Chapter 29 - Modern Physics I (Part - 1) - Physics, Solution by D C Pandey, NEET NEET Notes | EduRev

DC Pandey (Questions & Solutions) of Physics: NEET

Created by: Ciel Knowledge

NEET : Chapter 29 - Modern Physics I (Part - 1) - Physics, Solution by D C Pandey, NEET NEET Notes | EduRev

 Page 1


29. Modern Physics I
Introductory Exercise 29.1
1. The po si tron has same mass m as the
elec tron. The reduced mass of elec tron
po si tron atom is
m =
´
+
=
m m
m m
1
2
m
R
m e
c h
H
=
4
0
2 3
8 e
Þ   R
R
P
H
=
2
1 1
2
1
3
2 2
l
H
H
R = -
æ
è
ç
ö
ø
÷
1 1
2
1
3
2 2
l
P
P
R = -
æ
è
ç
ö
ø
÷
Þ  
l
l
P
H
H
P
R
R
= = 2
Þ l l
p H
= = ´ 2 2 6563 Å = 13126 Å
     = 1.31 m m
1 1
2
1
3
2 2
2
l
He
= -
æ
è
ç
ö
ø
÷
× R z
H
Þ   
1 1
2
l l
He H
= × z  Þ    l
l
He
H
Å
= =
2
6563
2
2 2
Þ    l
He
nm = 164
2.
1 1
2
1
2 2
l
= -
é
ë
ê
ù
û
ú
R
n
 for larg est wave length n = 3
Þ   
1 1
4
1
9 l
= -
é
ë
ê
ù
û
ú
R
Þ l = =
´ ´
36
5
36
5 10
7
R 1.097
Þ  l = 656 nm
3. For H-atom r
n h
me
n
=
e
p
0
2 2
2
u
e
nh
n
=
e
2
0
2
T
r
u
n h nh
me e
n
n
n
= =
´
´
2
2
2
2 2
0
2 2
p
p
e e
p
0
     =
4
0
3 3
4
e n h
me
r
T
me
n h
n
n
= =
1
4
4
0
2 3 3
e
Þ r
me
h
1
4
0
2 3
4
=
e
Þ n
1
31 19 4
12 2 3
10 1 6 10
4 10 10
=
´ ´ ´
´ ´ ´ ´
- -
- -
9.1
8.85 6.6
( . )
( ) (
4 3
)
Þ n
1
15
1 0 = ´ 6.58 Hz
n
n n
2
1
3
1
15
2 8
10
8
= = =
´ 6.58
     = ´ 0.823 10
15
 Hz
(b)   
1 1
1
1
2
2 2
l
= -
é
ë
ê
ù
û
ú
R
Þ  n
l
= = ´ ´
é
ë
ê
ù
û
ú
c
R 3 10
3
4
8
Þ n =
´ ´ ´ 9 10 10
4
8 7
1.097
        = ´ 2.46 Hz 10
15
(c) Number of revolutions
= ´ = ´ ´ ´
-
v T
2
15 8
10 1 10 0.823
= ´ 8.23 10
6
 revolution
Page 2


