Atoms and Nuclei: JEE Main Previous Year Questions (2021-2026)

 Page 1


JEE Main Previous Year Questions (2021-2026): 
Atoms and Nuclei  
 
(January 2026) 
 
Q1: A nucleus has mass number a and radius R
a
. Another nucleus has mass number ß 
and radius R
ß
.  
If ß = 8a then R
a
 / R
ß
 is :  
(a) 1  
(b) 2  
(c) 8  
(d) 0.5  
Ans: (d) 
Sol: 
The nucleus is assumed as a sphere where the volume is proportional to the total number of 
nucleons (mass number, A).  
Because the density of nuclear matter is approximately constant, the radius R of a nucleus is 
related to its mass number A by the following formula :  
R = R
0
 A
1/3
  
Where R is the nuclear radius.  
R
0
 is a constant (approximately 1.2 × 10
-15
 m or 1.2 fm).  
A is the mass number. 
Page 2


JEE Main Previous Year Questions (2021-2026): 
Atoms and Nuclei  
 
(January 2026) 
 
Q1: A nucleus has mass number a and radius R
a
. Another nucleus has mass number ß 
and radius R
ß
.  
If ß = 8a then R
a
 / R
ß
 is :  
(a) 1  
(b) 2  
(c) 8  
(d) 0.5  
Ans: (d) 
Sol: 
The nucleus is assumed as a sphere where the volume is proportional to the total number of 
nucleons (mass number, A).  
Because the density of nuclear matter is approximately constant, the radius R of a nucleus is 
related to its mass number A by the following formula :  
R = R
0
 A
1/3
  
Where R is the nuclear radius.  
R
0
 is a constant (approximately 1.2 × 10
-15
 m or 1.2 fm).  
A is the mass number. 
 
Therefore, the ratio of the radius of the first nucleus to the second is 0.5.  
Hence, the correct option is (D). 
 
Q2: An atom  is bombarded by shower of fundamental particles and in 10 s this 
atom absorbed 10 electrons, 10 protons and 9 neutrons. The percentage growth in the 
surface area of the nucleons is recorded by :  
(a) 150%  
(b) 900%  
(c) 125%  
(d) 225% 
Ans: (c) 
Sol: 
The mass number of initial nucleus is A
1
 = 8  
The radius of a nucleus is proportional to A
1/3
 (R = R
0
 A
1/3
).  
So, the nuclear radius of initial nucleus is R
1
 = R
0
 (8)
1/3
 = 2R
0
  
Hence, initial surface area 
Page 3


JEE Main Previous Year Questions (2021-2026): 
Atoms and Nuclei  
 
(January 2026) 
 
Q1: A nucleus has mass number a and radius R
a
. Another nucleus has mass number ß 
and radius R
ß
.  
If ß = 8a then R
a
 / R
ß
 is :  
(a) 1  
(b) 2  
(c) 8  
(d) 0.5  
Ans: (d) 
Sol: 
The nucleus is assumed as a sphere where the volume is proportional to the total number of 
nucleons (mass number, A).  
Because the density of nuclear matter is approximately constant, the radius R of a nucleus is 
related to its mass number A by the following formula :  
R = R
0
 A
1/3
  
Where R is the nuclear radius.  
R
0
 is a constant (approximately 1.2 × 10
-15
 m or 1.2 fm).  
A is the mass number. 
 
Therefore, the ratio of the radius of the first nucleus to the second is 0.5.  
Hence, the correct option is (D). 
 
Q2: An atom  is bombarded by shower of fundamental particles and in 10 s this 
atom absorbed 10 electrons, 10 protons and 9 neutrons. The percentage growth in the 
surface area of the nucleons is recorded by :  
(a) 150%  
(b) 900%  
(c) 125%  
(d) 225% 
Ans: (c) 
Sol: 
The mass number of initial nucleus is A
1
 = 8  
The radius of a nucleus is proportional to A
1/3
 (R = R
0
 A
1/3
).  
So, the nuclear radius of initial nucleus is R
1
 = R
0
 (8)
1/3
 = 2R
0
  
Hence, initial surface area 
 
The atom absorbs 10 protons and 9 neutrons.  
Total nucleons in final state of nucleus is 10 protons +9 neutrons = 19 nucleons.  
So, new mass number is A
2
 = A
1
 + 19 = 8 + 19 = 27  
The final radius, 
 
So, final surface area is 
 
The question has asked about the percentage growth recorded, that means final surface area is 
what percentage 
So, the percentage growth is, 
 
 
Q3: The binding energy for the following nuclear reactions are expressed in MeV. 
 
