Chapter 29 - Modern Physics I (Part - 2) - Physics, Solution by D C Pandey, NEET NEET Notes | EduRev

DC Pandey (Questions & Solutions) of Physics: NEET

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NEET : Chapter 29 - Modern Physics I (Part - 2) - Physics, Solution by D C Pandey, NEET NEET Notes | EduRev

 Page 1


25. Here E
1
= - 15.6 eV
(a) Hence ionization potential 
= - = E
1
15.6 eV
(b) We have l =
hc
E D
 for short wavelength 
DE is maximum
 Þ l =
´ ´ ´
- - ´ ´
-
-
-
6.6
5.3 1.6
Å
10 3 10
0 10
2335
34 8
19
( )
~
(c) Excitation potential for n = 3 state is
= - = - + = E E
3 1
3.08 15.6 12.52 V
(d) From n = 3 to n = 1 
DE E E = - = - + =
3 1
3.08 15.6 12.52eV
We know l =
hc
E D
Þ 
1 10
10 3 10
19
34 8
l
= =
´ ´
´ ´ ´
-
-
DE
hc
12.52 1.6
6.6
    = ´ 1.01 10
7
 ( ) m
- 1
26. (a) E
1
= - 6.52 eV
l = = 860 8600 nm Å
Energy of this photon 
= =
12375
8600
(eV)
1.44 eV
hence internal energy of atom after 
absorbing this photon is given by 
E E
i
= + = - +
1
1.44 eV 6.52 1.44
     = - 5.08 eV
(b) l
2
12375
4200
= =
(eV)
2.95 eV
hence internal energy of the atom after 
emission of this  photon is given by
E E
i
= - = - -
1
2.95 eV 2.68 2.95 eV ( )
Þ E
i
= - 5.63 eV
27. Heve U m r =
1
2
2 2
w Þ  F
dU
dr
m r = =
2 2
w
But 
mv
r
m r
2
2 2
= w
Þ v m r
2 2 2
= w
But by Bohr’s postulate mvr
nh
=
2p
Þ m v r
n h
2 2 2
2 2
2
4
=
p
Þ m r
n h
3 2 4
2 2
2
4
w
p
=
Þ          r
n h
m
4
2 2
2 3 2
4
=
p w
Þ             r n µ
28.
1
1 1
1
2
1
2
2
0
l l
a
K
R z = - -
é
ë
ê
ù
û
ú
= ( )
1
1 1
1
3
2
2
l
b
K
R z = - -
é
ë
ê
ù
û
ú
( )
Þ 
l
l
b
a
K
K
=
´
´
=
3 9
4 8
27
32
Þ l l l
b a
K K
= =
27
32
27
32
0
29. l
0
=
hc
eV
Þ l
0
34 8
19 3 1
10 3 10
10 25 10
=
´ ´ ´
´ ´ ´
-
-
6 .6
1 . 6
Þ l
0
1
= 49.5 p m
Þ l l
0 0
2 1
2 2 5 = = ´ = 49. pm 99 pm
[ ] 1 pm m =
-
10
12
30. f z
Ka
= ´ - ( ) ( ) 2.48 Hz 10 1
15 2
Þ 
3 10
10 2 1
8
15 2
´
= ´ -
l
a K
2.48 ( )
Þ 
3 10
10 10
1
8
10 15
2
´
´ ´ ´
= -
-
0.76 2.58
( ) z
( - - z 1 40 )
~
  Þ z = 41
31. l
i
hc
eV
=   Dl = 26 pm when V
f
= 1.5 V
Þ l l
f i
= - ( ) 26 pm
l l
f i
hc
e V
hc
eV
=
´
= = =
1.5 1.5 1.5 1.5
1 12 1
66
Page 2