29. Modern Physics I
Introductory Exercise 29.1
1. The po si tron has same mass m as the
elec tron. The reduced mass of elec tron
po si tron atom is
m =
´
+
=
m m
m m
1
2
m
R
m e
c h
H
=
4
0
2 3
8 e
Þ   R
R
P
H
=
2
1 1
2
1
3
2 2
l
H
H
R = -
æ
è
ç
ö
ø
÷
1 1
2
1
3
2 2
l
P
P
R = -
æ
è
ç
ö
ø
÷
Þ  
l
l
P
H
H
P
R
R
= = 2
Þ l l
p H
= = ´ 2 2 6563 Å = 13126 Å
     = 1.31 m m
1 1
2
1
3
2 2
2
l
He
= -
æ
è
ç
ö
ø
÷
× R z
H
Þ   
1 1
2
l l
He H
= × z  Þ    l
l
He
H
Å
= =
2
6563
2
2 2
Þ    l
He
nm = 164
2.
1 1
2
1
2 2
l
= -
é
ë
ê
ù
û
ú
R
n
 for larg est wave length n = 3
Þ   
1 1
4
1
9 l
= -
é
ë
ê
ù
û
ú
R
Þ l = =
´ ´
36
5
36
5 10
7
R 1.097
Þ  l = 656 nm
3. For H-atom r
n h
me
n
=
e
p
0
2 2
2
u
e
nh
n
=
e
2
0
2
T
r
u
n h nh
me e
n
n
n
= =
´
´
2
2
2
2 2
0
2 2
p
p
e e
p
0
     =
4
0
3 3
4
e n h
me
r
T
me
n h
n
n
= =
1
4
4
0
2 3 3
e
Þ r
me
h
1
4
0
2 3
4
=
e
Þ n
1
31 19 4
12 2 3
10 1 6 10
4 10 10
=
´ ´ ´
´ ´ ´ ´
- -
- -
9.1
8.85 6.6
( . )
( ) (
4 3
)
Þ n
1
15
1 0 = ´ 6.58 Hz
n
n n
2
1
3
1
15
2 8
10
8
= = =
´ 6.58
     = ´ 0.823 10
15
 Hz
(b)   
1 1
1
1
2
2 2
l
= -
é
ë
ê
ù
û
ú
R
Þ  n
l
= = ´ ´
é
ë
ê
ù
û
ú
c
R 3 10
3
4
8
Þ n =
´ ´ ´ 9 10 10
4
8 7
1.097
        = ´ 2.46 Hz 10
15
(c) Number of revolutions
= ´ = ´ ´ ´
-
v T
2
15 8
10 1 10 0.823
= ´ 8.23 10
6
 revolution
4.Reduce mass 
=
+
=
´
+
=
mm
mm
p
p
m
m
2071836
2071836
186
mm
mm ()
m
r
h
e
h
me
10
2
22
0
2
22
4
4
4
4
=
æ
è
ç
ç
ö
ø
÷
÷
=
æ
è
ç
ç
ö
ø
÷
÷
pe
pm
pe
p()186
Putting the value we get
r
1
13
10 = ´
-
2.55 m
E
e
h
1
4
0
2 2
8
2810 =
-
= -
m
e
eV
Ionization energy = - = E
1
2.81 keV
5. (a) l = =
´
´ ´
= ´
-
-
-
h
mv
6.6
4.8 m
10
46 10 30
10
34
3
34
(b) l =
´
´ ´
= ´
-
-
-
6.6
9.1
7.3  10 m
11
10
10 10
3 4
31 7
6. (a) After absorbing 12.3 eV the atom excited
to n = 3 state
1
1
1
2
l
L
R
n
= -
é
ë
ê
ù
û
ú
1
1
1
9
8
9
1
l
L
R
R
= -
é
ë
ê
ù
û
ú
=
Þ l
L
R
1
9
8
9
8 10
102
7
= =
´ ´
=
1.097
nm
1
1
1
4
2
l
L
R = -
é
ë
ê
ù
û
ú
Þ l
L
R
2
4
3
4
3 10
7
= =
´ ´
=
1.097
122 nm
1 1
2
1
3
5
36
2 2
l
B
R
R
= -
é
ë
ê
ù
û
ú
=
Þ l
B
R
=
36
5
Þ l
B
=
´ ´
=
36
5 10
7
1.097
653 nm
7. n n n
b a a
K K L
= + Þ l
l l
l l
a
a b
a b
L
K K
K K
=
´
-
c c c
K K L
l l l
b a a
= + Þ l
a
L
=
´
-
0.71 0.63
0.71 0.63
Þ 
1 1 1
l l l
a b a
L K K
= - l
a
L
= 5.59 nm
8. l =
hc
E D
Þ             l
a
K
E
=
´ ´ ´
- ´ ´
-
-
6.6
1.6
10 3 10
2870 10
34 8
1
19
( )
Þ 0.71
6.6
1.6
´ =
´ ´ ´
- ´ ´
-
-
-
10
10 3 10
2870 10
9
34 8
1
19
( ) E
Solving this we get 
     E
1
4613 = - eV
         l
K
B
E
=
´ ´ ´
- ´ ´
=
-
-
6.6
1.6
0.63
10 3 10
4613 10
34 8
3
19
( )
Solving this we get E
3
2650 = - eV
9. n
31
4
= =
E
h
f …(i)
n
21
3 3
4
4 3
4
= =
æ
è
ç
ö
ø
÷ =
E
h
E
h
f
n
32
4
= =
E
h
f
60
E
3
K
a
K
b
E = – 2870 eV
2
E
1
5 E
E
4 E
1
2
3
n = 3
n = 2
n = 1
Page 3


