JEE Main Previous Year Questions (2025): Atoms

 Page 1


JEE Main Previous Year Questions 
(2025): Atoms 
Q1: An electron in the hydrogen atom initially in the fourth excited state makes a 
transition to ?? th 
 energy state by emitting a photon of energy 2.86 eV . The integer 
value of ?? will be 
JEE Main 2025 (Online) 3rd April Evening Shift 
Ans: 2 
Solution: 
To find the integer value of ?? for which an electron transitions from the fourth excited state in a 
hydrogen atom, thus emitting a photon with an energy of 2.86 eV , we can follow these steps: 
We use the formula for the energy difference associated with electron transitions in a hydrogen 
atom: 
?? = 13.6 (
1
?? 1
2
-
1
?? 2
2
) 
However, in this Q, it seems there's a typo in the original explanation, as both indices are given 
as n
1
. To correct it, we should use this formula: 
?? = 13.6 (
1
?? 2
-
1
5
2
) 
where ?? 1
= 5 (the fifth energy level or fourth excited state) and ?? 2
= ?? (the state to which the 
electron transitions). 
Given the photon's energy is 2.86 eV , set up the equation: 
2.86 = 13.6 (
1
?? 2
-
1
25
) 
Solve for 
1
?? 2
 : 
1
?? 2
=
2.86
13.6
+
1
25
 
Calculate the value: 
1
?? 2
= 0.21 + 0.04 = 0.25 
Find ?? 2
 : 
?? 2
=
1
0.25
= 4 
Page 2


JEE Main Previous Year Questions 
(2025): Atoms 
Q1: An electron in the hydrogen atom initially in the fourth excited state makes a 
transition to ?? th 
 energy state by emitting a photon of energy 2.86 eV . The integer 
value of ?? will be 
JEE Main 2025 (Online) 3rd April Evening Shift 
Ans: 2 
Solution: 
To find the integer value of ?? for which an electron transitions from the fourth excited state in a 
hydrogen atom, thus emitting a photon with an energy of 2.86 eV , we can follow these steps: 
We use the formula for the energy difference associated with electron transitions in a hydrogen 
atom: 
?? = 13.6 (
1
?? 1
2
-
1
?? 2
2
) 
However, in this Q, it seems there's a typo in the original explanation, as both indices are given 
as n
1
. To correct it, we should use this formula: 
?? = 13.6 (
1
?? 2
-
1
5
2
) 
where ?? 1
= 5 (the fifth energy level or fourth excited state) and ?? 2
= ?? (the state to which the 
electron transitions). 
Given the photon's energy is 2.86 eV , set up the equation: 
2.86 = 13.6 (
1
?? 2
-
1
25
) 
Solve for 
1
?? 2
 : 
1
?? 2
=
2.86
13.6
+
1
25
 
Calculate the value: 
1
?? 2
= 0.21 + 0.04 = 0.25 
Find ?? 2
 : 
?? 2
=
1
0.25
= 4 
Consequently: 
?? = v4 = 2 
Thus, the electron transitions to the ?? = 2 energy state. 
Q2: During the transition of electron from state ?? to state ?? of a Bohr atom, the wavelength 
of emitted radiation is ???????? ?? °
 and it becomes ???????? ?? °
 when the electron jumps from state ?? 
to state ?? . Then the wavelength of the radiation emitted during the transition of electrons 
from state ?? to state ?? is 
JEE Main 2025 (Online) 24th January Morning Shift 
Options: 
A. 4000 ?? °
 
B. 6000?? °
 
C. 2000 ?? °
 
D. 3000 ?? °
 
Ans: D 
Solution: 
1
?? ????
=
?? ?? - ?? ?? h?? ,
1
?? ????
=
?? ?? - ?? ?? h?? 
Since the electron transitions from state ?? to state ?? in two steps ( ?? to ?? and then ?? to ?? ), the 
energy difference for the transition from A to B is given by 
?? ?? - ?? ?? = ( ?? ?? - ?? ?? )- ( ?? ?? - ?? ?? ) 
Expressing these energy differences in terms of wavelengths, 
h?? ?? ????
=
h?? ?? ????
-
h?? ?? ????
 
Cancelling out h?? , we obtain 
1
?? ????
=
1
?? ????
-
1
?? ????
 
