JEE Main Previous Year Questions: (2025): Structure of Atom | Chemistry for JEE Main & Advanced PDF Download

 Page 1


JEE Main Previous Year Questions 
(2025): Structure of Atom 
Q1: The energy of an electron in the first Bohr orbit of the H-atom is -13.6 eV . The 
magnitude of energy value of an electron in the first excited state of Be
3 +
 is  _ _ _ _ eV 
(nearest integer value). 
JEE Main 2025 (Online) 8th April Evening Shift 
Ans: 54 
Solution: 
To find the energy of an electron in the first excited state of a Be
3 +
 ion, we can use the formula 
for the energy of an electron in a hydrogen-like atom: 
E
T
= - 13 . 6
?? 2
?? 2
eV 
Given: 
Energy of the first orbit (ground state) of the hydrogen atom: E
1
= - 13 . 6 eV (where ?? = 1 and 
?? = 1 ). 
For Be
3 +
 : 
Atomic number ?? = 4. 
First excited state corresponds to ?? = 2. 
Calculation for Be
3 +
 : 
The energy ratio between the hydrogen atom and the Be
3 +
 ion can be written as: 
E
H
E
Be + 3
=
?? 1
2
?? 1
2
×
?? 2
2
?? 2
2
 
Substitute the known values: 
- 13 . 6
E
Be + 3
=
1
2
1
2
×
2
2
4
2
 
- 13 . 6
E
Be
+ 3
=
1
1
×
4
16
 
Solving this gives: 
E
Be
+ 3 = - 13 . 6 × 4 = - 54 . 4eV 
The magnitude of the energy of the electron in the first excited state of Be
3 +
 is | - 54 . 4 | =
54 . 4eV , which is approximately 54 eV when rounded to the nearest integer. 
 
Page 2


JEE Main Previous Year Questions 
(2025): Structure of Atom 
Q1: The energy of an electron in the first Bohr orbit of the H-atom is -13.6 eV . The 
magnitude of energy value of an electron in the first excited state of Be
3 +
 is  _ _ _ _ eV 
(nearest integer value). 
JEE Main 2025 (Online) 8th April Evening Shift 
Ans: 54 
Solution: 
To find the energy of an electron in the first excited state of a Be
3 +
 ion, we can use the formula 
for the energy of an electron in a hydrogen-like atom: 
E
T
= - 13 . 6
?? 2
?? 2
eV 
Given: 
Energy of the first orbit (ground state) of the hydrogen atom: E
1
= - 13 . 6 eV (where ?? = 1 and 
?? = 1 ). 
For Be
3 +
 : 
Atomic number ?? = 4. 
First excited state corresponds to ?? = 2. 
Calculation for Be
3 +
 : 
The energy ratio between the hydrogen atom and the Be
3 +
 ion can be written as: 
E
H
E
Be + 3
=
?? 1
2
?? 1
2
×
?? 2
2
?? 2
2
 
Substitute the known values: 
- 13 . 6
E
Be + 3
=
1
2
1
2
×
2
2
4
2
 
- 13 . 6
E
Be
+ 3
=
1
1
×
4
16
 
Solving this gives: 
E
Be
+ 3 = - 13 . 6 × 4 = - 54 . 4eV 
The magnitude of the energy of the electron in the first excited state of Be
3 +
 is | - 54 . 4 | =
54 . 4eV , which is approximately 54 eV when rounded to the nearest integer. 
 
Q2: Radius of the first excited state of Helium ion is given as : ?? ?? ? radius of first 
stationary state of hydrogen atom. 
JEE Main 2025 (Online) 22nd January Morning Shift 
Options: 
A. r =
a
0
4
 
B. r = 2 a
0
 
C. r = 4 a
0
 
D. r =
a
0
2
 
Ans: B 
Solution: 
?? ?? =
?? 2
?? 0
?? 
For the helium ion, which is hydrogen-like with a nuclear charge of ?? = 2, the first excited state 
corresponds to ?? = 2. Substituting these values: 
?? 2
=
( 2 )
2
?? 0
2
=
4 ?? 0
2
= 2 ?? 0
 
Thus, the radius of the first excited state of the helium ion is 2 ?? 0
. 
 
