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# Chapter 30 - Modern Physics II (Part - 1)- Physics, Solution by D C Pandey, NEET NEET Notes | EduRev

## DC Pandey (Questions & Solutions) of Physics: NEET

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## NEET : Chapter 30 - Modern Physics II (Part - 1)- Physics, Solution by D C Pandey, NEET NEET Notes | EduRev

``` Page 1

In tr o duc tor y Ex er cis e 30.2
1. D D E Mc =
2
Here, P = = ´ ´ 1 0 1 0 2 4 6 0
9 9
J / s J / d a y
Þ D M =
´ ´
´
= ´
-
1 0 2 4 6 0
3 1 0
1 0
9
8 2
4
( )
9 . 6 k g
2. Number of fis sion =
´ ´ ´
-
10
200 10 10
9
6 19
J/s
1.6
= ´ 3.125 10
19
3. Giv en re ac tion is
92
238
90
234
2
4
U T h H e ¾® +
D M = - ( 2 3 8 . 0 5 0 7 8 4 2 3 4 . 0 4 3 5 9 3
- 4 . 0 0 2 6 0 2 ) u
D M = 0 . 0 0 0 4 5 8 9 u
D D E M = ´ 9 3 1 . 5 M eV
= ´ 0.000 931.5 4589
= 4.27 MeV
4. Com plete reactions are
(a)
3
6
1
2
4
7
0
1
Li H Be + ® + n
(b)
17
35
1
1
16
32
2
4
Cl H S He + ® +
(c)
4
9
2
4
2 0
1
3 Be He He
4
+ ® + ( ) n
(d)
35
79
1
2
36
79
0
1
2 Br H Kr + ® + ( ) n
AIEEE Corner
¢ Subjective Questions (Level 1)
1. (a) Initially the rate of distingration is
-
æ
è
ç
ö
ø
÷ =
dN
dt
N
0
0
l
After 5 min
-
=
dN
dt
N l
Þ
N
N
dN
dt
dN
dt
0 0
4750
2700
=
æ
è
ç
ö
ø
÷
= =1.76
Now N N e
t
=
-
0
l
or l = -
æ
è
ç
ç
ö
ø
÷
÷
log
N
N
t
0
Þ l =
2.3026
10
0
t
N
N
log
=
2.3026
1.76
5
10
log ( )
Þ l = 0.113 /min
(b) Half-life = = =
0.693 0.693
0.113
6.132
l
min
2. We have A N = l
6 10
11
´ = ´ ´ l 1 10
15
Þ           l = ´
- -
6 10
4 1
s
T
1 2 4
6 10
1155
/
= =
´
=
-
0.693 0.693
l
s
= 19.25 min
3. A N = l
A = = ´ ´ 8 10
10
Ci 8 3.7 decay/s
l = =
0.693 0.693
5.3 yr T
1 2 /
=
´ ´ ´ ´
0.693
5.3 s 365 24 60 60
Þ N
A
=
l
=
´ ´ ´ ´ ´ ´ 8 10 365 24 3600
10
3.7 5.3
0.693
Þ N = ´ 7.2 10
19
83
Page 2

In tr o duc tor y Ex er cis e 30.2
1. D D E Mc =
2
Here, P = = ´ ´ 1 0 1 0 2 4 6 0
9 9
J / s J / d a y
Þ D M =
´ ´
´
= ´
-
1 0 2 4 6 0
3 1 0
1 0
9
8 2
4
( )
9 . 6 k g
2. Number of fis sion =
´ ´ ´
-
10
200 10 10
9
6 19
J/s
1.6
= ´ 3.125 10
19
3. Giv en re ac tion is
92
238
90
234
2
4
U T h H e ¾® +
D M = - ( 2 3 8 . 0 5 0 7 8 4 2 3 4 . 0 4 3 5 9 3
- 4 . 0 0 2 6 0 2 ) u
D M = 0 . 0 0 0 4 5 8 9 u
D D E M = ´ 9 3 1 . 5 M eV
= ´ 0.000 931.5 4589
= 4.27 MeV
4. Com plete reactions are
(a)
3
6
1
2
4
7
0
1
Li H Be + ® + n
(b)
17
35
1
1
16
32
2
4
Cl H S He + ® +
(c)
4
9
2
4
2 0
1
3 Be He He
4
+ ® + ( ) n
(d)
35
79
1
2
36
79
0
1
2 Br H Kr + ® + ( ) n
AIEEE Corner
¢ Subjective Questions (Level 1)
1. (a) Initially the rate of distingration is
-
æ
è
ç
ö
ø
÷ =
dN
dt
N
0
0
l
After 5 min
-
=
dN
dt
N l
Þ
N
N
dN
dt
dN
dt
0 0
4750
2700
=
æ
è
ç
ö
ø
÷
= =1.76
Now N N e
t
=
-
0
l
or l = -
æ
è
ç
ç
ö
ø
÷
÷
log
N
N
t
0
Þ l =
2.3026
10
0
t
N
N
log
=
2.3026
1.76
5
10
log ( )
Þ l = 0.113 /min
(b) Half-life = = =
0.693 0.693
0.113
6.132
l
min
2. We have A N = l
6 10
11
´ = ´ ´ l 1 10
15
Þ           l = ´
- -
6 10
4 1
s
T
1 2 4
6 10
1155
/
= =
´
=
-
0.693 0.693
l
s
= 19.25 min
3. A N = l
A = = ´ ´ 8 10
10
Ci 8 3.7 decay/s
l = =
0.693 0.693
5.3 yr T
1 2 /
=
´ ´ ´ ´
0.693
5.3 s 365 24 60 60
Þ N
A
=
l
=
´ ´ ´ ´ ´ ´ 8 10 365 24 3600
10
3.7 5.3
0.693
Þ N = ´ 7.2 10
19
83
6.023 ´ 10
23
nuclei = 60 g
1
60
10
23
=
´
g
6.023
Hence 7.2 ´ 10
19
nuclei
=
´ ´
´
60 10
10
19
23
7.2
6.023
= ´
-
7.11 10
3
g
4. Number of decay per second
m
M
N
A
´ ´ = ´ ´ l
1
238
6 10
23
´
´
0.693
4.5 yr 10
9
=
´ 6 10
238
23
´
´ ´ ´ ´ ´
0.693
4.5 s 10 365 24 60 60
9
= ´ 1.23 10
4
decay/s
5. Probability of decay
P e e
t t T
= - = -
- -
( ) ( )
/
1 1
l
mean
Þ P e e = - = - =
- -
( )
/
1 1
5 10 2
0.39
6. Since initially no Pb
206
nuclei is present and
after time t the ratio of
N
N
v
Pb
= 3
It means
1
4
of original U
238
nuclei decays.