29. Modern Physics I
Introductory Exercise 29.1
1. The po si tron has same mass m as the
elec tron. The reduced mass of elec tron
po si tron atom is
m =
´
+
=
m m
m m
1
2
m
R
m e
c h
H
=
4
0
2 3
8 e
Þ   R
R
P
H
=
2
1 1
2
1
3
2 2
l
H
H
R = -
æ
è
ç
ö
ø
÷
1 1
2
1
3
2 2
l
P
P
R = -
æ
è
ç
ö
ø
÷
Þ  
l
l
P
H
H
P
R
R
= = 2
Þ l l
p H
= = ´ 2 2 6563 Å = 13126 Å
     = 1.31 m m
1 1
2
1
3
2 2
2
l
He
= -
æ
è
ç
ö
ø
÷
× R z
H
Þ   
1 1
2
l l
He H
= × z  Þ    l
l
He
H
Å
= =
2
6563
2
2 2
Þ    l
He
nm = 164
2.
1 1
2
1
2 2
l
= -
é
ë
ê
ù
û
ú
R
n
 for larg est wave length n = 3
Þ   
1 1
4
1
9 l
= -
é
ë
ê
ù
û
ú
R
Þ l = =
´ ´
36
5
36
5 10
7
R 1.097
Þ  l = 656 nm
3. For H-atom r
n h
me
n
=
e
p
0
2 2
2
u
e
nh
n
=
e
2
0
2
T
r
u
n h nh
me e
n
n
n
= =
´
´
2
2
2
2 2
0
2 2
p
p
e e
p
0
     =
4
0
3 3
4
e n h
me
r
T
me
n h
n
n
= =
1
4
4
0
2 3 3
e
Þ r
me
h
1
4
0
2 3
4
=
e
Þ n
1
31 19 4
12 2 3
10 1 6 10
4 10 10
=
´ ´ ´
´ ´ ´ ´
- -
- -
9.1
8.85 6.6
( . )
( ) (
4 3
)
Þ n
1
15
1 0 = ´ 6.58 Hz
n
n n
2
1
3
1
15
2 8
10
8
= = =
´ 6.58
     = ´ 0.823 10
15
 Hz
(b)   
1 1
1
1
2
2 2
l
= -
é
ë
ê
ù
û
ú
R
Þ  n
l
= = ´ ´
é
ë
ê
ù
û
ú
c
R 3 10
3
4
8
Þ n =
´ ´ ´ 9 10 10
4
8 7
1.097
        = ´ 2.46 Hz 10
15
(c) Number of revolutions
= ´ = ´ ´ ´
-
v T
2
15 8
10 1 10 0.823
= ´ 8.23 10
6
 revolution
4.Reduce mass 
=
+
=
´
+
=
mm
mm
p
p
m
m
2071836
2071836
186
mm
mm ()
m
r
h
e
h
me
10
2
22
0
2
22
4
4
4
4
=
æ
è
ç
ç
ö
ø
÷
÷
=
æ
è
ç
ç
ö
ø
÷
÷
pe
pm
pe
p()186
Putting the value we get
r
1
13
10 = ´
-
2.55 m
E
e
h
1
4
0
2 2
8
2810 =
-
= -
m
e
eV
Ionization energy = - = E
1
2.81 keV
5. (a) l = =
´
´ ´
= ´
-
-
-
h
mv
6.6
4.8 m
10
46 10 30
10
34
3
34
(b) l =
´
´ ´
= ´
-
-
-
6.6
9.1
7.3  10 m
11
10
10 10
3 4
31 7
6. (a) After absorbing 12.3 eV the atom excited
to n = 3 state
1
1
1
2
l
L
R
n
= -
é
ë
ê
ù
û
ú
1
1
1
9
8
9
1
l
L
R
R
= -
é
ë
ê
ù
û
ú
=
Þ l
L
R
1
9
8
9
8 10
102
7
= =
´ ´
=
1.097
nm
1
1
1
4
2
l
L
R = -
é
ë
ê
ù
û
ú
Þ l
L
R
2
4
3
4
3 10
7
= =
´ ´
=
1.097
122 nm
1 1
2
1
3
5
36
2 2
l
B
R
R
= -
é
ë
ê
ù
û
ú
=
Þ l
B
R
=
36
5
Þ l
B
=
´ ´
=
36
5 10
7
1.097
653 nm
7. n n n
b a a
K K L
= + Þ l
l l
l l
a
a b
a b
L
K K
K K
=
´
-
c c c
K K L
l l l
b a a
= + Þ l
a
L
=
´
-
0.71 0.63
0.71 0.63
Þ 
1 1 1
l l l
a b a
L K K
= - l
a
L
= 5.59 nm
8. l =
hc
E D
Þ             l
a
K
E
=
´ ´ ´
- ´ ´
-
-
6.6
1.6
10 3 10
2870 10
34 8
1
19
( )
Þ 0.71
6.6
1.6
´ =
´ ´ ´
- ´ ´
-
-
-
10
10 3 10
2870 10
9
34 8
1
19
( ) E
Solving this we get 
     E
1
4613 = - eV
         l
K
B
E
=
´ ´ ´
- ´ ´
=
-
-
6.6
1.6
0.63
10 3 10
4613 10
34 8
3
19
( )
Solving this we get E
3
2650 = - eV
9. n
31
4
= =
E
h
f …(i)
n
21
3 3
4
4 3
4
= =
æ
è
ç
ö
ø
÷ =
E
h
E
h
f
n
32
4
= =
E
h
f
60
E
3
K
a
K
b
E = – 2870 eV
2
E
1
5 E
E
4 E
1
2
3
n = 3
n = 2
n = 1
Introductory Exercise 29.2
1. eV
hc
W
0
= -
l
Þ eV
0
34 8
7 19
10 3 10
2 10 10
=
´ ´ ´
´ ´ ´
-
-
- -
6.6
1.6
4.3eV
Þ eV
0
= - = 6.2eV 4.3eV 1.9eV
Þ V
0
=1.9 volt
2.P=1.5mW = ´
-
1.5 W 10
3
Energy of each photon 
=
´ ´ ´
´
-
-
6.62 10 3 10
4 10
34 8
7
= ´
-
4.96 10
19
 J
Number of photons incident per second
= =
´
´
-
-
P
Energy of each photon
1.5
4.96
10
10
3
19
             
~
- ´ 3 10
15
The number of photoelectrons produce 
= ´ ´ 0.1% 3 10
15
 = ´ 3 10
12
Current i = ´ ´ ´
-
3 10 10
12 19
1.6 A
= ´
-
4.8 A 10
7
 = 0.48 A m
3. K hf W hf hf
max
= - = -
0
Þ K f f
max
( ) µ -
0
4. K
hc
W
max
= -
l
=
´ ´ ´
´ ´ ´
-
é
ë
ê
ù
û
ú
-
- -
6.62
1.6
eV
10 3 10
2 10 10
3
34 8
7 19
= - = [ ] 6.20 eV 3.20 eV 3
The minimum kinetic energy = 0.
5.                  K h f f
max
[ ] = -
0
 
1.2 eV = - h f f [ ]
0
 …(i)
4.2 eV 1.5 = - h f f [ ]
0
 …(ii)
Dividing Eq. (i) and Eq. (ii)
       
124
42
0
0
=
-
-
f f
f f 1.5
Þ 3 2 7 7
0 0
f f f f - = -
Þ         5 4
0
f f =
 Þ  f f f
0
4
5
= = 0.8
Þ 1.2 1.6 6.62
0.8
´ ´ = ´ -
é
ë
ê
ù
û
ú
- -
10 10
19 34 0
0
f
f
Þ 
1.2 1.6
6.62
´ ´
´
= =
-
-
10
10
2
8 4
19
34
0 0
f f
Þ f
0
15
10 = ´ 1.16 Hz
Subjective Questions (Level I)
1. Here l = = ´
-
280 28 10
8
nm m
E
hc
= =
´ ´ ´
´
-
-
l
6.6
J
10 3 10
28 10
34 8
8
E =
´ ´
=
´
- -
19.8
J J
10 10
28
198 10
28
34 16 19
E =
´
´ ´
=
´
-
-
-
198 10
28 10
198
28
19
19
1.6
eV
1.6
4.6 eV
~
We have E mc =
2
 Þ m
E
c
= =
´
´ ´
-
2
19
16
198 10
28 9 10
Þ m = ´
-
8.2 kg 10
36
Momentum p mc = = ´ ´ ´
-
8.2 10 3 10
36 8
      = ´
-
2.46 kg-m/s 10
27
2. Intensity of light at a dis tance 2 m
From the source =
´
=
1
4 2
1
16
2
2
p p ( )
/ W m
 61
Page 4


