 0,
i.e.
so output acrosN R, Is given by as
A-, 8=0 or A8-1 shown in tigure
14 (c) Here. R,
800S2. R, 1922, 19. (a) The V-I characteristie tor a solar cell
urrent gain. f 0 , 9 is s shown in the tigure.
Voltage gain- Current gain x Resistance gain
0. (Open clecuit voltage)
Vo
Power gain-Current gain) |Valtage gain]
x

09643.84
sc (Short creult current)
15. (d)
Here, the p-n junetion diode is
trwand biased. hence it ofters zero resistance. 20. (d): "The barrier potential depends on type
of semiconductor (Por St, V, 0.7 V and for
AN V4N-(6) A =10
AGe,V,0.3 V), amount of doping and also on
RAB ks 1000
the temperature.
16 (b): Here.lnput signal. V, = 20os|15+ 2 21. (c): Voltage gain
and voltag* gain, A, 150 Current gain x Resistanee gain

As A, A,

Transconductance, Sm or R
Outputsignal. \A, Sm
Sune CE amplifier gives a phase ditlerence of ASRt
For tirst case, A, =G, g 0.03 mho,
-I80°) between input and output sigals, p25
G 0.03 Rut
()
For second case, A, Gg 0.02 mho, f20
0.02R, (ii)
Divide (ii) by (), we
get
G
 Semiconductor Electronics Materials, Devices and Simple Circuits 331
electrons Region (II) - Active region
22. (a): In n-type semiconductor, Region (11) - Saturation region
carriers and holes are
are majority charge
carriers and pentavalent atoms Using transistor as a switch it is used in cutoff
minority charge region or saturation region.
are dopants.
Using transistor as an amplifier it is used in
23. (a): R A.B active region.
circuit is 31. (a): The truth table ofthe given waveform
The output of the given logic
is as shown in the table.
X = A.B = A.B
Time interval Input A Input B Output C |
0 to t 0
24. (a) , to t2 0

The output of the given logic gate is to l

t to t4 0
C A-B=A B t to ts
It is the Boolean expression of AND gate. ts to t
Hence, the resulting gate is a AND gate.
The logic circuit is OR gate.
25. (b): The emitter injects electrons into the
base region of the n-p-n transistor and holes 32. (a): Here,
into the base region of p-n-p transistor. Inputresistance, R, =100 2
concentration is in Change in base current, Alz uA = 40
26. (b): The higher hole
p-region than that in n-region. Change in collector current, Ac= 2 mA
Load resistance, R = 4 k2 = 4 x 10* 2
27. (d): given circuit the upper diode
In the
= c 2 mA 2x10
A
D, is forward biased and the lower diode D, is gain, B 50
Current 40 LA 40x 10*° A
reverse biased. So, the current supplied by the AlB
battery is Voltage gain of the amplifier is
5V
Ay =B= 50x 4xl0
2000
I00A
102 2 =0.5 A 100

2 k2 2 x 10 2 33. (b):
28. (d): Here, R = =

A
1 k2 1 x 10' 2, B = 100
V=2 V, R, = =

Output voltage, V, = IcRc B

2V C
or c x10 =10 A =ImA
The Boolean expression of the given circuit is
Rc 2x10° Q2
Y= (A + B)-C
As B=or I= or Ig= 10A10
10s AA The table truth of the given inputs is as shown
100
in the table.
IpRg (10-5 A) (1 x 10* s2)
Input voltage, V, = =

Inputs Output
= 102 V= 10 mV
B C Y=(A+ B) C
Electronic configuration of carbon
29. (c) :
0
(C) is Is 2s 2p. The electronic configuration
0
of silicon (,,Si) is 1s 2s 2ps 3s 3p?
0
Hence, the four electrons of C and Si
bonding
orbit. 0
respectively lie in second and third
30. (b): In the given graph, From the above truth table it is clear that Y = 1,
when A = 1, B = 0 and C = 1
Region () -Cutoff region
332 wbG NEET-AIPMT Chapterwise Solutions Physics
The current through the zener diode is
Al
34. (a): Current gain. 5= I, I - I = (20 15)mA =5 mA

39. (c) 1p-n junction is said to be forward

(20-10)mA 10x10A biased when p side is at high potential than
30
(300- 100) uA 200x 10 A nside. It ia for circuit (A) and (C).
35. (d): 1n forward the positive
biasing, 40. (a):p-1ype semiconductor is obtained
terminal of the battery is connected to p-side when Si or Ge is doped with a trivalent impurity
and the negative terminal to n-side of p-n like aluminium (A), boron (B), indium (In) etc,
junction. The forward bias voltage opposes Here, n 1 . 5 x 10'" m', n, = 4.5 x I102 m '

the potential barrier. Due to this, the depletion As n,n, 1

region becomes thin.
(1.5x10 m = 5x10 m
36. (d): (i) 4.5x1022 m3
41. (b): In a n-type semiconductors, electrons
lt represents logic symbol of OR gate. are majority carriers and holes are minority
carriers.
(ii)
42. (a)
lt represents logic symbol of AND gate. 43. (b): Thedevice that can act as a complete
circuit is integrated circuit (IC).
(ii) 44. (c): Here, Voltage gain = 50
Input resistance, R = 100 2
It represents the logic symbol of NOT gate. Output resistance, R, = 200 2
(iv) 200 2
Resistance gain Ro.
R 100
=

100 2
2
2

It represents the logic symbol of NAND gate.