Hence                        N N e
t
=
-
0
l
N
N
N e
0
0
0
4
-
æ
è
ç
ö
ø
÷ =
- l

3
4
=
-
e
t l
Þ       t =
- log log 4 3
l
Þ t =
-
´ ´
(log log ) 4 3
10
9
0.693
4.5 yr
Þ t = ´ 1.88 yr 10
9
7. ( ) R N
1 0 1 0
32
4
P
= l
( ) R N
2 0 2 0
33
P
= l
R R R
0 1 0 2 0
32 33
= + ( ) ( )
P P
= + = N
0 1 2
4 3 ( ) l l mCi
( ) R N N e
t
t
1 1 1 0
32
1
4
P
= =
-
l l
l
( )
log
R N e
t 1 1 0
2
14
60 365
32
4
P
= ´
- ´ ´
l
( )
log
R N e
t 2 2 0
2
25
6 0 3 6 5
33
P
= ´
- ´ ´
l
R R R t
t
= + ( ) ( )
1 2
32 33
P P
= +
é
ë
ê
ê
ù
û
ú
ú
- ´ ´ - ´ ´
N e e
0 1
2
14
60 365
2
2
25
60 365
4 l l
log log
=
+
3
4
1 2
mCi
( ) l l
4
1
2 60 365
14
2
2
25
60 365
l l e e
-
´ ´
- ´ ´
+
é
ë
ê
ê
ù
û
ú
ú
log log
=
´
+
é
ë
ê
ù
û
ú
3
4 2
14
2
25
mCi
log log
4 2
14
2
25
2 60 365
14
2
25
60 365
log log
log log
e e
-
´ ´
- ´ ´
+ ´
é
ë
ê
ê
ù
û
ú
ú
» 0.205mCi
8. Complete reactions are
(a)
88
226
86
222
Ra RN ¾¾® + a
(b)
8
19
9
19
O F ¾¾® + + e n
(c)
13
25
12
25
Al Mg ¾¾® + +
+
e n
9. Only reaction (b) is possible.
10. DE = ´ + ´ - ( ) 7 7 1.000783 1.00867 14.00307
´ 931.5 MeV
Þ DE = 104.72 MeV
11. DE m m m O
p n
= + - ´ [ ( )] 8 8
8
16
931.5
= ´ + ´ - ( . 8 825 8 1 008665 1.007 15.994915)
´ 931.5  = 127.6 MeV
12. (a) Number of nuclei in kg
=
´
´
6.023 10
235
1
23
Energy
=
´
´ ´
6.023 10
235
200 10
23
6
´ ´
-
1.6 10
19
= ´ 8.09 J 10
13
84
Page 3

In tr o duc tor y Ex er cis e 30.2
1. D D E Mc =
2
Here, P = = ´ ´ 1 0 1 0 2 4 6 0
9 9
J / s J / d a y
Þ D M =
´ ´
´
= ´
-
1 0 2 4 6 0
3 1 0
1 0
9
8 2
4
( )
9 . 6 k g
2. Number of fis sion =
´ ´ ´
-
10
200 10 10
9
6 19
J/s
1.6
= ´ 3.125 10
19
3. Giv en re ac tion is
92
238
90
234
2
4
U T h H e ¾® +
D M = - ( 2 3 8 . 0 5 0 7 8 4 2 3 4 . 0 4 3 5 9 3
- 4 . 0 0 2 6 0 2 ) u
D M = 0 . 0 0 0 4 5 8 9 u
D D E M = ´ 9 3 1 . 5 M eV
= ´ 0.000 931.5 4589
= 4.27 MeV
4. Com plete reactions are
(a)
3
6
1
2
4
7
0
1
Li H Be + ® + n
(b)
17
35
1
1
16
32
2
4
Cl H S He + ® +
(c)
4
9
2
4
2 0
1
3 Be He He
4
+ ® + ( ) n
(d)
35
79
1
2
36
79
0
1
2 Br H Kr + ® + ( ) n
AIEEE Corner
¢ Subjective Questions (Level 1)
1. (a) Initially the rate of distingration is
-
æ
è
ç
ö
ø
÷ =
dN
dt
N
0
0
l
After 5 min
-
=
dN
dt
N l
Þ
N
N
dN
dt
dN
dt
0 0
4750
2700
=
æ
è
ç
ö
ø
÷
= =1.76
Now N N e
t
=
-
0
l
or l = -
æ
è
ç
ç
ö
ø
÷
÷
log
N
N
t
0
Þ l =
2.3026
10
0
t
N
N
log
=
2.3026
1.76
5
10
log ( )
Þ l = 0.113 /min
(b) Half-life = = =
0.693 0.693
0.113
6.132
l
min
2. We have A N = l
6 10
11
´ = ´ ´ l 1 10
15
Þ           l = ´
- -
6 10
4 1
s
T
1 2 4
6 10
1155
/
= =
´
=
-
0.693 0.693
l
s
= 19.25 min
3. A N = l
A = = ´ ´ 8 10
10
Ci 8 3.7 decay/s
l = =