Q3: Given below are two statements : 
Statement (I): A spectral line will be observed for a ?? ?? ?? ? ?? ?? ?? transition. 
Statement (II): ?? ?? ?? and ?? ?? ?? are degenerate orbitals. 
In the light of the above statements, choose the correct answer from the options 
given below: 
JEE Main 2025 (Online) 22nd January Evening Shift 
Options: 
A. Statement I is false but Statement II is true 
B. Both Statement I and Statement II are false 
C. Statement I is true but Statement II is false 
D. Both Statement I and Statement II are true 
Ans: A 
Solution: 
Let us analyze the two statements in the context of atomic orbitals and electronic transitions: 
Page 3


JEE Main Previous Year Questions 
(2025): Structure of Atom 
Q1: The energy of an electron in the first Bohr orbit of the H-atom is -13.6 eV . The 
magnitude of energy value of an electron in the first excited state of Be
3 +
 is  _ _ _ _ eV 
(nearest integer value). 
JEE Main 2025 (Online) 8th April Evening Shift 
Ans: 54 
Solution: 
To find the energy of an electron in the first excited state of a Be
3 +
 ion, we can use the formula 
for the energy of an electron in a hydrogen-like atom: 
E
T
= - 13 . 6
?? 2
?? 2
eV 
Given: 
Energy of the first orbit (ground state) of the hydrogen atom: E
1
= - 13 . 6 eV (where ?? = 1 and 
?? = 1 ). 
For Be
3 +
 : 
Atomic number ?? = 4. 
First excited state corresponds to ?? = 2. 
Calculation for Be
3 +
 : 
The energy ratio between the hydrogen atom and the Be
3 +
 ion can be written as: 
E
H
E
Be + 3
=
?? 1
2
?? 1
2
×
?? 2
2
?? 2
2
 
Substitute the known values: 
- 13 . 6
E
Be + 3
=
1
2
1
2
×
2
2
4
2
 
- 13 . 6
E
Be
+ 3
=
1
1
×
4
16
 
Solving this gives: 
E
Be
+ 3 = - 13 . 6 × 4 = - 54 . 4eV 
The magnitude of the energy of the electron in the first excited state of Be
3 +
 is | - 54 . 4 | =
54 . 4eV , which is approximately 54 eV when rounded to the nearest integer. 
 
Q2: Radius of the first excited state of Helium ion is given as : ?? ?? ? radius of first 
stationary state of hydrogen atom. 
JEE Main 2025 (Online) 22nd January Morning Shift 
Options: 
A. r =
a
0
4
 
B. r = 2 a
0
 
C. r = 4 a
0
 
D. r =
a
0
2
 
Ans: B 
Solution: 
?? ?? =
?? 2
?? 0
?? 
For the helium ion, which is hydrogen-like with a nuclear charge of ?? = 2, the first excited state 
corresponds to ?? = 2. Substituting these values: 
?? 2
=
( 2 )
2
?? 0
2
=
4 ?? 0
2
= 2 ?? 0
 
Thus, the radius of the first excited state of the helium ion is 2 ?? 0
. 
 
