Structure of Atom JEE Advanced previous year questions with solutions

5 solved JEE Advanced questions on Structure of Atom, free to read — no sign-in needed. The full chapter has 31 questions; sign in to attempt the remaining 26 in the exam simulator.

1. Q1JEE Advanced Adv 2026 (Paper 2)
   $\text{X}^{a+}$ and $\text{Y}^{b+}$ are hydrogen-like species. The wavelength of light absorbed during the transition between the states with principal quantum numbers $n = 1$ and $n = 2$ of $\text{X}^{a+}$ is $\lambda$. The wavelength of light absorbed during the transition between the states with principal quantum numbers $n = 2$ and $n = 4$ of $\text{Y}^{b+}$ is $9\lambda$. The lowest possible value of $(a + b)$ is ____.
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   Answer: 3

   For a hydrogen-like species, the wavelength of the absorbed light during a transition from $n_1$ to $n_2$ is given by the Rydberg formula: $\dfrac{1}{\lambda} = R Z^2 \left( \dfrac{1}{n_1^2} - \dfrac{1}{n_2^2} \right)$ For $\text{X}^{a+}$, the transition is from $n = 1$ to $n = 2$: $\dfrac{1}{\lambda} = R Z_X^2 \left( \dfrac{1}{1^2} - \dfrac{1}{2^2} \right) = \dfrac{3}{4} R Z_X^2 \Rightarrow \lambda = \dfrac{4}{3 R Z_X^2}$ For $\text{Y}^{b+}$, the transition is from $n = 2$ to $n = 4$: $\dfrac{1}{9\lambda} = R Z_Y^2 \left( \dfrac{1}{2^2} - \dfrac{1}{4^2} \right) = R Z_Y^2 \left( \dfrac{1}{4} - \dfrac{1}{16} \right) = \dfrac{3}{16} R Z_Y^2 \Rightarrow 9\lambda = \dfrac{16}{3 R Z_Y^2}$ Substituting the expression for $\lambda$ from the first equation into the second: $9 \left( \dfrac{4}{3 R Z_X^2} \right) = \dfrac{16}{3 R Z_Y^2}$ $\dfrac{12}{Z_X^2} = \dfrac{16}{3 Z_Y^2}$ $36 Z_Y^2 = 16 Z_X^2$ $9 Z_Y^2 = 4 Z_X^2 \Rightarrow 3 Z_Y = 2 Z_X$ $\dfrac{Z_X}{Z_Y} = \dfrac{3}{2}$ For the lowest possible integer values of atomic numbers, we get $Z_X = 3$ and $Z_Y = 2$. Since $\text{X}^{a+}$ and $\text{Y}^{b+}$ are hydrogen-like species, they must contain exactly $1$ electron. Therefore, the charge on the ion is $(Z - 1)$. For $\text{X}^{a+}$ ($Z_X = 3$): $a = 3 - 1 = 2$ For $\text{Y}^{b+}$ ($Z_Y = 2$): $b = 2 - 1 = 1$ The lowest possible value of $(a + b)$ is $2 + 1 = 3$. Answer: $3$
2. Q2JEE Advanced Adv 2026 (Paper 1)
   The 2s and the 2p orbital energies of hydrogen atom are $E_{2s}(\text{H})$ and $E_{2p}(\text{H})$, respectively. The 2s and the 2p orbital energies of lithium atom are $E_{2s}(\text{Li})$ and $E_{2p}(\text{Li})$, respectively. The correct option(s) about the orbital energies is(are)
   1. A.$E_{2s}(\text{Li}) \lt E_{2p}(\text{Li})$
   2. B.$E_{2s}(\text{H}) = E_{2p}(\text{H})$
   3. C.$E_{2p}(\text{H}) \lt E_{2s}(\text{Li})$
   4. D.$E_{2s}(\text{H}) \gt E_{2s}(\text{Li})$
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   Answer: A,B,D