25. Here E
1
= - 15.6 eV
(a) Hence ionization potential 
= - = E
1
15.6 eV
(b) We have l =
hc
E D
 for short wavelength 
DE is maximum
 Þ l =
´ ´ ´
- - ´ ´
-
-
-
6.6
5.3 1.6
Å
10 3 10
0 10
2335
34 8
19
( )
~
(c) Excitation potential for n = 3 state is
= - = - + = E E
3 1
3.08 15.6 12.52 V
(d) From n = 3 to n = 1 
DE E E = - = - + =
3 1
3.08 15.6 12.52eV
We know l =
hc
E D
Þ 
1 10
10 3 10
19
34 8
l
= =
´ ´
´ ´ ´
-
-
DE
hc
12.52 1.6
6.6
    = ´ 1.01 10
7
 ( ) m
- 1
26. (a) E
1
= - 6.52 eV
l = = 860 8600 nm Å
Energy of this photon 
= =
12375
8600
(eV)
1.44 eV
hence internal energy of atom after 
absorbing this photon is given by 
E E
i
= + = - +
1
1.44 eV 6.52 1.44
     = - 5.08 eV
(b) l
2
12375
4200
= =
(eV)
2.95 eV
hence internal energy of the atom after 
emission of this  photon is given by
E E
i
= - = - -
1
2.95 eV 2.68 2.95 eV ( )
Þ E
i
= - 5.63 eV
27. Heve U m r =
1
2
2 2
w Þ  F
dU
dr
m r = =
2 2
w
But 
mv
r
m r
2
2 2
= w
Þ v m r
2 2 2
= w
But by Bohr’s postulate mvr
nh
=
2p
Þ m v r
n h
2 2 2
2 2
2
4
=
p
Þ m r
n h
3 2 4
2 2
2
4
w
p
=
Þ          r
n h
m
4
2 2
2 3 2
4
=
p w
Þ             r n µ
28.
1
1 1
1
2
1
2
2
0
l l
a
K
R z = - -
é
ë
ê
ù
û
ú
= ( )
1
1 1
1
3
2
2
l
b
K
R z = - -
é
ë
ê
ù
û
ú
( )
Þ 
l
l
b
a
K
K
=
´
´
=
3 9
4 8
27
32
Þ l l l
b a
K K
= =
27
32
27
32
0
29. l
0
=
hc
eV
Þ l
0
34 8
19 3 1
10 3 10
10 25 10
=
´ ´ ´
´ ´ ´
-
-
6 .6
1 . 6
Þ l
0
1
= 49.5 p m
Þ l l
0 0
2 1
2 2 5 = = ´ = 49. pm 99 pm
[ ] 1 pm m =
-
10
12
30. f z
Ka
= ´ - ( ) ( ) 2.48 Hz 10 1
15 2
Þ 
3 10
10 2 1
8
15 2
´
= ´ -
l
a K
2.48 ( )
Þ 
3 10
10 10
1
8
10 15
2
´
´ ´ ´
= -
-
0.76 2.58
( ) z
( - - z 1 40 )
~
  Þ z = 41
31. l
i
hc
eV
=   Dl = 26 pm when V
f
= 1.5 V
Þ l l
f i
= - ( ) 26 pm
l l
f i
hc
e V
hc
eV
=
´
= = =
1.5 1.5 1.5 1.5
1 12 1
66
Þ ( ) l l
i i
- = 26
2
3
Þ 3 26 3 2 l l
i i
- ´ =
Þ l
i
= 78 pm
Þ 78 10
10 3 10
10
12
34 8
19
´ =
´ ´ ´
´ ´
-
-
-
6.6
1.6 V
Þ V =
´ ´
´ ´ ´
-
- -
6.6
1.6
3 10
78 10 10
26
12 19
Þ V = 15865 volt
32. V a z b = - ( ) 
Þ 
c
a z b
l
= -
2 2
( )
Þ    
1
2
2
l
= -
a
c
z b ( )
Þ 
1
887
13
2
2
pm
= -
a
c
b ( ) …(i)
and  
1
146
30
2
2
pm
= -
a
z
b ( ) …(ii)
Dividing Eq. (i) and Eq. (ii)
146
887
13
30
2
=
-
-
é
ë
ê
ù
û
ú
( )
( )
b
b
Þ 2.5 =
-
-
30
13
b
b
Þ 32.5 2.5 - = - b b 30
2.5 1.5 = b
   b =
5
3
1
26
5
3
26
2 2
l
= -
æ
è
ç
ö
ø
÷
a
c
1
26
5
3
887 13
5
3
26
2
2
l
=
f -
æ
è
ç
ö
ø
÷
´ -
æ
è
ç
ö
ø
÷
pm
Þ l
26
2
2
887 13
5
3
26
5
3
=
´ -
æ
è
ç
ö
ø
÷
-
æ
è
ç
ö
ø
÷
pm
= ´ 887
34
73
2
2
pm
( )
~
- 198 pm
33. f
RC
z = -
3
4
1 ( )
 4.2
1.1
´ =
´ ´ ´ ´
10
3 10 3 10
4
18
7 8
 ( ) z - 1
2
Þ
4.2
1.1
´ ´
´ ´
= -
10 4
9 10
1
18
15
2
( ) z 
Þ ( ) z - = 1 41 Þ z = 42
34. P Vi = = ´ = 40 10 400 kW mA W
% of P =
´
=
400 1
100
4 W
(a) Total power of X-rays = 4 W 
(b) Heat produced per second
= - = 400 4 396 J/s
Photoelectric effect
35. Einstein photo elec tric equation is
K h W
max
= - n
Þ eV
hc
W
0
= -
l
  Q K eV
max
=
0
Þ 10.4 eV
Å
1.7 eV = -
12375
l( )
Þ l( ) Å
12.1
Å = =
12375
1022
For H-atom l =
hc
E D
 Þ DE = =
12375
1022
12.1 eV
 67
Page 3


25. Here E
1
= - 15.6 eV
(a) Hence ionization potential 
= - = E
1
15.6 eV
(b) We have l =
hc
E D
 for short wavelength 
DE is maximum
 Þ l =
´ ´ ´
- - ´ ´
-
-
-
6.6
5.3 1.6
Å
10 3 10
0 10
2335
34 8
19
( )
~
(c) Excitation potential for n = 3 state is
= - = - + = E E
3 1
3.08 15.6 12.52 V
(d) From n = 3 to n = 1 
DE E E = - = - + =
3 1
3.08 15.6 12.52eV
We know l =
hc
E D
Þ 
1 10
10 3 10
19
34 8
l
= =
´ ´
´ ´ ´
-
-
DE
hc
12.52 1.6
6.6
    = ´ 1.01 10
7
 ( ) m
- 1
26. (a) E
1
= - 6.52 eV
l = = 860 8600 nm Å
Energy of this photon 
= =
12375
8600
(eV)
1.44 eV
hence internal energy of atom after 
absorbing this photon is given by 
E E
i
= + = - +
1
1.44 eV 6.52 1.44
     = - 5.08 eV
(b) l
2
12375
4200
= =
(eV)
2.95 eV
hence internal energy of the atom after 
emission of this  photon is given by
E E
i
= - = - -
1
2.95 eV 2.68 2.95 eV ( )
Þ E
i
= - 5.63 eV
27. Heve U m r =
1
2
2 2
w Þ  F
dU
dr
m r = =
2 2
w
But 
mv
r
m r
2
2 2
= w
Þ v m r
2 2 2
= w
But by Bohr’s postulate mvr
nh
=
2p
Þ m v r
n h
2 2 2
2 2
2
4
=
p
Þ m r
n h
3 2 4
2 2
2
4
w
p
=
Þ          r
n h
m
4
2 2
2 3 2
4
=
p w
Þ             r n µ
28.
1
1 1
1
2
1
2
2
0
l l
a
K
R z = - -
é
ë
ê
ù
û
ú
= ( )
1
1 1
1
3
2
2
l
b
K
R z = - -
é
ë
ê
ù
û
ú
( )
Þ 
l
l
b
a
K
K
=
´
´
=
3 9
4 8
27
32
Þ l l l
b a
K K
= =
27
32
27
32
0
29. l
0
=
hc
eV
Þ l
0
34 8
19 3 1
10 3 10
10 25 10
=
´ ´ ´
´ ´ ´
-
-
6 .6
1 . 6
Þ l
0
1
= 49.5 p m
Þ l l
0 0
2 1
2 2 5 = = ´ = 49. pm 99 pm
[ ] 1 pm m =
-
10
12
30. f z
Ka
= ´ - ( ) ( ) 2.48 Hz 10 1
15 2
Þ 
3 10
10 2 1
8
15 2
´
= ´ -
l
a K
2.48 ( )
Þ 
3 10
10 10
1
8
10 15
2
´
´ ´ ´
= -
-
0.76 2.58
( ) z
( - - z 1 40 )
~
  Þ z = 41
31. l
i
hc
eV
=   Dl = 26 pm when V
f
= 1.5 V
Þ l l
f i
= - ( ) 26 pm
l l
f i
hc
e V
hc
eV
=
´
= = =
1.5 1.5 1.5 1.5
1 12 1
66
Þ ( ) l l
i i
- = 26
2
3
Þ 3 26 3 2 l l
i i
- ´ =
Þ l
i
= 78 pm
Þ 78 10
10 3 10
10
12
34 8
19
´ =
´ ´ ´
´ ´
-
-
-
6.6
1.6 V
Þ V =
´ ´
´ ´ ´
-
- -
6.6
1.6
3 10
78 10 10
26
12 19
Þ V = 15865 volt
32. V a z b = - ( ) 
Þ 
c
a z b
l
= -
2 2
( )
Þ    
1
2
2
l
= -
a
c
z b ( )
Þ 
1
887
13
2
2
pm
= -
a
c
b ( ) …(i)
and  
1
146
30
2
2
pm
= -
a
z
b ( ) …(ii)
Dividing Eq. (i) and Eq. (ii)
146
887
13
30
2
=
-
-
é
ë
ê
ù
û
ú
( )
( )
b
b
Þ 2.5 =
-
-
30
13
b
b
Þ 32.5 2.5 - = - b b 30
2.5 1.5 = b
   b =
5
3
1
26
5
3
26
2 2
l
= -
æ
è
ç
ö
ø
÷
a
c
1
26
5
3
887 13
5
3
26
2
2
l
=
f -
æ
è
ç
ö
ø
÷
´ -
æ
è
ç
ö
ø
÷
pm
Þ l
26
2
2
887 13
5
3
26
5
3
=
´ -
æ
è
ç
ö
ø
÷
-
æ
è
ç
ö
ø
÷
pm
= ´ 887
34
73
2
2
pm
( )
~
- 198 pm
33. f
RC
z = -
3
4
1 ( )
 4.2
1.1
´ =
´ ´ ´ ´
10
3 10 3 10
4
18
7 8
 ( ) z - 1
2
Þ
4.2
1.1
´ ´
´ ´
= -
10 4
9 10
1
18
15
2
( ) z 
Þ ( ) z - = 1 41 Þ z = 42
34. P Vi = = ´ = 40 10 400 kW mA W
% of P =
´
=
400 1
100
4 W
(a) Total power of X-rays = 4 W 
(b) Heat produced per second
= - = 400 4 396 J/s
Photoelectric effect
35. Einstein photo elec tric equation is
K h W
max
= - n
Þ eV
hc
W
0
= -
l
  Q K eV
max
=
0
Þ 10.4 eV
Å
1.7 eV = -
12375
l( )
Þ l( ) Å
12.1
Å = =
12375
1022
For H-atom l =
hc
E D
 Þ DE = =
12375
1022
12.1 eV
 67
This difference equal to n = 3 ® = n 1
transition.
36. K h W
max
= - n
Þ K
max
=
´ ´ ´
´
-
-
-
6.6 1.5
1.6
3.7
10 10
10
34 15
19
Þ K
max
= - = 6.18 3.7 2.48 eV
37. Here work func tion 
       W(in eV)
Å
2.475 eV = =
12375
5000
K eV
max
= =
0
3 eV
K
hc
W
max
= -
l
Þ 3
12375
= -
l( ) in Å
2.475
Þ l = =
12375
2260
5.475
Å
38. Comparing the given graph with
K h W
max
= - n
 f = ´ 1 10
14
 Hz
(a) n
o
= threshold frequency q = ´ 10 10
14
Hz
= 10
15
Hz
(b) W = 4 eV 
(c) h = slope of the graph 
= =
´
CD
AD
8
20 10
14
eV
Hz
Þ h =
´ ´
´
= ´
-
-
8 10
2 10
10
19
15
34
1.6
6.4 J-s
39. Here 
v
u
1
2
3
1
(max)
) (max
=
Using Einstein equation, K
hc
W
max
, = -
l
we get
1
2
2
mv
hc
W
max
= -
l
where m is the mass of photoelectron
Þ 
1
2
1
2
1
m v
hc
W [ ]
( ) max
= -
l
 …(i)
and 
1
2
2
2
2
m v
hc
W [ ]
(max)
= -
l
 …(ii)
Dividing Eq. (i) and Eq. (ii), we get
v
v
hc
W
hc
W
1
2
2
1
2
max
max
æ
è
ç
ç
ö
ø
÷
÷
=
-
-
l
l
 Þ ( ) 3
2 1
2
=
-
-
hc
W
hc
W
l
l
Þ 
9
8
2 1
hc hc
W
l l
- =
Þ hc W
9
6000
1
3000
8 -
é
ë
ê
ù
û
ú
=
Þ 
6.6
1.6
´ ´ ´ ´
´ ´ ´
=
-
- -
10 3 10 7
6000 10 10
8
34 8
10 19
W
Þ W = 1.81 eV
Putting the value of W in Eq. (i)
  