Substitute the given values: 
1
?? ????
=
1
2000
-
1
6000
 
Finding a common denominator: 
1
?? ????
=
3 - 1
6000
=
2
6000
=
1
3000
 
Thus, the wavelength for the A to B transition is 
Page 3


JEE Main Previous Year Questions 
(2025): Atoms 
Q1: An electron in the hydrogen atom initially in the fourth excited state makes a 
transition to ?? th 
 energy state by emitting a photon of energy 2.86 eV . The integer 
value of ?? will be 
JEE Main 2025 (Online) 3rd April Evening Shift 
Ans: 2 
Solution: 
To find the integer value of ?? for which an electron transitions from the fourth excited state in a 
hydrogen atom, thus emitting a photon with an energy of 2.86 eV , we can follow these steps: 
We use the formula for the energy difference associated with electron transitions in a hydrogen 
atom: 
?? = 13.6 (
1
?? 1
2
-
1
?? 2
2
) 
However, in this Q, it seems there's a typo in the original explanation, as both indices are given 
as n
1
. To correct it, we should use this formula: 
?? = 13.6 (
1
?? 2
-
1
5
2
) 
where ?? 1
= 5 (the fifth energy level or fourth excited state) and ?? 2
= ?? (the state to which the 
electron transitions). 
Given the photon's energy is 2.86 eV , set up the equation: 
2.86 = 13.6 (
1
?? 2
-
1
25
) 
Solve for 
1
?? 2
 : 
1
?? 2
=
2.86
13.6
+
1
25
 
Calculate the value: 
1
?? 2
= 0.21 + 0.04 = 0.25 
Find ?? 2
 : 
?? 2
=
1
0.25
= 4 
Consequently: 
?? = v4 = 2 
Thus, the electron transitions to the ?? = 2 energy state. 
Q2: During the transition of electron from state ?? to state ?? of a Bohr atom, the wavelength 
of emitted radiation is ???????? ?? °
 and it becomes ???????? ?? °
 when the electron jumps from state ?? 
to state ?? . Then the wavelength of the radiation emitted during the transition of electrons 
from state ?? to state ?? is 
JEE Main 2025 (Online) 24th January Morning Shift 
Options: 
A. 4000 ?? °
 
B. 6000?? °
 
C. 2000 ?? °
 
D. 3000 ?? °
 
Ans: D 
Solution: 
1
?? ????
=
?? ?? - ?? ?? h?? ,
1
?? ????
=
?? ?? - ?? ?? h?? 
Since the electron transitions from state ?? to state ?? in two steps ( ?? to ?? and then ?? to ?? ), the 
energy difference for the transition from A to B is given by 
?? ?? - ?? ?? = ( ?? ?? - ?? ?? )- ( ?? ?? - ?? ?? ) 
Expressing these energy differences in terms of wavelengths, 
h?? ?? ????
=
h?? ?? ????
-
h?? ?? ????
 
Cancelling out h?? , we obtain 
1
?? ????
=
1
?? ????
-
1
?? ????
 
Substitute the given values: 
1
?? ????
=
1
2000
-
1
6000
 
Finding a common denominator: 
1
?? ????
=
3 - 1
6000
=
2
6000
=
1
3000
 
Thus, the wavelength for the A to B transition is 
?? ????
= 3000 Å. 
This matches Option D. 
Q3: The energy ?? and momentum ?? of a moving body of mass ?? are related by some 
equation. Given that ?? represents the speed of light, identify the correct equation 
JEE Main 2025 (Online) 24th January Evening Shift 
Options: 
A. E
2
= pc
2
+ m
2
c
4
 
B. E
2
= pc
2
+ m
2
c
2
 
C. E
2
= p
2
c
2
+ m
2
c
4
 
D. E
2
= p
2
c
2
+ m
2
c
2
 
Ans: C 
Solution: 
[?? ] = ?? 1
?? 2
?? -2
[???? ] = ?? 1
?? 1
?? -1
· ?? 1
?? -1
= ?? 1
?? 2
?? -2
[mc
2
] = M
1
?? 2
?? -2
?? 2
= M
1
 L
2
 T
-2
?? 2
= P
2
c
2
+ m
2
c
4
 
 
 
Q4: The frequency of revolution of the electron in Bohr's orbit varies with ?? , the 
principal quantum number as: 
JEE Main 2025 (Online) 28th January Evening Shift 
Options: 
A. 
1
?? 4
 
B. 
1
?? 2
 
C. 
1
?? 3
 
D. 
1
?? 
Ans: C 
Solution: 
Frequency of revolution ?
1
n
3
 
 
 