0.693 0.693
5.3 yr T
1 2 /
=
´ ´ ´ ´
0.693
5.3 s 365 24 60 60
Þ N
A
=
l
=
´ ´ ´ ´ ´ ´ 8 10 365 24 3600
10
3.7 5.3
0.693
Þ N = ´ 7.2 10
19
83
6.023 ´ 10
23
nuclei = 60 g
1
60
10
23
=
´
g
6.023
Hence 7.2 ´ 10
19
nuclei
=
´ ´
´
60 10
10
19
23
7.2
6.023
= ´
-
7.11 10
3
g
4. Number of decay per second
m
M
N
A
´ ´ = ´ ´ l
1
238
6 10
23
´
´
0.693
4.5 yr 10
9
=
´ 6 10
238
23
´
´ ´ ´ ´ ´
0.693
4.5 s 10 365 24 60 60
9
= ´ 1.23 10
4
decay/s
5. Probability of decay
P e e
t t T
= - = -
- -
( ) ( )
/
1 1
l
mean
Þ P e e = - = - =
- -
( )
/
1 1
5 10 2
0.39
6. Since initially no Pb
206
nuclei is present and
after time t the ratio of
N
N
v
Pb
= 3
It means
1
4
of original U
238
nuclei decays.
Hence                        N N e
t
=
-
0
l
N
N
N e
0
0
0
4
-
æ
è
ç
ö
ø
÷ =
- l

3
4
=
-
e
t l
Þ       t =
- log log 4 3
l
Þ t =
-
´ ´
(log log ) 4 3
10
9
0.693
4.5 yr
Þ t = ´ 1.88 yr 10
9
7. ( ) R N
1 0 1 0
32
4
P
= l
( ) R N
2 0 2 0
33
P
= l
R R R
0 1 0 2 0
32 33
= + ( ) ( )
P P
= + = N
0 1 2
4 3 ( ) l l mCi
( ) R N N e
t
t
1 1 1 0
32
1
4
P
= =
-
l l
l
( )
log
R N e
t 1 1 0
2
14
60 365
32
4
P
= ´
- ´ ´
l
( )
log
R N e
t 2 2 0
2
25
6 0 3 6 5
33
P
= ´
- ´ ´
l
R R R t
t
= + ( ) ( )
1 2
32 33
P P
= +
é
ë
ê
ê
ù
û
ú
ú
- ´ ´ - ´ ´
N e e
0 1
2
14
60 365
2
2
25
60 365
4 l l
log log
=
+
3
4
1 2
mCi
( ) l l
4
1
2 60 365
14
2
2
25
60 365
l l e e
-
´ ´
- ´ ´
+
é
ë
ê
ê
ù
û
ú
ú
log log
=
´
+
é
ë
ê
ù
û
ú
3
4 2
14
2
25
mCi
log log
4 2
14
2
25
2 60 365
14
2
25
60 365
log log
log log
e e
-
´ ´
- ´ ´
+ ´
é
ë
ê
ê
ù
û
ú
ú
» 0.205mCi
8. Complete reactions are
(a)
88
226
86
222
Ra RN ¾¾® + a
(b)
8
19
9
19
O F ¾¾® + + e n
(c)
13
25
12
25
Al Mg ¾¾® + +
+
e n
9. Only reaction (b) is possible.
10. DE = ´ + ´ - ( ) 7 7 1.000783 1.00867 14.00307
´ 931.5 MeV
Þ DE = 104.72 MeV
11. DE m m m O
p n
= + - ´ [ ( )] 8 8
8
16
931.5
= ´ + ´ - ( . 8 825 8 1 008665 1.007 15.994915)
´ 931.5  = 127.6 MeV
12. (a) Number of nuclei in kg
=
´
´
6.023 10
235
1
23
Energy
=
´
´ ´
6.023 10
235
200 10
23
6
´ ´
-
1.6 10
19
= ´ 8.09 J 10
13
84
(b) Mass =
´
´
8.09 J
J/g
10
30 10
13
3
=
´
´
8.09
g
10
30 10
13
3
= ´ ´
8.09
kg
3
10
1
10
9
3
= ´ 2.7 kg 10
6
13. Applying conservation of momentum
M v M v
a a
=
Ti Ti
Þ             v
M v
M
Ti
Ti
=
a a
K M v
a a a
= =
1
2
2
6.802 MeV
K M v M
M v
M
L
Ti Ti Ti Ti
Ti
= =
1
2
1
2
2
2
2
a a
Þ K
M
M
M v
Ti
Ti
=
æ
è
ç
ö
ø
÷
a
a a
1
2
2
= ´
4
208
6.802 MeV
= ´ =
1
52
6.802 0.1308 MeV
14. Power = = = 100 10 10
8 8
MW W J/s
=
´
=
-
10
10
10
8
13
21
1.6 J/MeV 1.6
MeV
s
Energy per fission = 185 MeV
Hence number of fissions =
´
10
185
21
1.6
/ s.