1
2 3000 10
1
2
10
m u
hc
( )
max
=
´
-
- ´ ´
-
181 16 10
19
. .
 
1
2
662 10 3 10
3 10
1
2
34 8
7
m u ( )
.
max
=
´ ´ ´
´
=
- ´
-
2 896 10
19
.
 
1
2
6 62 10 2 896 10
1
2 19 19
m u ( ) . .
m a x
= ´ - ´
- -
1
2
3 724 10
1
2 19
m u ( ) .
max
= ´
-
68
C
D A
– 2
– 4
0
2
4
6
8
K (eV)
max
10 20 30
Page 4


25. Here E
1
= - 15.6 eV
(a) Hence ionization potential 
= - = E
1
15.6 eV
(b) We have l =
hc
E D
 for short wavelength 
DE is maximum
 Þ l =
´ ´ ´
- - ´ ´
-
-
-
6.6
5.3 1.6
Å
10 3 10
0 10
2335
34 8
19
( )
~
(c) Excitation potential for n = 3 state is
= - = - + = E E
3 1
3.08 15.6 12.52 V
(d) From n = 3 to n = 1 
DE E E = - = - + =
3 1
3.08 15.6 12.52eV
We know l =
hc
E D
Þ 
1 10
10 3 10
19
34 8
l
= =
´ ´
´ ´ ´
-
-
DE
hc
12.52 1.6
6.6
    = ´ 1.01 10
7
 ( ) m
- 1
26. (a) E
1
= - 6.52 eV
l = = 860 8600 nm Å
Energy of this photon 
= =
12375
8600
(eV)
1.44 eV
hence internal energy of atom after 
absorbing this photon is given by 
E E
i
= + = - +
1
1.44 eV 6.52 1.44
     = - 5.08 eV
(b) l
2
12375
4200
= =
(eV)
2.95 eV
hence internal energy of the atom after 
emission of this  photon is given by
E E
i
= - = - -
1
2.95 eV 2.68 2.95 eV ( )
Þ E
i
= - 5.63 eV
27. Heve U m r =
1
2
2 2
w Þ  F
dU
dr
m r = =
2 2
w
But 
mv
r
m r
2
2 2
= w
Þ v m r
2 2 2
= w
But by Bohr’s postulate mvr
nh
=
2p
Þ m v r
n h
2 2 2
2 2
2
4
=
p
Þ m r
n h
3 2 4
2 2
2
4
w
p
=
Þ          r
n h
m
4
2 2
2 3 2
4
=
p w
Þ             r n µ
28.
1
1 1
1
2
1
2
2
0
l l
a
K
R z = - -
é
ë
ê
ù
û
ú
= ( )
1
1 1
1
3
2
2
l
b
K
R z = - -
é
ë
ê
ù
û
ú
( )
Þ 
l
l
b
a
K
K
=
´
´
=
3 9
4 8
27
32
Þ l l l
b a
K K
= =
27
32
27
32
0
29. l
0
=
hc
eV
Þ l
0
34 8
19 3 1
10 3 10
10 25 10
=
´ ´ ´
´ ´ ´
-
-
6 .6
1 . 6
Þ l
0
1
= 49.5 p m
Þ l l
0 0
2 1
2 2 5 = = ´ = 49. pm 99 pm
[ ] 1 pm m =
-
10
12
30. f z
Ka
= ´ - ( ) ( ) 2.48 Hz 10 1
15 2
Þ 
3 10
10 2 1
8
15 2
´
= ´ -
l
a K
2.48 ( )
Þ 
3 10
10 10
1
8
10 15
2
´
´ ´ ´
= -
-
0.76 2.58
( ) z
( - - z 1 40 )
~
  Þ z = 41
31. l
i
hc
eV
=   Dl = 26 pm when V
f
= 1.5 V
Þ l l
f i
= - ( ) 26 pm
l l
f i
hc
e V
hc
eV
=
´
= = =
1.5 1.5 1.5 1.5
1 12 1
66
Þ ( ) l l
i i
- = 26
2
3
Þ 3 26 3 2 l l
i i
- ´ =
Þ l
i
= 78 pm
Þ 78 10
10 3 10
10
12
34 8
19
´ =
´ ´ ´
´ ´
-
-
-
6.6
1.6 V
Þ V =
´ ´
´ ´ ´
-
- -
6.6
1.6
3 10
78 10 10
26
12 19
Þ V = 15865 volt
32. V a z b = - ( ) 
Þ 
c
a z b
l
= -
2 2
( )
Þ    
1
2
2
l
= -
a
c
z b ( )
Þ 
1
887
13
2
2
pm
= -
a
c
b ( ) …(i)
and  
1
146
30
2
2
pm
= -
a
z
b ( ) …(ii)
Dividing Eq. (i) and Eq. (ii)
146
887
13
30
2
=
-
-
é
ë
ê
ù
û
ú
( )
( )
b
b
Þ 2.5 =
-
-
30
13
b
b
Þ 32.5 2.5 - = - b b 30
2.5 1.5 = b
   b =
5
3
1
26
5
3
26
2 2
l
= -
æ
è
ç
ö
ø
÷
a
c
1
26
5
3
887 13
5
3
26
2
2
l
=
f -
æ
è
ç
ö
ø
÷
´ -
æ
è
ç
ö
ø
÷
pm
Þ l
26
2
2
887 13
5
3
26
5
3
=
´ -
æ
è
ç
ö
ø
÷
-
æ
è
ç
ö
ø
÷
pm
= ´ 887
34
73
2
2
pm
( )
~
- 198 pm
33. f
RC
z = -
3
4
1 ( )
 4.2
1.1
´ =
´ ´ ´ ´
10
3 10 3 10
4
18
7 8
 ( ) z - 1
2
Þ
4.2
1.1
´ ´
´ ´
= -
10 4
9 10
1
18
15
2
( ) z 
Þ ( ) z - = 1 41 Þ z = 42
34. P Vi = = ´ = 40 10 400 kW mA W
% of P =
´
=
400 1
100
4 W
(a) Total power of X-rays = 4 W 
(b) Heat produced per second
= - = 400 4 396 J/s
Photoelectric effect
35. Einstein photo elec tric equation is
K h W
max
= - n
Þ eV
hc
W
0
= -
l
  Q K eV
max
=
0
Þ 10.4 eV
Å
1.7 eV = -
12375
l( )
Þ l( ) Å
12.1
Å = =
12375
1022
For H-atom l =
hc
E D
 Þ DE = =
12375
1022
12.1 eV
 67
This difference equal to n = 3 ® = n 1
transition.
36. K h W
max
= - n
Þ K
max
=
´ ´ ´
´
-
-
-
6.6 1.5
1.6
3.7
10 10
10
34 15
19
Þ K
max
= - = 6.18 3.7 2.48 eV
37. Here work func tion 
       W(in eV)
Å
2.475 eV = =
12375
5000
K eV
max
= =
0
3 eV
K
hc
W
max
= -
l
Þ 3
12375
= -
l( ) in Å
2.475
Þ l = =
12375
2260
5.475
Å
38. Comparing the given graph with
K h W
max
= - n
 f = ´ 1 10
14
 Hz
(a) n
o
= threshold frequency q = ´ 10 10
14
Hz
= 10
15
Hz
(b) W = 4 eV 
(c) h = slope of the graph 
= =
´
CD
AD
8
20 10
14
eV
Hz
Þ h =
´ ´
´
= ´
-
-
8 10
2 10
10
19
15
34
1.6
6.4 J-s
39. Here 
v
u
1
2
3
1
(max)
) (max
=
Using Einstein equation, K
hc
W
max
, = -
l
we get
1
2
2
mv
hc
W
max
= -
l
where m is the mass of photoelectron
Þ 
1
2
1
2
1
m v
hc
W [ ]
( ) max
= -
l
 …(i)
and 
1
2
2
2
2
m v
hc
W [ ]
(max)
= -
l
 …(ii)
Dividing Eq. (i) and Eq. (ii), we get
v
v
hc
W
hc
W
1
2
2
1
2
max
max
æ
è
ç
ç
ö
ø
÷
÷
=
-
-
l
l
 Þ ( ) 3
2 1
2
=
-
-
hc
W
hc
W
l
l
Þ 
9
8
2 1
hc hc
W
l l
- =
Þ hc W
9
6000
1
3000
8 -
é
ë
ê
ù
û
ú
=
Þ 
6.6
1.6
´ ´ ´ ´
´ ´ ´
=
-
- -
10 3 10 7
6000 10 10
8
34 8
10 19
W
Þ W = 1.81 eV
Putting the value of W in Eq. (i)
  
