Atomic Structure: JEE Main Previous Year Questions (2021-2026)

 Page 1


 
JEE Main Previous Year Questions (2021-2026): 
Structure of Atom  
 
(January 2026) 
Q1: The wavelength of photon A is 400 nm. The frequency of photon B is 10¹6 s?¹. The wave 
number of photon C is 104 cm?¹. 
The correct order of energy of these photons is: 
A: A > C > B 
B: C > B > A 
C: B > A > C 
D: A > B > C 
Answer: C 
Explanation 
 
So, smaller wavelength means higher energy. Therefore: E
B
 > E
A
 > E
C
  
 
Q2: The wave numbers of three spectral lines of H atom are considered. Identify the set of 
spectral lines belonging to Balmer series. 
(R = Rydberg constant) 
A:  
B:  
C:  
Page 2


 
JEE Main Previous Year Questions (2021-2026): 
Structure of Atom  
 
(January 2026) 
Q1: The wavelength of photon A is 400 nm. The frequency of photon B is 10¹6 s?¹. The wave 
number of photon C is 104 cm?¹. 
The correct order of energy of these photons is: 
A: A > C > B 
B: C > B > A 
C: B > A > C 
D: A > B > C 
Answer: C 
Explanation 
 
So, smaller wavelength means higher energy. Therefore: E
B
 > E
A
 > E
C
  
 
Q2: The wave numbers of three spectral lines of H atom are considered. Identify the set of 
spectral lines belonging to Balmer series. 
(R = Rydberg constant) 
A:  
B:  
C:  
D:  
Answer: A 
Explanation:  
The Balmer series corresponds to transitions ending at n
1
 = 2. 
For hydrogen, the wave number is given by 
 
Now compute the ?rst few Balmer lines: 
For 
 
 
Q3:  
Page 3


 
JEE Main Previous Year Questions (2021-2026): 
Structure of Atom  
 
(January 2026) 
Q1: The wavelength of photon A is 400 nm. The frequency of photon B is 10¹6 s?¹. The wave 
number of photon C is 104 cm?¹. 
The correct order of energy of these photons is: 
A: A > C > B 
B: C > B > A 
C: B > A > C 
D: A > B > C 
Answer: C 
Explanation 
 
So, smaller wavelength means higher energy. Therefore: E
B
 > E
A
 > E
C
  
 
Q2: The wave numbers of three spectral lines of H atom are considered. Identify the set of 
spectral lines belonging to Balmer series. 
(R = Rydberg constant) 
A:  
B:  
C:  
D:  
Answer: A 
Explanation:  
The Balmer series corresponds to transitions ending at n
1
 = 2. 
For hydrogen, the wave number is given by 
 
Now compute the ?rst few Balmer lines: 
For 
 
 
Q3:  
 
Figure 1. electron probability density for 2 s orbital 
 
Figure 2. wave function for 2s orbital 
Which of the following points in Figure 2 most accurately represents the nodal surface as 
shown in Figure 1? 
A: D 
B: B 
C: C 
D: A 
Answer: B 
Explanation: 
The nodal surface (as seen in Figure 1) is the place where the probability density becomes zero. 
For an s-orbital, this nodal surface is a spherical node (a spherical shell around the nucleus). 
At a spherical (radial) node, the radial wave function becomes zero. 
Page 4


 
JEE Main Previous Year Questions (2021-2026): 
Structure of Atom  
 
(January 2026) 
Q1: The wavelength of photon A is 400 nm. The frequency of photon B is 10¹6 s?¹. The wave 
number of photon C is 104 cm?¹. 
The correct order of energy of these photons is: 
A: A > C > B 
B: C > B > A 
C: B > A > C 
D: A > B > C 
Answer: C 
Explanation 
 
So, smaller wavelength means higher energy. Therefore: E
B
 > E
A
 > E
C
  
 
Q2: The wave numbers of three spectral lines of H atom are considered. Identify the set of 
spectral lines belonging to Balmer series. 
(R = Rydberg constant) 
A:  
B:  
C:  
D:  
Answer: A 
Explanation:  
The Balmer series corresponds to transitions ending at n
1
 = 2. 
For hydrogen, the wave number is given by 
 
Now compute the ?rst few Balmer lines: 
For 
 
 
Q3:  
 
Figure 1. electron probability density for 2 s orbital 
 
Figure 2. wave function for 2s orbital 
Which of the following points in Figure 2 most accurately represents the nodal surface as 
shown in Figure 1? 
A: D 
B: B 
C: C 
D: A 
Answer: B 
Explanation: 
The nodal surface (as seen in Figure 1) is the place where the probability density becomes zero. 
For an s-orbital, this nodal surface is a spherical node (a spherical shell around the nucleus). 
At a spherical (radial) node, the radial wave function becomes zero. 
 
