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JEE Main Previous Year Questions (2025): Nuclei

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 Page 1


JEE Main Previous Year Qs (2025): 
Nuclei 
Q1: A radioactive nucleus ?? ?? has 3 times the decay constant as compared to the decay 
constant of another radioactive nucleus ?? ?? . If initial number of both nuclei are the same, 
what is the ratio of number of nuclei of ?? ?? to the number of nuclei of ?? ?? , after one half-life of 
?? ?? ? 
JEE Main 2025 (Online) 23rd January Morning Shift 
Options: 
A. 1 / 4 
B. 1 / 8 
C. 4 
D. 8 
Ans: A 
Solution: 
Let's work through the problem step by step. 
The decay of a radioactive nucleus is given by the formula: 
?? ( ?? ) = ?? 0
?? - ????
, 
where: 
?? ( ?? ) is the number of nuclei remaining at time ?? , 
?? 0
 is the initial number of nuclei, 
?? is the decay constant. 
For nucleus ?? 1
 with decay constant ?? 1
, its half-life ?? 1 / 2
 is: 
?? 1 / 2
=
ln ? 2
?? 1
. 
After one half-life of ?? 1
, the number of ?? 1
 nuclei remaining is: 
?? 1
= ?? 0
?? - ?? 1
?? 1 / 2
= ?? 0
?? - ?? 1
ln ? 2
?? 1
= ?? 0
?? - ln ? 2
=
?? 0
2
. 
Nucleus ?? 2
 has a decay constant that is 3 times that of ?? 1
 : 
?? 2
= 3 ?? 1
. 
For nucleus ?? 2
, the number of nuclei remaining after one half-life of ?? 1
 (which is ?? =
ln ? 2
?? 1
 ) is: 
?? 2
= ?? 0
?? - ?? 2
?? = ?? 0
?? - 3 ?? 1
ln ? 2
?? 1
= ?? 0
?? - 3 l n ? 2
= ?? 0
·
1
2
3
=
?? 0
8
. 
Now, we find the ratio of ?? 2
 nuclei to ?? 1
 nuclei after this time: 
?? 2
?? 1
=
?? 0
8
?? 0
2
=
1 / 8
1 / 2
=
1
4
. 
Therefore, the correct ratio is 
1
4
, which corresponds to Option A. 
Q2: Given below are two statements. One is labelled as Assertion (A) and the other is labelled 
as Reason (R). 
Assertion (A): The binding energy per nucleon is found to be practically independent of the 
atomic number ?? , for nuclei with mass numbers between 30 and 170 . 
Page 2