1
2 3000 10
1
2
10
m u
hc
( )
max
=
´
-
- ´ ´
-
181 16 10
19
. .
 
1
2
662 10 3 10
3 10
1
2
34 8
7
m u ( )
.
max
=
´ ´ ´
´
=
- ´
-
2 896 10
19
.
 
1
2
6 62 10 2 896 10
1
2 19 19
m u ( ) . .
m a x
= ´ - ´
- -
1
2
3 724 10
1
2 19
m u ( ) .
max
= ´
-
68
C
D A
– 2
– 4
0
2
4
6
8
K (eV)
max
10 20 30
Þ  ( )
.
.
max
u
1
2
19
31
3 724 10 2
9 1 10
=
´ ´
´
-
-
- ´ ´
-
1.81 16 10
19
.  Q     m
e
= ´
-
9 1 10
31
.
Þ  u
1
5
9 10
max
= ´ m/s
and v v
2 1
5
1
3
3 10
max max
= = ´ m/s
40. Here in ten si ty I = 2
2
W m / and
Area A = ´
-
1 10
4 2
m
Energy incident per unit time on the metal
surface
E IA = = ´
-
2 10
4
 W
= ´ =
´
´
-
-
-
2 10
2 10
10
4
4
19
J/s
1.6
eV
s
  