If X
3
, X
4
, X
5
 denote the stability of 
2
He
3
, 
2
He
4
 and 
2
He
5
, respectively, then the correct order 
is: 
(a) X
4
 > X
5
 > X
3 
(b) X
4
 < X
5
 < X
3 
(c) X
4
 > X
5
 < X
3 
(d) X
4
 = X
5
 = X
3 
Ans: (a) 
Sol:  
A nucleus is more stable if it has a higher binding energy i.e., is in a lower energy state.  
In exothermic reaction energy is released, ? Q > 0. The products are more stable than the 
reactants.  
Page 4


JEE Main Previous Year Questions (2021-2026): 
Atoms and Nuclei  
 
(January 2026) 
 
Q1: A nucleus has mass number a and radius R
a
. Another nucleus has mass number ß 
and radius R
ß
.  
If ß = 8a then R
a
 / R
ß
 is :  
(a) 1  
(b) 2  
(c) 8  
(d) 0.5  
Ans: (d) 
Sol: 
The nucleus is assumed as a sphere where the volume is proportional to the total number of 
nucleons (mass number, A).  
Because the density of nuclear matter is approximately constant, the radius R of a nucleus is 
related to its mass number A by the following formula :  
R = R
0
 A
1/3
  
Where R is the nuclear radius.  
R
0
 is a constant (approximately 1.2 × 10
-15
 m or 1.2 fm).  
A is the mass number. 
 
Therefore, the ratio of the radius of the first nucleus to the second is 0.5.  
Hence, the correct option is (D). 
 
Q2: An atom  is bombarded by shower of fundamental particles and in 10 s this 
atom absorbed 10 electrons, 10 protons and 9 neutrons. The percentage growth in the 
surface area of the nucleons is recorded by :  
(a) 150%  
(b) 900%  
(c) 125%  
(d) 225% 
Ans: (c) 
Sol: 
The mass number of initial nucleus is A
1
 = 8  
The radius of a nucleus is proportional to A
1/3
 (R = R
0
 A
1/3
).  
So, the nuclear radius of initial nucleus is R
1
 = R
0
 (8)
1/3
 = 2R
0
  
Hence, initial surface area 
 
The atom absorbs 10 protons and 9 neutrons.  
Total nucleons in final state of nucleus is 10 protons +9 neutrons = 19 nucleons.  
So, new mass number is A
2
 = A
1
 + 19 = 8 + 19 = 27  
The final radius, 
 
So, final surface area is 
 
The question has asked about the percentage growth recorded, that means final surface area is 
what percentage 
So, the percentage growth is, 
 
 
Q3: The binding energy for the following nuclear reactions are expressed in MeV. 
 
If X
3
, X
4
, X
5
 denote the stability of 
2
He
3
, 
2
He
4
 and 
2
He
5
, respectively, then the correct order 
is: 
(a) X
4
 > X
5
 > X
3 
(b) X
4
 < X
5
 < X
3 
(c) X
4
 > X
5
 < X
3 
(d) X
4
 = X
5
 = X
3 
Ans: (a) 
Sol:  
A nucleus is more stable if it has a higher binding energy i.e., is in a lower energy state.  
In exothermic reaction energy is released, ? Q > 0. The products are more stable than the 
reactants.  
In endothermic reaction energy is absorbed, ? Q < 0. The products are less stable than the 
reactants.  
The given two nuclear reactions are, 
0
 
Since energy is released, the product  is significantly more stable than the reactant 
. 
Therefore, X
4
 > X
3
. 
 
Since energy must be supplied to this reaction, the product  is less stable than the 
reactant .  
Therefore, X
4
 > X
5
.  
From reaction 1, the jump in stability from X
3
 to X
4
 is very large (20 MeV).  
From reaction 2, the drop in stability from X
4
 to X
5
 is very small (0.9 MeV).  
Because X
5
 is only slighty less stable than X
4
 , but X
3
 is significantly less stable than X
4
, we can 
conclude that X
5
 is still more stable than X
3
.  
Therefore, the order of stability is X
4
 > X
5
 > X
3
.  
Hence, the correct option is (a). 
 