JEE Main Previous Year Qs (2025): 
Nuclei 
Q1: A radioactive nucleus ?? ?? has 3 times the decay constant as compared to the decay 
constant of another radioactive nucleus ?? ?? . If initial number of both nuclei are the same, 
what is the ratio of number of nuclei of ?? ?? to the number of nuclei of ?? ?? , after one half-life of 
?? ?? ? 
JEE Main 2025 (Online) 23rd January Morning Shift 
Options: 
A. 1 / 4 
B. 1 / 8 
C. 4 
D. 8 
Ans: A 
Solution: 
Let's work through the problem step by step. 
The decay of a radioactive nucleus is given by the formula: 
?? ( ?? ) = ?? 0
?? - ????
, 
where: 
?? ( ?? ) is the number of nuclei remaining at time ?? , 
?? 0
 is the initial number of nuclei, 
?? is the decay constant. 
For nucleus ?? 1
 with decay constant ?? 1
, its half-life ?? 1 / 2
 is: 
?? 1 / 2
=
ln ? 2
?? 1
. 
After one half-life of ?? 1
, the number of ?? 1
 nuclei remaining is: 
?? 1
= ?? 0
?? - ?? 1
?? 1 / 2
= ?? 0
?? - ?? 1
ln ? 2
?? 1
= ?? 0
?? - ln ? 2
=
?? 0
2
. 
Nucleus ?? 2
 has a decay constant that is 3 times that of ?? 1
 : 
?? 2
= 3 ?? 1
. 
For nucleus ?? 2
, the number of nuclei remaining after one half-life of ?? 1
 (which is ?? =
ln ? 2
?? 1
 ) is: 
?? 2
= ?? 0
?? - ?? 2
?? = ?? 0
?? - 3 ?? 1
ln ? 2
?? 1
= ?? 0
?? - 3 l n ? 2
= ?? 0
·
1
2
3
=
?? 0
8
. 
Now, we find the ratio of ?? 2
 nuclei to ?? 1
 nuclei after this time: 
?? 2
?? 1
=
?? 0
8
?? 0
2
=
1 / 8
1 / 2
=
1
4
. 
Therefore, the correct ratio is 
1
4
, which corresponds to Option A. 
Q2: Given below are two statements. One is labelled as Assertion (A) and the other is labelled 
as Reason (R). 
Assertion (A): The binding energy per nucleon is found to be practically independent of the 
atomic number ?? , for nuclei with mass numbers between 30 and 170 . 
Reason ( ?? ) : Nuclear force is long range. In the light of the above statements, choose the 
correct answer from the options given below : 
JEE Main 2025 (Online) 23rd January Evening Shift 
Options: 
A. (A) is true but (R) is false 
B. Both (A) and (R) are true and (R) is the correct explanation of (A) 
C. (A) is false but (R) is true 
D. Both (A) and (R) are true but (R) is NOT the correct explanation of (A) 
Ans: A 
Solution: 
Let's analyze both statements: 
Assertion (A): "The binding energy per nucleon is found to be practically independent of the atomic 
number A, for nuclei with mass numbers between 30 and 170." 
In fact, for nuclei in this mass range, the binding energy per nucleon is nearly constant (around 7 -
8MeV per nucleon). This is because the attractive nuclear force saturates, leading to a relatively uniform 
binding energy per nucleon in medium to heavy nuclei. 
Therefore, Assertion (A) is true. 
Reason (R): "Nuclear force is long range." 
The nuclear force (or strong force) that holds the nucleons (protons and neutrons) together is actually 
very short-ranged. Its effective range is approximately 1 to 3 femtometers (fm), after which the force 
rapidly decreases. 
Hence, Reason (R) is false. 
Since the assertion is correct and the reason is incorrect, the correct answer is: 
Option A: (A) is true but (R) is false. 
Q3: Choose the correct nuclear process from the below options [ ?? : proton, ?? : neutron, ?? -
: 
electron, ?? +
: positron, ?? : neutrino, ?? ? : antineutrino] 
JEE Main 2025 (Online) 28th January Morning Shift 
Options: 
A. ?? ? ?? + ?? +
+ ?? 
B. n ? p + e
-
+ ?? ? 
C. n ? p + e
+
+ ?? ? 
D. n ? p + e
-
+ ?? 
Ans: B 
Solution: 
We know, theoretical equation for ?? -
decay, 
Page 3