Q3: Given below are two statements : 
Statement (I): A spectral line will be observed for a ?? ?? ?? ? ?? ?? ?? transition. 
Statement (II): ?? ?? ?? and ?? ?? ?? are degenerate orbitals. 
In the light of the above statements, choose the correct answer from the options 
given below: 
JEE Main 2025 (Online) 22nd January Evening Shift 
Options: 
A. Statement I is false but Statement II is true 
B. Both Statement I and Statement II are false 
C. Statement I is true but Statement II is false 
D. Both Statement I and Statement II are true 
Ans: A 
Solution: 
Let us analyze the two statements in the context of atomic orbitals and electronic transitions: 
Statement (I) 
"A spectral line will be observed for a 2 ?? ?? ? 2 ?? ?? transition." 
For an emission (or absorption) line to be observed, there must be a difference in energy 
between the initial and final states. 
In a typical hydrogen-like or many-electron atom (without additional external fields or splitting 
effects), the three 2 ?? orbitals ( 2 ?? ?? , 2 ?? ?? , 2 ?? ?? ) are degenerate-i.e., they all have the same 
energy. 
Consequently, a transition from 2 ?? ?? to 2 ?? ?? (both having the same energy) would involve no 
energy change. Hence, no photon is emitted or absorbed for this "transition." So you would not 
observe a spectral line for such a transition. 
Statement (II) 
" 2 ?? ?? and 2 ?? ?? are degenerate orbitals." 
Orbitals within the same subshell (e.g., 2 ?? subshell) are typically degenerate (same energy) in an 
isolated atom (especially a hydrogen-like atom). 
Thus, 2 ?? ?? and 2 ?? ?? (and 2 ?? ?? ) do indeed have the same energy. 
Therefore, Statement (II) is true. 
Conclusion 
Statement (I): False 
Statement (II): True 
Hence, the correct choice is: 
Option A: Statement I is false but Statement II is true. 
 
Q4: Heat treatment of muscular pain involves radiation of wavelength of about 900 
nm . Which spectral line of ?? atom is suitable for this? 
Given : Rydberg constant 
?? ?? = ????
?? ????
- ?? , ?? = ?? . ?? × ????
- ????
 ?? ?? , ?? = ?? × ????
?? ?? /?? ) 
JEE Main 2025 (Online) 23rd January Morning Shift 
Options: 
A. Lyman series, 8 ? 1 
B. Balmer series, 8 ? 2 
C. Paschen series, 5 ? 3 
D. Paschen series, 8 ? 3 
Ans: D 
Solution: To determine which spectral line of the hydrogen atom suits the wavelength for heat 
treatment of muscular pain ( 900 nm ), we begin by applying the Rydberg equation: 
Page 4


JEE Main Previous Year Questions 
(2025): Structure of Atom 
Q1: The energy of an electron in the first Bohr orbit of the H-atom is -13.6 eV . The 
magnitude of energy value of an electron in the first excited state of Be
3 +
 is  _ _ _ _ eV 
(nearest integer value). 
JEE Main 2025 (Online) 8th April Evening Shift 
Ans: 54 
Solution: 
To find the energy of an electron in the first excited state of a Be
3 +
 ion, we can use the formula 
for the energy of an electron in a hydrogen-like atom: 
E
T
= - 13 . 6
?? 2
?? 2
eV 
Given: 
Energy of the first orbit (ground state) of the hydrogen atom: E
1
= - 13 . 6 eV (where ?? = 1 and 
?? = 1 ). 
For Be
3 +
 : 
Atomic number ?? = 4. 
First excited state corresponds to ?? = 2. 
Calculation for Be
3 +
 : 
The energy ratio between the hydrogen atom and the Be
3 +
 ion can be written as: 
E
H
E
Be + 3
=
?? 1
2
?? 1
2
×
?? 2
2
?? 2
2
 
Substitute the known values: 
- 13 . 6
E
Be + 3
=
1
2
1
2
×
2
2
4
2
 
- 13 . 6
E
Be
+ 3
=
1
1
×
4
16
 
Solving this gives: 
E
Be
+ 3 = - 13 . 6 × 4 = - 54 . 4eV 
The magnitude of the energy of the electron in the first excited state of Be
3 +
 is | - 54 . 4 | =
54 . 4eV , which is approximately 54 eV when rounded to the nearest integer. 
 
Q2: Radius of the first excited state of Helium ion is given as : ?? ?? ? radius of first 
stationary state of hydrogen atom. 
JEE Main 2025 (Online) 22nd January Morning Shift 
Options: 
A. r =
a
0
4
 
B. r = 2 a
0
 
C. r = 4 a
0
 
D. r =
a
0
2
 
Ans: B 
Solution: 
?? ?? =
?? 2
?? 0
?? 
For the helium ion, which is hydrogen-like with a nuclear charge of ?? = 2, the first excited state 
corresponds to ?? = 2. Substituting these values: 
?? 2
=
( 2 )
2
?? 0
2
=
4 ?? 0
2
= 2 ?? 0
 
Thus, the radius of the first excited state of the helium ion is 2 ?? 0
. 
 