Power gain = (Voltage gain)
50x 50
37. = 1250
(c): When a small amount of antimony Resistance gain 2
(pentavalent) is added to germanium (tetravalent)
45. (d): It is clear from
crystal, then crystal becomes n-type semiconductor.
given logic circuit, that out A BY
In n-type semiconductor electrons are the
put Y is low when both the
majority charge carriers and the holes are the inputs are high, otherwise 0 0
minority charge carriers. it is high. Thus logic circuit
38. (a): +0 www is NAND gate.
I2502 Yz 46. (c)
20 V 15V IkQ
47. (c): Band gap 2.5 eV
The wavelength corresponding to 2.5 eV
The voltage drop across 1 k2 = Vz = 15 V
12400 eV A
The current through I k2 is = 4960A.
2.5 eV
15 V
I'= =15x10
110Q
A =15 mA 4000 A can excite this.
48. (b): (0)
The voltage drop across 250 2 20 V
=
- 15 V
= 5V
(ii)
The current
through 250 2 is
5V (iii)
I=
250 2 0.02 A =
=
20 mA
(iv)
 333
Semiconductor Electronics Materials, Devices and Simple Circuits
OR gate. NOT gate and NAND gates
are (iv),. A BYY
(ii) and (i) respectively. 0 0
49. (c) : For conmmon emitter, the current gain 0
1S
0 0
55. (c) : In a cubic crystal structure

i.C., at a given potential difference of CE a =b =c, a = ß = Y = 90°.

(10x10-5x10)A 5x10 = 50.

56. (d): p-type semiconductor.
B=
(200x10 -100x 10°) A 100x10 57. (b): The truth table corresponding to

50. (a) : Distance between nearest atoms in waveform is given by

body centred cubic lattice (bcc), d = * a

AB C

Given d 3.7 Å, a =3.7x2

3
4.3 Å 0 1 0

51. (d): Band gap = 2 eV. 10 0

this 0 0 0
Wavelength of radiation corresponding to
The given logic circuit gate is AND gate.
energy
58. (d): Current gain, ß = Alc /Al
hc12400 evA== 6200 A
6200 Å
E 2 eV (10-5) mA 5x10 = 100.
The frequency of this radiation
(150-100) uA 50x106
3x10 m/s 59. (c) : Resistivity of a semiconductor decreases
U = 5 x 10 Hz.
6200 x 101 m/s with increase in the temperature.
The atomic radius in f.c.c. crystal is
52. (b) : a 60. (c) : Zener diode is used for stabilisation
a
while p-n junction diode is used for rectification.
22
61. (d)
where a is the length of the edge of the crystal.
62. (a) : Band gap of carbon is 5.5 eV while
3.6 A
Atomic radius = - =1.27 A that of silicon is 1.1 eV.
2/2
53. (b): One applies negative feedback, which (Ec>(E)s
63. (b): Lattice constant for (f.c.c.)
reduces the output but makes it very stable. For
value of =a= interatomic spacing x v2 = 3.59 Å
voltage amplification amplifiers the
output voltage without the negative feedback 64. (b): In photocell, photoelectromotive force,
could be very high. The maximum value shown is the force that stimulates the emission of
here is 100. an electric current when photovoltaic action
54. (c) creates a potential difference between two
Y
points and the electric current depends on the
Y intensity of incident light.
B
65. (a): In semiconductors at room
Y'=A+ B. Y = A+ B = A+ B. temperature the electrons get enough energy so
that they are able to over come the
Truth table of the forbidden
given circuit is given by gap. Thus at room
temperature the valence
 mbG NEET-AIPMT Chapterwise Solutions Physics
334
band is partially empty and conduction Therefore number of atom per unit cell
band is partially filled. Conduction band in
at 0 K.
x8+1 =2.
semiconductor is completely enpty only
74. (d): The current gain ofa common emitter
66. (b): Vde = y transistor (B) is defined as the ratio of collector
Input current () to the base current ()
10V Also, I = Ip + Ici Ic/I = 0.96 (given)

:B=_
10V
Output
I-Ic
Now, E _ 1 - _0.04
67. (b): The truth table of OR gate is c 0.96 Ic 0.96 0.96

_ 0.96
B=C_ 24
A BY
0 0 0 Y=A +B
- I c 0.04
75. (a) :

P
Hole Electron
From truth table we can observe that if either of Electron
input is one then output is one. Also if both the
Hole
inputs are one then also output is one.
68. (c): A diode is said to be reverse biased if Depletion layer
p-type semiconductor of p-n junction is at low A p-n junction is shown in the figure. On
potential with respect to n-type semiconductor account of difference in concentration of charge
ofp-n junction. It is so for circuit (c). carriers in the two sections of p-n junction,
69. (b): n reverse biasing, the conduction the electrons from n-region diffuse through
across the p-n junction takes place due to the junction into p-region and the holes from
minority carriers, therefore the size of depletion p-region diffuse into n region.
region (potential barrier) rises. Since the hole is a vacancy of an electron.
70. (b): A n-p-n transistor conducts when when an electron from n region diffuses
emitter-base junction is forward biased while into the p-region, the electron falls into the
collector-base junction is reverse biased. vacancy, i.e., it completes the covalent bond.
71. (d): In full wave rectifier the fundamental Due to migration of charge carriers across
frequency in ripple is twice of input frequency. the junction, the n-region of the junction will
72. (a) Barrier
potential depends upon have its electrons neutralized by holes from
temperature, doping density and forward biasing the p-region, leaving only ionised donor atoms
73. (c):In body-centred cubic (b.c.c.) lattice (positive charges) which are bound and cannot
there are eight atoms at the corners of the cube move. Similarly, the p region of the junction will
and one at the centre as shown in the figure. have ionised acceptor atoms (negative charges)
which are immobile.
The accumulation of electric charges of opposite
polarities in the two regions of the junction gives
rise to an electric field between these regions
is connected across the
as
if a fictitious battery
connected
Junction with its positive terminal
connected
b.c.c. structure to n region and negative terminal
Semiconductor Electronics Materials, Devices and Simple Circuits 335
to p region. Theretore, in a p-n junction high
(b): B 41
potential is at n side and low potential is at p side.
82. (b): P =-_- -11-(0,11,)1-
76. (a): NAND gate is a combination of AND
83. (d): As a p-n junction diode conducts
and NOT gate.
ABY in forward bias and does not conduct in
0 0 0 reverse bias (current is practically zero), thus
A 0
Y 0 undirectional property leads to application of
B
diode in rectifiers.
AND gate
Truth table of AND gate 84. (a): Atomic radius for body centered

