Structure of Atom JEE Main previous year questions with solutions

5 solved JEE Main questions on Structure of Atom, free to read — no sign-in needed. The full chapter has 233 questions; sign in to attempt the remaining 228 in the exam simulator.

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1. Q1JEE Main 2026 (06 Apr, Shift 2)Bohr's model
   The Bohr radius of a hydrogen like species is $70.53$ pm. The species and the stationary state $(n)$ are respectively (Given : Hydrogen atom Bohr radius is $52.9$ pm)
   1. A.$\text{Li}^{2+}, 3$
   2. B.$\text{He}^+, 3$
   3. C.$\text{He}^+, 2$
   4. D.$\text{Li}^{2+}, 2$
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   Answer: (D)

   The radius of the $n$-th Bohr orbit for a hydrogen-like species is given by the formula: $r_n = a_0 \dfrac{n^2}{Z}$ where $a_0$ is the Bohr radius of the hydrogen atom ($52.9$ pm), $n$ is the principal quantum number, and $Z$ is the atomic number. Given $r_n = 70.53$ pm, we can write: $70.53 = 52.9 \times \dfrac{n^2}{Z}$ $\dfrac{n^2}{Z} = \dfrac{70.53}{52.9} = 1.333 = \dfrac{4}{3}$ Checking the given options: For $\text{Li}^{2+}$ and $n = 3$, $Z = 3 \Rightarrow \dfrac{n^2}{Z} = \dfrac{9}{3} = 3$ For $\text{He}^+$ and $n = 3$, $Z = 2 \Rightarrow \dfrac{n^2}{Z} = \dfrac{9}{2} = 4.5$ For $\text{He}^+$ and $n = 2$, $Z = 2 \Rightarrow \dfrac{n^2}{Z} = \dfrac{4}{2} = 2$ For $\text{Li}^{2+}$ and $n = 2$, $Z = 3 \Rightarrow \dfrac{n^2}{Z} = \dfrac{4}{3}$ The ratio matches for $\text{Li}^{2+}$ and $n = 2$. Answer: $\text{Li}^{2+}, 2$
2. Q2JEE Main 2025 (03 Apr, Shift 2)Dual Behaviour of Matter and Heisenberg Uncertainty Principle
   For electron in ' 2 s ' and ' 2 p ' orbitals, the orbital angular momentum values, respectively are :
   1. A.$\sqrt{2} \frac{\mathrm{~h}}{2 \pi}$ and 0
   2. B.$\frac{\mathrm{h}}{2 \pi}$ and $\sqrt{2} \frac{\mathrm{~h}}{2 \pi}$
   3. C.0 and $\sqrt{6} \frac{\mathrm{~h}}{2 \pi}$
   4. D.0 and $\sqrt{2} \frac{\mathrm{~h}}{2 \pi}$
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   Answer: (D)

   $\begin{aligned} & \text { Orbital angular momentum }=\sqrt{\ell(\ell+1)} \frac{\mathrm{h}}{2 \pi} \\ & \therefore \quad \text { For } 2 \mathrm{~s} \text { orbital }: \ell=0 \\ & \text { Orbital angular momentum }=0 \\ & \therefore \quad \text { For } 2 \mathrm{p} \text { orbital }: \ell=1 \\ & \quad \begin{aligned} \text { Orbital angular momentum } & =\sqrt{1(1+2)} \frac{\mathrm{h}}{2 \pi} \\ & =\sqrt{2} \frac{\mathrm{~h}}{2 \pi}\end{aligned}\end{aligned}$
3. Q3JEE Main 2024 (09 Apr, Shift 1)Electronic configuration
   Compare the energies of following sets of quantum numbers for multielectron system. (A) $\mathrm{n}=4,1=1$ (B) $\mathrm{n}=4, \mathrm{l}=2$ (C) $\mathrm{n}=3,1=1$ (D) $\mathrm{n}=3,1=2$ (E) $\mathrm{n}=4,1=0$ Choose the correct answer from the options given below :
   1. A.$\text { (B) } > \text { (A) } > \text { (C) } > \text { (E) } > \text { (D) }$
   2. B.$\text { (E) } < (\text { C }) < \text { (D) } < (\text { A }) < (\text { B })$
   3. C.$\text { (E) } > (\text { C }) > (\text { A }) > (\text { D }) > (B)$
   4. D.$\text { (C) } < \text { (E) } < \text { (D) } < \text { (A) } < \text { (B) }$
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   Answer: (D)

   Energy level can be determined by comparing ( $\mathrm{n}+\ell$ ) values (A) $\mathrm{n}=4, \ell=1 \Rightarrow(\mathrm{n}+\ell)=5$ (B) $\mathrm{n}=4, \ell=2 \Rightarrow(\mathrm{n}+\ell)=6$ (C) $\mathrm{n}=3, \ell=1 \Rightarrow(\mathrm{n}+\ell)=4$ (D) $\mathrm{n}=3, \ell=2 \Rightarrow(\mathrm{n}+\ell)=5$ (E) $\mathrm{n}=4, \ell=0 \Rightarrow(\mathrm{n}+\ell)=4$ For same value of $(n+\ell)$, orbital having higher value of $n$, will have more energy. (B) $\gt (\mathrm{A})\gt (\mathrm{D})\gt (\mathrm{E})\gt (\mathrm{C})$
4. Q4JEE Main 2023 (15 Apr, Shift 1)Atomic Mass and Atomic Size
   The total number of isoelectronic species from the given set is ________. ${O}^{2-},{F}^{–},Al,{Mg}^{2+},{Na}^{+},{O}^{+},Mg,{Al}^{3+},F$
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   Answer: 5

   Isoelectronic are those which have the same number of electrons. Let us write the number of electrons from the given species. Species No. of electrons ${O}^{–2}$ $10$ ${F}^{-}$ $10$ $Al$ $13$ ${Mg}^{2+}$ $10$ ${Na}^{+}$ $10$ ${O}^{+}$ $7$ $Mg$ $12$ ${Al}^{3+}$ $10$ $F$ $9$ Hence five species among the given are isoelectronic species.
5. Q5JEE Main 2022 (25 Jul, Shift 2)Hydrogen spectrum
   When the excited electron of a $H$ atom from $n=5$ drops to the ground state, the maximum number of emission lines observed are
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   Answer: 10

   Maximum number of emission lines $=\dfrac{\Delta n\left(\Delta n+1\right)}{2} \Delta n={n}_{2}-{n}_{1} {n}_{2}=5 {n}_{1}=1$ Maximum number of emission lines=$\dfrac{4\left(4+1\right)}{2}=\dfrac{20}{2}=10$ These lines correspond to the lyman series of the hydrogen spectrum