JEE Main Previous Year Qs (2025): 
Nuclei 
Q1: A radioactive nucleus ?? ?? has 3 times the decay constant as compared to the decay 
constant of another radioactive nucleus ?? ?? . If initial number of both nuclei are the same, 
what is the ratio of number of nuclei of ?? ?? to the number of nuclei of ?? ?? , after one half-life of 
?? ?? ? 
JEE Main 2025 (Online) 23rd January Morning Shift 
Options: 
A. 1 / 4 
B. 1 / 8 
C. 4 
D. 8 
Ans: A 
Solution: 
Let's work through the problem step by step. 
The decay of a radioactive nucleus is given by the formula: 
?? ( ?? ) = ?? 0
?? - ????
, 
where: 
?? ( ?? ) is the number of nuclei remaining at time ?? , 
?? 0
 is the initial number of nuclei, 
?? is the decay constant. 
For nucleus ?? 1
 with decay constant ?? 1
, its half-life ?? 1 / 2
 is: 
?? 1 / 2
=
ln ? 2
?? 1
. 
After one half-life of ?? 1
, the number of ?? 1
 nuclei remaining is: 
?? 1
= ?? 0
?? - ?? 1
?? 1 / 2
= ?? 0
?? - ?? 1
ln ? 2
?? 1
= ?? 0
?? - ln ? 2
=
?? 0
2
. 
Nucleus ?? 2
 has a decay constant that is 3 times that of ?? 1
 : 
?? 2
= 3 ?? 1
. 
For nucleus ?? 2
, the number of nuclei remaining after one half-life of ?? 1
 (which is ?? =
ln ? 2
?? 1
 ) is: 
?? 2
= ?? 0
?? - ?? 2
?? = ?? 0
?? - 3 ?? 1
ln ? 2
?? 1
= ?? 0
?? - 3 l n ? 2
= ?? 0
·
1
2
3
=
?? 0
8
. 
Now, we find the ratio of ?? 2
 nuclei to ?? 1
 nuclei after this time: 
?? 2
?? 1
=
?? 0
8
?? 0
2
=
1 / 8
1 / 2
=
1
4
. 
Therefore, the correct ratio is 
1
4
, which corresponds to Option A. 
Q2: Given below are two statements. One is labelled as Assertion (A) and the other is labelled 
as Reason (R). 
Assertion (A): The binding energy per nucleon is found to be practically independent of the 
atomic number ?? , for nuclei with mass numbers between 30 and 170 . 
Reason ( ?? ) : Nuclear force is long range. In the light of the above statements, choose the 
correct answer from the options given below : 
JEE Main 2025 (Online) 23rd January Evening Shift 
Options: 
A. (A) is true but (R) is false 
B. Both (A) and (R) are true and (R) is the correct explanation of (A) 
C. (A) is false but (R) is true 
D. Both (A) and (R) are true but (R) is NOT the correct explanation of (A) 
Ans: A 
Solution: 
Let's analyze both statements: 
Assertion (A): "The binding energy per nucleon is found to be practically independent of the atomic 
number A, for nuclei with mass numbers between 30 and 170." 
In fact, for nuclei in this mass range, the binding energy per nucleon is nearly constant (around 7 -
8MeV per nucleon). This is because the attractive nuclear force saturates, leading to a relatively uniform 
binding energy per nucleon in medium to heavy nuclei. 
Therefore, Assertion (A) is true. 
Reason (R): "Nuclear force is long range." 
The nuclear force (or strong force) that holds the nucleons (protons and neutrons) together is actually 
very short-ranged. Its effective range is approximately 1 to 3 femtometers (fm), after which the force 
rapidly decreases. 
Hence, Reason (R) is false. 
Since the assertion is correct and the reason is incorrect, the correct answer is: 
Option A: (A) is true but (R) is false. 
Q3: Choose the correct nuclear process from the below options [ ?? : proton, ?? : neutron, ?? -
: 
electron, ?? +
: positron, ?? : neutrino, ?? ? : antineutrino] 
JEE Main 2025 (Online) 28th January Morning Shift 
Options: 
A. ?? ? ?? + ?? +
+ ?? 
B. n ? p + e
-
+ ?? ? 
C. n ? p + e
+
+ ?? ? 
D. n ? p + e
-
+ ?? 
Ans: B 
Solution: 
We know, theoretical equation for ?? -
decay, 
 
Hence, option 2 is correct. 
Q4: Energy released when two deuterons ( ?
?? ?? ?? ) fuse to form a helium nucleus ( ?
?? ????
?? ) is 
:(Given : Binding energy per nucleon of ?
?? ?? ?? = ?? . ?????? ?? and binding energy per nucleon of 
?
?? ????
?? = ?? . ?????? ?? ) 
JEE Main 2025 (Online) 2nd April Evening Shift 
Options: 
A. 26.8 MeV 
B. 8.1 MeV 
C. 23.6 MeV 
D. 5.9 MeV 
Ans: C 
Solution: 
To calculate the energy released when two deuterons ( ?
1
H
2
) fuse to form a helium nucleus ( ?
2
He
4
), 
consider the following: 
The reaction is represented as: 
?
1
?? 2
+ ?
1
?? 2
? ?
2
?? ?? 4
 