   For hydrogen atom (a single-electron system), the energy of an orbital depends only on the principal quantum number $n$. Thus, the 2s and 2p orbitals are degenerate. $E_{2s}(\text{H}) = E_{2p}(\text{H})$ For lithium atom (a multi-electron system), the energy depends on both $n$ and $l$. Due to the penetration effect, the 2s electron penetrates closer to the nucleus and experiences a higher effective nuclear charge than the 2p electron. Thus, the 2s orbital has lower energy than the 2p orbital. $E_{2s}(\text{Li}) \lt E_{2p}(\text{Li})$ The energy of an orbital is given by $E \propto -\dfrac{Z_{\text{eff}}^2}{n^2}$. For hydrogen, $Z_{\text{eff}} = 1$. For the 2s electron in lithium, the nuclear charge is $Z=3$, and it is shielded by the $1s^2$ electrons. The effective nuclear charge $Z_{\text{eff}}$ for the 2s electron in Li is greater than 1 (approximately 1.3). Since $Z_{\text{eff}}(\text{Li}, 2s) \gt Z_{\text{eff}}(\text{H}, 2s)$, the energy of the 2s orbital in Li is more negative than that in H. $E_{2s}(\text{Li}) \lt E_{2s}(\text{H}) \Rightarrow E_{2s}(\text{H}) \gt E_{2s}(\text{Li})$ Since $E_{2p}(\text{H}) = E_{2s}(\text{H})$, we also have $E_{2p}(\text{H}) \gt E_{2s}(\text{Li})$. Answer: $E_{2s}(\text{Li}) \lt E_{2p}(\text{Li})$; $E_{2s}(\text{H}) = E_{2p}(\text{H})$; $E_{2s}(\text{H}) \gt E_{2s}(\text{Li})$
3. Q3JEE Advanced Adv 2024 (Paper 2)
   According to Bohr's model, the highest kinetic energy is associated with the electron in the
   1. A.first orbit of H atom
   2. B.first orbit of He+
   3. C.second orbit of He+
   4. D.second orbit of Li2+
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   Answer: (B)

   $K E=+13.6 \times \frac{Z^2}{n^2}$ (A) $\mathrm{KE}_{1,11}=+13.6 \times \frac{1^2}{1^2}=13.6 \mathrm{eV}$ (B) $\mathrm{KE}_{1, \mathrm{Hc}}=+13.6 \times \frac{2^2}{1^2}=13.6 \times 4 \mathrm{eV}$ (C) $\mathrm{KE}_{2, \mathrm{He}^{+}}=+13.6 \times \frac{2^2}{2^2}=13.6 \mathrm{eV}$ (D) $\mathrm{KE}_{2, \mathrm{Li}}{ }^{\mathrm{i}^{+}}=+13.6 \times \frac{3^2}{2^2}=13.6 \times \frac{9}{4} \mathrm{eV}$
4. Q4JEE Advanced Adv 2024 (Paper 1)
   Among the following, the correct statement(s) for electrons in an atom is(are)
   1. A.Uncertainty principle rules out the existence of definite paths for electrons.
   2. B.The energy of an electron in $2 s$ orbital of an atom is lower than the energy of an electron that is infinitely far away from the nucleus.
   3. C.According to Bohr's model, the most negative energy value for an electron is given by $\mathrm{n}=1$, which corresponds to the most stable orbit.
   4. D.According to Bohr's model, the magnitude of velocity of electrons increases with increase in values of $n$.
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   Answer: A,B,C

   (1) Uncertainity principle talks about probability of finding electrons in different regions around the nucleus rather than definite paths. (2) With increase in distance of electron from the nucleus, its energy increases. (3) Energy of electron $E_n=-13.6 \times \frac{\mathrm{Z}^2}{\mathrm{n}^2} \mathrm{eV} /$ atom. (4) Velocity of electron $V_n=2.19 \times 10^6 \times \frac{Z}{n} \mathrm{~m} / \mathrm{sec}$.
5. Q5JEE Advanced Adv 2023 (Paper 2)
   For ${He}^{+}$, a transition takes place from the orbit of radius $105.8pm$ to the orbit of radius $26.45pm$. The wavelength (in $nm$ ) of the emitted photon during the transition is Bohr radius, $a=52.9pm$ Rydberg constant, ${R}_{H}=2.2\times {10}^{-18}J$ Planck's constant, $h=6.6\times {10}^{-34}Js$ Speed of light, $c=3\times {10}^{8}{ms}^{-1}$ ]
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   Answer: 30

   The radius of the nth orbit can be represented as, $r=52.9\times \dfrac{{n}^{2}}{z}pm$ $∴105.8=\dfrac{52.9\times {n}^{2}}{2}∴{n}_{2}=2$ and $26.45=52.9\times \dfrac{{n}^{2}}{2}∴{n}_{1}=1$ $∵\Delta E={R}_{H}hC\times {z}^{2}\left[\dfrac{1}{{n}_{1}^{2}}-\dfrac{1}{{n}_{2}^{2}}\right]$ $\dfrac{hc}{\lambda }={R}_{H}hC\times {z}^{2}\left[\dfrac{1}{{n}_{1}^{2}}-\dfrac{1}{{n}_{2}^{2}}\right]$ $\dfrac{6.6\times {10}^{-34}\times 3\times {10}^{8}}{\lambda }=2.2\times {10}^{-18}\times 4\times \left[\dfrac{1}{1}-\dfrac{1}{4}\right]$ $\dfrac{6.6\times {10}^{-34}\times 3\times {10}^{8}}{\lambda }=2.2\times {10}^{-18}\times 4\times \dfrac{3}{4}$ $∴\lambda =300A$ $∴\lambda =30nm$