Nuclei Practice Questions - DPP for JEE

 Page 1


1. (c) Energy is released in a process when total binding energy (BE) of products is more than the
reactants. By calculations we can see that this happens in option (c).
Given W = 2Y
BE of reactants = 120 × 7.5 = 900 MeV
BE of products = 2 × (60 × 8.5) = 1020 MeV.
2. (c)
If a and B are emitted simultaneously.
3. (c) No. of nuclide at time t is given by  N = N
o
e
–?t
Where  N
o
 = initial nuclide
This equation is equivalent to y = ae
–kx
Thus correct graph is
4. (b) By conservation of energy,
,
where v is the speed of the daughter nuclei
5. (d) Radioactivity at T
1
 , R
1
 = ? N
1
Radioactivity at T
2
, R
2
 = ? N
2
?  Number of atoms decayed in time
(T
1
 – T
2
) = (N
1
 –N
2
)
=  
Page 2


1. (c) Energy is released in a process when total binding energy (BE) of products is more than the
reactants. By calculations we can see that this happens in option (c).
Given W = 2Y
BE of reactants = 120 × 7.5 = 900 MeV
BE of products = 2 × (60 × 8.5) = 1020 MeV.
2. (c)
If a and B are emitted simultaneously.
3. (c) No. of nuclide at time t is given by  N = N
o
e
–?t
Where  N
o
 = initial nuclide
This equation is equivalent to y = ae
–kx
Thus correct graph is
4. (b) By conservation of energy,
,
where v is the speed of the daughter nuclei
5. (d) Radioactivity at T
1
 , R
1
 = ? N
1
Radioactivity at T
2
, R
2
 = ? N
2
?  Number of atoms decayed in time
(T
1
 – T
2
) = (N
1
 –N
2
)
=  
6. (d)
 ;  ; R
1
 : R
2
 = 1 : 2
1/3
7. (c) The range of energy of ß-particles is from zero to some maximum
value.
8. (a) Mass defect = ZM
p
 + (A –Z)M
n
–M(A,Z)
or, = ZM
p
 + (A–Z) M
n
–M(A,Z)
? M (A, Z) = ZM
p
 + (A–Z)M
n
–
9. (a) As we know,  R = R
0
 (A)
1/3
where A = mass number
R
AI
 = R
0
 (27)
1/3
 = 3R
0
R
Te
 = R
0
 (125)
1/3
 = 5R
0
 = R
AI
10. (c)
11. (d)
12. (c) We assume that mass number of  nucleus when it was at rest = A
 Mass number of -particle = 4
 Mass number of remaining nucleus = A – 4
As there is no external force, so momentum of  the  system will remain
conserved.
 
negative sign represents  that  direction is opposite to the direction of
motion of -particle.
13. (d) Because radioactivity is a spontaneous phenomenon.
Page 3


1. (c) Energy is released in a process when total binding energy (BE) of products is more than the
reactants. By calculations we can see that this happens in option (c).
Given W = 2Y
BE of reactants = 120 × 7.5 = 900 MeV
BE of products = 2 × (60 × 8.5) = 1020 MeV.
2. (c)
If a and B are emitted simultaneously.
3. (c) No. of nuclide at time t is given by  N = N
o
e
–?t
Where  N
o
 = initial nuclide
This equation is equivalent to y = ae
–kx
Thus correct graph is
4. (b) By conservation of energy,
,
where v is the speed of the daughter nuclei
5. (d) Radioactivity at T
1
 , R
1
 = ? N
1
Radioactivity at T
2
, R
2
 = ? N
2
?  Number of atoms decayed in time
(T
1
 – T
2
) = (N
1
 –N
2
)
=  
6. (d)
 ;  ; R
1
 : R
2
 = 1 : 2
1/3
7. (c) The range of energy of ß-particles is from zero to some maximum
value.
8. (a) Mass defect = ZM
p
 + (A –Z)M
n
–M(A,Z)
or, = ZM
p
 + (A–Z) M
n
–M(A,Z)
? M (A, Z) = ZM
p
 + (A–Z)M
n
–
9. (a) As we know,  R = R
0
 (A)
1/3
where A = mass number
R
AI
 = R
0
 (27)
1/3
 = 3R
0
R
Te
 = R
0
 (125)
1/3
 = 5R
0
 = R
AI
10. (c)
11. (d)
12. (c) We assume that mass number of  nucleus when it was at rest = A
 Mass number of -particle = 4
 Mass number of remaining nucleus = A – 4
As there is no external force, so momentum of  the  system will remain
conserved.
 
