Nuclei Practice Questions - DPP for NEET

 Page 1


1. (c). The energy produced per second is
= 1000 × 10
3
 J = 
6
19
10
1.6 10
-
´
 eV  6.25 × 10
24
 eV
The number of fissions should be, thus
number = 
24
6
6.25 10
200 10
´
´
 = 3.125 × 10
16
2. (b). No. of atoms in 2kg 
  92
U
235
 = 
2
235
 × N
A
= 
2
235
 × (6.02 × 10
26
) = 5.12 × 10
24
Fission rate = 
24
5.12 10
30 246060
´
´´´
 = 1.975 × 10
18
 per sec
Usable energy per fission = 185 MeV
\ Power output
= (185 × 10
6
)(1.975 × 10
18
)(1.6 × 10
–19
) watt
= 58.4 × 10
6
 watt = 58.46 MW
3. (d). \ 6 gm of 
6
C
12
 contains atoms = 
23
6 10
2
´
 and each
atom of 
6
C
12
 contains electron, protons  and neutrons
= 6, 6, 6
\ No. of electron, protons and neutron in 6 gm of
6
C
12
 = 18 × 10
23
, 18 × 10
23
, 18 × 10
23
4. (c). Use r = Mass/volume
= 
27
15
1.66 10 16
(4/3)(310)
-
-
´´
p´
 = 2.35 × 10
17
 kg m
-3
5. (a). Mass defect Dm = M (Ra 226) – M(Rn 222) – M (a)
    = 226.0256 – 222.0175 – 4.00026 = 0.0053 u.
6. (a). E = mc
2 
= (1.66 × 10
–27
) (3 × 10
8
)
2
 J
  = 1.49 × 10
–10
 J
= 
10
13
1.49 10
1.6 10
-
-
´
´
 MeV = 931.49 MeV
7. (b). E = mc
2
       = (9.1 × 10
–31
) (3 × 10
8
)
2
 J = 0.51 MeV
8. (c). DE = D mc
2
Dm = 
0.5
100
kg = 0.005 kg
c = 3 × 10
8
 m/s
DE = 0.005 × (3 × 10
8
)
2
DE = 4.5 × 10
14
 J or watt-sec
DE = 
14
4.5 10
60 60
´
´
 = 1.25 × 10
111
 watt hour
DE = 1.25 × 10
8
 kWH
9. (b). By the forumula N = N
0
e
–lt
Given 
0
N
N
 = 
1
20
 and l = 
0.6931
3.8
 Þ 20 = 
0.631×
3.8
t
e
Taking log of both sides
or log 20 = 10
0.6931 ×
log
3.8
t
e
or 1.3010 = 
0.6931 × 0.4343
3.8
t ´
 Þ t = 16.5 days
10. (b). A = 238 - 4 = 234, Z = 92 - 2 = 90
11. (a). Dm = 0.03 a.m.u., A = 4
Þ DE = 
m 931
A
D´
Þ DE = 
0.03 931
4
´
 = 7 MeV
12. (a). Q DE = Dm × 931 MeV
Þ Dm = 
E 2.23
931 931
D
= = 0.0024 a.m.u.
13. (a). 
1/3
Al
1/3
Te
R (27) 36
R 5 10
(125)
= ==
14. (d). R = R
0
 A
1/3
  = 1.2 × 10
-15
 × (64)
1/3
        = 1.2 × 10
–15
 × 4 = 4.8 fm
15. (b). Number of protrons in nucleus = atomic number = 11
Number of electrons = number of protons = 11.
Number of neutrons = mass number A – atomic number Z
N = 24 – 11 = 13
16. (d). Q equivalent mass of each photon = 1/2000 amu
Q 1 amu = 931 MeV
\ Energy of each photon = 
931
2000
 = 0.465 MeV
17. (c). Deuterium, the isotope of hydrogen consits of one proton
and neutron. Therefore mass of nuclear constituents of
deuterium = mass of proton + mass of neutron
    = 1.00759 + 1.00898 = 2.01657 amu.
mass of nucleus of deuterium = 2.01470 amu.
Mass defect = 2.01657 – 2.01470 = 0.00187 amu.
Binding energy = DE = 0.00187 × 931 MeV = 1.741 MeV .