Q4: Two electrons are moving in orbits of two hydrogen like atoms with speeds 3 × 10
5
 
m/s and 2.5 × 10
5
 m/s respectively. If the radii of these orbits are nearly same then the 
possible order of energy states are ______ respectively.  
(a) 8 and 10  
(b) 10 and 12  
(c) 9 and 8  
Page 5


JEE Main Previous Year Questions (2021-2026): 
Atoms and Nuclei  
 
(January 2026) 
 
Q1: A nucleus has mass number a and radius R
a
. Another nucleus has mass number ß 
and radius R
ß
.  
If ß = 8a then R
a
 / R
ß
 is :  
(a) 1  
(b) 2  
(c) 8  
(d) 0.5  
Ans: (d) 
Sol: 
The nucleus is assumed as a sphere where the volume is proportional to the total number of 
nucleons (mass number, A).  
Because the density of nuclear matter is approximately constant, the radius R of a nucleus is 
related to its mass number A by the following formula :  
R = R
0
 A
1/3
  
Where R is the nuclear radius.  
R
0
 is a constant (approximately 1.2 × 10
-15
 m or 1.2 fm).  
A is the mass number. 
 
Therefore, the ratio of the radius of the first nucleus to the second is 0.5.  
Hence, the correct option is (D). 
 
Q2: An atom  is bombarded by shower of fundamental particles and in 10 s this 
atom absorbed 10 electrons, 10 protons and 9 neutrons. The percentage growth in the 
surface area of the nucleons is recorded by :  
(a) 150%  
(b) 900%  
(c) 125%  
(d) 225% 
Ans: (c) 
Sol: 
The mass number of initial nucleus is A
1
 = 8  
The radius of a nucleus is proportional to A
1/3
 (R = R
0
 A
1/3
).  
So, the nuclear radius of initial nucleus is R
1
 = R
0
 (8)
1/3
 = 2R
0
  
Hence, initial surface area 
 
The atom absorbs 10 protons and 9 neutrons.  
Total nucleons in final state of nucleus is 10 protons +9 neutrons = 19 nucleons.  
So, new mass number is A
2
 = A
1
 + 19 = 8 + 19 = 27  
The final radius, 
 
So, final surface area is 
 
The question has asked about the percentage growth recorded, that means final surface area is 
what percentage 
So, the percentage growth is, 
 
 
Q3: The binding energy for the following nuclear reactions are expressed in MeV. 
 
If X
3
, X
4
, X
5
 denote the stability of 
2
He
3
, 
2
He
4
 and 
2
He
5
, respectively, then the correct order 
is: 
(a) X
4
 > X
5
 > X
3 
(b) X
4
 < X
5
 < X
3 
(c) X
4
 > X
5
 < X
3 
(d) X
4
 = X
5
 = X
3 
Ans: (a) 
Sol:  
A nucleus is more stable if it has a higher binding energy i.e., is in a lower energy state.  
In exothermic reaction energy is released, ? Q > 0. The products are more stable than the 
reactants.  
In endothermic reaction energy is absorbed, ? Q < 0. The products are less stable than the 
reactants.  
The given two nuclear reactions are, 
0
 
Since energy is released, the product  is significantly more stable than the reactant 
. 
Therefore, X
4
 > X
3
. 
 
Since energy must be supplied to this reaction, the product  is less stable than the 
reactant .  
Therefore, X
4
 > X
5
.  
From reaction 1, the jump in stability from X
3
 to X
4
 is very large (20 MeV).  
From reaction 2, the drop in stability from X
4
 to X
5
 is very small (0.9 MeV).  
Because X
5
 is only slighty less stable than X
4
 , but X
3
 is significantly less stable than X
4
, we can 
conclude that X
5
 is still more stable than X
3
.  
Therefore, the order of stability is X
4
 > X
5
 > X
3
.  
Hence, the correct option is (a). 
 
Q4: Two electrons are moving in orbits of two hydrogen like atoms with speeds 3 × 10
5
 
m/s and 2.5 × 10
5
 m/s respectively. If the radii of these orbits are nearly same then the 
possible order of energy states are ______ respectively.  
(a) 8 and 10  
(b) 10 and 12  
(c) 9 and 8  
(d) 6 and 5 
Ans: (d) 
Sol: 
 
The velocity of an electron in the n
th
 orbit is proportional to the atomic number Z and inversely 
proportional to the principal quantum number n. 
 
Where k
v
 is constant for the velocity.  
The radius of the n
th
 orbit is proportional to the square of the principal quantum number n and 
inversely proportional to the atomic number Z.