Number of nuclei in 1 kg
U
6.023
235
26
10
235
=
´
Hence t =
´ ´ ´
´
6.023 1.6 10 185
235 10
26
21
= 8.78 days
15. (a) The given reaction is
1
2
1
2
1 1
1
H H H H
3
+ ¾® +
DM m m m = - - [ ( ) ( ) ( )] 2
1
2
1
3
1
1
H H H
= - - [ ( ] 2 7825 2.014102) 3.016049 1.00 u
DM = 0.000433 u
Q M = ´ = ´ D 931.5 0.000433 931.5
» 4.05 MeV
(b) DM = ´ - (2 2.014102 3.016049
- ´ 1.008665) 931.5
DM = 3.25 MeV
(c) DM = (2.014102 3.016049 4.002603 + -
- ´ 1.008665 931.5 )
D M = 1 7 . 5 7 M eV
16. The given reaction is
He He Be
4 4 8
+ ¾®
DM = ´ - ( ) 2 4.0026 8.0053 u
= - (8.0052 8.0053) u
DM = - (0.0001) u
Q DM is negative this reaction is not
energetically favourable
D D E M = ´ 931.5
= - ´ ´
-
1 10
4
931.5 MeV
= - 93.15 keV
17. Number of nuclei in 1 kg water
=
´ 6.023 10
18
26
Heavy water =
´
´
´
-
6.023 1.5 10
18
10
100
26 2

=
´
´
6.023 1.5
18
10
22
Energy realesed per fission
= ´ - - ´ ( ) 2 2.014102 3.016049 1.007825 931.5
= ´ ´ ´ ´ ´
- -
4.33 931.5 1.6 J 10 10 10
3 6 19
Hence total energy =
´ ´ 6.023 1.5 10
18
22
´ ´ ´ ´
-
4.33 931.5 1.6 10
16
= 3200 MJ
85
Page 4

In tr o duc tor y Ex er cis e 30.2
1. D D E Mc =
2
Here, P = = ´ ´ 1 0 1 0 2 4 6 0
9 9
J / s J / d a y
Þ D M =
´ ´
´
= ´
-
1 0 2 4 6 0
3 1 0
1 0
9
8 2
4
( )
9 . 6 k g
2. Number of fis sion =
´ ´ ´
-
10
200 10 10
9
6 19
J/s
1.6
= ´ 3.125 10
19
3. Giv en re ac tion is
92
238
90
234
2
4
U T h H e ¾® +
D M = - ( 2 3 8 . 0 5 0 7 8 4 2 3 4 . 0 4 3 5 9 3
- 4 . 0 0 2 6 0 2 ) u
D M = 0 . 0 0 0 4 5 8 9 u
D D E M = ´ 9 3 1 . 5 M eV
= ´ 0.000 931.5 4589
= 4.27 MeV
4. Com plete reactions are
(a)
3
6
1
2
4
7
0
1
Li H Be + ® + n
(b)
17
35
1
1
16
32
2
4
Cl H S He + ® +
(c)
4
9
2
4
2 0
1
3 Be He He
4
+ ® + ( ) n
(d)
35
79
1
2
36
79
0
1
2 Br H Kr + ® + ( ) n
AIEEE Corner
¢ Subjective Questions (Level 1)
1. (a) Initially the rate of distingration is
-
æ
è
ç
ö
ø
÷ =
dN
dt
N
0
0
l
After 5 min
-
=
dN
dt
N l
Þ
N
N
dN
dt
dN
dt
0 0
4750
2700
=
æ
è
ç
ö
ø
÷
= =1.76
Now N N e
t
=
-
0
l
or l = -
æ
è
ç
ç
ö
ø
÷
÷
log
N
N
t
0
Þ l =
2.3026
10
0
t
N
N
log
=
2.3026
1.76
5
10
log ( )
Þ l = 0.113 /min
(b) Half-life = = =
0.693 0.693
0.113
6.132
l
min
2. We have A N = l
6 10
11
´ = ´ ´ l 1 10
15
Þ           l = ´
- -
6 10
4 1
s
T
1 2 4
6 10
1155
/
= =
´
=
-
0.693 0.693
l
s
= 19.25 min
3. A N = l
A = = ´ ´ 8 10
10
Ci 8 3.7 decay/s
l = =
0.693 0.693
5.3 yr T
1 2 /
=
´ ´ ´ ´
0.693
5.3 s 365 24 60 60
Þ N
A
=
l
=
´ ´ ´ ´ ´ ´ 8 10 365 24 3600
10
3.7 5.3
0.693
Þ N = ´ 7.2 10
19
83
6.023 ´ 10
23
nuclei = 60 g
1
60
10
23
=
´
g
6.023
Hence 7.2 ´ 10
19
nuclei
=
´ ´
´
60 10
10
19
23
7.2
6.023
= ´
-
7.11 10
3
g
4. Number of decay per second
m
M
N
A
´ ´ = ´ ´ l
1
238
6 10
23
´
´
0.693
4.5 yr 10
9
=
´ 6 10
238
23
´
´ ´ ´ ´ ´
0.693
4.5 s 10 365 24 60 60
9
= ´ 1.23 10
4
decay/s
5. Probability of decay
P e e
t t T
= - = -
- -
( ) ( )
/
1 1
l
mean
Þ P e e = - = - =
- -
( )
/
1 1
5 10 2
0.39
6. Since initially no Pb
206
nuclei is present and
after time t the ratio of
N
N
v
Pb
= 3
It means
1
4
of original U
238
nuclei decays.
