DC Pandey Solutions: Modern Physics II- 2

# DC Pandey Solutions: Modern Physics II- 2 | DC Pandey Solutions for NEET Physics PDF Download

``` Page 1

5. Here assertion is wrong since b-decay
process is n p e v ® + +
–
but reason is true hence correct option is (d).
6. Here assertion is true but reason is false.
Correct option is (c).
7. Here both assertion and reason are true and
reason may or may not be true. Correct
option is (a, b)
8. Both assertion and reason are true but
reason is not correct explanation of
assertion. Hence correct option is (b).
9. Here reason is true but assertion is false
Q 1 amu = 931.5 MeV
Correct option is (d)
10. Both assertion and reason are true but
reason does not correctly explain assertion.
Hence correct option is (b).
11. Here both assertion and reason are true and
reason may or may not be correct
explanation of assertion .
Hence correct options are (a, b).
Objective Questions (Level 2)
¢ Single option correct
1. Let initially substance have N
i
nuclei then
N N e
i
t
=
- l
dN
dt
N e
i
t
= -
-
l
l
At      t t =
we get
dN
dt
N e N
t t
i
t
æ
è
ç
ö
ø
÷ = - =
=
-
l
l
0
…(i)
At t t = 4
dN
dt
N e
N
t t
i
t
æ
è
ç
ö
ø
÷ = - =
=
-
4
4 0
16
l
l
…(ii)
Dividing Eq. (i) and Eq. (ii) we get
e
t 3
16
l
= …(iii)
Now at t t =
æ
è
ç
ö
ø
÷
11
2
dN
dt
N e
t
t
i
t
æ
è
ç
ö
ø
÷ = -
=
-
11
2
11
2
l
l
= - ´
- -
l
l l
N e e
i
t t 8
2
3
2
=
-
=
´
-
l
l
l
N e
e
N
i
t
t
4
3
0
16 16
From Eqs. (ii) and (iii)
=
N
0
64
Hence, correct option is (b).
2. We have l l l = + = +
1 2
2
30
2
60
log log
Þ     l =
log 2
20
Now    N N e
t
=
-
0
l
Þ
N
N e
t
0
0
2
20
4
=
- log
Þ  log log 4
20
2 =
t
Þ 2 2
20
2 log log =
t
Þ          t = 40 yr
Hence correct option is (c).
3. From graph it is clear that number of
nucleons in X is N
3
and binding energy per
nucleon is E
3
for Y nucleon is N
2
and BE per
nucleon is E
2
.
Hence X Y E N E N + = +
3 3 2 2
Similerly W E N =
1 1
88
Page 2

5. Here assertion is wrong since b-decay
process is n p e v ® + +
–
but reason is true hence correct option is (d).
6. Here assertion is true but reason is false.
Correct option is (c).
7. Here both assertion and reason are true and
reason may or may not be true. Correct
option is (a, b)
8. Both assertion and reason are true but
reason is not correct explanation of
assertion. Hence correct option is (b).
9. Here reason is true but assertion is false
Q 1 amu = 931.5 MeV
Correct option is (d)
10. Both assertion and reason are true but
reason does not correctly explain assertion.
Hence correct option is (b).
11. Here both assertion and reason are true and
reason may or may not be correct
explanation of assertion .
Hence correct options are (a, b).
Objective Questions (Level 2)
¢ Single option correct
1. Let initially substance have N
i
nuclei then
N N e
i
t
=
- l
dN
dt
N e
i
t
= -
-
l
l
At      t t =
we get
dN
dt
N e N
t t
i
t
æ
è
ç
ö
ø
÷ = - =
=
-
l
l
0
…(i)
At t t = 4
dN
dt
N e
N
t t
i
t
æ
è
ç
ö
ø
÷ = - =
=
-
4
4 0
16
l
l
…(ii)
Dividing Eq. (i) and Eq. (ii) we get
e
t 3
16
l
= …(iii)
Now at t t =
æ
è
ç
ö
ø
÷
11
2
dN
dt
N e
t
t
i
t
æ
è
ç
ö
ø
÷ = -
=
-
11
2
11
2
l
l
= - ´
- -
l
l l
N e e
i
t t 8
2
3
2
=
-
=
´
-
l
l
l
N e
e
N
i
t
t
4
3
0
16 16
From Eqs. (ii) and (iii)
=
N
0
64
Hence, correct option is (b).
2. We have l l l = + = +
1 2
2
30
2
60
log log
Þ     l =
log 2
20
Now    N N e
t
=
-
0
l
Þ
N
N e
t
0
0
2
20
4
=
- log
Þ  log log 4
20
2 =
t
Þ 2 2
20
2 log log =
t
Þ          t = 40 yr
Hence correct option is (c).
3. From graph it is clear that number of
nucleons in X is N
3
and binding energy per
nucleon is E
3
for Y nucleon is N
2
and BE per
nucleon is E
2
.
Hence X Y E N E N + = +
3 3 2 2
Similerly W E N =
1 1
88
The reaction is W X Y ® +
The energy released is
( ) E N E N E N
3 3 2 2 1 1
+ -
Hence Correct option is (b).
4. Energy = ´ + ´ - ´ ( ) 110 200 8.2 90 8.2 7.4
= ´ - 200 (8.2 7.4)
= ´ = 200 160 0.8 MeV
Hence correct option is (d).