=
´ 2 10
15
1.6
eV/s
Energy of each photon = 10.6 eV
Number of photons incident on surface 
=
´
´
2 10
106
15
1.6 .
Number of photoelectrons emitted 
= ´
´
´
0.53
1.6 100
2 10
106
15
.
= ´ 6.25 10
11
 per second
Minimum KE = 0
Maximum KE 10.6 5.6 eV eV = - = ( ) 5
41. K
hc
W
max
= -
l
Þ 
K
max
=
´ ´ ´
´
- ´ ´
-
-
6.6
1.6
10 3 10
180 10
2 10
34 8
9
19
1
2
3 10
18
2 10
2
18
19
m v
e m ax
=
´ ´
- ´ ´
-
-
6.6
1.6
1
2
11 10 10
2 19 19
m v
e max
= ´ - ´
- -
3.2
            = ´
-
7.2 J 10
19
v
max
=
´ ´
´
= ´
-
-
2 10
10
10
19
31
6
7.2
9.1
1.25 m/s
r
mv
eB
= =
´ ´ ´
´ ´ ´
-
- -
max
9.1 1.25
1.6
10 10
10 5 10
3 1 6
1 9 5
r = 0.148 m
42. The given equation is 
E t = ´ ( ) [sin (( ) 100 5 10
15
v / m
+ ´ sin ( ) ] 8 10
15
t
Light consist of two different frequencies.
Maximum frequency 
=
´
= ´
8 10
2
10
15
15
p
1.27 Hz
For maximum KE we will use Einstein’s
equation
( )
max
KE = - h W n
=
´ ´ ´
´
-
-
-
1.27 6.62
1.6
10 10
10
2
15 34
19
Þ ( )
max
KE 3.27 eV = 
43. Here E E x ct = ´ -
0
7
10 sin ( ) 1.57 frequency
of the wave
n =
´ ´
´
= ´
1.57
3.14
0.75
3 10
2
10
15
15
We have 
eV
0
34 15
19
10 10
10
=
´ ´ ´
´
-
æ
è
ç
ç
ö
ø
÷
÷
-
-
6.6 0.75
1.8
1.9 eV
Þ V
0
= 1.2 V 
¢ Objective Questions (Level-1)
1. Einstein photo elec tric equa tion is
K h W
max
= - n
its slope = = h planck constant which is
same for all metals and independent of
intensity of radiation.
Hence correct option is (d).
2. Since cur rent is di rectly pro por tional to
in ten sity there fore as cur rent is in creased
 69
Page 5