AY cubic structure is r = * av3

A Y 4
4r 4(3.7/2) = 4.3 A.
NOT gate or, a
Truth table of NOT gate 3 1.732
A|B Y 85. (c)
86. (b): In forward biasing, the conduction
B
across p-n junction takes place due to migration
of majority carriers (i.e., electrons from n-side
NAND gate
Truth table of NAND gate to p-side and holes from p-side to n-side), hence
Hence the given truth table is of a NAND gate the size of depletion region decreases.
77. (a): In forward biasing, the resistance of 87. (d) 88. (d)
p-n junction diode is very low to the flow of 89. (c): Because P (phosphorus) is pentavalent.
current. So practically all the voltage will be
dropped across the resistance R, i.e., voltage 90. (d) 91. (a)
across R will be V. n_0.01
In reverse biasing, the resistance of p-n junction 92. (d): I,
diode is very high. So the voltage drop across R Rin 10B
0.01 = 5x10 A = 500 mA.
is zero. =Bl, =50x
78. (b): D,> reverse biased 10
93. (c):On reversing the polarity of the
and D, forward biased.
battery, the p-n junction is reverse biased. As a
Equivalent circuit is
20 2 result of which its resistance becomes high and
current through the junction drops to almost
30 Q
zero.
W K ww-
30 2 94. (c): Voltage drop across diode (Vp) = 0.5 V;

Maximum power rating of diode (P) = 100 mW

ww E -ww F
20 2 5V 20 2 5V = 100 x 103 W
5V and source voltage (V,) =1.5 V.
A.
(30+20)Q 50 The resistance of diode (Rp)

79. (a):C=a=0.98, =B== 49 Vp (0.5 =2.59

E P 100x10
80. (b)
And current in diode (I,) ===0.22
81. (a): A diode is said to be forward biased Rp 2.5
if p-type semiconductor of p-n junction is
Therefore total resistance in circuit (R)
at positive potential with respect to n-type
semiconductor of p-n junction. It is so for
Circuit (a)
V,5-7.592
D 0.2
 330
mtG NEET-AIPMT Chapterwise Solutions Physic
And the value of the series
resistor 105.(a): The function of emitter is to
=
Total resistance of the circuit supplv
-

Resistance the majority carriers. So, it is heavily doped.

of diode
7.5-2.5= 5 Q. 106.(c): van der Waals bonding is the weaket
bonding in solids.
95. (a): Current
gain (B) o
1-ar B-Bo= o
=

107.(b): The amplifying action of a triode

is
based on the fact that a small change in
grid
or.
B=aBa =a(l+B) or a =
P voltage causes a large change in plate current,
1+B The AC input signal which is to be
96. (b): For NOR amplified is
A +B.gate, Y = superimposed on the grid potential.
Therefore from the given truth table 108. (a): According to Child's Law, I =KV3/2
A BA+B Y=A+B Thus Io V3/2

0 109.(d): Copper is a conductor so its resistance

10 decreases on decreasing temperature as thermal
agitation decreases whereas germanium is semi-
0
conductor which on decreasing
0 0 0 1
temperature
resistance increases.
97. (a): In p typegermanium semiconductor, 110. (b) : The
it must be
doped with a trivalent impurity atom. depletion
layer in thep-n junction
Since indium is a third region is caused by diffusion of charge carriers.
group member, therefore
germanium must be doped in indium. 111. (b): This truth table is
of identity,
98. (c): In Y= A+ B, hence
n-p-n transistor, the electrons are OR gate.
majority carriers in emitter, which from 112. (a): To use transistor
base collector while using n-p-n transistor
to
move as an
emitter base junction is forward bias while
amplifier the
as the
an amplifier.
collector base junction is reverse biased.
99. (b): Arsenic is pentavalent,
when added with silicon it leaves one electron
therefore 113. (a): The phase difference between output
voltage and input signal voltage in common
as a free electron. In
this case the conduction of base transistor or circuit is zero.
electricity is due to motion of electrons, so the
114. (b): Voltage
resulting material isn-type semiconductor. gain of an amplifier
100. (b) Output voltage HR
101.(c): According to figure Y= AB. Input voltage RL +Tp
The negative sign indicates that
Therefore it is NAND gate. the output
102.(c): a = 4.225 Á
voltage differs in phase from the input voltage
by 180° (7T). This holds for a pure resistive load.
For bcc cubic cell, 4r = V3 xa
115.(d): Due to heating, when a free electron
Therefore 2r =
V5 Xa_ 1.732 x4.225s 3.66 A
is produced then
simultaneously a hole is also
2 2 produced.
103. (b): As soon as
the p-n junction formed, is
116. (d): Radiowaves of constant
there is an immediate diffusion of the charge can be amplitude
carriers across the junction due to thermal produced by using oscillator with proper
feedback.
agitation. After diffusion, these charge carriers
combine with their 117.(a): For forward biasing of p-n
counterparts and neutralise
each other. Therefore correct direction of flow the positive terminal of junction,
external battery is to
of carriers is be connected to
depicted in figure (b). p semiconductor and negative
terminal of battery to the n
104.(b): Diamond is very hard due to large semiconductor.
cohesive energy. 118.(c): Semiconductors
temperature.
are
insulators at roonm
Aloms and Muclei 305