So, in Figure 2, the correct point is where the 2s wave function curve cuts the horizontal axis 
(changes sign). That point represents the spherical node shown in Figure 1. 
 
Q4: The wavelength of spectral line obtained in the spectrum of Li
2+
 ion, when the transition 
takes place between two levels whose sum is 4 and difference is 2 , is 
A: 1.14 × 10?6 cm 
B: 2.28 × 10?6 cm 
C: 1.14 × 10?7 cm 
D: 2.28 × 10?7 cm 
Answer: A 
Explanation:  
n1 ? lower energy level 
n2 ? higher energy level 
n1 + n2 = 4, n2 = 3 
n2 - n1 = 2, n1 = 1 
Rydberg's formula: 
 
 
Q5: The work functions of two metals (M
A
 and M
B
) are in the 1 : 2 ratio. When these metals are 
exposed to photons of energy 6 eV, the kinetic energy of liberated electrons of M
A
 : M
B
 is in the 
Page 5


 
JEE Main Previous Year Questions (2021-2026): 
Structure of Atom  
 
(January 2026) 
Q1: The wavelength of photon A is 400 nm. The frequency of photon B is 10¹6 s?¹. The wave 
number of photon C is 104 cm?¹. 
The correct order of energy of these photons is: 
A: A > C > B 
B: C > B > A 
C: B > A > C 
D: A > B > C 
Answer: C 
Explanation 
 
So, smaller wavelength means higher energy. Therefore: E
B
 > E
A
 > E
C
  
 
Q2: The wave numbers of three spectral lines of H atom are considered. Identify the set of 
spectral lines belonging to Balmer series. 
(R = Rydberg constant) 
A:  
B:  
C:  
D:  
Answer: A 
Explanation:  
The Balmer series corresponds to transitions ending at n
1
 = 2. 
For hydrogen, the wave number is given by 
 
Now compute the ?rst few Balmer lines: 
For 
 
 
Q3:  
 
Figure 1. electron probability density for 2 s orbital 
 
Figure 2. wave function for 2s orbital 
Which of the following points in Figure 2 most accurately represents the nodal surface as 
shown in Figure 1? 
A: D 
B: B 
C: C 
D: A 
Answer: B 
Explanation: 
The nodal surface (as seen in Figure 1) is the place where the probability density becomes zero. 
For an s-orbital, this nodal surface is a spherical node (a spherical shell around the nucleus). 
At a spherical (radial) node, the radial wave function becomes zero. 
 
So, in Figure 2, the correct point is where the 2s wave function curve cuts the horizontal axis 
(changes sign). That point represents the spherical node shown in Figure 1. 
 
Q4: The wavelength of spectral line obtained in the spectrum of Li
2+
 ion, when the transition 
takes place between two levels whose sum is 4 and difference is 2 , is 
A: 1.14 × 10?6 cm 
B: 2.28 × 10?6 cm 
C: 1.14 × 10?7 cm 
D: 2.28 × 10?7 cm 
Answer: A 
Explanation:  
n1 ? lower energy level 
n2 ? higher energy level 
n1 + n2 = 4, n2 = 3 
n2 - n1 = 2, n1 = 1 
Rydberg's formula: 
 
 
Q5: The work functions of two metals (M
A
 and M
B
) are in the 1 : 2 ratio. When these metals are 
exposed to photons of energy 6 eV, the kinetic energy of liberated electrons of M
A
 : M
B
 is in the 
ratio 2.642 : 1. 
The work functions (in eV) of M
A
 and M
B
 respectively are: 
A: 3.1, 6.2 
B: 1.5, 3.0 
C: 2.3, 4.6 
D: 1.4, 2.8 
Answer: C 
Explanation: