Structure of Atom JEE Main PYQs (2020–2026)

Solved Structure of Atom PYQs

3 recent questions with answers. The full set of 162 is free to practise.

1. JEE Main 2026EasyDual Nature of Electromagnetic Radiation & Photoelectric Effect
   Q1

   Consider two radiations of wavelengths
   
   $${\lambda }_1=2000\mathring{A}$$
   
   $${\lambda }_2=6000\mathring{A}$$
   
   The ratio of the energies of these two radiations $\left(\frac{E_1}{E_2}\right)$ is $\_\_\_\_$ (Nearest integer).

   Numerical answer type

   Show answer & solution

   Correct answer: 3

   We know that the energy of a photon is given by the formula
   
   $$E=\frac{hc}{\lambda }$$
   
   where
   
   $h$ = Planck’s constant
   
   $c$ = velocity of light
   
   $\lambda$ = wavelength of the radiation.
   
   For the two radiations,
   
   $$\frac{E_1}{E_2}=\frac{\frac{hc}{{\lambda }_1}}{\frac{hc}{{\lambda }_2}}=\frac{{\lambda }_2}{{\lambda }_1}$$
   
   $${\lambda }_1=2000\text{A˚},{\lambda }_2=6000\text{A˚}$$
   
   $$\frac{E_1}{E_2}=\frac{6000}{2000}$$
   
   $$\frac{E_1}{E_2}=3$$
   
   Final Answer
   
   $$\boxed{\frac{E_1}{E_2}=3}$$
   
   Hence, the ratio of energies of the two radiations is 3 (nearest integer).
2. JEE Main 2026MediumBohr's Model of Hydrogen Atom & Line Spectrum
   Q2

   The hydrogen spectrum consists of several spectral lines in Lyman series $(L_1,L_2$ , $L_3\dots ;L_1$ has lowest energy among Lyman series). Similarly it consists of several spectral lines in Balmer series $(B_1,B_2,B_3\dots ;B_1$ has lowest energy among Balmer lines). The energy of $L_1$ is $x$ times the energy of $B_1$ . The value of $x$ is $\_\_\_\_$ $\times 10^{-1}$
   
   . (Nearest integer)

   Numerical answer type

   Show answer & solution

   Correct answer: 54

   For hydrogen, the energy difference for a spectral line is given by
   
   $$\Delta E=13.6\times Z^2\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)$$
   
   Step 1: Identify the transitions
   
   In the Lyman series, electrons fall to $n_1=1$ . The first line $L_1$ is the transition $2\to 1$ .
   
   In the Balmer series, electrons fall to $n_1=2$ . The first line $B_1$ is the transition $3\to 2$ .
   
   Step 2: Find energy of $L_1$
   
   $$\begin{array}{rl}& \Delta E\left(L_1\right)=13.6\times Z^2\left(\frac{1}{1^2}-\frac{1}{2^2}\right)=13.6Z^2\times \frac{3}{4}\end{array}$$
   
   Step 3: Find energy of $B_1$
   
   $$\begin{array}{rl}& \Delta E\left(B_1\right)=13.6\times Z^2\times \left(\frac{1}{2^2}-\frac{1}{3^2}\right)=13.6\times Z^2\times \frac{5}{4\times 9}\end{array}$$
   
   Step 4: Take the ratio to get $x$
   
   The factor $13.6Z^2$ cancels in the ratio.
   
   $$\begin{array}{rl}& \frac{\Delta E\left(L_1\right)}{\Delta E\left(B_1\right)}=\frac{3}{5}\times 9=\frac{27}{5}=x\end{array}$$
   
   Step 5: Write in the asked form
   
   $$\begin{array}{rl}& x=\left(\frac{27}{5}\times 10\right)\times 10^{-1}=54\times 10^{-1}\end{array}$$
3. JEE Main 2026EasyElectronic Configuration of Atoms (Aufbau, Pauli, Hund's Rule)
   Q3

   Arrange the following atomic orbitals of multi electron atoms in order of increasing energy.
   
   A. $n=3,l=2,m=+1$
   
   B. $n=4,l=0,m=0$
   
   C. $n=6,l=1,m=0$
   
   D. $n=5,l=1,m=+1$
   
   E. $n=2,l=1,m=+1$
   
   Choose the correct answer from the options given below:

   • A$C<D<B<A<E$
   • B$B<A<E<C<D$
   • C$E<C<D<B<A$
   • D$E<B<A<D<C$

   Show answer & solution

   Correct answer: (D)

   For multi-electron atoms, the energy of an orbital is decided by the $n+l$ rule.
   
   The orbital with smaller value of $n+l$ has lower energy.
   
   If two orbitals have the same value of $n+l$, then the orbital with smaller $n$ has lower energy.
   
   Also, energy does not depend on $m$.
   
   Now let us identify each orbital:
   
   A: $n=3,\text{mspace}6mul=2$
   
   This is $3d$
   
   $$n+l=3+2=5$$
   
   B: $n=4,\text{mspace}6mul=0$
   
   This is $4s$
   
   $$n+l=4+0=4$$
   
   C: $n=6,\text{mspace}6mul=1$
   
   This is $6p$
   
   $$n+l=6+1=7$$
   
   D: $n=5,\text{mspace}6mul=1$
   
   This is $5p$
   
   $$n+l=5+1=6$$
   
   E: $n=2,\text{mspace}6mul=1$
   
   This is $2p$
   
   $$n+l=2+1=3$$
   
   Now arrange in increasing energy:
   
   $$2p<4s<3d<5p<6p$$
   
   So,
   
   $$E<B<A<D<C$$
   
   Hence, the correct answer is:
   
   $$\boxed{\text{Option D}}$$

Topic-wise question count

Where JEE Main has asked from Structure of Atom, most frequent first.

• Bohr's Model of Hydrogen Atom & Line Spectrum47
• Quantum Mechanical Model: Quantum Numbers & Orbitals27
• Dual Nature of Electromagnetic Radiation & Photoelectric Effect24
• Dual Nature of Matter (de Broglie) & Heisenberg Uncertainty Principle21
• Electronic Configuration of Atoms (Aufbau, Pauli, Hund's Rule)17
• Shapes of Atomic Orbitals & Radial/Angular Nodes17
• Subatomic Particles & Atomic Models (Thomson, Rutherford)9

Year-wise Structure of Atom questions in JEE Main

Number of questions from this chapter in each year's paper(s).