18. (a). E = 
E m 931
AA
D D´
= MeV
Dm = (3m
p
 + 4m
n
) – mass of Li
7
Dm = (3 × 1.00759 + 4 × 1.00898) – 7.01653
Dm = 0.04216 a.m.u.
DE = 
0.04216 931 39.25
77
´
= = 5.6 MeV
Page 2


1. (c). The energy produced per second is
= 1000 × 10
3
 J = 
6
19
10
1.6 10
-
´
 eV  6.25 × 10
24
 eV
The number of fissions should be, thus
number = 
24
6
6.25 10
200 10
´
´
 = 3.125 × 10
16
2. (b). No. of atoms in 2kg 
  92
U
235
 = 
2
235
 × N
A
= 
2
235
 × (6.02 × 10
26
) = 5.12 × 10
24
Fission rate = 
24
5.12 10
30 246060
´
´´´
 = 1.975 × 10
18
 per sec
Usable energy per fission = 185 MeV
\ Power output
= (185 × 10
6
)(1.975 × 10
18
)(1.6 × 10
–19
) watt
= 58.4 × 10
6
 watt = 58.46 MW
3. (d). \ 6 gm of 
6
C
12
 contains atoms = 
23
6 10
2
´
 and each
atom of 
6
C
12
 contains electron, protons  and neutrons
= 6, 6, 6
\ No. of electron, protons and neutron in 6 gm of
6
C
12
 = 18 × 10
23
, 18 × 10
23
, 18 × 10
23
4. (c). Use r = Mass/volume
= 
27
15
1.66 10 16
(4/3)(310)
-
-
´´
p´
 = 2.35 × 10
17
 kg m
-3
5. (a). Mass defect Dm = M (Ra 226) – M(Rn 222) – M (a)
    = 226.0256 – 222.0175 – 4.00026 = 0.0053 u.
6. (a). E = mc
2 
= (1.66 × 10
–27
) (3 × 10
8
)
2
 J
  = 1.49 × 10
–10
 J
= 
10
13
1.49 10
1.6 10
-
-
´
´
 MeV = 931.49 MeV
7. (b). E = mc
2
       = (9.1 × 10
–31
) (3 × 10
8
)
2
 J = 0.51 MeV
8. (c). DE = D mc
2
Dm = 
0.5
100
kg = 0.005 kg
c = 3 × 10
8
 m/s
DE = 0.005 × (3 × 10
8
)
2
DE = 4.5 × 10
14
 J or watt-sec
DE = 
14
4.5 10
60 60
´
´
 = 1.25 × 10
111
 watt hour
DE = 1.25 × 10
8
 kWH
9. (b). By the forumula N = N
0
e
–lt
Given 
0
N
N
 = 
1
20
 and l = 
0.6931
3.8
 Þ 20 = 
0.631×
3.8
t
e
Taking log of both sides
or log 20 = 10
0.6931 ×
log
3.8
t
e
or 1.3010 = 
0.6931 × 0.4343
3.8
t ´
 Þ t = 16.5 days
10. (b). A = 238 - 4 = 234, Z = 92 - 2 = 90
11. (a). Dm = 0.03 a.m.u., A = 4
Þ DE = 
m 931
A
D´
Þ DE = 
0.03 931
4
´
 = 7 MeV
12. (a). Q DE = Dm × 931 MeV
Þ Dm = 
E 2.23
931 931
D
= = 0.0024 a.m.u.
13. (a). 
1/3
Al
1/3
Te
R (27) 36
R 5 10
(125)
= ==
14. (d). R = R
0
 A
1/3
  = 1.2 × 10
-15
 × (64)
1/3
        = 1.2 × 10
–15
 × 4 = 4.8 fm
15. (b). Number of protrons in nucleus = atomic number = 11
Number of electrons = number of protons = 11.
Number of neutrons = mass number A – atomic number Z
N = 24 – 11 = 13
16. (d). Q equivalent mass of each photon = 1/2000 amu
Q 1 amu = 931 MeV
\ Energy of each photon = 
931
2000
 = 0.465 MeV
17. (c). Deuterium, the isotope of hydrogen consits of one proton
and neutron. Therefore mass of nuclear constituents of
deuterium = mass of proton + mass of neutron
    = 1.00759 + 1.00898 = 2.01657 amu.
mass of nucleus of deuterium = 2.01470 amu.
