Atomic Structure Previous Year Questions (JEE Main 2026)
Solved JEE Main 2026 previous year questions on Atomic Structure with step-by-step solutions covering Bohr model, hydrogen spectrum, quantum numbers, radial nodes, ionization energy and the photoelectric effect.
A curated set of JEE Main 2026 previous-year questions on Atomic Structure, each solved step by step so you can check both the final answer and the reasoning.
Solutions are AI-generated and pending review.
Solution
Wave number: $\bar{\nu} = R_H Z^2 \left(\dfrac{1}{n_1^2} - \dfrac{1}{n_2^2}\right)$, with $Z = 1$ for H.
Balmer first line ($n_1 = 2,\ n_2 = 3$):
$$\bar{\nu}_B = R_H\left(\dfrac{1}{2^2} - \dfrac{1}{3^2}\right) = R_H\left(\dfrac{1}{4} - \dfrac{1}{9}\right) = R_H \cdot \dfrac{5}{36}$$Brackett first line ($n_1 = 4,\ n_2 = 5$):
$$\bar{\nu}_{Br} = R_H\left(\dfrac{1}{4^2} - \dfrac{1}{5^2}\right) = R_H\left(\dfrac{1}{16} - \dfrac{1}{25}\right) = R_H \cdot \dfrac{9}{400}$$Ratio:
$$\dfrac{\bar{\nu}_B}{\bar{\nu}_{Br}} = \dfrac{5/36}{9/400} = \dfrac{5}{36}\cdot\dfrac{400}{9} = \dfrac{2000}{324} = 6.17$$Writing as $5 : x$: $\ x = \dfrac{5}{6.17} = 0.81$, so the ratio is $5 : 0.81$.
Answer: B
Solution
Filling orbitals strictly by the Aufbau order (as the question demands, ignoring the Cr anomaly):
$$1s^2\ 2s^2\ 2p^6\ 3s^2\ 3p^6\ 4s^2\ 3d^4$$Cumulative electron counts: $1s^2 \to 2$, up to $2p^6 \to 10$, up to $3p^6 \to 18$. Therefore electrons 19 and 20 go into the $4s$ subshell.
The 19th electron is the first electron entering $4s$:
$$n = 4,\quad l = 0\ (s\text{-subshell}),\quad m = 0,\quad s = +\tfrac{1}{2}$$Answer: D
Solution
Shortest wavelength, Lyman (H): transition $n = \infty \to 1$, $Z = 1$:
$$\dfrac{1}{x} = R\cdot 1^2\left(\dfrac{1}{1^2} - 0\right) = R \quad\Rightarrow\quad R = \dfrac{1}{x}$$Longest wavelength, Balmer ($\text{He}^+$): smallest energy gap $n = 3 \to 2$, $Z = 2$:
$$\dfrac{1}{\lambda} = R\cdot 2^2\left(\dfrac{1}{2^2} - \dfrac{1}{3^2}\right) = 4R\left(\dfrac{1}{4} - \dfrac{1}{9}\right) = 4R\cdot\dfrac{5}{36} = \dfrac{5R}{9}$$Therefore:
$$\lambda = \dfrac{9}{5R} = \dfrac{9}{5}\cdot\dfrac{1}{R} = \dfrac{9x}{5}$$Answer: $\dfrac{9x}{5}$
Solution
Use radial nodes $= n - l - 1$ and nodal planes (angular nodes) $= l$.
| Orbital | $n$ | $l$ | Radial nodes $=n-l-1$ | Nodal planes $=l$ | LIST-II |
|---|---|---|---|---|---|
| A. 2s | 2 | 0 | $1$ | $0$ | IV |
| B. 3s | 3 | 0 | $2$ | $0$ | III |
| C. 3p | 3 | 1 | $1$ | $1$ | II |
| D. 4d | 4 | 2 | $1$ | $2$ | I |
Matching: A-IV, B-III, C-II, D-I.
Answer: D
Solution
The process $\text{Li}(g) \to \text{Li}^{2+}(g)$ removes two electrons, so the molar energy is the sum of the first two ionization enthalpies:
$$\Delta E = IE_1 + IE_2 = 520 + 7297 = 7817\ \text{kJ mol}^{-1}$$The requested unit is kJ mol$^{-1}$ (a per-mole quantity), so the given mass of 3.5 mg does not change the molar answer.