ANSWER KEY

(d) 2. (b) 3. (c) 4. (d) 5. (a) 6. (b) 7. (None) 8. (b)

1.
9. (c) 10. (d) 11. (a) 12. (c) 13. (c) 14. (d) 15. (c) 16. (d)
17. (c) 18. (c) 19. (a) 20. (d) 21. (c) 22. (d) 23. (b) 24. (d)
25. (c) 26. (d) 27. (a) 28. (d) 29. (b) 30. (d) 31. (b) 32. (c)
33. (b) 34. (c) 35. (c) 36. (c) 37. (c) 38. (c) 39. (b) 40. (d)
41. (a) 42. (c) 43. (c) 44. (a) 45. (d) 46. (c) 47. (a) 48. (c)
49. (c) 50. (b) 51. (a) 52. (a) 53. (c) 54. (a) 55. (a) 56. (a)
57. (d) 58. (d) 59. (c) 60. (b) 61. (c) 62. (a) 63. (c) 64. (a)
65. (a) 66. (c) 67. (b) 68. (d) 69. (b) 70. (c) 71. (a) 72 (d)
73. (d) 74. (c) 75. (d) 76. (a) 77. (a) 78. (c) 79. (c) 80. (a)
81. (a) 82. (c) 83. (d) 84. (a) 85. (a) 86. (b) 87. (b) 88. (a)
89. (a) 90. (a) 91. (c) 92. (a) 93. (a) 94 (a) 95. (a) 96. (b)
97. (d) 98. (b.d) 99. (c) 100. (c) 101. (c) 102. (a) 103. (d) 104. (a)
106. (d) 107. (c) 108. (b) 109. (b) 110. (b) 111. (b) 112. (d)
105. (c)
114. (b) 115 (b) 116. (a) 117. (d) 118. (b) 119. (b) 120. (c)
113. (d)
123. (b) 124. (b) 125. (b) 126. (a) 127. (a) 128. (d)
121. (b) 122. (b)
130. (a) 131. (d) 132. (c) 133. (a) 134. (d) 135. (a) 136. (a)
129. (c)
140. (a) 141. (c) 142. (a) 143. (b) 144. (d)
137. (c) 138. (c) 139. (d)
146. (c) 147. (a) 148. (b) 149. (d) 150. (d) 151. (d) 152. (d)
145. (a)

EXPLANATIONS

1. (d): Total energy of electron in nh orbit, and its ground state energy, E,= -13.6 eV

-13.622
Charge of muon =charge of electron
eV Mass of muon = 207 x (mass of electron)
E,
n
Kinetic energy of electron in nh orbit, Therefore, when electron is replaced by muon
0.51A
K.E,=3.622
then,first Bohr radius, =
?ev
207
2.56 x 10 m
Potential energy of electron in n orbit, and ground state energy, Ej = -13.6 x 207

= -2815.2 eV = -2.815 keV

PE=27.222

PE= -eV
4. (d): Given, 2 = 2.2 x 10 s

Ihus, total energy ofelectron, E, = -K.E. =

and rate of radioactive disintegration,
2

'K.E. = 3.4 eV Given E,= -3.4 eV] dN10s

=10°
dt
P.E. = 2x -3.4 = - 6.8 eV. 0.693 0.693
2.2x109 *°.15x 10-10
(b): Alpha particle is a positively charged
the nucleus of the
1/2
particle. It is identical to
helium (,He) atom, so it contains 2 protons
Now, we know that, N = Noea
dN
and 2 neutrons. -ÀN,eA = -AN
for dt
c):Given, radius of first Bohr orbit
Ciectron in a hydrogen atom, r = 0.51 A 100=3.15 x 10-l"x N=N= 3.17 x 10
 mtG NEET-AIPMT Chapterwise Solutions PhySics
5. (a): Number of nuclei
= 600
remaining. 9. (c): When electron jumps from higher
- 450 150 orbit to lower orbit then, wavelength of emitted
According to the law of radioactive decay photon is given by,

where N, is the number of nuclei

mtial.

2 where T =half life. and

-9F =
7
27T,=2x 10 minutes =20 minutes 10. (d): No = Nuclei at time t = 0
b ) : In a Bohr orbit
of the hydrogen atom, N =
Remaining nuclei after 40% decay
Ametik
Su
enengy= (Total energy)
-
=
(1 -

0.4) N, =
0.6 No%
Ainctic energy: Total energy =1:-1| N, =
Remaining nuclei after 85% decay
C)The number of =
(1 - 0.85) N, 0.15 No
radioactive nuclei 'N
=

t any time t is given as N2_0.15No

N)= Noe N
where A is number of 0.6No
radioactive nuclei Hence,
samyple at some arbitrary time t 0 and Ainis the two half life is
required between 40%
radioactive decay constant. the
=
decay and 85% decay of a radioactive
substance.
Given: Time taken 2t/2 2 x 30 min =

A 8A, Ag à, Naa NoB No

=
= = =
11.
=
=
60 min
(a): Here, R 10 m =

The wave number of the

last line of the Balmer
S e series in
hydrogen spectrum is given by
=S=?h 10=0.25 x10' ml
-1= 7h or t= 12. (c): Distance of
closest
7n
Negative value of time is not possible. O-particle of mass m movingapproach when an
with velocity is
bombarded on a heavy nucleus of v