Hence                        N N e
t
=
-
0
l
N
N
N e
0
0
0
4
-
æ
è
ç
ö
ø
÷ =
- l

3
4
=
-
e
t l
Þ       t =
- log log 4 3
l
Þ t =
-
´ ´
(log log ) 4 3
10
9
0.693
4.5 yr
Þ t = ´ 1.88 yr 10
9
7. ( ) R N
1 0 1 0
32
4
P
= l
( ) R N
2 0 2 0
33
P
= l
R R R
0 1 0 2 0
32 33
= + ( ) ( )
P P
= + = N
0 1 2
4 3 ( ) l l mCi
( ) R N N e
t
t
1 1 1 0
32
1
4
P
= =
-
l l
l
( )
log
R N e
t 1 1 0
2
14
60 365
32
4
P
= ´
- ´ ´
l
( )
log
R N e
t 2 2 0
2
25
6 0 3 6 5
33
P
= ´
- ´ ´
l
R R R t
t
= + ( ) ( )
1 2
32 33
P P
= +
é
ë
ê
ê
ù
û
ú
ú
- ´ ´ - ´ ´
N e e
0 1
2
14
60 365
2
2
25
60 365
4 l l
log log
=
+
3
4
1 2
mCi
( ) l l
4
1
2 60 365
14
2
2
25
60 365
l l e e
-
´ ´
- ´ ´
+
é
ë
ê
ê
ù
û
ú
ú
log log
=
´
+
é
ë
ê
ù
û
ú
3
4 2
14
2
25
mCi
log log
4 2
14
2
25
2 60 365
14
2
25
60 365
log log
log log
e e
-
´ ´
- ´ ´
+ ´
é
ë
ê
ê
ù
û
ú
ú
» 0.205mCi
8. Complete reactions are
(a)
88
226
86
222
Ra RN ¾¾® + a
(b)
8
19
9
19
O F ¾¾® + + e n
(c)
13
25
12
25
Al Mg ¾¾® + +
+
e n
9. Only reaction (b) is possible.
10. DE = ´ + ´ - ( ) 7 7 1.000783 1.00867 14.00307
´ 931.5 MeV
Þ DE = 104.72 MeV
11. DE m m m O
p n
= + - ´ [ ( )] 8 8
8
16
931.5
= ´ + ´ - ( . 8 825 8 1 008665 1.007 15.994915)
´ 931.5  = 127.6 MeV
12. (a) Number of nuclei in kg
=
´
´
6.023 10
235
1
23
Energy
=
´
´ ´
6.023 10
235
200 10
23
6
´ ´
-
1.6 10
19
= ´ 8.09 J 10
13
84
(b) Mass =
´
´
8.09 J
J/g
10
30 10
13
3
=
´
´
8.09
g
10
30 10
13
3
= ´ ´
8.09
kg
3
10
1
10
9
3
= ´ 2.7 kg 10
6
13. Applying conservation of momentum
M v M v
a a
=
Ti Ti
Þ             v
M v
M
Ti
Ti
=
a a
K M v
a a a
= =
1
2
2
6.802 MeV
K M v M
M v
M
L
Ti Ti Ti Ti
Ti
= =
1
2
1
2
2
2
2
a a
Þ K
M
M
M v
Ti
Ti
=
æ
è
ç
ö
ø
÷
a
a a
1
2
2
= ´
4
208
6.802 MeV
= ´ =
1
52
6.802 0.1308 MeV
14. Power = = = 100 10 10
8 8
MW W J/s
=
´
=
-
10
10
10
8
13
21
1.6 J/MeV 1.6
MeV
s
Energy per fission = 185 MeV
Hence number of fissions =
´
10
185
21
1.6
/ s.
Number of nuclei in 1 kg
U
6.023
235
26
10
235
=
´
Hence t =
´ ´ ´
´
6.023 1.6 10 185
235 10
26
21
= 8.78 days
15. (a) The given reaction is
1
2
1
2
1 1
1
H H H H
3
+ ¾® +
DM m m m = - - [ ( ) ( ) ( )] 2
1
2
1
3
1
1
H H H
= - - [ ( ] 2 7825 2.014102) 3.016049 1.00 u
DM = 0.000433 u
Q M = ´ = ´ D 931.5 0.000433 931.5
» 4.05 MeV
(b) DM = ´ - (2 2.014102 3.016049
- ´ 1.008665) 931.5
DM = 3.25 MeV
(c) DM = (2.014102 3.016049 4.002603 + -
- ´ 1.008665 931.5 )
D M = 1 7 . 5 7 M eV
16. The given reaction is
He He Be
4 4 8
+ ¾®
DM = ´ - ( ) 2 4.0026 8.0053 u
= - (8.0052 8.0053) u
DM = - (0.0001) u
Q DM is negative this reaction is not
energetically favourable
D D E M = ´ 931.5
= - ´ ´
-
1 10
4
931.5 MeV
= - 93.15 keV
17. Number of nuclei in 1 kg water
=
´ 6.023 10
18
26
Heavy water =
´
´
´
-
6.023 1.5 10
18
10
100
26 2

=
´
´
6.023 1.5
18
10
22
Energy realesed per fission
= ´ - - ´ ( ) 2 2.014102 3.016049 1.007825 931.5
= ´ ´ ´ ´ ´
- -
4.33 931.5 1.6 J 10 10 10
3 6 19
Hence total energy =
´ ´ 6.023 1.5 10
18
22
´ ´ ´ ´
-
4.33 931.5 1.6 10
16
= 3200 MJ
85
¢ Objective Questions (Level-1)
1. Since during b
-
decay a neutron in the
nucleus is transformed into a proton, an
electron and an antineutrino as
n P e ® + +
-
g.
Hence Correct option is (c).
2. Since nuclear force is same for all nucleons.
Hence  F F F
1 2 3
= =
Correct option is (a)..