Given the binding energy per nucleon: 
For ?
1
H
2
: 1 . 1MeV 
For ?
2
He
4
: 7 . 0MeV 
Calculating Binding Energy 
Binding Energy for Reactants: 
Each deuteron ( ?
1
H
2
 ) has a binding energy of 1.1 MeV per nucleon. 
Since there are two nucleons in a deuteron, the total binding energy for one deuteron is 1 . 1 × 2 =
2 . 2MeV. 
Therefore, for two deuterons: 2 . 2 × 2 = 4 . 4MeV. 
Binding Energy for Product: 
For ?
2
He
4
 : Each of the four nucleons has a binding energy of 7.0 MeV . 
Total binding energy for the helium nucleus: 7 . 0 × 4 = 28 . 0M eV. 
Energy Released (Q) 
The energy released, Q , is the difference between the binding energy of the products and the reactants: 
Q = BE
product 
- BE
reactants 
= 28 . 0MeV - 4 . 4Me V = 23 . 6MeV 
Thus, the energy released when two deuterons fuse to form a helium nucleus is ???? . ?? ?????? . 
 
 
Page 4


JEE Main Previous Year Qs (2025): 
Nuclei 
Q1: A radioactive nucleus ?? ?? has 3 times the decay constant as compared to the decay 
constant of another radioactive nucleus ?? ?? . If initial number of both nuclei are the same, 
what is the ratio of number of nuclei of ?? ?? to the number of nuclei of ?? ?? , after one half-life of 
?? ?? ? 
JEE Main 2025 (Online) 23rd January Morning Shift 
Options: 
A. 1 / 4 
B. 1 / 8 
C. 4 
D. 8 
Ans: A 
Solution: 
Let's work through the problem step by step. 
The decay of a radioactive nucleus is given by the formula: 
?? ( ?? ) = ?? 0
?? - ????
, 
where: 
?? ( ?? ) is the number of nuclei remaining at time ?? , 
?? 0
 is the initial number of nuclei, 
?? is the decay constant. 
For nucleus ?? 1
 with decay constant ?? 1
, its half-life ?? 1 / 2
 is: 
?? 1 / 2
=
ln ? 2
?? 1
. 
After one half-life of ?? 1
, the number of ?? 1
 nuclei remaining is: 
?? 1
= ?? 0
?? - ?? 1
?? 1 / 2
= ?? 0
?? - ?? 1
ln ? 2
?? 1
= ?? 0
?? - ln ? 2
=
?? 0
2
. 
Nucleus ?? 2
 has a decay constant that is 3 times that of ?? 1
 : 
?? 2
= 3 ?? 1
. 
For nucleus ?? 2
, the number of nuclei remaining after one half-life of ?? 1
 (which is ?? =
ln ? 2
?? 1
 ) is: 
?? 2
= ?? 0
?? - ?? 2
?? = ?? 0
?? - 3 ?? 1
ln ? 2
?? 1
= ?? 0
?? - 3 l n ? 2
= ?? 0
·
1
2
3
=
?? 0
8
. 
Now, we find the ratio of ?? 2
 nuclei to ?? 1
 nuclei after this time: 
?? 2
?? 1
=
?? 0
8
?? 0
2
=
1 / 8
1 / 2
=
1
4
. 
Therefore, the correct ratio is 
1
4
, which corresponds to Option A. 
Q2: Given below are two statements. One is labelled as Assertion (A) and the other is labelled 
as Reason (R). 
Assertion (A): The binding energy per nucleon is found to be practically independent of the 
atomic number ?? , for nuclei with mass numbers between 30 and 170 . 
Reason ( ?? ) : Nuclear force is long range. In the light of the above statements, choose the 
correct answer from the options given below : 
JEE Main 2025 (Online) 23rd January Evening Shift 
Options: 
A. (A) is true but (R) is false 
B. Both (A) and (R) are true and (R) is the correct explanation of (A) 
C. (A) is false but (R) is true 
D. Both (A) and (R) are true but (R) is NOT the correct explanation of (A) 
Ans: A 
Solution: 
Let's analyze both statements: 
Assertion (A): "The binding energy per nucleon is found to be practically independent of the atomic 
number A, for nuclei with mass numbers between 30 and 170." 
In fact, for nuclei in this mass range, the binding energy per nucleon is nearly constant (around 7 -
8MeV per nucleon). This is because the attractive nuclear force saturates, leading to a relatively uniform 
binding energy per nucleon in medium to heavy nuclei. 
Therefore, Assertion (A) is true. 
Reason (R): "Nuclear force is long range." 
The nuclear force (or strong force) that holds the nucleons (protons and neutrons) together is actually 
very short-ranged. Its effective range is approximately 1 to 3 femtometers (fm), after which the force 
rapidly decreases. 
Hence, Reason (R) is false. 
Since the assertion is correct and the reason is incorrect, the correct answer is: 
Option A: (A) is true but (R) is false. 
Q3: Choose the correct nuclear process from the below options [ ?? : proton, ?? : neutron, ?? -
: 
electron, ?? +
: positron, ?? : neutrino, ?? ? : antineutrino] 
JEE Main 2025 (Online) 28th January Morning Shift 
Options: 
A. ?? ? ?? + ?? +
+ ?? 
B. n ? p + e
-
+ ?? ? 
C. n ? p + e
+
+ ?? ? 
D. n ? p + e
-
+ ?? 
Ans: B 
Solution: 
We know, theoretical equation for ?? -
decay, 
 