So given ratio in question should be given by charge Ze, is

=e
NA Ze
TEgmv 'To
8. (b): The wavelength of last line of Balmer
SeTies
13. (c): If
Prh and PHe are the momenta of
thorium and helium
Rc according
nuclei
to law of respectively, then
The wavelength of last line of Lyman series momentum conservation of linear
0 PTh t PHe or

sign shows that bothPTh=-PHe

Re -ve
are
directions. moving in opposite
But in magnitude
PihPHe
Aoms and Nuclei
307
I( m and mn are the masses of thorium and
16. (d): Binding energy of Li nucleus
nuelei respectively, then
helium
energy of thorium nucleus 7x 5.60 MeV 39.2 MeV
Kinctic 1S

nucleus
Pih and that of helium nucleus is Binding energy of He
Th 2myh 4x 7.06 MeV 28.24 MeV
The reaction is
Kae 2mie
Piic ATh-Pth | e li+ H2(He)+Q
KHc Pte h
ButPh Pie and mHe mTh Q= 2(BE of He)-(BE of Li)
Th Kue or Kiue Kh
28.24 MeV 39.2 MeV
Thus the helium nucleus has more kinetic =2 x

= 56.48 MeV 39.2 MeV = 17.28 MeV

the thorium nucleus.
energy than 17. (c): X Y
14. (d): Energy of electron in He* 3r orbit 0
Number of nuclei at t = 0 No
Number of nuclei after time t N-
E=-13.6xeV=-13.6xev
(As per question)
= -13.6 xx1.6x101J=-9.7x107 Na-
As per Bohr's model.
7No 7x =x or x=No
Kinetic energy of electron in the 3rd orbit =-E 8
Remaining nuclei of isotope X*
9.7 x 101 = m,v?

2 x9.7 x 1o-19
9.1x10-31.6 x10° ms
=
No-x = No N--)
So three half lives would have been passed.
t= nT =3 x 1.4 x 10 years
15. (c): Energy of the photon, E=" = 4.2 x 10 years
Hence, the age of the rock is 4.2 x 10 years.
E=
6.63 x107 x3x10, 18. (): The wavelength of different spectral
975 x10l0
lines of Lyman series is given by
6.63 x10x3x10 1 eV =12.75 eV
975x10x1.6X101 wheren=2,3,4,..
After absorbing a photon of energy 12.75 eV,
where subscript L refers to Lyman.
the electron will reach to third excited state
For longest wavelength, n = 2
of energy -0.85 eV, since energy difference
l and n 4 is 12.75 eV.
corresponding to n = =
--
..()
Number of spectral lines emitted
longest
(n)n-1)_(4)(4-1) The wavelength of different spectral series of
2 Balmer series is given by
-0.85 eV
-1.5l eV
n=
where n =
3, 4,5,.
-3.4 eV
where subscript B reters to Balmer
For longest wavelength, n = 3

- 13.6 eV
Bongs
()
 Solutions Physics
30 mbG NEET-AIPMT Chapterwise

4
n= 3, n,
=

Divide (i) by (i), we get In first case,

R )
36 3 27 3
In second case, n, =
2, n, =

19. (a) As H+ He R.(i)

Here, AM 0.02866 u Divide (ii) by (i), we get
The energy liberated per u is

AM931 MeV 20
4 36 7
25. (c): Nuclear radius, R = R,A
(0.02866 931 MeV= 26.7MeV=6.675 MeV
4 constant and A is the mass number
where R, is a
20. (d): Y RAL_(27) 3
Initial number of atoms No 0 Cu (64)!/3 4
Number of atoms after N No-N
time or RcuXRA=x3.6fermi =4.8 fermi
As per question
26. (d): Let after t s amount of the A, and Az
N
7N= Ng - N or 8N =No will become equal in the mixture.
Ng-N 7
N
As N=No
No where n is the number of half-lives
where n is the no. of half lives
A /20
For A, N=Nou

-C)- For A. N=Noz

n 3
According to question, N =N;
orf=n7/2 3x20 years=60 years 40 160
2/20 t/10
Hence, the age of rock is 60 years. 20= 4(2/20) or 2t/10 = 22 2t/20
21. (c) 22. (d)
23. (b): For a given energy,y-rays has highest 210 =220
penetrating power and o-particles has least
penetrating power. 10 20
2 or 10 =2
20
24. (d)
or =2 or t= 40 sS
n-4(3d excitedstate) 20
n 32nd excited state)
27. (a): According to Rydberg formula
n-2(1t excitedstate)

=I(excited stale)
Acording to Rydberg formula Here, n= 1, n, = 5
 ingtnconerrvationof linear momentum
31 (b):
stamentum o photon - Momentum of atom
where n is umber of half live
24R 24hR
7 1 O1 1
191A
-(- or m-4

Let the age of rock be t years

( According to radioactive decay law

where N Number of radioactive nuclei at

or f= nT
T2 = 4 x 50 years 200years
timc
-Number of radioactive nucdei left undecayed 32. (c):According to Einstein's mass energy
a tany time
relation
decay constant E =m or m=
At time of the sample had decayed
Mass decay per second
10 Ww
N
No=No AEP 1000
x
Am I
.() =

Ar A (3x 10 m/s
At time ofthesample had decayed, 10
9 x 1 0 1 68 / s

NNo NoNo ..(ii)

Mass decay per hour
Divide (i) by (1i), we get 10

-lt2-) x60x60
Ar
60=|
9x 10
kg/s (3600 )

= 4x 10 kg =40 x 10g= 40 mg
l-1)= In2 33. (b): Momentum of emitted photon
hu
In2 In 2 In2 =Pphoton C
In 2 From the law of conservation of linear
Ti/2 momentum, Momentum of recoil nucleus
=
T =

50 days Paucleus Pphoton

30. (d): The wavelength of the first line of hu
. Mv=
Vman series for hydrogen atom is
where v is the recoil speed of the nucleus
hu
or v=F-
Ihe wavelength of the second line of Balmer Mc
series for hydrogen like ion is The recoil energy of the nucleus

-MM-K 2.M (Usingii))

*
According to questionà =

alpha
34. (c):When an
particleHei
emitted, the mass number and the atomic
number of the daughter nucleus decreases by
32 four and two respectively. Whei a beta partile
or Z= 4 or Z = 2
 310 mtG NEET-AIPMT Chapterwise Solutions Physics
(B) is emitted. the atomic number of the Number of nuclei of R
daughter nucleus increases by one but the mass 4N M, INo ,9No
2
number remains the same.