5. The reaction is
1
2
1
2
2
4
H H He energy + ® +
Energy = ´ - ´ (4 7 2 1.1) MeV
= - = (28 4.4) 23.6 MeV
Hence Correct option is (b).
6. Total energy released per second
= ´ 16 10
6
W
= ´ 16 10
6
J/s
Energy per fission = 200 MeV
= ´ ´ ´
-
200 10 10
6 19
1.6
= ´ ´
-
2 10
11
1.6 J
Q Efficiency = 50%
Hence power (energy converted per second)
= ´ ´ ´ = ´
- -
2 10
50
100
10
11 11
1.6 1.6 J
Number of fission =
´
´
=
-
16 10
10
10
6
11
18
1.6
/s
Hence correct option is (d).
7.
dN
dt
A N = -l
Q After time N become
conservation  Þ
dN
dt
= 0
Þ      N
A A
T
AT
= = =
l
log
log
2
2
Hence correct option is (d).
8. By conservation of momentum
M v M M
H H H
= + ¢ ( ) n
Þ       v
v
¢ =
2
Let initial KE of H-atom = K
Final KE of each H-atom =
K
2
For excitation
K
E E
2 4
13 6
2 1
= - =
-
+
æ
è
ç
ö
ø
÷
13.6
eV .
Þ
K
2
= 10.2 eV
Þ K = ´ ´ ´
-
2 10
19
10.2 1.6
Þ
1
2
2 10
2 19
M u
H
10.2 1.2 1.6 = ´ ´ ´ ´
-
Þ u
H
1.2 10.2 1.6
1.673
=
´ ´ ´ ´ ´
´
-
2 2 10
10
19
27
= ´ 6.25 m/s 10
4
Hence correct option is (c).
9. Let us suppose just before the death no
original activity A
0
is given as
A N
0 0
= l …(i)
exponentially ie,
A
dN
dt
N N e
t
= = =
-
l l
l
0
…(ii)
Dividing eq. (ii) by eq. (i) we get
A
A
e
t
0
=
- l
or lt
A
A
= log
0
or t
A
A
=
1
0
l
log
Now A
0
15 = decay/min/gram
A =
375
200
decay/min/g
but  l =
0.693
yr 5730
89
Page 3

5. Here assertion is wrong since b-decay
process is n p e v ® + +
–
but reason is true hence correct option is (d).
6. Here assertion is true but reason is false.
Correct option is (c).
7. Here both assertion and reason are true and
reason may or may not be true. Correct
option is (a, b)
8. Both assertion and reason are true but
reason is not correct explanation of
assertion. Hence correct option is (b).
9. Here reason is true but assertion is false
Q 1 amu = 931.5 MeV
Correct option is (d)
10. Both assertion and reason are true but
reason does not correctly explain assertion.
Hence correct option is (b).
11. Here both assertion and reason are true and
reason may or may not be correct
explanation of assertion .
Hence correct options are (a, b).
Objective Questions (Level 2)
¢ Single option correct
1. Let initially substance have N
i
nuclei then
N N e
i
t
=
- l
dN
dt
N e
i
t
= -
-
l
l
At      t t =
we get
dN
dt
N e N
t t
i
t
æ
è
ç
ö
ø
÷ = - =
=
-
l
l
0
…(i)
At t t = 4
dN
dt
N e
N
t t
i
t
æ
è
ç
ö
ø
÷ = - =
=
-
4
4 0
16
l
l
…(ii)
Dividing Eq. (i) and Eq. (ii) we get
e
t 3
16
l
= …(iii)
Now at t t =
æ
è
ç
ö
ø
÷
11
2
dN
dt
N e
t
t
i
t
æ
è
ç
ö
ø
÷ = -
=
-
11
2
11
2
l
l
= - ´
- -
l
l l
N e e
i
t t 8
2
3
2
=
-
=
´
-
l
l
l
N e
e
N
i
t
t
4
3
0
16 16
From Eqs. (ii) and (iii)
=
N
0
64
Hence, correct option is (b).
2. We have l l l = + = +
1 2
2
30
2
60
log log
Þ     l =
log 2
20
Now    N N e
t
=
-
0
l
Þ
N
N e
t
0
0
2
20
4
=
- log
Þ  log log 4
20
2 =
t
Þ 2 2
20
2 log log =
t
Þ          t = 40 yr
Hence correct option is (c).
3. From graph it is clear that number of
nucleons in X is N
3
and binding energy per
nucleon is E
3
for Y nucleon is N
2
and BE per
nucleon is E
2
.
Hence X Y E N E N + = +
3 3 2 2
Similerly W E N =
1 1
88
The reaction is W X Y ® +
The energy released is
( ) E N E N E N
3 3 2 2 1 1
+ -
Hence Correct option is (b).
4. Energy = ´ + ´ - ´ ( ) 110 200 8.2 90 8.2 7.4
= ´ - 200 (8.2 7.4)
= ´ = 200 160 0.8 MeV
Hence correct option is (d).
5. The reaction is
1
2
1
2
2
4
H H He energy + ® +
Energy = ´ - ´ (4 7 2 1.1) MeV
= - = (28 4.4) 23.6 MeV
Hence Correct option is (b).