29. Modern Physics I
Introductory Exercise 29.1
1. The po si tron has same mass m as the
elec tron. The reduced mass of elec tron
po si tron atom is
m =
´
+
=
m m
m m
1
2
m
R
m e
c h
H
=
4
0
2 3
8 e
Þ   R
R
P
H
=
2
1 1
2
1
3
2 2
l
H
H
R = -
æ
è
ç
ö
ø
÷
1 1
2
1
3
2 2
l
P
P
R = -
æ
è
ç
ö
ø
÷
Þ  
l
l
P
H
H
P
R
R
= = 2
Þ l l
p H
= = ´ 2 2 6563 Å = 13126 Å
     = 1.31 m m
1 1
2
1
3
2 2
2
l
He
= -
æ
è
ç
ö
ø
÷
× R z
H
Þ   
1 1
2
l l
He H
= × z  Þ    l
l
He
H
Å
= =
2
6563
2
2 2
Þ    l
He
nm = 164
2.
1 1
2
1
2 2
l
= -
é
ë
ê
ù
û
ú
R
n
 for larg est wave length n = 3
Þ   
1 1
4
1
9 l
= -
é
ë
ê
ù
û
ú
R
Þ l = =
´ ´
36
5
36
5 10
7
R 1.097
Þ  l = 656 nm
3. For H-atom r
n h
me
n
=
e
p
0
2 2
2
u
e
nh
n
=
e
2
0
2
T
r
u
n h nh
me e
n
n
n
= =
´
´
2
2
2
2 2
0
2 2
p
p
e e
p
0
     =
4
0
3 3
4
e n h
me
r
T
me
n h
n
n
= =
1
4
4
0
2 3 3
e
Þ r
me
h
1
4
0
2 3
4
=
e
Þ n
1
31 19 4
12 2 3
10 1 6 10
4 10 10
=
´ ´ ´
´ ´ ´ ´
- -
- -
9.1
8.85 6.6
( . )
( ) (
4 3
)
Þ n
1
15
1 0 = ´ 6.58 Hz
n
n n
2
1
3
1
15
2 8
10
8
= = =
´ 6.58
     = ´ 0.823 10
15
 Hz
(b)   
1 1
1
1
2
2 2
l
= -
é
ë
ê
ù
û
ú
R
Þ  n
l
= = ´ ´
é
ë
ê
ù
û
ú
c
R 3 10
3
4
8
Þ n =
´ ´ ´ 9 10 10
4
8 7
1.097
        = ´ 2.46 Hz 10
15
(c) Number of revolutions
= ´ = ´ ´ ´
-
v T
2
15 8
10 1 10 0.823
= ´ 8.23 10
6
 revolution
4.Reduce mass 
=
+
=
´
+
=
mm
mm
p
p
m
m
2071836
2071836
186
mm
mm ()
m
r
h
e
h
me
10
2
22
0
2
22
4
4
4
4
=
æ
è
ç
ç
ö
ø
÷
÷
=
æ
è
ç
ç
ö
ø
÷
÷
pe
pm
pe
p()186
Putting the value we get
r
1
13
10 = ´
-
2.55 m
E
e
h
1
4
0
2 2
8
2810 =
-
= -
m
e
eV
Ionization energy = - = E
1
2.81 keV
5. (a) l = =
´
´ ´
= ´
-
-
-
h
mv
6.6
4.8 m
10
46 10 30
10
34
3
34
(b) l =
´
´ ´
= ´
-
-
-
6.6
9.1
7.3  10 m
11
10
10 10
3 4
31 7
6. (a) After absorbing 12.3 eV the atom excited
to n = 3 state
1
1
1
2
l
L
R
n
= -
é
ë
ê
ù
û
ú
1
1
1
9
8
9
1
l
L
R
R
= -
é
ë
ê
ù
û
ú
=
Þ l
L
R
1
9
8
9
8 10
102
7
= =
´ ´
=
1.097
nm
1
1
1
4
2
l
L
R = -
é
ë
ê
ù
û
ú
Þ l
L
R
2
4
3
4
3 10
7
= =
´ ´
=
1.097
122 nm
1 1
2
1
3
5
36
2 2
l
B
R
R
= -
é
ë
ê
ù
û
ú
=
Þ l
B
R
=
36
5
Þ l
B
=
´ ´
=
36
5 10
7
1.097
653 nm
7. n n n
b a a
K K L
= + Þ l
l l
l l
a
a b
a b
L
K K
K K
=
´
-
c c c
K K L
l l l
b a a
= + Þ l
a
L
=
´
-
0.71 0.63
0.71 0.63
Þ 
1 1 1
l l l
a b a
L K K
= - l
a
L
= 5.59 nm
8. l =
hc
E D
Þ             l
a
K
E
=
´ ´ ´
- ´ ´
-
-
6.6
1.6
10 3 10
2870 10
34 8
1
19
( )
Þ 0.71
6.6
1.6
´ =
´ ´ ´
- ´ ´
-
-
-
10
10 3 10
2870 10
9
34 8
1
19
( ) E
Solving this we get 
     E
1
4613 = - eV
         l
K
B
E
=
´ ´ ´
- ´ ´
=
-
-
6.6
1.6
0.63
10 3 10
4613 10
34 8
3
19
( )
Solving this we get E
3
2650 = - eV
9. n
31
4
= =
E
h
f …(i)
n
21
3 3
4
4 3
4
= =
æ
è
ç
ö
ø
÷ =
E
h
E
h
f
n
32
4
= =
E
h
f
60
E
3
K
a
K
b
E = – 2870 eV
2
E
1
5 E
E
4 E
1
2
3
n = 3
n = 2
n = 1
Introductory Exercise 29.2
1. eV
hc
W
0
= -
l
Þ eV
0
34 8
7 19
10 3 10
2 10 10
=
´ ´ ´
´ ´ ´
-
-
- -
6.6
1.6
4.3eV
Þ eV
0
= - = 6.2eV 4.3eV 1.9eV
Þ V
0
=1.9 volt
2.P=1.5mW = ´
-
1.5 W 10
3
Energy of each photon 
=
´ ´ ´
´
-
-
6.62 10 3 10
4 10
34 8
7
= ´
-
4.96 10
19
 J
Number of photons incident per second
= =
´
´
-
-
P
Energy of each photon
1.5
4.96
10
10
3
19
             