Mass defect = 2.01657 – 2.01470 = 0.00187 amu.
Binding energy = DE = 0.00187 × 931 MeV = 1.741 MeV .
18. (a). E = 
E m 931
AA
D D´
= MeV
Dm = (3m
p
 + 4m
n
) – mass of Li
7
Dm = (3 × 1.00759 + 4 × 1.00898) – 7.01653
Dm = 0.04216 a.m.u.
DE = 
0.04216 931 39.25
77
´
= = 5.6 MeV
DPP/ P 56
156
19. (d). The sun radiates energy in all directions in a sphere. At
a distance R, the energy received per unit area per
second is 1.4 KJ (given). Therefore the energy released
in area 4pR
2
 per sec is = 1400 × 4pR
2
 Joule the energy
released per day      = 1400 × 4pR
2
 × 86400J
where   R = 1.5 × 10
11
 m, Thus
DE = 1400×4 × 3.14 × (1.5 × 10
11
)
2
 × 86400
The equivalent mass is
Dm = DE/c
2
Dm = 
112
16
1400 4 3.14 (1.5 10 ) 86400
9 10
´´ ´ ´´
´
Dm = 3.8 × 10
14
 kg
20. (b) 
Fission
+
+
Fusion
A
.. BE
A
21. (c) 
/ 10/5
0
11
50000
22
tT
t
NN
æö æö
==
ç÷ ç÷
èø èø
= 12500
22. (d). Power received from the reactor,
P = 1000 KW = 1000 × 1000 W = 10
6
 J/s
P = 
6
19
10
1.6 10
-
´
 eV/sec.
P = 6.25 × 10
18
 MeV/sec
\ number of atoms disintegrated per sec
  = 
18
6.25 10
200
´
 = 3.125 × 10
16
Energy released per hour = 10
6
 × 60 × 60 Joule
Mass decay per hour = Dm = 
2
E
c
D
Þ Dm = 
6
82
10 60 60
(3 10)
´´
´
Þ Dm = 4 × 10
–8
 kg
23. (a)
24. (a) In fusion two lighter nuclei combines, it is not the
radioactive decay.
25. (c) The number of 
12
C atoms in 1g of carbon,
23
6.022 10
1
12 12
A
N
N mN
´
=´Þ=´
     = 5.02 × 10
22
 atoms.
The ratio of 
14
C/
12
C atoms = 1.3 × 10
–12
 (Given)
\ Number of 
14
C atoms   = 5.02 × 10
22
 × 1.3 × 10
–
12
      = 6.5 × 10
10
\  Rate of decay R
0
 = lN
0
 = 0
1/2
0.693
N
T
\  R
0
 = 
10
0.693 6.5 10
5730 365 24 3600
´´
´ ´´
       0.25 Bq 0.25 ==  (decays/s)
26. (c) For 10g sample, number of decays = 0.5 per second.
i.e. R = 0.05 and R
0
 = 0.25 for each gram of 
14
C
00
0 1/2
ln(/ ) ln(/ ) 1
1 (0.693/ )
t
RR RR R
et
RT
-l
=Þ==
l
Þ 
5730 years 0.25
ln
0.693 0.05
æö
=´
ç÷
èø
t = 13310 years
27. (d) If there are no other radioactive ingredients, the sample
is very recent. But the error of measurement must be
high unless the statistical error itself is large. In any
case, for an old sample, the activity will not be higher
than that of a recent one.
28. (d) The penetrating power is maximum in case of gamma
rays because gamma rays are an electromagnetic
radiation of very small wavelength.
29. (b) b-particles, being emitted with very high velocity (up
to 0.99 c). So, according to Einstein's theory of
relatively, the mass of a b-particle is much higher
compared to is its rest mass (m
0
). The velocity of
electrons obtained by other means is very small
compared to c (V elocity of light). So its mass remains
nearly m
0
. But b-particle and electron both are similar
particles.
30. (b) Electron capture occurs more often than positron
emission in heavy elements. This is because if positron
emission is energetically allowed, electron capture is
necessarily allowed, but the reverse is not true i.e. when
electron caputre is energetically allowed, positron
emission is not necessarily allowed.