6. Total energy released per second
= ´ 16 10
6
W
= ´ 16 10
6
J/s
Energy per fission = 200 MeV
= ´ ´ ´
-
200 10 10
6 19
1.6
= ´ ´
-
2 10
11
1.6 J
Q Efficiency = 50%
Hence power (energy converted per second)
= ´ ´ ´ = ´
- -
2 10
50
100
10
11 11
1.6 1.6 J
Number of fission =
´
´
=
-
16 10
10
10
6
11
18
1.6
/s
Hence correct option is (d).
7.
dN
dt
A N = -l
Q After time N become
conservation  Þ
dN
dt
= 0
Þ      N
A A
T
AT
= = =
l
log
log
2
2
Hence correct option is (d).
8. By conservation of momentum
M v M M
H H H
= + ¢ ( ) n
Þ       v
v
¢ =
2
Let initial KE of H-atom = K
Final KE of each H-atom =
K
2
For excitation
K
E E
2 4
13 6
2 1
= - =
-
+
æ
è
ç
ö
ø
÷
13.6
eV .
Þ
K
2
= 10.2 eV
Þ K = ´ ´ ´
-
2 10
19
10.2 1.6
Þ
1
2
2 10
2 19
M u
H
10.2 1.2 1.6 = ´ ´ ´ ´
-
Þ u
H
1.2 10.2 1.6
1.673
=
´ ´ ´ ´ ´
´
-
2 2 10
10
19
27
= ´ 6.25 m/s 10
4
Hence correct option is (c).
9. Let us suppose just before the death no
original activity A
0
is given as
A N
0 0
= l …(i)
exponentially ie,
A
dN
dt
N N e
t
= = =
-
l l
l
0
…(ii)
Dividing eq. (ii) by eq. (i) we get
A
A
e
t
0
=
- l
or lt
A
A
= log
0
or t
A
A
=
1
0
l
log
Now A
0
15 = decay/min/gram
A =
375
200
decay/min/g
but  l =
0.693
yr 5730
89
t = ´
´
= ´
5730 15 200
375
5730 200
25 0.693 0.693
log log
Þ t = ´
5730
8
0.693
log
= ´ = ´
5730
3 2 5730 3
0.693
log
Þ t = 17190 yr
Hence correct option is (c).
10. N N e
P
t t
=
- +
0
1
l( )
…(i)
and          N N e
Q
t
=
-
0
l
Now A N
P P
= l and A N
Q Q
= l
Þ
A
A
N
N
e
P
Q
P
Q
t
= =
- l
1
Þ   lt
A
A
Q
P
1
=
æ
è
ç
ç
ö
ø
÷
÷
log
Þ t
A
A
T
A
A
Q
P
Q
P
1
1
=
æ
è
ç
ç
ö
ø
÷
÷
=
æ
è
ç
ç
ö
ø
÷
÷
l
log log
Hence correct option is (b).
11. The given reactions are
1
2
1
2
1
3
H H H + ® + p
1
2
1 2
4
H H He
3
+ ® + n
Þ 3 H He
1
2
2
4
® + + n p
Mass defect
Dm = ´ - - - ( ) 3 2.014 4.001 1.007 1.008
Dm = 0.026 amu
Energy Released
= ´ = ´
-
0.026 MeV 3.87 J 931 10
12
This energy produced by the three
deutronatoms. Total energy released by
10
40
deutrons
= ´ ´ = ´
-
10
3
10 10
40
12 28
3.87 J 1.29 J
The average power P = = 10 10
16 16
W J/s
Therefore total time to exhaust all
deutrons of the star will be
t =
´
»
1.29
s s.
10
10
10
28
16
12
Hence correct option is (c).
12. N N e N e
t t
t
1 0 0
2
1 1
= =
-
-
l
log
…(i)
N N e N e
t t
t
2 0 0
2
2 2
= =
-
-
l
log
…(ii)
R N
1 1 1
= l …(iii)
and R N
2 2 2 2
= = l l …(iv)
Let after time t, R R
1 2
= then
Þ
R
R
1
1
1 =
Þ
l
l
1
2
1
2
1 ´ =
N
N
Þ
l
l
1
2
2
1
2
2 1
1 2
= =
- é
ë
ê
ù
û
ú
´
N
N
e
t t
t t
t log
Þ          log log
t
t
t
t t
t t
2
1
2 1
1 2
2 = ´ ´
- æ
è
ç
ç
ö
ø
÷
÷
Þ   t
t t
t t
t
t
1
1 2
2 1
2
1
=
-
´
0.693 ( )
log
Hence Correct option is (a).
13. The given reaction is
Z
A
X Y
232
90
¾® + a
Þ
Z
A
X Y
232
90 2
¾® + He
4
Þ Z = 92 and A = 228
Q Initially X is in rest hence momentum of
a-particle after decay will be equal and
opposite of Y.
Þ M v M v
Y Y
=
a a
Þ         v
M
M
v
Y
Y
=
a
a
Total kinetic energy
K M v M v
T Y Y
= +
1
2
2 2
( )
a a
Þ K M v M
M v
M
T y
Y
= +
é
ë
ê
ù
û
ú
1
2
2
2 2
2
a a
a a
90
Page 4

5. Here assertion is wrong since b-decay
process is n p e v ® + +
–
but reason is true hence correct option is (d).
6. Here assertion is true but reason is false.
Correct option is (c).