X "'x 38. (c):The energy of nth orbit of hydrogen

35. (c) Extremely high temperature needed atom is given as

for fusion make kinetic energy large enough to 13.6

E- eV
Overcome coulomb repulsion between nuclei.
36. (c): Here, Stopping potential, Vo = 10V 13.6
Work function, W = 2.75 eV E=-13.6eV; E =- 22 =-3.4 eV
According to Einstein's photoelectric equation 13.6 13.6
eV= hu - W or hv = evo + W E- =-1.5 eV:E, == =-0.85 eV
= 10 eV+2.75 eV = 12.75 eV
When an electron in the hydrogen atom makes E E -1.5 (-3.4) 1.9 eV
-

= - =

E -E = -0.85 (-1.5) = 0.65 eV

a transition from excited state n to the
gound
state (n = 1), then the frequency (v) of the 39. (b): For Li nucleus,
emitted photon is given by Mass defect, AM = 0.042 u
hu=E,-E = hv= lu=931.5 MeVIc?
AM = 0.042 x 931.5 MeV/e = 39.1 MeV/c
13.6 Binding energy, E, = AMe
For hydrogen atom, E, =

According to given problem -39.1 -39.1 Mev

13.6 Binding energy per nucleon,
+13.6 =12.75 (Using(0)
E 39.1Mev
5.6 MeV
=0.85 = 0.85
=16 A
40. (d): According to activity law
or n=4 R= Roe
37. (c) P Q According to given problem,
No. of nuclei, at t = 0 4N No Ro = No counts per minute
Half-life 1 min 2 min
No. of nuclei after Np No R counts per minute
time t
t= 5 minutes
Let after t min the number of nuclei of P and Q
are equa. Substituting these values in equation (), we get

Np =4N
/l

Ng -No[;)
/2
No-No3
el=eA
As Np =
No
5A=lor A= per minute

2 Att= Tu the activity R reduces to

/2 or 4 / 2 or 4 22 or 22 2!2 2
where T2 half lite of a
=
radioactive sample
or 2 or f = 4 min
From equation (i), we get
After 4 minutes, both P and Q have
number of nuclei.
equal RoRe
2
/2 or = 2
AANINE ANY NiAY 311

laking natuual logaritlhms ot both sides ot above 49. (c): ZM, + (A - Z)M, - M(A, Z)

-
B.E
mass defect =

lg,log Slog, 2 minutes B.E

M(A. Z) =ZM, +(A - Z)M,
()
-

50. (b): A, : A, = 1:33

4. (a) 42. ()
Their radii will be in the ratio
4 ()A,AN, at time f A, =AN, at time t
R A : RoA,U =1:3
therd umber ot uclei decayed during
A
tineinterval ( t , ) is Density =

44 TR
3

Ihe indig eneryy per ucleon of a deuteron PAPA24 R, 3

3
1.1 2.2 MeV Their nuclear densities will be the
tal binding energy =2x = same.

he nding eneryr er nucleon of a helium n=2

51. (a)
ei AMel
8 MeV
tal binding energr= 4 x7 =

ctte eneryr roleasd n= -E, = - 13.6 eV

H atom
A(8- 2)= 23.6 MeV
Ist excitation energy E, E, -

=(-3.4 + 13.6)
= 10.2 eV

ö,52 (a): X Noe"; =

X, =Noe^z
First ienars dy emission emitting
atewutritto sitmultanaoush: Yemits a resulting
tt te ewitai level ot B
which in turn ennits a y

Yis the answer.

A(5À-A) 4
5h. Ag A
isotope ot the (c): Given: = =

t e ultant iaugdter is an 53.

Xtal punnt nadeus Att=0(Na) = (No)s

a Energv odtheproirtileis the potential

eergy at iwt aqypnuh.
According to radioactive decay,
o

-130 NA=A .(i)

. (ii)
(Ns
- 18se Divide (i) by (ii), we get
emtted here
=AAAg or., A=-5À-À)¥
N3
 312 WbG NEET-AIPMT Chapterwise Solutions Physics

For first excited state, n = 2

C (C)-C" E =-13.6-3.4
22
eV
Kinetic energy of an electron in the first excited
or, 4Ar2
4A
state is
54. (a): In beta nminus (B) decay, a neutron K= -E, = 3.4 eV.
is transtormed into a proton and an electron
59. (c): lonisation potential of hydrogen atom
is emitted with the nucleus along with an
is 13.6 eV.
antineutrino.
Energy required for exciting the hydrogen atom
+c +, where i
is the antineutrino. in the ground state to orbit n is given by

55. (a): In mass spectrometer when ions are

E= E,-E
accelerated through potential V, i.e., 12.1- +13.6
m=qV .(i)

where m is the mass of ion, q is the charge of

or, -1.5=*13.65
or, = = 9 or, n=3
n 1.5
the ion. Number of spectral lines emitted
As the magnetic field curves the path of the ions
in a semicircular orbit n(n-1)3x2 -3.
2

Bq r= Bgk (ii) 60. (b): According to activity law, R = Roe

R
R = Roe a n d R = Rge2
Substituting (ii) in (i), we get

q or, 1-_2V
Ra Re -12
F

BR
Since V, B are constants, or, R = R,e"Ali-12)
charge on the ion
Or, -

mass of the ion

61. (c): H + }H> He+ energy
R
56. (a) Energy released
57. (d): Nuclear radii
R (R,)A3 =
=
B.E. of He-2(B.E. of H)
where A is the mass number. 28 2(2.2) = 28 - 4.4 23.6 MeV.