Page 4


JEE Main Previous Year Questions 
(2025): Atoms 
Q1: An electron in the hydrogen atom initially in the fourth excited state makes a 
transition to ?? th 
 energy state by emitting a photon of energy 2.86 eV . The integer 
value of ?? will be 
JEE Main 2025 (Online) 3rd April Evening Shift 
Ans: 2 
Solution: 
To find the integer value of ?? for which an electron transitions from the fourth excited state in a 
hydrogen atom, thus emitting a photon with an energy of 2.86 eV , we can follow these steps: 
We use the formula for the energy difference associated with electron transitions in a hydrogen 
atom: 
?? = 13.6 (
1
?? 1
2
-
1
?? 2
2
) 
However, in this Q, it seems there's a typo in the original explanation, as both indices are given 
as n
1
. To correct it, we should use this formula: 
?? = 13.6 (
1
?? 2
-
1
5
2
) 
where ?? 1
= 5 (the fifth energy level or fourth excited state) and ?? 2
= ?? (the state to which the 
electron transitions). 
Given the photon's energy is 2.86 eV , set up the equation: 
2.86 = 13.6 (
1
?? 2
-
1
25
) 
Solve for 
1
?? 2
 : 
1
?? 2
=
2.86
13.6
+
1
25
 
Calculate the value: 
1
?? 2
= 0.21 + 0.04 = 0.25 
Find ?? 2
 : 
?? 2
=
1
0.25
= 4 
Consequently: 
?? = v4 = 2 
Thus, the electron transitions to the ?? = 2 energy state. 
Q2: During the transition of electron from state ?? to state ?? of a Bohr atom, the wavelength 
of emitted radiation is ???????? ?? °
 and it becomes ???????? ?? °
 when the electron jumps from state ?? 
to state ?? . Then the wavelength of the radiation emitted during the transition of electrons 
from state ?? to state ?? is 
JEE Main 2025 (Online) 24th January Morning Shift 
Options: 
A. 4000 ?? °
 
B. 6000?? °
 
C. 2000 ?? °
 
D. 3000 ?? °
 
Ans: D 
Solution: 
1
?? ????
=
?? ?? - ?? ?? h?? ,
1
?? ????
=
?? ?? - ?? ?? h?? 
Since the electron transitions from state ?? to state ?? in two steps ( ?? to ?? and then ?? to ?? ), the 
energy difference for the transition from A to B is given by 
?? ?? - ?? ?? = ( ?? ?? - ?? ?? )- ( ?? ?? - ?? ?? ) 
Expressing these energy differences in terms of wavelengths, 
h?? ?? ????
=
h?? ?? ????
-
h?? ?? ????
 
Cancelling out h?? , we obtain 
1
?? ????
=
1
?? ????
-
1
?? ????
 
Substitute the given values: 
1
?? ????
=
1
2000
-
1
6000
 
Finding a common denominator: 
1
?? ????
=
3 - 1
6000
=
2
6000
=
1
3000
 
Thus, the wavelength for the A to B transition is 
?? ????
= 3000 Å. 
This matches Option D. 
Q3: The energy ?? and momentum ?? of a moving body of mass ?? are related by some 
equation. Given that ?? represents the speed of light, identify the correct equation 
JEE Main 2025 (Online) 24th January Evening Shift 
Options: 
A. E
2
= pc
2
+ m
2
c
4
 
B. E
2
= pc
2
+ m
2
c
2
 
C. E
2
= p
2
c
2
+ m
2
c
4
 
D. E
2
= p
2
c
2
+ m
2
c
2
 
Ans: C 
Solution: 
[?? ] = ?? 1
?? 2
?? -2
[???? ] = ?? 1
?? 1
?? -1
· ?? 1
?? -1
= ?? 1
?? 2
?? -2
[mc
2
] = M
1
?? 2
?? -2
?? 2
= M
1
 L
2
 T
-2
?? 2
= P
2
c
2
+ m
2
c
4
 
 
 
Q4: The frequency of revolution of the electron in Bohr's orbit varies with ?? , the 
principal quantum number as: 
JEE Main 2025 (Online) 28th January Evening Shift 
Options: 
A. 
1
?? 4
 
B. 
1
?? 2
 
C. 
1
?? 3
 
D. 
1
?? 
Ans: C 
Solution: 
Frequency of revolution ?
1
n
3
 
 
 