7. Here both assertion and reason are true and
reason may or may not be true. Correct
option is (a, b)
8. Both assertion and reason are true but
reason is not correct explanation of
assertion. Hence correct option is (b).
9. Here reason is true but assertion is false
Q 1 amu = 931.5 MeV
Correct option is (d)
10. Both assertion and reason are true but
reason does not correctly explain assertion.
Hence correct option is (b).
11. Here both assertion and reason are true and
reason may or may not be correct
explanation of assertion .
Hence correct options are (a, b).
Objective Questions (Level 2)
¢ Single option correct
1. Let initially substance have N
i
nuclei then
N N e
i
t
=
- l
dN
dt
N e
i
t
= -
-
l
l
At      t t =
we get
dN
dt
N e N
t t
i
t
æ
è
ç
ö
ø
÷ = - =
=
-
l
l
0
…(i)
At t t = 4
dN
dt
N e
N
t t
i
t
æ
è
ç
ö
ø
÷ = - =
=
-
4
4 0
16
l
l
…(ii)
Dividing Eq. (i) and Eq. (ii) we get
e
t 3
16
l
= …(iii)
Now at t t =
æ
è
ç
ö
ø
÷
11
2
dN
dt
N e
t
t
i
t
æ
è
ç
ö
ø
÷ = -
=
-
11
2
11
2
l
l
= - ´
- -
l
l l
N e e
i
t t 8
2
3
2
=
-
=
´
-
l
l
l
N e
e
N
i
t
t
4
3
0
16 16
From Eqs. (ii) and (iii)
=
N
0
64
Hence, correct option is (b).
2. We have l l l = + = +
1 2
2
30
2
60
log log
Þ     l =
log 2
20
Now    N N e
t
=
-
0
l
Þ
N
N e
t
0
0
2
20
4
=
- log
Þ  log log 4
20
2 =
t
Þ 2 2
20
2 log log =
t
Þ          t = 40 yr
Hence correct option is (c).
3. From graph it is clear that number of
nucleons in X is N
3
and binding energy per
nucleon is E
3
for Y nucleon is N
2
and BE per
nucleon is E
2
.
Hence X Y E N E N + = +
3 3 2 2
Similerly W E N =
1 1
88
The reaction is W X Y ® +
The energy released is
( ) E N E N E N
3 3 2 2 1 1
+ -
Hence Correct option is (b).
4. Energy = ´ + ´ - ´ ( ) 110 200 8.2 90 8.2 7.4
= ´ - 200 (8.2 7.4)
= ´ = 200 160 0.8 MeV
Hence correct option is (d).
5. The reaction is
1
2
1
2
2
4
H H He energy + ® +
Energy = ´ - ´ (4 7 2 1.1) MeV
= - = (28 4.4) 23.6 MeV
Hence Correct option is (b).
6. Total energy released per second
= ´ 16 10
6
W
= ´ 16 10
6
J/s
Energy per fission = 200 MeV
= ´ ´ ´
-
200 10 10
6 19
1.6
= ´ ´
-
2 10
11
1.6 J
Q Efficiency = 50%
Hence power (energy converted per second)
= ´ ´ ´ = ´
- -
2 10
50
100
10
11 11
1.6 1.6 J
Number of fission =
´
´
=
-
16 10
10
10
6
11
18
1.6
/s
Hence correct option is (d).
7.
dN
dt
A N = -l
Q After time N become
conservation  Þ
dN
dt
= 0
Þ      N
A A
T
AT
= = =
l
log
log
2
2
Hence correct option is (d).
8. By conservation of momentum
M v M M
H H H
= + ¢ ( ) n
Þ       v
v
¢ =
2
Let initial KE of H-atom = K
Final KE of each H-atom =
K
2
For excitation
K
E E
2 4
13 6
2 1
= - =
-
+
æ
è
ç
ö
ø
÷
13.6
eV .
Þ
K
2
= 10.2 eV
Þ K = ´ ´ ´
-
2 10
19
10.2 1.6
Þ
1
2
2 10
2 19
M u
H
10.2 1.2 1.6 = ´ ´ ´ ´
-
Þ u
H
1.2 10.2 1.6
1.673
=
´ ´ ´ ´ ´
´
-
2 2 10
10
19
27
= ´ 6.25 m/s 10
4
Hence correct option is (c).
9. Let us suppose just before the death no
original activity A
0
is given as
A N
0 0
= l …(i)
exponentially ie,
A
dN
dt
N N e
t
= = =
-
l l
l
0
…(ii)
Dividing eq. (ii) by eq. (i) we get
A
A
e
t
0
=
- l
or lt
A
A
= log
0
or t
A
A
=
1
0
l
log
Now A
0
15 = decay/min/gram
A =
375
200
decay/min/g
but  l =
0.693
yr 5730
89
t = ´
´
= ´
5730 15 200
375
5730 200
25 0.693 0.693
log log
Þ t = ´
5730
8
0.693
log
= ´ = ´
5730
3 2 5730 3
0.693
log
Þ t = 17190 yr
Hence correct option is (c).
10. N N e
P
t t
=
- +
0
1
l( )
…(i)
and          N N e
Q
t
=
-
0
l
Now A N
P P
= l and A N
Q Q
= l
Þ
A
A
N
N
e
P
Q
P
Q
t
= =
- l
1
Þ   lt
A
A
Q
P
1
=
æ
è
ç
ç
ö
ø
÷
÷
log
Þ t
A
A
T
A
A
Q
P
Q
P
1
1
=
æ
è
ç
ç
ö
ø
÷
÷
=
æ
è
ç
ç
ö
ø
÷
÷
l
log log
Hence correct option is (b).