62. (a): Nuclear radii R =

R,(A)3,
where R 1.2 fm
RA or R« (A)/3
or. Re = x 3.6 =6 fm. (9)/3
3 / or, RBe (9
(Given Ra =3.6 fm) RGe (A/3 2RBe (A)V3
58. (d): Energy of h orbit of hydrogen atom Cgiven RGe = 2R)
or, (A) = 2 x (9)!/
is given by 3
or, A =2 x9 = 8 x 9 = 72.
-13.6 The number of nucleons in Ge is
E
- N = 1
72
63. (c): Energy released, E = (Am) x 931 MeV
For ground state, n
=1 Am =mass of product -

mass ot reactant
-13.6 -13.6eV Am = c- a - b
E (Am) x 931 or, E
=

=(c -

a -

b).
 AIDmS and Nuclei 313
74. (c)
64. (a): K.E=P.E
75. (d):
is ncgative.
But PE. Volume of atom
Total energy 3 = 105
Volume of nucleus
r(10
E-PE--.4eV.
K.E. =+3.4 eV. 76. (a): Let, f = 0, M, = 10 g

Isotones means number of neutron

(a):
65.
rmains same.
=
2t =

2 (given)
66. (c)
Then from, M = Moe = 10ce

E-E) (Ec Eg) =

-

+ (Ep -

E)
-1o-1.356
hc hc
hc 77. (a): Radius offirst orbit, r «

2 for doubly ionized lithium Z (= 3) will be

ionized lithium,
or maximum, hence for doubly r

will be minimum.

68. (d): For nuclei A> 56 binding

having 78. (c): Mass defect =
2Mp + 2MN MHe -

= 2x 1.0073 + 2 x 1.0087 4.0015 = 0.03005

nucleon gradually decreases.
energy per
is the Binding energy (931 = x mass defect) MeV
69. (b): Z is number of protons and A = 931 x 0.0305 MeV = 28.4 MeV
total number of protons and neutrons. (1 amu = 931 Mev).
70. (c): In nuclear fusion the mass of end 79. (c): Mass number = atomic number + n0.
product or resultant is always less than the sum of neutrons
of initial product, the rest is liberated in the For hydrogen, number of neutrons = 0
form of energy, like in Sun energy is liberated
So, mass number = Atomic number.
due to fusion of two hydrogen atoms.
Hence mass number is sometimes equal too
71. (a) atomic number.
80. (a) : B-decay.
72 (d): Using N =No
81. (a): The nuclei of light elements have a
lower binding energy than that for the elements
n = 2. of intermediate mass. They are therefore less
stable; consequently the fusion of the light
The total time in which radium change to 25 g is elements results in more stable nucleus.
2 x 1600= 3200 yr. 82. (c): Number ofinitial active nuclei
73. (d): The charge on hydrogen nucleus = 4 x 10lo

9 te Number of decayed nuclei after 10 days (half

charge on electron, G2= -e life)
Coulomb force, F =K 92 =Kte-e) 4x106
4X102x
2
Ke Ke Remaining number of nuclei after 10 days
r. = 4x 10'6 - 2 x 10o = 2 x 10o
2
S14 mtG NEET-AIPMT Chapterwise Sokuions

Number ot deravad nuclei in next 10 days 10x +11(100-x)

= 10.81 x= 19
100
96 of B is 100 19 =81. Ratio is 19 81
102. (a): Centripetal force = force of attraction
in next 10
Similarly number of decayed nuclei of nucdeus on electron
davs=0.5 10
Total number of nucdei decaved atter 30

10+Ix 10 +0.5 x10ls =3.5 x 10s

103.(d): Energy = hu
3 (d): The nuclear reaction is
H N+ He hc 6.625 x 10*x3x10
So when a deuteron is bombarded on os 21x10
nuceus then an a-particle (,He ) is emitted and = 0.9464 x 10-24 J=lx 10-24 J.
the product nucleus is -N4 104. (a) For A. 80 min. = 4 half lives
$4 (a): a-rays are positively charged particles.
No. of atoms left =
5.
(a):B+ He+ 3li For B, 80 min. = 2 half lives
16

s6. (b): = ; no.of decays Ratio 1:4.

No. of atoms left =
. =

105.(c)
=8
106.(d): In nuclear fission, the chain reaction
is controlled in such way that only one neutron,
Time for S8 half life = 100 hours.
produced in each fission, causes further fission.
s7. (b): 2dsino = tà : (sino)max = 1 Therefore some moderator is used to slow
ic., a
2

down the neutrons. Heavy water is used for this

mas =2x 28 x 10* =5.6 x 10 nm. purpose

88. (a): E
107.(c): Energy of the ground electronic state
-13.6 eV
ofhydrogen atom E =

We know that energy of the first excited state for

89. (a) 90. (a) 91. (c)
second orbit (where n =2)

92. (a):u E
13.6
n)
13.63.4
(2
ev.
its
93. (a) 94. (a) 95. (a) 108.(b): When a hydrogen atom is in
excited level, then n = 2.
96. (b): Jump to second orbit leads to Balmer
Theretore radius of hydrogen atom in its irst
series. The jump from 4th orbit shall give rise to
excited level (r) * n*r,« (2) = 4 r
second line of Balmer series.
97. (d) 109.(b): Y-ray are most penetrating radiations
98. (b,d): la reduce the mass number by 110.(b): By the law of photo-electric etfect
4units and atomic number by 2 units, while 1ß
=eV or
only increase the atomic number byl unit.