Q5: The number of spectral lines emitted by atomic hydrogen that is in the 4th energy 
level, is 
JEE Main 2025 (Online) 29th January Evening Shift 
Options: 
A. 3 
B. 6 
C. 1 
D. 0 
Ans: B 
Solution: 
When an e transition from n
th 
 to ground state then ? ( ?? - 1) spectral lines are produced. 
Given, ?? = 4 
So, ? ( 4 - 1) = ? 3 = 3 + 2 + 1 = 6 
Hence, 6 spectral lines are emitted by atomic Hydrogen that is in the 4
th 
 energy level 
 
Q6: Considering Bohr's atomic model for hydrogen atom : 
(A) the energy of ?? atom in ground state is same as energy of ????
+
 ion in its first 
excited state. 
(B) the energy of ?? atom in ground state is same as that for ????
++
ion in its second 
excited state. 
(C) the energy of ?? atom in its ground state is same as that of ????
+
 ion for its ground 
state. 
(D) the energy of ????
+
ion in its first excited state is same as that for ????
++
ion in its 
ground state. 
Choose the correct answer from the options given below: 
JEE Main 2025 (Online) 2nd April Morning Shift 
Options: 
A. (A), (B) only 
B. (A), (D) only 
C. (A), (C) only 
D. (B), (D) only 
Page 5


JEE Main Previous Year Questions 
(2025): Atoms 
Q1: An electron in the hydrogen atom initially in the fourth excited state makes a 
transition to ?? th 
 energy state by emitting a photon of energy 2.86 eV . The integer 
value of ?? will be 
JEE Main 2025 (Online) 3rd April Evening Shift 
Ans: 2 
Solution: 
To find the integer value of ?? for which an electron transitions from the fourth excited state in a 
hydrogen atom, thus emitting a photon with an energy of 2.86 eV , we can follow these steps: 
We use the formula for the energy difference associated with electron transitions in a hydrogen 
atom: 
?? = 13.6 (
1
?? 1
2
-
1
?? 2
2
) 
However, in this Q, it seems there's a typo in the original explanation, as both indices are given 
as n
1
. To correct it, we should use this formula: 
?? = 13.6 (
1
?? 2
-
1
5
2
) 
where ?? 1
= 5 (the fifth energy level or fourth excited state) and ?? 2
= ?? (the state to which the 
electron transitions). 
Given the photon's energy is 2.86 eV , set up the equation: 
2.86 = 13.6 (
1
?? 2
-
1
25
) 
Solve for 
1
?? 2
 : 
1
?? 2
=
2.86
13.6
+
1
25
 
Calculate the value: 
1
?? 2
= 0.21 + 0.04 = 0.25 
Find ?? 2
 : 
?? 2
=
1
0.25
= 4 
Consequently: 
?? = v4 = 2 
Thus, the electron transitions to the ?? = 2 energy state. 
Q2: During the transition of electron from state ?? to state ?? of a Bohr atom, the wavelength 
of emitted radiation is ???????? ?? °
 and it becomes ???????? ?? °
 when the electron jumps from state ?? 
to state ?? . Then the wavelength of the radiation emitted during the transition of electrons 
from state ?? to state ?? is 
JEE Main 2025 (Online) 24th January Morning Shift 
Options: 
A. 4000 ?? °
 
B. 6000?? °
 
C. 2000 ?? °
 
D. 3000 ?? °
 
Ans: D 
Solution: 
1
?? ????
=
?? ?? - ?? ?? h?? ,
1
?? ????
=
?? ?? - ?? ?? h?? 
Since the electron transitions from state ?? to state ?? in two steps ( ?? to ?? and then ?? to ?? ), the 
energy difference for the transition from A to B is given by 
?? ?? - ?? ?? = ( ?? ?? - ?? ?? )- ( ?? ?? - ?? ?? ) 
Expressing these energy differences in terms of wavelengths, 
h?? ?? ????
=
h?? ?? ????
-
h?? ?? ????
 
Cancelling out h?? , we obtain 
1
?? ????
=
1
?? ????
-
1
?? ????
 