~
- ´ 3 10
15
The number of photoelectrons produce 
= ´ ´ 0.1% 3 10
15
 = ´ 3 10
12
Current i = ´ ´ ´
-
3 10 10
12 19
1.6 A
= ´
-
4.8 A 10
7
 = 0.48 A m
3. K hf W hf hf
max
= - = -
0
Þ K f f
max
( ) µ -
0
4. K
hc
W
max
= -
l
=
´ ´ ´
´ ´ ´
-
é
ë
ê
ù
û
ú
-
- -
6.62
1.6
eV
10 3 10
2 10 10
3
34 8
7 19
= - = [ ] 6.20 eV 3.20 eV 3
The minimum kinetic energy = 0.
5.                  K h f f
max
[ ] = -
0
 
1.2 eV = - h f f [ ]
0
 …(i)
4.2 eV 1.5 = - h f f [ ]
0
 …(ii)
Dividing Eq. (i) and Eq. (ii)
       
124
42
0
0
=
-
-
f f
f f 1.5
Þ 3 2 7 7
0 0
f f f f - = -
Þ         5 4
0
f f =
 Þ  f f f
0
4
5
= = 0.8
Þ 1.2 1.6 6.62
0.8
´ ´ = ´ -
é
ë
ê
ù
û
ú
- -
10 10
19 34 0
0
f
f
Þ 
1.2 1.6
6.62
´ ´
´
= =
-
-
10
10
2
8 4
19
34
0 0
f f
Þ f
0
15
10 = ´ 1.16 Hz
Subjective Questions (Level I)
1. Here l = = ´
-
280 28 10
8
nm m
E
hc
= =
´ ´ ´
´
-
-
l
6.6
J
10 3 10
28 10
34 8
8
E =
´ ´
=
´
- -
19.8
J J
10 10
28
198 10
28
34 16 19
E =
´
´ ´
=
´
-
-
-
198 10
28 10
198
28
19
19
1.6
eV
1.6
4.6 eV
~
We have E mc =
2
 Þ m
E
c
= =
´
´ ´
-
2
19
16
198 10
28 9 10
Þ m = ´
-
8.2 kg 10
36
Momentum p mc = = ´ ´ ´
-
8.2 10 3 10
36 8
      = ´
-
2.46 kg-m/s 10
27
2. Intensity of light at a dis tance 2 m
From the source =
´
=
1
4 2
1
16
2
2
p p ( )
/ W m
 61
Let plate area is A
Energy incident on unit time is
E A
1
1
16
=
p
w
Energy of each photon 
=
´ ´ ´
´
-
-
6.6
4.8
10 3 10
10
34 8
7
Number of photons striking per unit area
n
A
A A
=
´ ´ ´
´ ´ ´ ´ ´
-
-
1
16
10
10 3 10
7
34 8
p
4.8
6.6
= ´ 4.82 10
16
 per m
2
 s
3. Here p = ´
-
8.24 kg-m/s 10
28
(a) Energy of photon E pc =
Þ E = ´ ´ ´
-
8.24 10 3 10
28 8
Þ E = ´
-
2.47 J 10
19
Energy in eV
in joule
1.6
=
´
-
E
10
19
          =
´
´
-
-
2.47
1.6
10
10
19
19
Energy (in eV) = 1.54 eV
(b) Wavelength 
l = =
´
´
=
-
-
h
p
6.6
8.24
nm
10
10
804
34
28
This wavelength in Infrared region.
4. We have c f = l Þ f
c
= =
´
´
-
l
3 10
6 10
8
7
m/s
m
Þ f = ´ 5 10
14
Hz
We have p
E
t
= Þ E pt = power per sec =
energy
Þ P E =
Þ 75 = ´ ´ ( ) h v n
Þ =
´ ´ ´
-
75
10 5 10
34 14
6.6
Þ n = ´ 2.3 10
20
 photons/sec
5. (a) E = = ´ ´ ´
-
2.45 MeV 2.45 1.6 10 J
19 6
10
E h = n Þ n = =
´
´
-
-
E
h
3.92
6.6
10
10
13
34
Þ n = ´ 5.92 Hz 10
20
(b) We have c = nl Þ l
n
=
c
Þ l =
´
´
= ´
-
3 10
10
8
20
5.92
5.06  10 m
13
6. We have    p mK = 2
Þ p mK
1 1
2 = and p mK
2 2
2 =
Þ 
p
p
K
K
1
2
1
2
=
Þ   
1
2
1
2
=
K
K
 [ ] Q p p
2 1
2 =
Þ K K
2 1
4 =
(b) E p c
1 1
= and E p c
2 2
= Þ E p c
2 1
2 =
Þ E E
2 1
2 =
7. (a) Since power = energy per unit line
              let n be the number of photons
Þ P nE = Þ 10 = ´ n
nc
l
Þ n
hc
=
´
=
´ ´
´ ´ ´
-
-
10 10 500 10
10 3 10
9
34 8
l
6.6
Þ n = ´ 2.52 10
19
(b) Force exerted on that surface
F
P
c
= =
´
= ´
-
10
3 10
8
3.33  10 N
8
8. Absorbing (power) light =70% of in ci dent
light
Þ P
a
=70% of 10 7 W W =
Refractive power = 30% of 10 W
Þ P
R
= 3 W
The force exerted = +
P
c
P
c
a R
2
=
+ ´
=
´
7 2 3 13
3 10
8
c
= ´
-
4.3 N 10
8
62
Page 5


29. Modern Physics I
Introductory Exercise 29.1
1. The po si tron has same mass m as the
elec tron. The reduced mass of elec tron
po si tron atom is
m =
´
+
=
m m
m m
1
2
m
R
m e
c h
H
=
4
0
2 3
8 e
Þ   R
R
P
H
=
2
1 1
2
1
3
2 2
l
H
H
R = -
æ
è
ç
ö
ø
÷
1 1
2
1
3
2 2
l
P
P
R = -
æ
è
ç
ö
ø
÷
Þ  
l
l
P
H
H
P
R
R
= = 2
Þ l l
p H
= = ´ 2 2 6563 Å = 13126 Å
     = 1.31 m m
1 1
2
1
3
2 2
2
l
He
= -
æ
è
ç
ö
ø
÷
× R z
H
Þ   
1 1
2
l l
He H
= × z  Þ    l
l
He
H
Å
= =
2
6563
2
2 2
Þ    l
He
nm = 164
2.
1 1
2
1
2 2
l
= -
é
ë
ê
ù
û
ú
R
n
 for larg est wave length n = 3
Þ   
1 1
4
1
9 l
= -
é
ë
ê
ù
û
ú
R
Þ l = =
´ ´
36
5
36
5 10
7
R 1.097
Þ  l = 656 nm
3. For H-atom r
n h
me
n
=
e
p
0
2 2
2
u
e
nh
n
=
e
2
0
2
T
r
u
n h nh
me e
n
n
n
= =
´
´
2
2
2
2 2
0
2 2
p
p
e e
p
0
     =
4
0
3 3
4
e n h
me
r
T
me
n h
n
n
= =
1
4
4
0
2 3 3
e
Þ r
me
h
1
4
0
2 3
4
=
e
Þ n
1
31 19 4
12 2 3
10 1 6 10
4 10 10
=
´ ´ ´
´ ´ ´ ´
- -
- -
9.1
8.85 6.6
( . )
( ) (
4 3
)
Þ n
1
15
1 0 = ´ 6.58 Hz
n
n n
2
1
3
1
15
2 8
10
8
= = =
´ 6.58
     = ´ 0.823 10
15
 Hz
(b)   
1 1
1
1
2
2 2
l
= -
é
ë
ê
ù
û
ú
R
Þ  n
l
= = ´ ´
é
ë
ê
ù
û
ú
c
R 3 10
3
4
8
Þ n =
´ ´ ´ 9 10 10
4
8 7
1.097
        = ´ 2.46 Hz 10
15
(c) Number of revolutions
= ´ = ´ ´ ´
-
v T
2
15 8
10 1 10 0.823
= ´ 8.23 10
6
 revolution
4.Reduce mass 
=
+
=
´
+
=
mm
mm
p
p
m
m
2071836
2071836
186
mm
mm ()
m
r
h
e
h
me
10
2
22
0
2
22
4
4
4
4
=
æ
è
ç
ç
ö
ø
÷
÷
=
æ
è
ç
ç
ö
ø
÷
÷
pe
pm
pe
p()186
Putting the value we get
r
1
13
10 = ´
-
2.55 m
E
e
h
1
4
0
2 2
8
2810 =
-
= -
m
e
eV
Ionization energy = - = E
1
2.81 keV
5. (a) l = =
´
´ ´
= ´
-
-
-
h
mv
6.6
4.8 m
10
46 10 30
10
34
3
34
(b) l =
´
´ ´
= ´
-
-
-
6.6
9.1
7.3  10 m
11
10
10 10
3 4
31 7
6. (a) After absorbing 12.3 eV the atom excited
to n = 3 state
1
1
1
2
l
L
R
n
= -
é
ë
ê
ù
û
ú
1
1
1
9
8
9
1
l
L
R
R
= -
é
ë
ê
ù
û
ú
=
Þ l
L
R
1
9
8
9
8 10
102
7
= =
´ ´
=
1.097
nm
1
1
1
4
2
l
L
R = -
é
ë
ê
ù
û
ú
Þ l
L
R
2
4
3
4
3 10
7
= =
´ ´
=
1.097
122 nm
1 1
2
1
3
5
36
2 2
l
B
R
R
= -
é
ë
ê
ù
û
ú
=
Þ l
B
R
=
36
5
Þ l
B
=
´ ´
=
36
5 10
7
1.097
653 nm
7. n n n
b a a
K K L
= + Þ l
l l
l l
a
a b
a b
L
K K
K K
=
´
-
c c c
K K L
l l l
b a a
= + Þ l
a
L
=
´
-
0.71 0.63
0.71 0.63
Þ 
1 1 1
l l l
a b a
L K K
= - l
a
L
= 5.59 nm
8. l =
hc
E D
Þ             l
a
K
E
=
´ ´ ´
- ´ ´
-
-
6.6
1.6
10 3 10
2870 10
34 8
1
19
( )
Þ 0.71
6.6
1.6
´ =
´ ´ ´
- ´ ´
-
-
-
10
10 3 10
2870 10
9
34 8
1
19
( ) E
Solving this we get 
     E
1
4613 = - eV
         l
K
B
E
=
´ ´ ´
- ´ ´
=
-
-
6.6
1.6
0.63
10 3 10
4613 10
34 8
3
19
( )
Solving this we get E
3
2650 = - eV
9. n
31
4
= =
E
h
f …(i)
n
21
3 3
4
4 3
4
= =
æ
è
ç
ö
ø
÷ =
E
h
E
h
f
n
32
4
= =
E
h
f
60
E
3
K
a
K
b
E = – 2870 eV
2
E
1
5 E
E
4 E
1
2
3
n = 3
n = 2
n = 1
Introductory Exercise 29.2
1. eV
hc
W
0
= -
l
Þ eV
0
34 8
7 19
10 3 10
2 10 10
=
´ ´ ´
´ ´ ´
-
-
- -
6.6
1.6
4.3eV
Þ eV
0
= - = 6.2eV 4.3eV 1.9eV
Þ V
0
=1.9 volt
2.P=1.5mW = ´
-
1.5 W 10
3
Energy of each photon 
=
´ ´ ´
´
-
-
6.62 10 3 10
4 10
34 8
7
= ´
-
4.96 10
19
 J
Number of photons incident per second
= =
´
´
-
-
P
Energy of each photon
1.5
4.96
10
10
3
19
             
~
- ´ 3 10
15
The number of photoelectrons produce 
= ´ ´ 0.1% 3 10
15
 = ´ 3 10
12
Current i = ´ ´ ´
-
3 10 10
12 19
1.6 A
= ´
-
4.8 A 10
7
 = 0.48 A m
3. K hf W hf hf
max
= - = -
0
Þ K f f
max
( ) µ -
0
4. K
hc
W
max
= -
l
=
´ ´ ´
´ ´ ´
-
é
ë
ê
ù
û
ú
-
- -
6.62
1.6
eV
10 3 10
2 10 10
3
34 8
7 19
= - = [ ] 6.20 eV 3.20 eV 3
The minimum kinetic energy = 0.
5.                  K h f f
max
[ ] = -
0
 
1.2 eV = - h f f [ ]
0
 …(i)
4.2 eV 1.5 = - h f f [ ]
0
 …(ii)
Dividing Eq. (i) and Eq. (ii)
       
124
42
0
0
=
-
-
f f
f f 1.5
Þ 3 2 7 7
0 0
f f f f - = -
Þ         5 4
0
f f =
 Þ  f f f
0
4
5
= = 0.8
Þ 1.2 1.6 6.62
0.8
´ ´ = ´ -
é
ë
ê
ù
û
ú
- -
10 10
19 34 0
0
f
f
Þ 
1.2 1.6
6.62
´ ´
´
= =
-
-
10
10
2
8 4
19
34
0 0
f f
Þ f
0
15
10 = ´ 1.16 Hz
Subjective Questions (Level I)
1. Here l = = ´
-
280 28 10
8
nm m
E
hc
= =
´ ´ ´
´
-
-
l
6.6
J
10 3 10
28 10
34 8
8
E =
´ ´
=
´
- -
19.8
J J
10 10
28
198 10
28
34 16 19
E =
´
´ ´
=
´
-
-
-
198 10
28 10
198
28
19
19
1.6
eV
1.6
4.6 eV
~
We have E mc =
2
 Þ m
E
c
= =
´
´ ´
-
2
19
16
198 10
28 9 10
Þ m = ´
-
8.2 kg 10
36
Momentum p mc = = ´ ´ ´
-
8.2 10 3 10
36 8
      = ´
-
2.46 kg-m/s 10
27
2. Intensity of light at a dis tance 2 m
From the source =
´
=
1
4 2
1
16
2
2
p p ( )
/ W m
 61
Let plate area is A
Energy incident on unit time is
E A
1
1
16
=
p
w
Energy of each photon 
=
´ ´ ´
´
-
-
6.6
4.8
10 3 10
10
34 8
7
Number of photons striking per unit area
n
A
A A
=
´ ´ ´
´ ´ ´ ´ ´
-
-
1
16
10
10 3 10
7
34 8
p
4.8
6.6
= ´ 4.82 10
16
 per m
2
 s
3. Here p = ´
-
8.24 kg-m/s 10
28
(a) Energy of photon E pc =
Þ E = ´ ´ ´
-
8.24 10 3 10
28 8
Þ E = ´
-
2.47 J 10
19
Energy in eV
in joule
1.6
=
´
-
E
10
19
          =
´
´
-
-
2.47
1.6
10
10
19
19
Energy (in eV) = 1.54 eV
(b) Wavelength 
l = =
´
´
=
-
-
h
p
6.6
8.24
nm
10
10
804
34
28
This wavelength in Infrared region.
4. We have c f = l Þ f
c
= =
´
´
-
l
3 10
6 10
8
7
m/s
m
Þ f = ´ 5 10
14
Hz
We have p
E
t
= Þ E pt = power per sec =
energy
Þ P E =
Þ 75 = ´ ´ ( ) h v n
Þ =
´ ´ ´
-
75
10 5 10
34 14
6.6
Þ n = ´ 2.3 10
20
 photons/sec
5. (a) E = = ´ ´ ´
-
2.45 MeV 2.45 1.6 10 J
19 6
10
E h = n Þ n = =
´
´
-
-
E
h
3.92
6.6
10
10
13
34
Þ n = ´ 5.92 Hz 10
20
(b) We have c = nl Þ l
n
=
c
Þ l =
´
´
= ´
-
3 10
10
8
20
5.92
5.06  10 m
13
6. We have    p mK = 2
Þ p mK
1 1
2 = and p mK
2 2
2 =
Þ 
p
p
K
K
1
2
1
2
=
Þ   
1
2
1
2
=
K
K
 [ ] Q p p
2 1
2 =
Þ K K
2 1
4 =
(b) E p c
1 1
= and E p c
2 2
= Þ E p c
2 1
2 =
Þ E E
2 1
2 =
7. (a) Since power = energy per unit line
              let n be the number of photons
Þ P nE = Þ 10 = ´ n
nc
l
Þ n
hc
=
´
=
´ ´
´ ´ ´
-
-
10 10 500 10
10 3 10
9
34 8
l
6.6
Þ n = ´ 2.52 10
19
(b) Force exerted on that surface
F
P
c
= =
´
= ´
-
10
3 10
8
3.33  10 N
8
8. Absorbing (power) light =70% of in ci dent
light
Þ P
a
=70% of 10 7 W W =
Refractive power = 30% of 10 W
Þ P
R
= 3 W
The force exerted = +
P
c
P
c
a R
2
=
+ ´
=
´
7 2 3 13
3 10
8
c
= ´
-
4.3 N 10
8
62
9. Force = rate of change of mo men tum
F
p
t
=
D
D
 here Dt = 1 s
F p = ° 2 60 cos
F p =
Þ F
nh
=
l
 
