DC Pandey Solutions: Modern Physics I- 1

# DC Pandey Solutions: Modern Physics I- 1 | DC Pandey Solutions for NEET Physics PDF Download

``` 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
```

122 docs

## FAQs on DC Pandey Solutions: Modern Physics I- 1 - DC Pandey Solutions for NEET Physics

 1. What is the importance of studying modern physics?
Ans. Studying modern physics is important as it helps us understand the fundamental laws and principles that govern the behavior of matter and energy at the atomic and subatomic levels. It allows us to comprehend the theories and concepts behind technologies such as lasers, semiconductors, nuclear power, and quantum computing.
 2. How can I prepare for the DC Pandey Modern Physics I-1 exam?
Ans. To prepare for the DC Pandey Modern Physics I-1 exam, it is recommended to thoroughly understand the concepts and theories covered in the book. Regularly practice the numerical problems and solved examples given in the book to enhance problem-solving skills. Additionally, taking mock tests and solving previous years' question papers can help in familiarizing oneself with the exam pattern and time management.
 3. What are the major topics covered in DC Pandey Modern Physics I-1?
Ans. The major topics covered in DC Pandey Modern Physics I-1 include the photoelectric effect, the dual nature of radiation and matter, the Bohr's model of the atom, radioactive decay, nuclear physics, and elementary particles. These topics delve into the behavior of particles at the microscopic level and their interactions.
 4. Are there any recommended reference materials for studying modern physics?
Ans. Apart from DC Pandey Modern Physics I-1, some recommended reference materials for studying modern physics include "Concepts of Modern Physics" by Arthur Beiser, "Modern Physics for Scientists and Engineers" by John R. Taylor, and "Introduction to Quantum Mechanics" by David J. Griffiths. These books provide in-depth explanations and additional practice problems to supplement the understanding of modern physics.
 5. How can I improve my understanding of quantum mechanics and its applications?
Ans. Improving understanding of quantum mechanics and its applications can be done by reading books specifically dedicated to the subject, such as "Quantum Mechanics: Concepts and Applications" by Nouredine Zettili or "Principles of Quantum Mechanics" by R. Shankar. Additionally, online resources like video lectures and tutorials can provide visual explanations and demonstrations of quantum mechanical concepts. Regularly practicing problems related to quantum mechanics can also help solidify the understanding of its applications.

## DC Pandey Solutions for NEET Physics

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