Hence, option 2 is correct. 
Q4: Energy released when two deuterons ( ?
?? ?? ?? ) fuse to form a helium nucleus ( ?
?? ????
?? ) is 
:(Given : Binding energy per nucleon of ?
?? ?? ?? = ?? . ?????? ?? and binding energy per nucleon of 
?
?? ????
?? = ?? . ?????? ?? ) 
JEE Main 2025 (Online) 2nd April Evening Shift 
Options: 
A. 26.8 MeV 
B. 8.1 MeV 
C. 23.6 MeV 
D. 5.9 MeV 
Ans: C 
Solution: 
To calculate the energy released when two deuterons ( ?
1
H
2
) fuse to form a helium nucleus ( ?
2
He
4
), 
consider the following: 
The reaction is represented as: 
?
1
?? 2
+ ?
1
?? 2
? ?
2
?? ?? 4
 
Given the binding energy per nucleon: 
For ?
1
H
2
: 1 . 1MeV 
For ?
2
He
4
: 7 . 0MeV 
Calculating Binding Energy 
Binding Energy for Reactants: 
Each deuteron ( ?
1
H
2
 ) has a binding energy of 1.1 MeV per nucleon. 
Since there are two nucleons in a deuteron, the total binding energy for one deuteron is 1 . 1 × 2 =
2 . 2MeV. 
Therefore, for two deuterons: 2 . 2 × 2 = 4 . 4MeV. 
Binding Energy for Product: 
For ?
2
He
4
 : Each of the four nucleons has a binding energy of 7.0 MeV . 
Total binding energy for the helium nucleus: 7 . 0 × 4 = 28 . 0M eV. 
Energy Released (Q) 
The energy released, Q , is the difference between the binding energy of the products and the reactants: 
Q = BE
product 
- BE
reactants 
= 28 . 0MeV - 4 . 4Me V = 23 . 6MeV 
Thus, the energy released when two deuterons fuse to form a helium nucleus is ???? . ?? ?????? . 
 