25. Here E
1
= - 15.6 eV
(a) Hence ionization potential 
= - = E
1
15.6 eV
(b) We have l =
hc
E D
 for short wavelength 
DE is maximum
 Þ l =
´ ´ ´
- - ´ ´
-
-
-
6.6
5.3 1.6
Å
10 3 10
0 10
2335
34 8
19
( )
~
(c) Excitation potential for n = 3 state is
= - = - + = E E
3 1
3.08 15.6 12.52 V
(d) From n = 3 to n = 1 
DE E E = - = - + =
3 1
3.08 15.6 12.52eV
We know l =
hc
E D
Þ 
1 10
10 3 10
19
34 8
l
= =
´ ´
´ ´ ´
-
-
DE
hc
12.52 1.6
6.6
    = ´ 1.01 10
7
 ( ) m
- 1
26. (a) E
1
= - 6.52 eV
l = = 860 8600 nm Å
Energy of this photon 
= =
12375
8600
(eV)
1.44 eV
hence internal energy of atom after 
absorbing this photon is given by 
E E
i
= + = - +
1
1.44 eV 6.52 1.44
     = - 5.08 eV
(b) l
2
12375
4200
= =
(eV)
2.95 eV
hence internal energy of the atom after 
emission of this  photon is given by
E E
i
= - = - -
1
2.95 eV 2.68 2.95 eV ( )
Þ E
i
= - 5.63 eV
27. Heve U m r =
1
2
2 2
w Þ  F
dU
dr
m r = =
2 2
w
But 
mv
r
m r
2
2 2
= w
Þ v m r
2 2 2
= w
But by Bohr’s postulate mvr
nh
=
2p
Þ m v r
n h
2 2 2
2 2
2
4
=
p
Þ m r
n h
3 2 4
2 2
2
4
w
p
=
Þ          r
n h
m
4
2 2
2 3 2
4
=
p w
Þ             r n µ
28.
1
1 1
1
2
1
2
2
0
l l
a
K
R z = - -
é
ë
ê
ù
û
ú
= ( )
1
1 1
1
3
2
2
l
b
K
R z = - -
é
ë
ê
ù
û
ú
( )
Þ 
l
l
b
a
K
K
=
´
´
=
3 9
4 8
27
32
Þ l l l
b a
K K
= =
27
32
27
32
0
29. l
0
=
hc
eV
Þ l
0
34 8
19 3 1
10 3 10
10 25 10
=
´ ´ ´
´ ´ ´
-
-
6 .6
1 . 6
Þ l
0
1
= 49.5 p m
Þ l l
0 0
2 1
2 2 5 = = ´ = 49. pm 99 pm
[ ] 1 pm m =
-
10
12
30. f z
Ka
= ´ - ( ) ( ) 2.48 Hz 10 1
15 2
Þ 
3 10
10 2 1
8
15 2
´
= ´ -
l
a K
2.48 ( )
Þ 
3 10
10 10
1
8
10 15
2
´
´ ´ ´
= -
-
0.76 2.58
( ) z
( - - z 1 40 )
~
  Þ z = 41
31. l
i
hc
eV
=   Dl = 26 pm when V
f
= 1.5 V
Þ l l
f i
= - ( ) 26 pm
l l
f i
hc
e V
hc
eV
=
´
= = =
1.5 1.5 1.5 1.5
1 12 1
66
Þ ( ) l l
i i
- = 26
2
3
Þ 3 26 3 2 l l
i i
- ´ =
Þ l
i
= 78 pm
Þ 78 10
10 3 10
10
12
34 8
19
´ =
´ ´ ´
´ ´
-
-
-
6.6
1.6 V
Þ V =
´ ´
´ ´ ´
-
- -
6.6
1.6
3 10
78 10 10
26
12 19
Þ V = 15865 volt
32. V a z b = - ( ) 
Þ 
c
a z b
l
= -
2 2
( )
Þ    
1
2
2
l
= -
a
c
z b ( )
Þ 
1
887
13
2
2
pm
= -
a
c
b ( ) …(i)
and  
1
146
30
2
2
pm
= -
a
z
b ( ) …(ii)
Dividing Eq. (i) and Eq. (ii)
146
887
13
30
2
=
-
-
é
ë
ê
ù
û
ú
( )
( )
b
b
Þ 2.5 =
-
-
30
13
b
b
Þ 32.5 2.5 - = - b b 30
2.5 1.5 = b
   b =
5
3
1
26
5
3
26
2 2
l
= -
æ
è
ç
ö
ø
÷
a
c
1
26
5
3
887 13
5
3
26
2
2
l
=
f -
æ
è
ç
ö
ø
÷
´ -
æ
è
ç
ö
ø
÷
pm
Þ l
26
2
2
887 13
5
3
26
5
3
=
´ -
æ
è
ç
ö
ø
÷
-
æ
è
ç
ö
ø
÷
pm
= ´ 887
34
73
2
2
pm
( )
~
- 198 pm
33. f
RC
z = -
3
4
1 ( )
 4.2
1.1
´ =
´ ´ ´ ´
10
3 10 3 10
4
18
7 8
 ( ) z - 1
2
Þ
4.2
1.1
´ ´
´ ´
= -
10 4
9 10
1
18
15
2
( ) z 
Þ ( ) z - = 1 41 Þ z = 42
34. P Vi = = ´ = 40 10 400 kW mA W
% of P =
´
=
400 1
100
4 W
(a) Total power of X-rays = 4 W 
(b) Heat produced per second
= - = 400 4 396 J/s
Photoelectric effect
35. Einstein photo elec tric equation is
K h W
max
= - n
Þ eV
hc
W
0
= -
l
  Q K eV
max
=
0
Þ 10.4 eV
Å
1.7 eV = -
12375
l( )
Þ l( ) Å
12.1
Å = =
12375
1022
For H-atom l =
hc
E D
 Þ DE = =
12375
1022
12.1 eV
 67
This difference equal to n = 3 ® = n 1
transition.
36. K h W
max
= - n
Þ K
max
=
´ ´ ´
´
-
-
-
6.6 1.5
1.6
3.7
10 10
10
34 15
19
Þ K
max
= - = 6.18 3.7 2.48 eV
37. Here work func tion 
       W(in eV)
Å
2.475 eV = =
12375
5000
K eV
max
= =
0
3 eV
K
hc
W
max
= -
l
Þ 3
12375
= -
l( ) in Å
2.475
Þ l = =
12375
2260
5.475
Å
38. Comparing the given graph with
K h W
max
= - n
 f = ´ 1 10
14
 Hz
(a) n
o
= threshold frequency q = ´ 10 10
14
Hz
= 10
15
Hz
(b) W = 4 eV 
(c) h = slope of the graph 
= =
´
CD
AD
8
20 10
14
eV
Hz
Þ h =
´ ´
´
= ´
-
-
8 10
2 10
10
19
15
34
1.6
6.4 J-s
39. Here 
v
u
1
2
3
1
(max)
) (max
=
Using Einstein equation, K
hc
W
max
, = -
l
we get
1
2
2
mv
hc
W
max
= -
l
where m is the mass of photoelectron
Þ 
1
2
1
2
1
m v
hc
W [ ]
( ) max
= -
l
 …(i)
and 
1
2
2
2
2
m v
hc
W [ ]
(max)
= -
l
 …(ii)
Dividing Eq. (i) and Eq. (ii), we get
v
v
hc
W
hc
W
1
2
2
1
2
max
max
æ
è
ç
ç
ö
ø
÷
÷
=
-
-
l
l
 Þ ( ) 3
2 1
2
=
-
-
hc
W
hc
W
l
l
Þ 
9
8
2 1
hc hc
W
l l
- =
Þ hc W
9
6000
1
3000
8 -
é
ë
ê
ù
û
ú
=
Þ 
6.6
1.6
´ ´ ´ ´
´ ´ ´
=
-
- -
10 3 10 7
6000 10 10
8
34 8
10 19
W
Þ W = 1.81 eV
Putting the value of W in Eq. (i)
  
1
2 3000 10
1
2
10
m u
hc
( )
max
=
´
-
- ´ ´
-
181 16 10
19
. .
 
1
2
662 10 3 10
3 10
1
2
34 8
7
m u ( )
.
max
=
´ ´ ´
´
=
- ´
-
2 896 10
19
.
 