Q3: Given below are two statements : 
Statement (I): A spectral line will be observed for a ?? ?? ?? ? ?? ?? ?? transition. 
Statement (II): ?? ?? ?? and ?? ?? ?? are degenerate orbitals. 
In the light of the above statements, choose the correct answer from the options 
given below: 
JEE Main 2025 (Online) 22nd January Evening Shift 
Options: 
A. Statement I is false but Statement II is true 
B. Both Statement I and Statement II are false 
C. Statement I is true but Statement II is false 
D. Both Statement I and Statement II are true 
Ans: A 
Solution: 
Let us analyze the two statements in the context of atomic orbitals and electronic transitions: 
Statement (I) 
"A spectral line will be observed for a 2 ?? ?? ? 2 ?? ?? transition." 
For an emission (or absorption) line to be observed, there must be a difference in energy 
between the initial and final states. 
In a typical hydrogen-like or many-electron atom (without additional external fields or splitting 
effects), the three 2 ?? orbitals ( 2 ?? ?? , 2 ?? ?? , 2 ?? ?? ) are degenerate-i.e., they all have the same 
energy. 
Consequently, a transition from 2 ?? ?? to 2 ?? ?? (both having the same energy) would involve no 
energy change. Hence, no photon is emitted or absorbed for this "transition." So you would not 
observe a spectral line for such a transition. 
Statement (II) 
" 2 ?? ?? and 2 ?? ?? are degenerate orbitals." 
Orbitals within the same subshell (e.g., 2 ?? subshell) are typically degenerate (same energy) in an 
isolated atom (especially a hydrogen-like atom). 
Thus, 2 ?? ?? and 2 ?? ?? (and 2 ?? ?? ) do indeed have the same energy. 
Therefore, Statement (II) is true. 
Conclusion 
Statement (I): False 
Statement (II): True 
Hence, the correct choice is: 
Option A: Statement I is false but Statement II is true. 
 
Q4: Heat treatment of muscular pain involves radiation of wavelength of about 900 
nm . Which spectral line of ?? atom is suitable for this? 
Given : Rydberg constant 
?? ?? = ????
?? ????
- ?? , ?? = ?? . ?? × ????
- ????
 ?? ?? , ?? = ?? × ????
?? ?? /?? ) 
JEE Main 2025 (Online) 23rd January Morning Shift 
Options: 
A. Lyman series, 8 ? 1 
B. Balmer series, 8 ? 2 
C. Paschen series, 5 ? 3 
D. Paschen series, 8 ? 3 
Ans: D 
Solution: To determine which spectral line of the hydrogen atom suits the wavelength for heat 
treatment of muscular pain ( 900 nm ), we begin by applying the Rydberg equation: 
1
?? = R
H
Z
2
(
1
?? 1
2
-
1
?? 2
2
) 
Given: 
?? = 900 nm = 9 × 10
- 5
 cm 
R
H
= 10
5
 cm
- 1
 
Hydrogen atom ( Z = 1 ) 
We proceed by solving the equation: 
1
?? × R
H
=
1
?? 1
2
-
1
?? 2
2
 
Substituting the known values: 
1
9 × 10
- 5
 cm × 10
5
 cm
- 1
=
1
?? 1
2
-
1
?? 2
2
 
This simplifies to: 
1
9
=
1
?? 1
2
-
1
?? 2
2
 
This condition is satisfied when ?? 1
= 3 and ?? 2
= 8. 
Therefore, the appropriate spectral transition is from ?? 2
= 8 to ?? 1
= 3, which corresponds to 
the Paschen series transition 8 ? 3. 
 