99. (c) 111.(b): Energy ot hydrogen atom in ground

state =-13.6 eV and quantum number (n) =2.
100.(c)
We know that energy of hydrogen atom
101.(c) Let ;B be present as x
and percentage of B ' = (100 x)
(E)=-3.0_13.6
= -3.4 eV
Average atomic coeticient (2
Atoms and Nuclei
315
112.(d): According to Bohrs principle, radius
118.(b):
of orbit (r) An Epx- n k Ze kTe
P.E.
4T me R
and
K.E. = mv=
2 2R
where n- principal quantum number. Therefore when a hydrogen atom is raised
113.(d): Velocity ratio(V,: v,) = 2: 1 from the ground, it increases the value of the
Mass (m) Volumeo r' radius R. As a result of this, both K.E. and PE.
According to law of conservation of momentum, decrease.
m,V m,V, 119. (b)
120.(c) Absorption spectrum involves only
Theretore m22
excitation of ground level to higher level.
V2 m Therefore spectral lines 1, 2, 3 will occur in the
absorption spectrum.
121. (b): Mass number of helium (Apie) = 4 and

or r T 1 :21/ mass number of sulphur (A5) = 32.

Radius ofnucleus, r = roA)"(A)". Therefore
114.(b):On emission ofone -particle, atomic
number decreases by 2 units and atomic mass
number decrease by 4 units. 2
Herc, decrease in mass number = 200 168 THe
32 122. (b): From binding energy curve, the curve
Number of o-particles = 32/4 = 8. reaches peak for 2,FeS6
While the emission of B-particle does not effect 123.(b): Neon street sign is a source of line
the mass number and atomic number increases emission spectrum.
by I unit.
Number of B-particles = 16 - 10 = 6. 124.(b): Fission rate
total power
115. (b): No. of nucleon on reactant side =
1.56x10-
Binding energy for one nucleon =x energy 200 x 1.6x10-13
Binding energy for 4 nucleons = 4x fission
Similarly on product side binding energy = 4x2
125.(b): n=4

Now Q= change in binding energy =

42- *). n= 3

116.(a) Half-life time =30 minutes; Rate of n = 2

decrease (N) = 5 per second and n = 1

total time = 2 hours = 120 minutes. Relation for

No. of spectral lines = n(n-1)_ 4 -6
initial and f1nal count rate 2 2
time/half-life 20/3
126. (a): As roc n', therefore, radius of 2nd
Bohr's orbit = 4a,
Ny
Therefore N, = 16 x N = 16 x 5 = 80 s
127.(a): Fusion reaction
Transition of hydrogen atom from
17.(d): 128.(d): E = E, - Es
orbit n = 4 to n = 2.
13.6 -0.85+1.51 =0.66 eV
Wave number -
129.(c): l a.m.u = 931 MeV

= l6/3R. 130.(a): u-particle =He'. It contains 2p and

2n.
 316 mtG NEET-AIPMT Chapterwise Solutions Physics
As some mass is converted into binding energy, | 140. (a) : As C and ,N4 have same no. of

therefore, mass of a particle is slightly less than neutrons

sum of the masses of 2p and 21. (13 6 = 7 for C and 14 7 = 7 for N), they are

131.(d) Clearly C and N* have same electronic isotones.

configuration as they are isoelectronic. 141.(c): Nuclear forces are short range forces
132. (c): The nuclear radius r varies with mass 142.(a): Ro« (A)'3 from R =R, AS
RA (27) and RT, o« (125)
number A according to the relation
r= To A"3> roc A'/3 or A oc RAL 5 110
Now density =
mass RTe
volume 143. (b):5,C2+o ,C">,N"+ -1B°+Energy
Further mass A and volume e r 6400
=8
mass
144.(d): Number of half lives, n== 800
= constant
volume
256
133.(a): Nucleus contains only neutrons and
From equation (ii), B has 2 units of
protons. 145. (a) :

6400/16000 charge more than C.

From equation (i), A loses 2 units of charge by
134.(d) :

135. (a): Z = 11 i.e., number of protons = 11,

)-6 emission of alpha particle. Hence, A and C are
their number is
isotopes as charge same.

A 23 146. (c) : Curie is a unit of radioactivity

Number of neutrons A Z 12= =

'Number in
-

of electron = 0 (No electron in nucleus)

147.(a): Average binding energy/nucleon
nuclei is of the order of 8 MeV.
Therefore 11, 12, 0. 148.(b): Bohr used quantisation of angular
136. (a) : Second excited state corresponds to
momentum. For stationary orbits, Angular
n =3 nh
momentum l0= where n = 1, 2, 3,...etc.
E= eV =1.51 eV 2T
32 149.(d)
but one has to ionise only from ground state.
150. (d): In two halflives, the activity becomes
Even if one has to excite an atom from n =
3, one one fourth.
has to excite from n = 1.
Activity on 1 - 8 - 91 was 2 micro curie
137. (c) : Nuclear force is the same between any
Activity before two months,
two nucleons. 4x 2 micro-Curie =8 micro curie
138. (c) For all first group elements, Na, K, Rb,
151.(d) : Two successive B decays increase the
Cs, Fr. They have one electron in the s subshell.
charge no. by 2.
circumference of an orbit in an
139. (d): The 152. (d): E« Z and Z for singly ionised helium
of wave associated
atom in terms of wavelength
is 2 (i.e., 2 protons in the nucleus)
with electron is given by the relation,
Circumference =
nA, where n =
1, 2, 3, ..
(E)He4 x 13.6 = 54.4 eV.