1
2
6 62 10 2 896 10
1
2 19 19
m u ( ) . .
m a x
= ´ - ´
- -
1
2
3 724 10
1
2 19
m u ( ) .
max
= ´
-
68
C
D A
– 2
– 4
0
2
4
6
8
K (eV)
max
10 20 30
Þ  ( )
.
.
max
u
1
2
19
31
3 724 10 2
9 1 10
=
´ ´
´
-
-
- ´ ´
-
1.81 16 10
19
.  Q     m
e
= ´
-
9 1 10
31
.
Þ  u
1
5
9 10
max
= ´ m/s
and v v
2 1
5
1
3
3 10
max max
= = ´ m/s
40. Here in ten si ty I = 2
2
W m / and
Area A = ´
-
1 10
4 2
m
Energy incident per unit time on the metal
surface
E IA = = ´
-
2 10
4
 W
= ´ =
´
´
-
-
-
2 10
2 10
10
4
4
19
J/s
1.6
eV
s
  
=
´ 2 10
15
1.6
eV/s
Energy of each photon = 10.6 eV
Number of photons incident on surface 
=
´
´
2 10
106
15
1.6 .
Number of photoelectrons emitted 
= ´
´
´
0.53
1.6 100
2 10
106
15
.
= ´ 6.25 10
11
 per second
Minimum KE = 0
Maximum KE 10.6 5.6 eV eV = - = ( ) 5
41. K
hc
W
max
= -
l
Þ 
K
max
=
´ ´ ´
´
- ´ ´
-
-
6.6
1.6
10 3 10
180 10
2 10
34 8
9
19
1
2
3 10
18
2 10
2
18
19
m v
e m ax
=
´ ´
- ´ ´
-
-
6.6
1.6
1
2
11 10 10
2 19 19
m v
e max
= ´ - ´
- -
3.2
            = ´
-
7.2 J 10
19
v
max
=
´ ´
´
= ´
-
-
2 10
10
10
19
31
6
7.2
9.1
1.25 m/s
r
mv
eB
= =
´ ´ ´
´ ´ ´
-
- -
max
9.1 1.25
1.6
10 10
10 5 10
3 1 6
1 9 5
r = 0.148 m
42. The given equation is 
E t = ´ ( ) [sin (( ) 100 5 10
15
v / m
+ ´ sin ( ) ] 8 10
15
t
Light consist of two different frequencies.
Maximum frequency 
=
´
= ´
8 10
2
10
15
15
p
1.27 Hz
For maximum KE we will use Einstein’s
equation
( )
max
KE = - h W n
=
´ ´ ´
´
-
-
-
1.27 6.62
1.6
10 10
10
2
15 34
19
Þ ( )
max
KE 3.27 eV = 
43. Here E E x ct = ´ -
0
7
10 sin ( ) 1.57 frequency
of the wave
n =
´ ´
´
= ´
1.57
3.14
0.75
3 10
2
10
15
15
We have 
eV
0
34 15
19
10 10
10
=
´ ´ ´
´
-
æ
è
ç
ç
ö
ø
÷
÷
-
-
6.6 0.75
1.8
1.9 eV
Þ V
0
= 1.2 V 
¢ Objective Questions (Level-1)
1. Einstein photo elec tric equa tion is
K h W
max
= - n
its slope = = h planck constant which is
same for all metals and independent of
intensity of radiation.
Hence correct option is (d).
2. Since cur rent is di rectly pro por tional to
in ten sity there fore as cur rent is in creased
 69
in ten si ty is in creased since l
min
µ
1
V
, if V is
de creased l
min
 is in creased.
Hence correct option is (c).
3. For hydrogen atom (Bohr’s model) nth
or bital speed v
e
n h
n
=
2
0
2e
For first orbit n = 1
Þ v
e
h
1
2
0
2
=
´ e
 
=
´
´ ´ ´ ´
-
- -
( ) 1.6
8.85 6.62
10
2 10 10
19 2
12 34
Þ v
c
~
-
æ
è
ç
ö
ø
÷
´ ´ =
1
137
3 10
137
8
Hence correct option is (c).
4.
86
22
3
80
210
4
84
210
A X B ¾® ¾®
a
b
Hence correct option is (b)
5. l
min
( )
(
~
in Å
in volt)
0.62 Å = =
´
-
12375 12375
20 1000 V
Hence correct option is (c).
6. We have 
1
2
2
m v
e max
= eV
Þ     v
eV
m
max
=
2
=
´ ´ ´ ´
´
-
-
2 10 18 1000
19
31
1.6
9.1 10
Þ    v
max
~
- ´ 8 10
7
m/s
Hence correct option is (a).
7. For hy dro gen atom v
e
n h
n
=
2
0
2e
Þ v
e
h
v
e
h
2
2
0
3
2
0
2 2 2 3
=
e ´
=
´
,
e
Þ 
v
v
2
3
3
2
=
Let l
2
 and l
3
 are the de-Broglie
wavelengths
Þ 
l
l
2
3
2
3
3
2
= =
h
mv
n
mv
v
v
Þ 
l
l
2
3
2
3
=
Hence correct option is (a).
8. For hy dro gen like atom
E
z
n
n
=
-
2
2
(13.6 eV)
For ground state n = 1
Þ E z
1
2
= - ´ 13.6 eV
But  E
1
= - 122.4 eV
Þ  - = - ´ 122.4 eV 13.6 eV z
2
Þ   z
2
9 =
Þ    z = 3 
Hence it is Li
2 +
The correct option is (c).
9. l
min
=
hc
eV
 Þ  
D D l
l
min
min
´ = - ´ 100 100
V
V
Percentage change in l
min
% = - 2
Hence l
min
 is decreased by 2%
correct option is (c)
10. E
z
n
n
=
-
2
2
(13.6 eV) for first ex cited state n = 2
Þ E
z
2
2
4
=
-
( ) 13.6 eV
Þ   - =
-
´ 13.6 eV 13.6 eV
z
2
4
 
Þ    z = 2
Hence it is He
+
Correct option is  (a).
11. l
min
( )
( )
in Å
in volt
=
12375
V
Þ V = = ´
12375
1
10
3
12.375 V
Þ V = 12.4 eV
70
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