Q5: Given below are two statements about X-ray spectra of elements : 
Statement (I): A plot of v ?? ( ?? = frequency of X-rays emitted) vs atomic mass is a 
straight line. 
Statement (II): A plot of ?? ( ?? = frequency of ?? -rays emitted) vs atomic number is a 
straight line.  
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. Statement I is false but Statement II is true 
B. Both Statement I and Statement II are true 
C. Statement I is true but Statement II is false 
D. Both Statement I and Statement II are false 
Ans: D 
Solution: 
 
Page 5


JEE Main Previous Year Questions 
(2025): Structure of Atom 
Q1: The energy of an electron in the first Bohr orbit of the H-atom is -13.6 eV . The 
magnitude of energy value of an electron in the first excited state of Be
3 +
 is  _ _ _ _ eV 
(nearest integer value). 
JEE Main 2025 (Online) 8th April Evening Shift 
Ans: 54 
Solution: 
To find the energy of an electron in the first excited state of a Be
3 +
 ion, we can use the formula 
for the energy of an electron in a hydrogen-like atom: 
E
T
= - 13 . 6
?? 2
?? 2
eV 
Given: 
Energy of the first orbit (ground state) of the hydrogen atom: E
1
= - 13 . 6 eV (where ?? = 1 and 
?? = 1 ). 
For Be
3 +
 : 
Atomic number ?? = 4. 
First excited state corresponds to ?? = 2. 
Calculation for Be
3 +
 : 
The energy ratio between the hydrogen atom and the Be
3 +
 ion can be written as: 
E
H
E
Be + 3
=
?? 1
2
?? 1
2
×
?? 2
2
?? 2
2
 
Substitute the known values: 
- 13 . 6
E
Be + 3
=
1
2
1
2
×
2
2
4
2
 
- 13 . 6
E
Be
+ 3
=
1
1
×
4
16
 
Solving this gives: 
E
Be
+ 3 = - 13 . 6 × 4 = - 54 . 4eV 
The magnitude of the energy of the electron in the first excited state of Be
3 +
 is | - 54 . 4 | =
54 . 4eV , which is approximately 54 eV when rounded to the nearest integer. 
 
Q2: Radius of the first excited state of Helium ion is given as : ?? ?? ? radius of first 
stationary state of hydrogen atom. 
JEE Main 2025 (Online) 22nd January Morning Shift 
Options: 
A. r =
a
0
4
 
B. r = 2 a
0
 
C. r = 4 a
0
 
D. r =
a
0
2
 
Ans: B 
Solution: 
?? ?? =
?? 2
?? 0
?? 
For the helium ion, which is hydrogen-like with a nuclear charge of ?? = 2, the first excited state 
corresponds to ?? = 2. Substituting these values: 
?? 2
=
( 2 )
2
?? 0
2
=
4 ?? 0
2
= 2 ?? 0
 
Thus, the radius of the first excited state of the helium ion is 2 ?? 0
. 
 
Q3: Given below are two statements : 
Statement (I): A spectral line will be observed for a ?? ?? ?? ? ?? ?? ?? transition. 
Statement (II): ?? ?? ?? and ?? ?? ?? are degenerate orbitals. 
In the light of the above statements, choose the correct answer from the options 
given below: 
JEE Main 2025 (Online) 22nd January Evening Shift 
Options: 
A. Statement I is false but Statement II is true 
B. Both Statement I and Statement II are false 
C. Statement I is true but Statement II is false 
D. Both Statement I and Statement II are true 
Ans: A 
Solution: 
Let us analyze the two statements in the context of atomic orbitals and electronic transitions: 
Statement (I) 
"A spectral line will be observed for a 2 ?? ?? ? 2 ?? ?? transition." 
For an emission (or absorption) line to be observed, there must be a difference in energy 
between the initial and final states. 
In a typical hydrogen-like or many-electron atom (without additional external fields or splitting 
effects), the three 2 ?? orbitals ( 2 ?? ?? , 2 ?? ?? , 2 ?? ?? ) are degenerate-i.e., they all have the same 
energy. 
Consequently, a transition from 2 ?? ?? to 2 ?? ?? (both having the same energy) would involve no 
energy change. Hence, no photon is emitted or absorbed for this "transition." So you would not 
observe a spectral line for such a transition. 
Statement (II) 
" 2 ?? ?? and 2 ?? ?? are degenerate orbitals." 
Orbitals within the same subshell (e.g., 2 ?? subshell) are typically degenerate (same energy) in an 
isolated atom (especially a hydrogen-like atom). 
Thus, 2 ?? ?? and 2 ?? ?? (and 2 ?? ?? ) do indeed have the same energy. 
Therefore, Statement (II) is true. 
Conclusion 
Statement (I): False 
Statement (II): True 
Hence, the correct choice is: 
Option A: Statement I is false but Statement II is true. 
 