where n is number of photons striking per
second
Þ F =
´ ´ ´
´
=
-
-
-
1 10 10
663 10
10
19 34
9
8
6.63
N
10. Here out put en ergy = 60 W/s
Pressure p =
´
´
= ´
-
2 60
3 10
8
4  10 N
7
de-Broglie wavelength
11. Here m = Þ = ´
-
5 5 10
3
g m kg,  v = 340 m/s
by de-Broglie hypothesis wavelength
l = =
´
´ ´
-
-
h
mv
6.62 10
5 10 340
34
3
Þ l = ´
-
3.9  10 m
34
Since l is too small. No wave like property
is exhibit.
12.(a) l
e
e
h
m v
= =
´
´ ´ ´
-
-
6.6
9.1 4.7
10
10 10
34
31 6
= ´
-
1.55 m 10
10
(b) l
p
=
´
´ ´ ´ ´
-
- -
6.6
9.1 4.7
10
1836 10 10
34
31 6
= ´
-
8.44 m 10
14
13. (a) l =
h
p
 Þ p
h
= =
´
´
-
-
l
6.6
2.8
10
10
34
10
Þ p = ´
-
2.37 kg - m /s 10
24
(b) Q p m K
e
2
2 = Þ K
p
m
e
=
2
2
Þ K =
´
´ ´
-
-
( ) 2.37
9.1
10
2 10
24 2
31
Þ K = ´
-
3.07 J 10
18
K
K
( ) in eV
in J
1.6
3.07
1.6
=
´
=
´
´
-
-
-
10
10
10
19
18
19
Þ K =19.2 eV
14. Here T = ° + ° = 273 20 293 K
v
RT
M
rms
=
3
 Þ l =
h
Mv
rms
Þ l =
h
MRT 3
=
´
´ ´ ´ ´ ´
-
-
6.6
9.1 8.31
10
3 1836 10 293
34
31
Þ l =1.04 Å
15. For hy dro gen like atom
E K = - Here E = - 3.4 eV
Þ K = = ´ ´
-
3.4 eV 3.4 1.6 J 10
19
l = =
h
p
h
m K
e
2
=
´
´ ´ ´ ´ ´
-
- -
6.6
9.1 3.4 1.6
10
2 10 10
34
31 19
Þ l = 6.663 Å
16. In Bohr model the ve loc ity of elec tron in nth
or bit is given by
U
e
nh
n
=
2
0
2e
Putting the values of e h , , e
0
 and n = 1, we
get
U
1
7
10 = ´ 2.19 m/s and U
4
6
10
=
´ 2.19
4
m/s
 63
60
p cos 60
p cos 60°
p sin 60°
p sin 60
p
Read More
Offer running on EduRev: Apply code STAYHOME200 to get INR 200 off on our premium plan EduRev Infinity!

Complete Syllabus of NEET

Dynamic Test

Content Category

Related Searches

Summary

,

MCQs

,

practice quizzes

,

Solution by D C Pandey

,

Chapter 29 - Modern Physics I (Part - 1) - Physics

,

NEET NEET Notes | EduRev

,

study material

,

Objective type Questions

,

Viva Questions

,

pdf

,

Chapter 29 - Modern Physics I (Part - 1) - Physics

,

NEET NEET Notes | EduRev

,

Extra Questions

,

Previous Year Questions with Solutions

,

past year papers

,

Solution by D C Pandey

,

mock tests for examination

,

Sample Paper

,

Semester Notes

,

Solution by D C Pandey

,

Free

,

ppt

,

Important questions

,

Chapter 29 - Modern Physics I (Part - 1) - Physics

,

shortcuts and tricks

,

NEET NEET Notes | EduRev

,

video lectures

,

Exam

;