MODERN PHYSICS- SOLUTION(3-1-16)
2
1.
(b) k  
 Z 1

1
 27  1 
 
 Ni   Co

2


Z

1
(Z  1)
 28  1 
Co
 Ni

2
2.
2
2
 26 
 26 
  Ni     Co    179  165 .9 pm  179 pm .
27
 
 27 
hc
hc
hc
(b) min 
 1 
and  2 
eV
eV1
eV 2
hc  1
1 
  . Given V2 = 1.5 V1

e  V2 V1 
on solving we get V1 = 16000 volt = 16 kV.
(c) Radius of circular path described by a charged particle in a magnetic field is given by
   2  1 
3.
q 2 B 2r 2  e  eB 2r 2
2mK
; where K = Kinetic energy of electron  K 
 
2m
qB
m  2
r
2

 1

1
 1 .7  10 11  1 .6  10 19  
 10  5   (1)2
2
 17

 8  10 20 J  0 .5 eV
By using E = W0 + Kmax
 12375 
 W0  E  K max  
 eV  0 . 5 eV  4 . 5 eV
 2475 
4.
5.
12375
 3 .09 eV Photoelectrons emits if energy of incident light > work function.
4000
hc
hc
 W0  E1 and
 W0  E 2
(c) E = W0 + Kmax
(c) E 
1
2
 hc  W0 1  E11 and hc  W0 2  E2 2
 W0 1  E11  W0 2  E2 2  W0 
6.
E1 1  E 2  2
.
(2  1 )
(b) vmax= 4  108 cm/sec = 4  106 m/sec.
1
1
2
 K max  mv max
  9  10  31  (4  10 6 )2
2
2
= 7.2  10–18 J = 45 eV.
K
45 eV
 45 volt .
Hence, stopping potential  V0  max 
e
e
e
(c) Since spot is same, hence
should be same i.e.,
m
As q1 : q2 : q3 = 1 : 3 : 5. Hence m1 : m2 : m3 = 1 : 3 : 5
0.5
h
199 
h
199 h
h
(c)     




p
200
p  p 200 p
100
p  p
200
p  p = 199 p
 p  p 
199
1
(b) For the incident electron mv 2  eV or p 2  2m eV
2
h
h
 de-Broglie wavelength 1  
p
2m eV
***
7.
8.
9.
Shortest X-ray wavelength 2 

10.
1
1

2
c
 V  e 
   
 2 m 
hc
eV
10 4
 1 .8  10 11
2
= 0.1
3  10 8
(a) Energy of the electron, when it comes out from the second plate = 200 eV – 100 eV = 100 eV
Hence accelerating potential difference = 100 V
12 .27 12 .27
Electron 

 1 .23 Å
V
100
11.
(a) (4n+2) series starts from U238 and it’s stable end product is Pb206.
12.
(a) Activity depends upon mass, but  doesn’t change.
13.
(b) The given reaction is a nuclear reaction, which can take place only if a proton (a hydrogen
nucleus) comes into contact with a lithium nucleus. If the hydrogen is in the atomic from,
the interaction between it's electron cloud and the electron cloud of a lithium atom keeps
the two nuclei from getting close to each other. Even if isolated protons are used, they must
be fired at the Li atom with enough kinetic energy to over come the electric repulsion
between the proton and Li atom.
14.
(c) N  N 0 e  t 
dN
 N 0 e  t
dt
dN
dt
Initially at t = 0,
 N0
t 0
where N0 = Initial number of undecayed atoms


15.
dN
dt
Mass of the sample
M
MN A


Mass of a single atom of X A / N A
A

t 0
MN A  0 . 693 MN A

A
AT
(b) In 10 sec, number of nuclei has been reduced to one fourth (25% to 6.25%).
Therefore it’s half life is T1/2 = 5 sec.
 Mean life T 
16.
T1 / 2
5

= 7.21 sec.
0.693 0.693
(b) By using R  R0 A1 / 3 
1/3

R
 A
 
R He  4 
R1  A1

R 2  A2
1/3




1/3
 A
 (14 )1 / 3   
4
 A = 56 so Z = 56 – 30 = 26.
17.
(c) Energy of an electron in ground state of an atom (Bohr’s hydrogen like atom) is given as
E  13 .6 Z 2 eV
(Z = atomic number of the atom)
2

18.
(Eion )H  Z H 
1
1
   

(Eion )Li  Z Li 
9
3
(c) By using E  
 – 0.544  
2
13 . 6
eV
n2
(for H2 atom)
13 . 6
 n2 = 25  n = 5
n2
 Angular momentum = n
19.
 Eionisation= 13.6 Z2
h
5h

.
2
2
(b) From conservation of momentum, two identical photons must travel in opposite directions
hc
with equal magnitude of momentum and energy

from conservation of energy
 
h
.
m0c
hc


hc

 m0c2  m0c2
20.
(a) (i)
16 S
32
 0 n1 15 p 32 1 H 1
(ii) 9 F19 1 H 1 2 He 4  8 O16
(iii)
7
N 14  0 n1 6 C 14 1 H 1
21.
(a) In sodium chloride the Na  and Cl  ions both have noble gas electron configuration
corresponding to completely filled bands. Since the bands do not overlap, there must be a
gap between the filled bands and the empty bands above them, so NaCl is an insulator.
22.
(b) Pure Cu is already an excellent conductor, since it has a partially filled conduction band,
furthermore, Cu forms a metallic crystal as opposed to the covalent crystals of silicon or
germanium, so the scheme of using an impurity to donate or accept an electron does not
work for copper. In fact adding impurities to copper decreases the conductivity because an
impurity tends to scatter electrons, impeding the flow of current.
23.
(b) V  VCE  IC RL
 15 = 7 + IC  2  103  iC = 4 mA
  
iC
4
 0 .04 mA
 iB 
100
iB
24.
(a) Charge carriers inside the P-type semiconductor are holes (mainly). Inside the conductor
charge carriers are electrons and for cell ions are the charge carriers.
25.
(a) The ionisation energy of an isolated atom is different from it’s value in crystalline lattice,
because in the latter case each bound electron is influenced by many atoms in the periodic
crystalline lattice.
26.
(a) Here the emitter base junction of N-P-N transistor is forward biased with battery VBB
through resistance R. When the value of R is reduced, then the emitter current ie will
increase. As a result the collector current will also increase. (ic = ie – ib). Due to increase in
ic, the potential difference across L increases and hence the reading of voltmeter will
increases.
27.
(c) Vdc  Vac 
28.
(c)
2 V0


0.6V
2  6 .28
 4 V.
3 .14
R
i
R
(2 .6  0 .6)
= 400 .
5  10  3
2.6V
29.
(c) According to the given figure A is at lower potential w.r.t. B. Hence both diodes are in
reverse biasing, so equivalent, circuit can be redrawn as follows.
 Equivalent resistance between A and B
R = 8 + 2 + 6 = 16 .
4
6
2
A
B
8
30.
12
(a) Diodes D1 and D3 are forward biased and D2 is reverse biased so the circuit can be redrawn
as follows.
i
E
R
R
R
R
i
E