3. Given reaction is
90
200
80
168
X Y ¾®
Difference in mass number = - = 200 168 32
Hence Number of a-particles = =
32
4
8
Difference in atomic number = 10
hence number of b -particles = 6
Hence correct option is (d).
4. The reaction is
92
235
0
1
54
138
38
94
0
1
3 U Xe Sr + ¾® + + n n ( )
The correct option is (b) three neutrons.
5. The reactions are A ® + b a and b b ® + C 2
After one a atomic number reduced by 2 and
after 2b atomic number increased by.
Hence A and C are isotopes Correct option
is (d).
6. Here m m
p n
= = 1.00785 u 1.00866 u ,
and m
a
= 4.00274 u
Dm m m m
p n
= + - 2( )
a
Þ Dm = + - [ ( ) ] 2 1.00785 1.00866 4.00274 u
Þ Dm = 00 . 3028 u
D D E m = ´ 931.5 MeV
= ´ = 0.3028 931.5 28.21 MeV
Hence correct option is (c).
7. N N N N = - =
0 0 0
7
8
1
8
But N N
n
=
æ
è
ç
ö
ø
÷ 0
1
2
Þ
1
8
1
2
0 0
N N
n
=
æ
è
ç
ö
ø
÷
Þ     n = 3
Þ  3 8
1 2
´ = T
/
s
Þ         T
1 2
8
3
/
= s
Hence correct option is (d).
8. N N e
t
=
-
0
l
for mean life t =
1
l
Þ N N e
N
e
= =
- ´
0
1
0
l
l
Hence the fraction disintegrated
=
-
= -
æ
è
ç
ö
ø
÷
N N
N e
0
0
1
1
Correct option is (b).
9. N N N N = - =
0 0 0
7
8
1
8
But N N
n
=
æ
è
ç
ö
ø
÷ 0
1
2
Þ
1
8
1
2
=
æ
è
ç
ö
ø
÷
n
Þn = 3
Hence half life T
1 2
15
3
5
/
= =
min
min
Correct option is (a).
10. Since radioactive substance loses half of its
activity in 4 days it means its half life
T
12
4
/
= days
Now A = 5% of A
0

Þ A A =
1
20
0
Þ
A
A
0
1
20
=
But    A A e
t
=
-
0
l
Þ
1
20
=
-
e
t l
Þ  log 20 = lt Þ t =
log 20
l
86
Page 5

In tr o duc tor y Ex er cis e 30.2
1. D D E Mc =
2
Here, P = = ´ ´ 1 0 1 0 2 4 6 0
9 9
J / s J / d a y
Þ D M =
´ ´
´
= ´
-
1 0 2 4 6 0
3 1 0
1 0
9
8 2
4
( )
9 . 6 k g
2. Number of fis sion =
´ ´ ´
-
10
200 10 10
9
6 19
J/s
1.6
= ´ 3.125 10
19
3. Giv en re ac tion is
92
238
90
234
2
4
U T h H e ¾® +
D M = - ( 2 3 8 . 0 5 0 7 8 4 2 3 4 . 0 4 3 5 9 3
- 4 . 0 0 2 6 0 2 ) u
D M = 0 . 0 0 0 4 5 8 9 u
D D E M = ´ 9 3 1 . 5 M eV
= ´ 0.000 931.5 4589
= 4.27 MeV
4. Com plete reactions are
(a)
3
6
1
2
4
7
0
1
Li H Be + ® + n
(b)
17
35
1
1
16
32
2
4
Cl H S He + ® +
(c)
4
9
2
4
2 0
1
3 Be He He
4
+ ® + ( ) n
(d)
35
79
1
2
36
79
0
1
2 Br H Kr + ® + ( ) n
AIEEE Corner
¢ Subjective Questions (Level 1)
1. (a) Initially the rate of distingration is
-
æ
è
ç
ö
ø
÷ =
dN
dt
N
0
0
l
After 5 min
-
=
dN
dt
N l
Þ
N
N
dN
dt
dN
dt
0 0
4750
2700
=
æ
è
ç
ö
ø
÷
= =1.76
Now N N e
t
=
-
0
l
or l = -
æ
è
ç
ç
ö
ø
÷
÷
log
N
N
t
0
Þ l =
2.3026
10
0
t
N
N
log
=
2.3026
1.76
5
10
log ( )
Þ l = 0.113 /min
(b) Half-life = = =
0.693 0.693
0.113
6.132
l
min
2. We have A N = l
6 10
11
´ = ´ ´ l 1 10
15
Þ           l = ´
- -
6 10
4 1
s
T
1 2 4
6 10
1155
/
= =
´
=
-
0.693 0.693
l
s
= 19.25 min
3. A N = l
A = = ´ ´ 8 10
10
Ci 8 3.7 decay/s
l = =
0.693 0.693
5.3 yr T
1 2 /
=
´ ´ ´ ´
0.693
5.3 s 365 24 60 60
Þ N
A
=
l
=
´ ´ ´ ´ ´ ´ 8 10 365 24 3600
10
3.7 5.3
0.693
Þ N = ´ 7.2 10
19
83
6.023 ´ 10
23
nuclei = 60 g
1
60
10
23
=
´
g
6.023
Hence 7.2 ´ 10
19
nuclei
=
´ ´
´
60 10
10
19
23
7.2
6.023
= ´
-
7.11 10
3
g
4. Number of decay per second
m
M
N
A
´ ´ = ´ ´ l
1
238
6 10
23
´
´
0.693
4.5 yr 10
9
=
´ 6 10
238
23
´
´ ´ ´ ´ ´
0.693
4.5 s 10 365 24 60 60
9
= ´ 1.23 10
4
decay/s
5. Probability of decay
P e e
t t T
= - = -
- -
( ) ( )
/
1 1
l
mean
Þ P e e = - = - =
- -
( )
/
1 1
5 10 2
0.39
6. Since initially no Pb
206
nuclei is present and
after time t the ratio of
N
N
v
Pb
= 3
It means
1
4
of original U
238
nuclei decays.