Substitute the given values: 
1
?? ????
=
1
2000
-
1
6000
 
Finding a common denominator: 
1
?? ????
=
3 - 1
6000
=
2
6000
=
1
3000
 
Thus, the wavelength for the A to B transition is 
?? ????
= 3000 Å. 
This matches Option D. 
Q3: The energy ?? and momentum ?? of a moving body of mass ?? are related by some 
equation. Given that ?? represents the speed of light, identify the correct equation 
JEE Main 2025 (Online) 24th January Evening Shift 
Options: 
A. E
2
= pc
2
+ m
2
c
4
 
B. E
2
= pc
2
+ m
2
c
2
 
C. E
2
= p
2
c
2
+ m
2
c
4
 
D. E
2
= p
2
c
2
+ m
2
c
2
 
Ans: C 
Solution: 
[?? ] = ?? 1
?? 2
?? -2
[???? ] = ?? 1
?? 1
?? -1
· ?? 1
?? -1
= ?? 1
?? 2
?? -2
[mc
2
] = M
1
?? 2
?? -2
?? 2
= M
1
 L
2
 T
-2
?? 2
= P
2
c
2
+ m
2
c
4
 
 
 
Q4: The frequency of revolution of the electron in Bohr's orbit varies with ?? , the 
principal quantum number as: 
JEE Main 2025 (Online) 28th January Evening Shift 
Options: 
A. 
1
?? 4
 
B. 
1
?? 2
 
C. 
1
?? 3
 
D. 
1
?? 
Ans: C 
Solution: 
Frequency of revolution ?
1
n
3
 
 
 
Q5: The number of spectral lines emitted by atomic hydrogen that is in the 4th energy 
level, is 
JEE Main 2025 (Online) 29th January Evening Shift 
Options: 
A. 3 
B. 6 
C. 1 
D. 0 
Ans: B 
Solution: 
When an e transition from n
th 
 to ground state then ? ( ?? - 1) spectral lines are produced. 
Given, ?? = 4 
So, ? ( 4 - 1) = ? 3 = 3 + 2 + 1 = 6 
Hence, 6 spectral lines are emitted by atomic Hydrogen that is in the 4
th 
 energy level 
 
Q6: Considering Bohr's atomic model for hydrogen atom : 
(A) the energy of ?? atom in ground state is same as energy of ????
+
 ion in its first 
excited state. 
(B) the energy of ?? atom in ground state is same as that for ????
++
ion in its second 
excited state. 
(C) the energy of ?? atom in its ground state is same as that of ????
+
 ion for its ground 
state. 
(D) the energy of ????
+
ion in its first excited state is same as that for ????
++
ion in its 
ground state. 
Choose the correct answer from the options given below: 
JEE Main 2025 (Online) 2nd April Morning Shift 
Options: 
A. (A), (B) only 
B. (A), (D) only 
C. (A), (C) only 
D. (B), (D) only 
Ans: A 
Solution: 
In Bohr's atomic model, the energy of an electron in an atom is given by the formula: 
?? ?
?? 2
?? 2
 
where ?? is the atomic number and ?? is the principal quantum number of the electron's orbit. 
Let's consider the implications: 
For hydrogen (H), ?? = 1. 
For helium ion ( He
+
) , ?? = 2. 
For lithium ion ( Li
++
) , ?? = 3. 
Now, identifying the relevant states: 
Ground State: This corresponds to ?? = 1. 
First Excited State: This corresponds to ?? = 2. 
Second Excited State: This corresponds to ?? = 3. 
Let's evaluate the energy comparisons: 
Hydrogen Atom in Ground State: ?? ?
1
2
1
2
= 1. 
Helium Ion ( He
+
) in the First Excited State: 
?? ?
2
2
2
2
= 1 
This matches the energy of hydrogen in its ground state. 
Lithium Ion ( Li
++
) in the Second Excited State: 
?? ?
3
2
3
2
= 1 
This also matches the energy of hydrogen in its ground state. 
From this analysis, we can conclude: 
The energy of a hydrogen atom in its ground state is equal to the energy of a He
+
ion in its first 
excited state. 
The energy of a hydrogen atom in its ground state also equals the energy of a Li
++
ion in its 
second excited state. 
Hence, statements (A) and (B) are correct. 
Q7: Assuming the validity of Bohr's atomic model for hydrogen like ions the radius of 
????
++
ion in its ground state is given by 
?? ?? ?? ?? , where ?? = ____ (Where ?? ?? is the first 
Bohr's radius.) 
JEE Main 2025 (Online) 2nd April Evening Shift 
Options: 
A. 2 
B. 9 
C. 1 
D. 3 
Ans: D 
Solution: