JEE Exam > JEE Notes > DC Pandey Solutions for JEE Physics > DC Pandey Solutions: Modern Physics I- 1
DC Pandey Solutions: Modern Physics I- 1 | DC Pandey Solutions for JEE Physics PDF Download
| Download, print and study this document offline |
Please wait while the PDF view is loading
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
FAQs on DC Pandey Solutions: Modern Physics I- 1 - DC Pandey Solutions for JEE 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.
About this Document
|
4.94/5 Rating |
|
Dec 23, 2024 Last updated |
Document Description: DC Pandey Solutions: Modern Physics I- 1 for JEE 2024 is part of DC Pandey Solutions for JEE Physics preparation.
The notes and questions for DC Pandey Solutions: Modern Physics I- 1 have been prepared according to the JEE exam syllabus. Information about DC Pandey Solutions: Modern Physics I- 1 covers topics
like and DC Pandey Solutions: Modern Physics I- 1 Example, for JEE 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises and tests below for DC Pandey Solutions: Modern Physics I- 1.
Introduction of DC Pandey Solutions: Modern Physics I- 1 in English is available as part of our DC Pandey Solutions for JEE Physics
for JEE & DC Pandey Solutions: Modern Physics I- 1 in Hindi for DC Pandey Solutions for JEE Physics course.
Download more important topics related with notes, lectures and mock test series for JEE
Exam by signing up for free. JEE: DC Pandey Solutions: Modern Physics I- 1 | DC Pandey Solutions for JEE Physics
Description
Full syllabus notes, lecture & questions for DC Pandey Solutions: Modern Physics I- 1 | DC Pandey Solutions for JEE Physics - JEE | Plus excerises question with solution to help you revise complete syllabus for DC Pandey Solutions for JEE Physics | Best notes, free PDF download
Information about DC Pandey Solutions: Modern Physics I- 1
In this doc you can find the meaning of DC Pandey Solutions: Modern Physics I- 1 defined & explained in the simplest way possible. Besides explaining types of
DC Pandey Solutions: Modern Physics I- 1 theory, EduRev gives you an ample number of questions to practice DC Pandey Solutions: Modern Physics I- 1 tests, examples and also practice JEE
tests
|
Explore Courses for JEE exam
|
|
Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
Related Searches
MCQs
,practice quizzes
,mock tests for examination
,Important questions
,study material
,Exam
,Summary
,Previous Year Questions with Solutions
,DC Pandey Solutions: Modern Physics I- 1 | DC Pandey Solutions for JEE Physics
,DC Pandey Solutions: Modern Physics I- 1 | DC Pandey Solutions for JEE Physics
,past year papers
,DC Pandey Solutions: Modern Physics I- 1 | DC Pandey Solutions for JEE Physics
,Free
,Extra Questions
,ppt
,video lectures
,Objective type Questions
,shortcuts and tricks
,Viva Questions
,Sample Paper
,Semester Notes
;
Additional Information about DC Pandey Solutions: Modern Physics I- 1 for JEE Preparation
DC Pandey Solutions: Modern Physics I- 1 Free PDF Download
The DC Pandey Solutions: Modern Physics I- 1 is an invaluable resource that delves deep into the core of the JEE exam.
These study notes are curated by experts and cover all the essential topics and concepts, making your preparation more efficient and effective.
With the help of these notes, you can grasp complex subjects quickly, revise important points easily,
and reinforce your understanding of key concepts. The study notes are presented in a concise and easy-to-understand manner,
allowing you to optimize your learning process. Whether you're looking for best-recommended books, sample papers, study material,
or toppers' notes, this PDF has got you covered. Download the DC Pandey Solutions: Modern Physics I- 1 now and kickstart your journey towards success in the JEE exam.
Importance of DC Pandey Solutions: Modern Physics I- 1
The importance of DC Pandey Solutions: Modern Physics I- 1 cannot be overstated, especially for JEE aspirants.
This document holds the key to success in the JEE exam.
It offers a detailed understanding of the concept, providing invaluable insights into the topic.
By knowing the concepts well in advance, students can plan their preparation effectively.
Utilize this indispensable guide for a well-rounded preparation and achieve your desired results.
DC Pandey Solutions: Modern Physics I- 1 Notes
DC Pandey Solutions: Modern Physics I- 1 Notes offer in-depth insights into the specific topic to help you master it with ease.
This comprehensive document covers all aspects related to DC Pandey Solutions: Modern Physics I- 1.
It includes detailed information about the exam syllabus, recommended books, and study materials for a well-rounded preparation.
Practice papers and question papers enable you to assess your progress effectively.
Additionally, the paper analysis provides valuable tips for tackling the exam strategically.
Access to Toppers' notes gives you an edge in understanding complex concepts.
Whether you're a beginner or aiming for advanced proficiency, DC Pandey Solutions: Modern Physics I- 1 Notes on EduRev are your ultimate resource for success.
DC Pandey Solutions: Modern Physics I- 1 JEE Questions
The "DC Pandey Solutions: Modern Physics I- 1 JEE Questions" guide is a valuable resource for all aspiring students preparing for the
JEE exam. It focuses on providing a wide range of practice questions to help students gauge
their understanding of the exam topics. These questions cover the entire syllabus, ensuring comprehensive preparation.
The guide includes previous years' question papers for students to familiarize themselves with the exam's format and difficulty level.
Additionally, it offers subject-specific question banks, allowing students to focus on weak areas and improve their performance.
Study DC Pandey Solutions: Modern Physics I- 1 on the App
Students of JEE can study DC Pandey Solutions: Modern Physics I- 1 alongwith tests & analysis from the EduRev app,
which will help them while preparing for their exam. Apart from the DC Pandey Solutions: Modern Physics I- 1,
students can also utilize the EduRev App for other study materials such as previous year question papers, syllabus, important questions, etc.
The EduRev App will make your learning easier as you can access it from anywhere you want.
The content of DC Pandey Solutions: Modern Physics I- 1 is prepared as per the latest JEE syllabus.
|
© EduRev
|
Education Revolution
|
|
Signup to see your scores
go up within 7 days!
Access 1000+ FREE Docs, Videos and Tests
Takes less than 10 seconds to signup