Hence                        N N e
t
=
-
0
l
N
N
N e
0
0
0
4
-
æ
è
ç
ö
ø
÷ =
- l

3
4
=
-
e
t l
Þ       t =
- log log 4 3
l
Þ t =
-
´ ´
(log log ) 4 3
10
9
0.693
4.5 yr
Þ t = ´ 1.88 yr 10
9
7. ( ) R N
1 0 1 0
32
4
P
= l
( ) R N
2 0 2 0
33
P
= l
R R R
0 1 0 2 0
32 33
= + ( ) ( )
P P
= + = N
0 1 2
4 3 ( ) l l mCi
( ) R N N e
t
t
1 1 1 0
32
1
4
P
= =
-
l l
l
( )
log
R N e
t 1 1 0
2
14
60 365
32
4
P
= ´
- ´ ´
l
( )
log
R N e
t 2 2 0
2
25
6 0 3 6 5
33
P
= ´
- ´ ´
l
R R R t
t
= + ( ) ( )
1 2
32 33
P P
= +
é
ë
ê
ê
ù
û
ú
ú
- ´ ´ - ´ ´
N e e
0 1
2
14
60 365
2
2
25
60 365
4 l l
log log
=
+
3
4
1 2
mCi
( ) l l
4
1
2 60 365
14
2
2
25
60 365
l l e e
-
´ ´
- ´ ´
+
é
ë
ê
ê
ù
û
ú
ú
log log
=
´
+
é
ë
ê
ù
û
ú
3
4 2
14
2
25
mCi
log log
4 2
14
2
25
2 60 365
14
2
25
60 365
log log
log log
e e
-
´ ´
- ´ ´
+ ´
é
ë
ê
ê
ù
û
ú
ú
» 0.205mCi
8. Complete reactions are
(a)
88
226
86
222
Ra RN ¾¾® + a
(b)
8
19
9
19
O F ¾¾® + + e n
(c)
13
25
12
25
Al Mg ¾¾® + +
+
e n
9. Only reaction (b) is possible.
10. DE = ´ + ´ - ( ) 7 7 1.000783 1.00867 14.00307
´ 931.5 MeV
Þ DE = 104.72 MeV
11. DE m m m O
p n
= + - ´ [ ( )] 8 8
8
16
931.5
= ´ + ´ - ( . 8 825 8 1 008665 1.007 15.994915)
´ 931.5  = 127.6 MeV
12. (a) Number of nuclei in kg
=
´
´
6.023 10
235
1
23
Energy
=
´
´ ´
6.023 10
235
200 10
23
6
´ ´
-
1.6 10
19
= ´ 8.09 J 10
13
84
(b) Mass =
´
´
8.09 J
J/g
10
30 10
13
3
=
´
´
8.09
g
10
30 10
13
3
= ´ ´
8.09
kg
3
10
1
10
9
3
= ´ 2.7 kg 10
6
13. Applying conservation of momentum
M v M v
a a
=
Ti Ti
Þ             v
M v
M
Ti
Ti
=
a a
K M v
a a a
= =
1
2
2
6.802 MeV
K M v M
M v
M
L
Ti Ti Ti Ti
Ti
= =
1
2
1
2
2
2
2
a a
Þ K
M
M
M v
Ti
Ti
=
æ
è
ç
ö
ø
÷
a
a a
1
2
2
= ´
4
208
6.802 MeV
= ´ =
1
52
6.802 0.1308 MeV
14. Power = = = 100 10 10
8 8
MW W J/s
=
´
=
-
10
10
10
8
13
21
1.6 J/MeV 1.6
MeV
s
Energy per fission = 185 MeV
Hence number of fissions =
´
10
185
21
1.6
/ s.
Number of nuclei in 1 kg
U
6.023
235
26
10
235
=
´
Hence t =
´ ´ ´
´
6.023 1.6 10 185
235 10
26
21
= 8.78 days
15. (a) The given reaction is
1
2
1
2
1 1
1
H H H H
3
+ ¾® +
DM m m m = - - [ ( ) ( ) ( )] 2
1
2
1
3
1
1
H H H
= - - [ ( ] 2 7825 2.014102) 3.016049 1.00 u
DM = 0.000433 u
Q M = ´ = ´ D 931.5 0.000433 931.5
» 4.05 MeV
(b) DM = ´ - (2 2.014102 3.016049
- ´ 1.008665) 931.5
DM = 3.25 MeV
(c) DM = (2.014102 3.016049 4.002603 + -
- ´ 1.008665 931.5 )
D M = 1 7 . 5 7 M eV
16. The given reaction is
He He Be
4 4 8
+ ¾®
DM = ´ - ( ) 2 4.0026 8.0053 u
= - (8.0052 8.0053) u
DM = - (0.0001) u
Q DM is negative this reaction is not
energetically favourable
D D E M = ´ 931.5
= - ´ ´
-
1 10
4
931.5 MeV
= - 93.15 keV
17. Number of nuclei in 1 kg water
=
´ 6.023 10
18
26
Heavy water =
´
´
´
-
6.023 1.5 10
18
10
100
26 2

=
´
´
6.023 1.5
18
10
22
Energy realesed per fission
= ´ - - ´ ( ) 2 2.014102 3.016049 1.007825 931.5
= ´ ´ ´ ´ ´
- -
4.33 931.5 1.6 J 10 10 10
3 6 19
Hence total energy =
´ ´ 6.023 1.5 10
18
22
´ ´ ´ ´
-
4.33 931.5 1.6 10
16
= 3200 MJ
85
¢ Objective Questions (Level-1)
1. Since during b
-
decay a neutron in the
nucleus is transformed into a proton, an
electron and an antineutrino as
n P e ® + +
-
g.