Q4: Heat treatment of muscular pain involves radiation of wavelength of about 900 
nm . Which spectral line of ?? atom is suitable for this? 
Given : Rydberg constant 
?? ?? = ????
?? ????
- ?? , ?? = ?? . ?? × ????
- ????
 ?? ?? , ?? = ?? × ????
?? ?? /?? ) 
JEE Main 2025 (Online) 23rd January Morning Shift 
Options: 
A. Lyman series, 8 ? 1 
B. Balmer series, 8 ? 2 
C. Paschen series, 5 ? 3 
D. Paschen series, 8 ? 3 
Ans: D 
Solution: To determine which spectral line of the hydrogen atom suits the wavelength for heat 
treatment of muscular pain ( 900 nm ), we begin by applying the Rydberg equation: 
1
?? = R
H
Z
2
(
1
?? 1
2
-
1
?? 2
2
) 
Given: 
?? = 900 nm = 9 × 10
- 5
 cm 
R
H
= 10
5
 cm
- 1
 
Hydrogen atom ( Z = 1 ) 
We proceed by solving the equation: 
1
?? × R
H
=
1
?? 1
2
-
1
?? 2
2
 
Substituting the known values: 
1
9 × 10
- 5
 cm × 10
5
 cm
- 1
=
1
?? 1
2
-
1
?? 2
2
 
This simplifies to: 
1
9
=
1
?? 1
2
-
1
?? 2
2
 
This condition is satisfied when ?? 1
= 3 and ?? 2
= 8. 
Therefore, the appropriate spectral transition is from ?? 2
= 8 to ?? 1
= 3, which corresponds to 
the Paschen series transition 8 ? 3. 
 
Q5: Given below are two statements about X-ray spectra of elements : 
Statement (I): A plot of v ?? ( ?? = frequency of X-rays emitted) vs atomic mass is a 
straight line. 
Statement (II): A plot of ?? ( ?? = frequency of ?? -rays emitted) vs atomic number is a 
straight line.  
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. Statement I is false but Statement II is true 
B. Both Statement I and Statement II are true 
C. Statement I is true but Statement II is false 
D. Both Statement I and Statement II are false 
Ans: D 
Solution: 
 
Q6: Given below are two statements : 
Statement (I): For a given shell, the total number of allowed orbitals is given by ?? ?? . 
Statement (II): For any subshell, the spatial orientation of the orbitals is given by - ?? to 
+ ?? values including zero. 
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. Both Statement I and Statement II are true 
B. Both Statement I and Statement II are false 
C. Statement I is false but Statement II is true 
D. Statement I is true but Statement II is false 
Ans: A 
Solution: 
For a shell total number of orbitals = n
2
 
Magnetic quantum number have values ( - l to + l ) including 0 . 
 
Q7: For hydrogen atom, the orbital/s with lowest energy is/are : 
(A) 4 s 
(B) ?? ?? ?? 
(C) ?? ?? ?? ?? - ?? ?? 
(D) ?? ?? ?? ?? 
(E) ?? ?? ?? 
Choose the correct answer from the options given below: 
JEE Main 2025 (Online) 24th January Evening Shift 
Options: 
A. (A) only 
B. (B) only 
C. (B), (C) and (D) only 
D. (A) and (E) only 
Ans: C