11. The given reactions are
1
2
1
2
1
3
H H H + ® + p
1
2
1 2
4
H H He
3
+ ® + n
Þ 3 H He
1
2
2
4
® + + n p
Mass defect
Dm = ´ - - - ( ) 3 2.014 4.001 1.007 1.008
Dm = 0.026 amu
Energy Released
= ´ = ´
-
0.026 MeV 3.87 J 931 10
12
This energy produced by the three
deutronatoms. Total energy released by
10
40
deutrons
= ´ ´ = ´
-
10
3
10 10
40
12 28
3.87 J 1.29 J
The average power P = = 10 10
16 16
W J/s
Therefore total time to exhaust all
deutrons of the star will be
t =
´
»
1.29
s s.
10
10
10
28
16
12
Hence correct option is (c).
12. N N e N e
t t
t
1 0 0
2
1 1
= =
-
-
l
log
…(i)
N N e N e
t t
t
2 0 0
2
2 2
= =
-
-
l
log
…(ii)
R N
1 1 1
= l …(iii)
and R N
2 2 2 2
= = l l …(iv)
Let after time t, R R
1 2
= then
Þ
R
R
1
1
1 =
Þ
l
l
1
2
1
2
1 ´ =
N
N
Þ
l
l
1
2
2
1
2
2 1
1 2
= =
- é
ë
ê
ù
û
ú
´
N
N
e
t t
t t
t log
Þ          log log
t
t
t
t t
t t
2
1
2 1
1 2
2 = ´ ´
- æ
è
ç
ç
ö
ø
÷
÷
Þ   t
t t
t t
t
t
1
1 2
2 1
2
1
=
-
´
0.693 ( )
log
Hence Correct option is (a).
13. The given reaction is
Z
A
X Y
232
90
¾® + a
Þ
Z
A
X Y
232
90 2
¾® + He
4
Þ Z = 92 and A = 228
Q Initially X is in rest hence momentum of
a-particle after decay will be equal and
opposite of Y.
Þ M v M v
Y Y
=
a a
Þ         v
M
M
v
Y
Y
=
a
a
Total kinetic energy
K M v M v
T Y Y
= +
1
2
2 2
( )
a a
Þ K M v M
M v
M
T y
Y
= +
é
ë
ê
ù
û
ú
1
2
2
2 2
2
a a
a a
90
Þ K M v
M
M
T
Y
= +
é
ë
ê
ù
û
ú
1
2
1
2
a a
a
Þ K K
T
= +
é
ë
ê
ù
û
ú
a
1
4
228
Þ K K
T a
=
232
228
Hence Correct option is (b).
14. Energy of emitted photon = 7 MeV
= ´ ´ ´
-
7 10 10
6 19
1.6 J
= ´
-
11.2 J 10
13
Momentum of photon =
´
´
-
11.2 J
10 m/s
10
3
13
8
= ´
-
11.2
kg-m/s
3
10
21
Q Initial nucleus is stationary
Applying conservation of momentum
principle
O = + Þ = -
® ® ® ®
P P P P nuc photon nuc photon
Þ | | | | P P
® ®
= - nuc photon
Þ P
nuc
11.2
kg- m/s = ´
-
3
10
21
Mass of nucleus = 24 amu
= ´ ´
-
24 10
27
1.66 kg
But    P mK
nuc nuc
2
2 =
Þ
K
P
m
nuc
nuc
11.2 11.2 10
1.66
Joule = =
´ ´
´ ´ ´ ´
-
-
2 42
27
2 9 2 24 10
Þ
K
c nuc
11.2
1.66 1.6
eV =
´ ´
´ ´ ´ ´ ´
-
- -
112 10
18 24 10 10
42
27 19
.
» 1.1 keV
Hence correct option is  (b).
15. Let time interval between two instants is t
1
then
N N e
t t
1 0
1
=
- + l( )
and N N e
t
2 0
2 =
-l
A N N e
t t
1 1 0
1
= =
- +
l l
l( )
A N N e
t
2 2 0
2 = =
-
l l
l
( )
Þ
A
A
e
t 1
2
1
2
1
= ´
-l
Þ
2
1
2
1
A
A
e
t
=
-l
Þ log
A
A
t
2
1
1
2
æ
è
ç
ç
ö
ø
÷
÷
= l
Þ t
A
A
1
2
1
1
2
=
æ
è
ç
ç
ö
ø
÷
÷
l
log
Þ t
T A
A
1
2
1
2 2
=
æ
è
ç
ç
ö
ø
÷
÷
log
log
Hence Correct option is (c).
16. The given reaction is
1
2
1
2
1
3
1
1
H H H H + ¾® +
DM m m m = - - [ ( ) ( ) ( )] 2
1
2
1
3
1
1
H H H
= ´ - - ( ) 2 2.014102 3.016049 1.007825 amu
= ´
-
4.33 amu 10
3
DE = ´ ´
-
4.33 931.5 MeV 10
3
= 4 MeV
Hence correct option is (c).
17. Number of fusion required to generate
1 kWh
=
´ ´
´ ´ ´
-
1 10 3600
4 10 10
3
6 19
1.6
=
´
= ´ »
36 10
10 10
18
18 18
6.4
5.6
Hence correct option is  (b).
18. The energy released = 4 MeV
This energy produced by two atoms.
Hence energy produced per atom
= = ´ ´
-
2 2 10
13
MeV 1.6 J
Hence number of atom fused to produced
1 kWJ
91
Page 5

5. Here assertion is wrong since b-decay
process is n p e v ® + +
–
but reason is true hence correct option is (d).
6. Here assertion is true but reason is false.
Correct option is (c).
7. Here both assertion and reason are true and
reason may or may not be true. Correct
option is (a, b)
8. Both assertion and reason are true but
reason is not correct explanation of
assertion. Hence correct option is (b).
9. Here reason is true but assertion is false
Q 1 amu = 931.5 MeV
Correct option is (d)
10. Both assertion and reason are true but
reason does not correctly explain assertion.
Hence correct option is (b).
11. Here both assertion and reason are true and
reason may or may not be correct
explanation of assertion .
Hence correct options are (a, b).
Objective Questions (Level 2)
¢ Single option correct
1. Let initially substance have N
i
nuclei then
N N e
i
t
=
- l
dN
dt
N e
i
t
= -
-
l
l
At      t t =
we get
dN
dt
N e N
t t
i
t
æ
è
ç
ö
ø
÷ = - =
=
-
l
l
0
…(i)
At t t = 4
dN
dt
N e
N
t t
i
t
æ
è
ç
ö
ø
÷ = - =
=
-
4
4 0
16
l
l
…(ii)
Dividing Eq. (i) and Eq. (ii) we get
e
t 3
16
l
= …(iii)
Now at t t =
æ
è
ç
ö
ø
÷
11
2
dN
dt
N e
t
t
i
t
æ
è
ç
ö
ø
÷ = -
=
-
11
2
11
2
l
l
= - ´
- -
l
l l
N e e
i
t t 8
2
3
2
=
-
=
´
-
l
l
l
N e
e
N
i
t
t
4
3
0
16 16
From Eqs. (ii) and (iii)
=
N
0
64
Hence, correct option is (b).
2. We have l l l = + = +
1 2
2
30
2
60
log log
Þ     l =
log 2
20
Now    N N e
t
=
-
0
l
Þ
N
N e
t
0
0
2
20
4
=
- log
Þ  log log 4
20
2 =
t
Þ 2 2
20
2 log log =
t
Þ          t = 40 yr
Hence correct option is (c).
3. From graph it is clear that number of
nucleons in X is N
3
and binding energy per
nucleon is E
3
for Y nucleon is N
2
and BE per
nucleon is E
2
.
Hence X Y E N E N + = +
3 3 2 2
Similerly W E N =
1 1
88
The reaction is W X Y ® +
The energy released is
( ) E N E N E N
3 3 2 2 1 1
+ -
Hence Correct option is (b).
4. Energy = ´ + ´ - ´ ( ) 110 200 8.2 90 8.2 7.4
= ´ - 200 (8.2 7.4)
= ´ = 200 160 0.8 MeV
Hence correct option is (d).
5. The reaction is
1
2
1
2
2
4
H H He energy + ® +
Energy = ´ - ´ (4 7 2 1.1) MeV
= - = (28 4.4) 23.6 MeV
Hence Correct option is (b).
6. Total energy released per second
= ´ 16 10
6
W
= ´ 16 10
6
J/s
Energy per fission = 200 MeV
= ´ ´ ´
-
200 10 10
6 19
1.6
= ´ ´
-
2 10
11
1.6 J
Q Efficiency = 50%
Hence power (energy converted per second)
= ´ ´ ´ = ´
- -
2 10
50
100
10
11 11
1.6 1.6 J
Number of fission =
´
´
=
-
16 10
10
10
6
11
18
1.6
/s
Hence correct option is (d).
7.
dN
dt
A N = -l
Q After time N become
conservation  Þ
dN
dt
= 0
Þ      N
A A
T
AT
= = =
l
log
log
2
2
Hence correct option is (d).
8. By conservation of momentum
M v M M
H H H
= + ¢ ( ) n
Þ       v
v
¢ =
2
Let initial KE of H-atom = K
Final KE of each H-atom =
K
2
For excitation
K
E E
2 4
13 6
2 1
= - =
-
+
æ
è
ç
ö
ø
÷
13.6
eV .
Þ
K
2
= 10.2 eV
Þ K = ´ ´ ´
-
2 10
19
10.2 1.6
Þ
1
2
2 10
2 19
M u
H
10.2 1.2 1.6 = ´ ´ ´ ´
-
Þ u
H
1.2 10.2 1.6
1.673
=
´ ´ ´ ´ ´
´
-
2 2 10
10
19
27
= ´ 6.25 m/s 10
4
Hence correct option is (c).
9. Let us suppose just before the death no
original activity A
0
is given as
A N
0 0
= l …(i)
exponentially ie,
A
dN
dt
N N e
t
= = =
-
l l
l
0
…(ii)
Dividing eq. (ii) by eq. (i) we get
A
A
e
t
0
=
- l
or lt
A
A
= log
0
or t
A
A
=
1
0
l
log
Now A
0
15 = decay/min/gram
A =
375
200
decay/min/g
but  l =
0.693
yr 5730
89
t = ´
´
= ´
5730 15 200
375
5730 200
25 0.693 0.693
log log
Þ t = ´
5730
8
0.693
log
= ´ = ´
5730
3 2 5730 3
0.693
log
Þ t = 17190 yr
Hence correct option is (c).
10. N N e
P
t t
=
- +
0
1
l( )
…(i)
and          N N e
Q
t
=
-
0
l
Now A N
P P
= l and A N
Q Q
= l
Þ
A
A
N
N
e
P
Q
P
Q
t
= =
- l
1
Þ   lt
A
A
Q
P
1
=
æ
è
ç
ç
ö
ø
÷
÷
log
Þ t
A
A
T
A
A
Q
P
Q
P
1
1
=
æ
è
ç
ç
ö
ø
÷
÷
=
æ
è
ç
ç
ö
ø
÷
÷
l
log log
Hence correct option is (b).
11. The given reactions are
1
2
1
2
1
3
H H H + ® + p
1
2
1 2
4
H H He
3
+ ® + n
Þ 3 H He
1
2
2
4
® + + n p
Mass defect
Dm = ´ - - - ( ) 3 2.014 4.001 1.007 1.008
Dm = 0.026 amu
Energy Released
= ´ = ´
-
0.026 MeV 3.87 J 931 10
12
This energy produced by the three
deutronatoms. Total energy released by
10
40
deutrons
= ´ ´ = ´
-
10
3
10 10
40
12 28
3.87 J 1.29 J
The average power P = = 10 10
16 16
W J/s
Therefore total time to exhaust all
deutrons of the star will be
t =
´
»
1.29
s s.
10
10
10
28
16
12
Hence correct option is (c).
12. N N e N e
t t
t
1 0 0
2
1 1
= =
-
-
l
log
…(i)
N N e N e
t t
t
2 0 0
2
2 2
= =
-
-
l
log
…(ii)
R N
1 1 1
= l …(iii)
and R N
2 2 2 2
= = l l …(iv)
Let after time t, R R
1 2
= then
Þ
R
R
1
1
1 =
Þ
l
l
1
2
1
2
1 ´ =
N
N
Þ
l
l
1
2
2
1
2
2 1
1 2
= =
- é
ë
ê
ù
û
ú
´
N
N
e
t t
t t
t log
Þ          log log
t
t
t
t t
t t
2
1
2 1
1 2
2 = ´ ´
- æ
è
ç
ç
ö
ø
÷
÷
Þ   t
t t
t t
t
t
1
1 2
2 1
2
1
=
-
´
0.693 ( )
log
Hence Correct option is (a).
13. The given reaction is
Z
A
X Y
232
90
¾® + a
Þ
Z
A
X Y
232
90 2
¾® + He
4
Þ Z = 92 and A = 228
Q Initially X is in rest hence momentum of
a-particle after decay will be equal and
opposite of Y.
Þ M v M v
Y Y
=
a a
Þ         v
M
M
v
Y
Y
=
a
a
Total kinetic energy
K M v M v
T Y Y
= +
1
2
2 2
( )
a a
Þ K M v M
M v
M
T y
Y
= +
é
ë
ê
ù
û
ú
1
2
2
2 2
2
a a
a a
90
Þ K M v
M
M
T
Y
= +
é
ë
ê
ù
û
ú
1
2
1
2
a a
a
Þ K K
T
= +
é
ë
ê
ù
û
ú
a
1
4
228
Þ K K
T a
=
232
228
Hence Correct option is (b).
14. Energy of emitted photon = 7 MeV
= ´ ´ ´
-
7 10 10
6 19
1.6 J
= ´
-
11.2 J 10
13
Momentum of photon =
´
´
-
11.2 J
10 m/s
10
3
13
8
= ´
-
11.2
kg-m/s
3
10
21
Q Initial nucleus is stationary
Applying conservation of momentum
principle
O = + Þ = -
® ® ® ®
P P P P nuc photon nuc photon
Þ | | | | P P
® ®
= - nuc photon
Þ P
nuc
11.2
kg- m/s = ´
-
3
10
21
Mass of nucleus = 24 amu
= ´ ´
-
24 10
27
1.66 kg
But    P mK
nuc nuc
2
2 =
Þ
K
P
m
nuc
nuc
11.2 11.2 10
1.66
Joule = =
´ ´
´ ´ ´ ´
-
-
2 42
27
2 9 2 24 10
Þ
K
c nuc
11.2
1.66 1.6
eV =
´ ´
´ ´ ´ ´ ´
-
- -
112 10
18 24 10 10
42
27 19
.
» 1.1 keV
Hence correct option is  (b).
15. Let time interval between two instants is t
1
then
N N e
t t
1 0
1
=
- + l( )
and N N e
t
2 0
2 =
-l
A N N e
t t
1 1 0
1
= =
- +
l l
l( )
A N N e
t
2 2 0
2 = =
-
l l
l
( )
Þ
A
A
e
t 1
2
1
2
1
= ´
-l
Þ
2
1
2
1
A
A
e
t
=
-l
Þ log
A
A
t
2
1
1
2
æ
è
ç
ç
ö
ø
÷
÷
= l
Þ t
A
A
1
2
1
1
2
=
æ
è
ç
ç
ö
ø
÷
÷
l
log
Þ t
T A
A
1
2
1
2 2
=
æ
è
ç
ç
ö
ø
÷
÷
log
log
Hence Correct option is (c).
16. The given reaction is
1
2
1
2
1
3
1
1
H H H H + ¾® +
DM m m m = - - [ ( ) ( ) ( )] 2
1
2
1
3
1
1
H H H
= ´ - - ( ) 2 2.014102 3.016049 1.007825 amu
= ´
-
4.33 amu 10
3
DE = ´ ´
-
4.33 931.5 MeV 10
3
= 4 MeV
Hence correct option is (c).
17. Number of fusion required to generate
1 kWh
=
´ ´
´ ´ ´
-
1 10 3600
4 10 10
3
6 19
1.6
=
´
= ´ »
36 10
10 10
18
18 18
6.4
5.6
Hence correct option is  (b).
18. The energy released = 4 MeV
This energy produced by two atoms.
Hence energy produced per atom
= = ´ ´
-
2 2 10
13
MeV 1.6 J
Hence number of atom fused to produced
1 kWJ
91
=
´
´ ´
= ´
-
36 10
2 10
18
10
5
13
18
1.6 1.6
Mass of deutrium which contain
18
10
18
1.6
atom ´
=
´
´ ´
= ´
-
18 10
10
10
18
23
5
1.6 6.02
3.7 kg.
Hence correct option is (c).
¢ More Than one Option is Correct
1. x N =
0
, y N = l
0
Þ
x
y
N
N
= =
0
0
1
l l
where l is decay constant
Hence
x
y
is constant throught.
Q
x
y
T
T
= = =
1 1
l
0.693
0.693
Þ
x
y
T > ,  xy N = l( )
0
2
For one half life N
N
=
0
2
Þ ( ) xy
N N xy
T
=
æ
è
ç
ö
ø
÷ = = l
l
0
2
0
2
2 4 4
Hence correct options are (a), (b) and (d).
2. The correct options are (a), (b), (c) and (d).
3. A nucleus in excited state emits a high
energy photon called as g-ray. The reaction
is
X X
*
¾® + g
Hence by gamma radiation atomic number
and mass number are not changed. Since
after emission of one a atomic number
reduced by 2
2
4
( ) a and after 2b atomic
number is increased by (2). Hence correct
options are (a), (b) and (c).
4. Here half lives are T and 2T and N N
x
=
0
,
N N
y
=
0
after 4T for first substance = 4 half
lives and after 4T the second substance = 2
Half lines.
Þ N N
N
x
=
æ
è
ç
ö
ø
÷
=
0
4
0
1
2 16
N N
N
y
=
æ
è
ç
ö
ø
÷
=
0
2
0
1
2 4
Þ     x
N
N
N
N
x
y
= = =
0
0
16
4
1
4
/
/
Let their activity are R
x
and R
y
.
Þ R N
x x x
= l and R N
y y y
= l
Þ y
R
R
N
N
T
T
x
y
x
y
x
y
= = ´ = ´
l
l
0.693
0.693
2
1
4
Þ
R
R
y
x
y
= =
1
2
Hence correct options are (b) and (c).
5. Since nuclear forces are vary short range
charge independent, no electromagnetic and
they exchange (n p ® orp n ® ). Hence the
correct options are (a),  (b), (c) and (d).
6. Q R R A =
0
1 3 /

r
p
p
= =
´ ´
-
M
R
A
R A
4 3
10
4
3
3
27
0
3 /
1.67 kg
Þ r is independent of A.
But r =
´
´ ´ ´
-
-
1.67 kg
3.14 1.3
10
4
3
10
27
15 3
( )
= ´ 1.8 kg/m 10
17 3
Hence correct options are (b) and (c).
92
```

122 docs

## FAQs on DC Pandey Solutions: Modern Physics II- 2 - DC Pandey Solutions for NEET Physics

 1. What is the concept of modern physics?
Ans. Modern physics is a branch of physics that deals with the study of the behavior of matter and energy at the atomic and subatomic levels. It includes topics such as quantum mechanics, relativity, particle physics, and nuclear physics.
 2. What is the importance of studying modern physics?
Ans. Studying modern physics is important as it helps us understand the fundamental nature of the universe and the behavior of matter and energy at the smallest scales. It has led to significant technological advancements and has been the basis for many scientific breakthroughs.
 3. What are the main topics covered in DC Pandey's book on Modern Physics II?
Ans. DC Pandey's book on Modern Physics II covers topics such as nuclear physics, radioactivity, X-rays, and semiconductors. It provides a comprehensive understanding of these concepts and their applications.
 4. How can I effectively prepare for the Modern Physics II exam using DC Pandey's book?
Ans. To effectively prepare for the Modern Physics II exam using DC Pandey's book, it is recommended to first read and understand the concepts thoroughly. Practice solving the numerical problems and review the solved examples given in the book. Additionally, solve previous years' question papers to get familiar with the exam pattern and improve time management skills.
 5. Are there any additional resources available to supplement the study of Modern Physics II?
Ans. Yes, apart from DC Pandey's book, there are several online resources available, such as video lectures, online tutorials, and practice tests, which can further enhance your understanding of Modern Physics II. These resources can provide additional explanations and examples to reinforce the concepts learned from the book.

## DC Pandey Solutions for NEET Physics

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