 
Q5: Match the LIST-I with LIST-II 
List - I List - II 
A. ?
?? ?? ?? + ?
????
??????
?? ? ?
????
??????
???? + ?
????
????
???? + ?? ?
?? ?? ?? I. Chemical reaction 
B. ?? ?? ?? + ?? ?? ? ?? ?? ?? ?? II. Fusion with +ve Q value 
C. ?
?? ?? ?? + ?
?? ?? ?? ? ?
?? ?? ???? + ?
?? ?? ?? III. Fission 
D. ?
?? ?? ?? + ?
?? ?? ?? ? ?
?? ?? ?? + ?
?? ?? ?? IV. Fusion with -ve ?? value 
Choose the correct answer from the options given below: 
JEE Main 2025 (Online) 3rd April Morning Shift 
Options: 
A. A-II, B-I, C-III, D-IV 
B. A-III, B-I, C-II, D-IV 
C. A-III, B-I, C-IV, D-II 
D. A-II, B-I, C-IV, D-III 
Ans: B 
Solution: 
Here's the correct matching: 
A. 
?
0
1
n + ?
92
235
U ? ?
54
140
Xe + ?
38
94
Sr + 2 ?
0
1
n 
? This is nuclear fission ? III. 
B. 
2 H
2
+ O
2
? 2 H
2
O 
? This is a chemical reaction ? I. 
C. 
?
1
2
H + ?
1
2
H ? ?
2
3
He + ?
0
1
n 
? This D-D fusion releases about 3 . 3MeV ? fusion with + v eQ ? II. 
D. 
?
1
1
H + ?
1
3
H ? ?
1
2
H + ?
1
2
H 
? The mass of products exceeds reactants ? requires energy ? fusion with -ve Q ? IV. 
So the answer is 
Option B: A-III, B-I, C-II, D-IV. 
Q6: A radioactive material ?? first decays into ?? and then ?? decays to non-radioactive 
material ?? . Which of the following figure represents time dependent mass of ?? , ?? and ?? ? 
JEE Main 2025 (Online) 4th April Evening Shift 
Options: 
Page 5


JEE Main Previous Year Qs (2025): 
Nuclei 
Q1: A radioactive nucleus ?? ?? has 3 times the decay constant as compared to the decay 
constant of another radioactive nucleus ?? ?? . If initial number of both nuclei are the same, 
what is the ratio of number of nuclei of ?? ?? to the number of nuclei of ?? ?? , after one half-life of 
?? ?? ? 
JEE Main 2025 (Online) 23rd January Morning Shift 
Options: 
A. 1 / 4 
B. 1 / 8 
C. 4 
D. 8 
Ans: A 
Solution: 
Let's work through the problem step by step. 
The decay of a radioactive nucleus is given by the formula: 
?? ( ?? ) = ?? 0
?? - ????
, 
where: 
?? ( ?? ) is the number of nuclei remaining at time ?? , 
?? 0
 is the initial number of nuclei, 
?? is the decay constant. 
For nucleus ?? 1
 with decay constant ?? 1
, its half-life ?? 1 / 2
 is: 
?? 1 / 2
=
ln ? 2
?? 1
. 
After one half-life of ?? 1
, the number of ?? 1
 nuclei remaining is: 
?? 1
= ?? 0
?? - ?? 1
?? 1 / 2
= ?? 0
?? - ?? 1
ln ? 2
?? 1
= ?? 0
?? - ln ? 2
=
?? 0
2
. 
Nucleus ?? 2
 has a decay constant that is 3 times that of ?? 1
 : 
?? 2
= 3 ?? 1
. 
For nucleus ?? 2
, the number of nuclei remaining after one half-life of ?? 1
 (which is ?? =
ln ? 2
?? 1
 ) is: 
?? 2
= ?? 0
?? - ?? 2
?? = ?? 0
?? - 3 ?? 1
ln ? 2
?? 1
= ?? 0
?? - 3 l n ? 2
= ?? 0
·
1
2
3
=
?? 0
8
. 
Now, we find the ratio of ?? 2
 nuclei to ?? 1
 nuclei after this time: 
?? 2
?? 1
=
?? 0
8
?? 0
2
=
1 / 8
1 / 2
=
1
4
. 
Therefore, the correct ratio is 
1
4
, which corresponds to Option A. 
Q2: Given below are two statements. One is labelled as Assertion (A) and the other is labelled 
as Reason (R). 
Assertion (A): The binding energy per nucleon is found to be practically independent of the 
atomic number ?? , for nuclei with mass numbers between 30 and 170 . 
Reason ( ?? ) : Nuclear force is long range. In the light of the above statements, choose the 
correct answer from the options given below : 
JEE Main 2025 (Online) 23rd January Evening Shift 
Options: 
A. (A) is true but (R) is false 
B. Both (A) and (R) are true and (R) is the correct explanation of (A) 
C. (A) is false but (R) is true 
D. Both (A) and (R) are true but (R) is NOT the correct explanation of (A) 
Ans: A 
Solution: 
Let's analyze both statements: 
Assertion (A): "The binding energy per nucleon is found to be practically independent of the atomic 
number A, for nuclei with mass numbers between 30 and 170." 
In fact, for nuclei in this mass range, the binding energy per nucleon is nearly constant (around 7 -
8MeV per nucleon). This is because the attractive nuclear force saturates, leading to a relatively uniform 
binding energy per nucleon in medium to heavy nuclei. 
Therefore, Assertion (A) is true. 
Reason (R): "Nuclear force is long range." 
The nuclear force (or strong force) that holds the nucleons (protons and neutrons) together is actually 
very short-ranged. Its effective range is approximately 1 to 3 femtometers (fm), after which the force 
rapidly decreases. 
Hence, Reason (R) is false. 
Since the assertion is correct and the reason is incorrect, the correct answer is: 
Option A: (A) is true but (R) is false. 
Q3: Choose the correct nuclear process from the below options [ ?? : proton, ?? : neutron, ?? -
: 
electron, ?? +
: positron, ?? : neutrino, ?? ? : antineutrino] 
JEE Main 2025 (Online) 28th January Morning Shift 
Options: 
A. ?? ? ?? + ?? +
+ ?? 
B. n ? p + e
-
+ ?? ? 
C. n ? p + e
+
+ ?? ? 
D. n ? p + e
-
+ ?? 
Ans: B 
Solution: 
We know, theoretical equation for ?? -
decay, 
 