Answer: 7817
Solution
$\text{Li}^{2+}$ is a single-electron (hydrogen-like) species with $Z = 3$, electron in the ground state ($n = 1$). Ionization energy scales as $Z^2$:
$$IE = 2.18\times10^{-18}\cdot Z^2 = 2.18\times10^{-18}\times 3^2$$$$= 2.18\times10^{-18}\times 9 = 1.962\times10^{-17}\ \text{J atom}^{-1}$$Answer: $1.962 \times 10^{-17}$
Solution
Bohr radius: $r_n = a_0\,\dfrac{n^2}{Z}$. Identical radii require equal $\dfrac{n^2}{Z}$.
| Species | $n$ | $Z$ | $n^2/Z$ |
|---|---|---|---|
| A. H (1st) | 1 | 1 | $1$ |
| B. He$^+$ (1st) | 1 | 2 | $0.5$ |
| C. He$^+$ (2nd) | 2 | 2 | $2$ |
| D. Li$^{2+}$ (1st) | 1 | 3 | $0.33$ |
| E. Be$^{3+}$ (2nd) | 2 | 4 | $1$ |
A and E both give $n^2/Z = 1$, so they have identical radii.
Answer: B
Solution
Photoelectric equation: $\dfrac{hc}{\lambda} = W_0 + KE$.
Work function in joules:
$$W_0 = 2.3\times 1.6\times10^{-19} = 3.68\times10^{-19}\ \text{J}$$Photon energy:
$$E = W_0 + KE = 3.68\times10^{-19} + 2.8\times10^{-20} = 3.96\times10^{-19}\ \text{J}$$Wavelength:
$$\lambda = \dfrac{hc}{E} = \dfrac{6.6\times10^{-34}\times 3.0\times10^{8}}{3.96\times10^{-19}} = \dfrac{1.98\times10^{-25}}{3.96\times10^{-19}} = 5.0\times10^{-7}\ \text{m}$$$$\lambda = 500\ \text{nm} = 5\times10^{2}\ \text{nm}$$So $x = 5$.
Answer: 5
Solution
Bohr radius: $r_n = 52.9\,\dfrac{n^2}{Z}$ pm. Set equal to $70.53$ pm and test the options.
$$\dfrac{n^2}{Z} = \dfrac{70.53}{52.9} = 1.333 = \dfrac{4}{3}$$This is satisfied by $n^2 = 4,\ Z = 3$, i.e. $n = 2$ and $Z = 3$ ($\text{Li}^{2+}$):
$$r = 52.9\times\dfrac{2^2}{3} = 52.9\times\dfrac{4}{3} = 70.53\ \text{pm}\ \checkmark$$Answer: D
Solution
For multi-electron atoms, energy order follows the $(n+l)$ rule; ties are broken by lower $n$.
| Label | Orbital | $n+l$ |
|---|---|---|
| E | 2p | $3$ |
| B | 4s | $4$ |
| A | 3d | $5$ |
| D | 5p | $6$ |
| C | 6p | $7$ |
Increasing energy: $E < B < A < D < C$.
Answer: D
Solution
A. True — Heisenberg’s uncertainty principle applies to microscopic particles such as electrons.
B. True — a higher principal quantum number means a larger orbital, so $2p_x < 3p_x$ in size.
C. False — orbital energies depend on nuclear charge. For a given shell, energy $\propto Z^2$ (H-like) or is shifted by $Z_{\text{eff}}$ in multi-electron atoms, so the $2s$ energy of Li is much lower (more negative) than that of H.
D. True — Cr shows the stable half-filled configuration $[\text{Ar}]\,3d^5\,4s^1$.
Correct statements: A, B and D.
Answer: B
Solution
A. True — for equal $(n+l)$, the orbital with lower $n$ has lower energy (a corollary of the Aufbau rule).
B. False — as atomic number increases, the increased nuclear attraction makes orbital energies more negative (they decrease, not increase).
C. True — $3p_x$ is larger than $2p_x$ due to the higher principal quantum number.
D. Radial nodes $= n - l - 1$:
$$5f:\ 5-3-1=1,\quad 6s:\ 6-0-1=5,\quad 4d:\ 4-2-1=1,\quad 5p:\ 5-1-1=3,\quad 5d:\ 5-2-1=2$$The $5d$ orbital has 2 radial nodes, so the claim “none have 2 radial nodes” is False.
Correct statements: A and C only.
Answer: B
Solution
Photon energy is inversely proportional to wavelength: $E = \dfrac{hc}{\lambda}$, so $E \propto \dfrac{1}{\lambda}$.
$$\dfrac{E_1}{E_2} = \dfrac{\lambda_2}{\lambda_1} = \dfrac{6000}{2000} = 3$$Answer: 3