negative sign represents  that  direction is opposite to the direction of
motion of -particle.
13. (d) Because radioactivity is a spontaneous phenomenon.
14. (a)
15. (d) Let at time t
1
 & t
2
, number of particles be N
1
 & N
2
. So,
16. (a) Iodine and Yttrium are medium sized nuclei and therefore, have
more binding energy per nucleon as compared to Uranium which
has a big nuclei and less B.E./nucleon.
17. (d) Nuclear force is not the same between any two nucleons.
18. (c) Binding energy per nucleon for fission products is higher relative to
Binding energy per nucleon for parent nucleus, i.e., more masses are
lost and are obtained as kinetic energy of fission products. So, the
given ratio < 1.
19. (d) For  is positive. This is because E
b
 for C is
greater than the E
b
 for A and B.
Again for  is positive. This is because E
b
 for D and E
is greater than E
b
 for F .
20. (a) = 
As v
2
 is zero, m
2
 > m
1
, v ‘
1
 is in the opposite direction.
m
1
 = 1, m
2
 = A.
?
The fraction of total energy retained is
Page 4


1. (c) Energy is released in a process when total binding energy (BE) of products is more than the
reactants. By calculations we can see that this happens in option (c).
Given W = 2Y
BE of reactants = 120 × 7.5 = 900 MeV
BE of products = 2 × (60 × 8.5) = 1020 MeV.
2. (c)
If a and B are emitted simultaneously.
3. (c) No. of nuclide at time t is given by  N = N
o
e
–?t
Where  N
o
 = initial nuclide
This equation is equivalent to y = ae
–kx
Thus correct graph is
4. (b) By conservation of energy,
,
where v is the speed of the daughter nuclei
5. (d) Radioactivity at T
1
 , R
1
 = ? N
1
Radioactivity at T
2
, R
2
 = ? N
2
?  Number of atoms decayed in time
(T
1
 – T
2
) = (N
1
 –N
2
)
=  
6. (d)
 ;  ; R
1
 : R
2
 = 1 : 2
1/3
7. (c) The range of energy of ß-particles is from zero to some maximum
value.
8. (a) Mass defect = ZM
p
 + (A –Z)M
n
–M(A,Z)
or, = ZM
p
 + (A–Z) M
n
–M(A,Z)
? M (A, Z) = ZM
p
 + (A–Z)M
n
–
9. (a) As we know,  R = R
0
 (A)
1/3
where A = mass number
R
AI
 = R
0
 (27)
1/3
 = 3R
0
R
Te
 = R
0
 (125)
1/3
 = 5R
0
 = R
AI
10. (c)
11. (d)
12. (c) We assume that mass number of  nucleus when it was at rest = A
 Mass number of -particle = 4
 Mass number of remaining nucleus = A – 4
As there is no external force, so momentum of  the  system will remain
conserved.
 
negative sign represents  that  direction is opposite to the direction of
motion of -particle.
13. (d) Because radioactivity is a spontaneous phenomenon.
14. (a)
15. (d) Let at time t
1
 & t
2
, number of particles be N
1
 & N
2
. So,
16. (a) Iodine and Yttrium are medium sized nuclei and therefore, have
more binding energy per nucleon as compared to Uranium which
has a big nuclei and less B.E./nucleon.
17. (d) Nuclear force is not the same between any two nucleons.
18. (c) Binding energy per nucleon for fission products is higher relative to
Binding energy per nucleon for parent nucleus, i.e., more masses are
lost and are obtained as kinetic energy of fission products. So, the
given ratio < 1.
19. (d) For  is positive. This is because E
b
 for C is
greater than the E
b
 for A and B.
Again for  is positive. This is because E
b
 for D and E
is greater than E
b
 for F .
20. (a) = 
As v
2
 is zero, m
2
 > m
1
, v ‘
1
 is in the opposite direction.
m
1
 = 1, m
2
 = A.
?
The fraction of total energy retained is
21. (69)
22. (0.511)
The mass defect during the process
 = 1.6725 × 10
–27
 – (1.6725 × 10
–27
+ 9 × 10
–
31
kg)
= – 9 × 10
–31
 kg
The energy released during the process
E = ?mc
2
E =  9 × 10
–31
× 9 × 10
16
 = 81 × 10
–15
 Joules
E = 
23. (78) The number of  a- particles released =8
Therefore the atomic number should decrease by 16
The number of ß
–
-particles released = 4
Therefore the atomic number should increase by 4.
Also the number of ß
+
 particles released is 2, which should decrease the
atomic number by 2.
Therefore the final atomic number
= Z –16 + 4 – 2 = Z –14
= 92 – 14  =  78
24. (3.125 × 10
13
)
25. (0.693) Average life of the nuclei is
Page 5