Hence, option 2 is correct. 
Q4: Energy released when two deuterons ( ?
?? ?? ?? ) fuse to form a helium nucleus ( ?
?? ????
?? ) is 
:(Given : Binding energy per nucleon of ?
?? ?? ?? = ?? . ?????? ?? and binding energy per nucleon of 
?
?? ????
?? = ?? . ?????? ?? ) 
JEE Main 2025 (Online) 2nd April Evening Shift 
Options: 
A. 26.8 MeV 
B. 8.1 MeV 
C. 23.6 MeV 
D. 5.9 MeV 
Ans: C 
Solution: 
To calculate the energy released when two deuterons ( ?
1
H
2
) fuse to form a helium nucleus ( ?
2
He
4
), 
consider the following: 
The reaction is represented as: 
?
1
?? 2
+ ?
1
?? 2
? ?
2
?? ?? 4
 
Given the binding energy per nucleon: 
For ?
1
H
2
: 1 . 1MeV 
For ?
2
He
4
: 7 . 0MeV 
Calculating Binding Energy 
Binding Energy for Reactants: 
Each deuteron ( ?
1
H
2
 ) has a binding energy of 1.1 MeV per nucleon. 
Since there are two nucleons in a deuteron, the total binding energy for one deuteron is 1 . 1 × 2 =
2 . 2MeV. 
Therefore, for two deuterons: 2 . 2 × 2 = 4 . 4MeV. 
Binding Energy for Product: 
For ?
2
He
4
 : Each of the four nucleons has a binding energy of 7.0 MeV . 
Total binding energy for the helium nucleus: 7 . 0 × 4 = 28 . 0M eV. 
Energy Released (Q) 
The energy released, Q , is the difference between the binding energy of the products and the reactants: 
Q = BE
product 
- BE
reactants 
= 28 . 0MeV - 4 . 4Me V = 23 . 6MeV 
Thus, the energy released when two deuterons fuse to form a helium nucleus is ???? . ?? ?????? . 
 