Hence Correct option is (c).
2. Since nuclear force is same for all nucleons.
Hence  F F F
1 2 3
= =
Correct option is (a)..
3. Given reaction is
90
200
80
168
X Y ¾®
Difference in mass number = - = 200 168 32
Hence Number of a-particles = =
32
4
8
Difference in atomic number = 10
hence number of b -particles = 6
Hence correct option is (d).
4. The reaction is
92
235
0
1
54
138
38
94
0
1
3 U Xe Sr + ¾® + + n n ( )
The correct option is (b) three neutrons.
5. The reactions are A ® + b a and b b ® + C 2
After one a atomic number reduced by 2 and
after 2b atomic number increased by.
Hence A and C are isotopes Correct option
is (d).
6. Here m m
p n
= = 1.00785 u 1.00866 u ,
and m
a
= 4.00274 u
Dm m m m
p n
= + - 2( )
a
Þ Dm = + - [ ( ) ] 2 1.00785 1.00866 4.00274 u
Þ Dm = 00 . 3028 u
D D E m = ´ 931.5 MeV
= ´ = 0.3028 931.5 28.21 MeV
Hence correct option is (c).
7. N N N N = - =
0 0 0
7
8
1
8
But N N
n
=
æ
è
ç
ö
ø
÷ 0
1
2
Þ
1
8
1
2
0 0
N N
n
=
æ
è
ç
ö
ø
÷
Þ     n = 3
Þ  3 8
1 2
´ = T
/
s
Þ         T
1 2
8
3
/
= s
Hence correct option is (d).
8. N N e
t
=
-
0
l
for mean life t =
1
l
Þ N N e
N
e
= =
- ´
0
1
0
l
l
Hence the fraction disintegrated
=
-
= -
æ
è
ç
ö
ø
÷
N N
N e
0
0
1
1
Correct option is (b).
9. N N N N = - =
0 0 0
7
8
1
8
But N N
n
=
æ
è
ç
ö
ø
÷ 0
1
2
Þ
1
8
1
2
=
æ
è
ç
ö
ø
÷
n
Þn = 3
Hence half life T
1 2
15
3
5
/
= =
min
min
Correct option is (a).
10. Since radioactive substance loses half of its
activity in 4 days it means its half life
T
12
4
/
= days
Now A = 5% of A
0

Þ A A =
1
20
0
Þ
A
A
0
1
20
=
But    A A e
t
=
-
0
l
Þ
1
20
=
-
e
t l
Þ  log 20 = lt Þ t =
log 20
l
86
But  l =
log
/
e
T
2
1 2

Þ = = ´ t T T
e
e
1 2 1 2
10
10
20
2
20
2
4
/ /
log
log
log
log
= 4.32
Þ t = 17.3 days
Hence correct option is (c).
11. Total energy released per sec
= = ´ 1.6 MW 1.6 J/s 10
6
Energy released per fission= 200 MeV
= ´ ´ ´
-
200 10 10
6 19
1.6 J
= ´ ´
-
2 10
11
1.6 J
Number of fission per second
=
´
´ ´
= ´
-
1.6
1.6
10
2 10
5 10
6
11
16
/s
Hence correct option is (a).
12. Q R R A =
0
1 3 /
Volume = =
4
3
4
3
3
0
3
p p R R A
Mass of nucleus = ´ ´
-
A 1.67 kg 10
27
Density r
p
= =
´ ´
-
mass
volume
1.67 kg A
R A
10
4
3
27
0
3
Þ          r
p
=
´
-
1.67 kg 10
4
3
27
0
3
R
Q r is independent of A, hence ratio of
densities
r
r
1
2
1 = .
Correct option is (d).
13.         N N e
t
=
0
â€“ l
Þ
N
N
e
t
0
=
â€“ l
Þ
N
N
e e
e
0
0 693
6 93
10
1
1
= = =
´ â€“
.
.
â€“
Fractional change
=
-
= -
æ
è
ç
ö
ø
÷
-
N N
N e
0
0
1
1
063
~
.
Hence correct option is (b).
6% it means N
N
=
0
16
We have    N N
n
=
æ
è
ç
ö
ø
÷
( )
0
1
2

Þ
1
16
1
2
=
æ
è
ç
ö
ø
÷
n
Þ n = 4
4 2
1 2
´ = T h
/
Þ T
1 2
30
/
= min
Hence correct option is (a).
15. Probability of a nucleus for survival of
time t.
p
N
N
N e
N
e
t
t
(survival)
= = =
-
-
0
0
0
l
l
For one mean life t =
1
l
Þ P
survival
= = =
- ´
-
e e
e
l
l
1
1
1
Hence Correct option is (a).
¢ Assertion and Reason
1. Here both assertion and reason are true but
reason does not explain assertion. Hence
correct option is (b).
2. Here assertion is false but reason is true
since for heavier nucleus binding energy per
nucleon is least.
Correct option is (d).
3. Here both assertion and reason are true but
reason is not correct explanation of
assertion. Hence correct option is (b).
4. Here a ssertion is true but reason is false
since electromagnetic waves are produced
by accelerating charge particles.
Correct option is (c).
87
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