1. (c) Energy is released in a process when total binding energy (BE) of products is more than the
reactants. By calculations we can see that this happens in option (c).
Given W = 2Y
BE of reactants = 120 × 7.5 = 900 MeV
BE of products = 2 × (60 × 8.5) = 1020 MeV.
2. (c)
If a and B are emitted simultaneously.
3. (c) No. of nuclide at time t is given by  N = N
o
e
–?t
Where  N
o
 = initial nuclide
This equation is equivalent to y = ae
–kx
Thus correct graph is
4. (b) By conservation of energy,
,
where v is the speed of the daughter nuclei
5. (d) Radioactivity at T
1
 , R
1
 = ? N
1
Radioactivity at T
2
, R
2
 = ? N
2
?  Number of atoms decayed in time
(T
1
 – T
2
) = (N
1
 –N
2
)
=  
6. (d)
 ;  ; R
1
 : R
2
 = 1 : 2
1/3
7. (c) The range of energy of ß-particles is from zero to some maximum
value.
8. (a) Mass defect = ZM
p
 + (A –Z)M
n
–M(A,Z)
or, = ZM
p
 + (A–Z) M
n
–M(A,Z)
? M (A, Z) = ZM
p
 + (A–Z)M
n
–
9. (a) As we know,  R = R
0
 (A)
1/3
where A = mass number
R
AI
 = R
0
 (27)
1/3
 = 3R
0
R
Te
 = R
0
 (125)
1/3
 = 5R
0
 = R
AI
10. (c)
11. (d)
12. (c) We assume that mass number of  nucleus when it was at rest = A
 Mass number of -particle = 4
 Mass number of remaining nucleus = A – 4
As there is no external force, so momentum of  the  system will remain
conserved.
 
negative sign represents  that  direction is opposite to the direction of
motion of -particle.
13. (d) Because radioactivity is a spontaneous phenomenon.
14. (a)
15. (d) Let at time t
1
 & t
2
, number of particles be N
1
 & N
2
. So,
16. (a) Iodine and Yttrium are medium sized nuclei and therefore, have
more binding energy per nucleon as compared to Uranium which
has a big nuclei and less B.E./nucleon.
17. (d) Nuclear force is not the same between any two nucleons.
18. (c) Binding energy per nucleon for fission products is higher relative to
Binding energy per nucleon for parent nucleus, i.e., more masses are
lost and are obtained as kinetic energy of fission products. So, the
given ratio < 1.
19. (d) For  is positive. This is because E
b
 for C is
greater than the E
b
 for A and B.
Again for  is positive. This is because E
b
 for D and E
is greater than E
b
 for F .
20. (a) = 
As v
2
 is zero, m
2
 > m
1
, v ‘
1
 is in the opposite direction.
m
1
 = 1, m
2
 = A.
?
The fraction of total energy retained is
21. (69)
22. (0.511)
The mass defect during the process
 = 1.6725 × 10
–27
 – (1.6725 × 10
–27
+ 9 × 10
–
31
kg)
= – 9 × 10
–31
 kg
The energy released during the process
E = ?mc
2
E =  9 × 10
–31
× 9 × 10
16
 = 81 × 10
–15
 Joules
E = 
23. (78) The number of  a- particles released =8
Therefore the atomic number should decrease by 16
The number of ß
–
-particles released = 4
Therefore the atomic number should increase by 4.
Also the number of ß
+
 particles released is 2, which should decrease the
atomic number by 2.
Therefore the final atomic number
= Z –16 + 4 – 2 = Z –14
= 92 – 14  =  78
24. (3.125 × 10
13
)
25. (0.693) Average life of the nuclei is
t
av
 = ....(i)
Half life of the nuclei
t
1/2
 = ....(ii)
from (i) and (ii)
t
av
 =