 
Q5: Match the LIST-I with LIST-II 
List - I List - II 
A. ?
?? ?? ?? + ?
????
??????
?? ? ?
????
??????
???? + ?
????
????
???? + ?? ?
?? ?? ?? I. Chemical reaction 
B. ?? ?? ?? + ?? ?? ? ?? ?? ?? ?? II. Fusion with +ve Q value 
C. ?
?? ?? ?? + ?
?? ?? ?? ? ?
?? ?? ???? + ?
?? ?? ?? III. Fission 
D. ?
?? ?? ?? + ?
?? ?? ?? ? ?
?? ?? ?? + ?
?? ?? ?? IV. Fusion with -ve ?? value 
Choose the correct answer from the options given below: 
JEE Main 2025 (Online) 3rd April Morning Shift 
Options: 
A. A-II, B-I, C-III, D-IV 
B. A-III, B-I, C-II, D-IV 
C. A-III, B-I, C-IV, D-II 
D. A-II, B-I, C-IV, D-III 
Ans: B 
Solution: 
Here's the correct matching: 
A. 
?
0
1
n + ?
92
235
U ? ?
54
140
Xe + ?
38
94
Sr + 2 ?
0
1
n 
? This is nuclear fission ? III. 
B. 
2 H
2
+ O
2
? 2 H
2
O 
? This is a chemical reaction ? I. 
C. 
?
1
2
H + ?
1
2
H ? ?
2
3
He + ?
0
1
n 
? This D-D fusion releases about 3 . 3MeV ? fusion with + v eQ ? II. 
D. 
?
1
1
H + ?
1
3
H ? ?
1
2
H + ?
1
2
H 
? The mass of products exceeds reactants ? requires energy ? fusion with -ve Q ? IV. 
So the answer is 
Option B: A-III, B-I, C-II, D-IV. 
Q6: A radioactive material ?? first decays into ?? and then ?? decays to non-radioactive 
material ?? . Which of the following figure represents time dependent mass of ?? , ?? and ?? ? 
JEE Main 2025 (Online) 4th April Evening Shift 
Options: 
A.  
B.  
C.  
D.  
Ans: B 
Solution: 
P ? Q ? R
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FAQs on JEE Main Previous Year Questions (2025): Nuclei

1. What are the basic properties of atomic nuclei that are important for understanding nuclear physics?
Ans. Atomic nuclei are characterized by their mass number (A), which is the total number of protons and neutrons, and their atomic number (Z), which indicates the number of protons. The stability of a nucleus depends on the ratio of neutrons to protons, where certain ratios lead to stable configurations while others may result in radioactivity. Additionally, nuclei exhibit properties such as binding energy, which is the energy required to disassemble the nucleus into its constituent protons and neutrons, and the concept of isotopes, which are variants of elements that have the same number of protons but different numbers of neutrons.
2. What is nuclear binding energy, and why is it significant?
Ans. Nuclear binding energy is the energy that holds the protons and neutrons together within an atomic nucleus. It is significant because it reflects the stability of the nucleus; a higher binding energy indicates a more stable nucleus. The binding energy can be calculated by determining the mass defect, which is the difference in mass between the individual nucleons and the mass of the nucleus. This concept is essential in nuclear fission and fusion processes, where the release or absorption of binding energy plays a crucial role.
3. How does the process of nuclear fission work?
Ans. Nuclear fission is the process by which a heavy nucleus splits into two or more lighter nuclei, along with the release of energy and additional neutrons. This process typically occurs when a nucleus absorbs a neutron, becomes unstable, and then breaks apart. The released neutrons can then induce further fission reactions in nearby nuclei, leading to a chain reaction. Fission is the principle behind nuclear reactors and atomic bombs, where controlled or uncontrolled chain reactions are utilized to produce energy or explosive force.
4. What are the differences between alpha, beta, and gamma radiation?
Ans. Alpha, beta, and gamma radiation are three types of radiation emitted during radioactive decay. Alpha radiation consists of helium nuclei (⁴He) and carries a +2 charge; it has low penetration power and can be stopped by a sheet of paper. Beta radiation consists of high-energy electrons (β⁻) or positrons (β⁺) and has moderate penetration ability, being stopped by a few millimeters of plastic or aluminum. Gamma radiation is electromagnetic radiation with no charge, possessing high penetration power and requiring dense materials like lead for protection. Understanding these differences is crucial for safety and applications in nuclear technology.
5. What role do isotopes play in medical applications and nuclear energy?
Ans. Isotopes play a significant role in both medical applications and nuclear energy. In medicine, isotopes such as ¹²³I and ¹⁸F are used in diagnostic imaging and cancer treatment. For instance, ¹²³I is utilized in thyroid scans, while ¹⁸F is used in positron emission tomography (PET) scans. In nuclear energy, isotopes like Uranium-235 (²³⁵U) are essential for fission reactions in nuclear reactors, providing a significant source of energy. The unique properties of isotopes enable their diverse applications across various fields, including healthcare and energy production.
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