Atoms NEET PYQs (2005–2026) with Solutions

10 solved Atoms PYQs

Real NEET and AIPMT questions from this chapter, newest first. Try each one, then open the answer and the step-by-step explanation.

1. Question 1 (NEET 2026)

   NEET 2026EasyAlpha-particle scattering experiment

   In Geiger-Marsden experiment, the number of scattered $\alpha$-particles $N(\theta)$ is plotted as a function of scattering angle $\theta$. Which of the following options represents the correct plot?

   1. AOption A: Graph A
   2. BOption B: Graph B
   3. COption C: Graph C
   4. DOption D: Graph D

   Show answer & explanationHide answer & explanation

   Correct answer: (C) Graph C

   Explanation

   The graphs (log-scale $N(\theta)$ from 10 to $10^7$ against $\theta$ from 0° to 180°): (A) $N$ increases steadily with $\theta$; (B) $N$ nearly flat, then falling at large angles; (C) $N$ very large (about $10^7$) at small angles, falling steeply and levelling off near 10 towards 180°; (D) bell-shaped curve peaking around 90°. In the Geiger–Marsden experiment most α-particles pass through the gold foil almost undeflected; the number scattered falls very rapidly as the angle increases, and only about 1 in 8000 is deflected by more than 90°. Rutherford's analysis gives $N(\theta) \propto \frac{1}{\sin^4(\theta/2)}$, a steeply decreasing function of $\theta$. This matches option (C).

2. Question 2 (NEET 2026)

   NEET 2026MediumBohr model: orbit radius and speed

   Consider that an electron is revolving in an excited state of Hydrogen atom with velocity $\sqrt{25.6} \times 10^5\ \mathrm{ms^{-1}}$. The radius of the orbit is $x \times 10^{-9}\ \mathrm{m}$. The value of $x$ is: [Take the mass of electron to the $9 \times 10^{-31}\ \mathrm{kg}$, charge of electron = $-1.6 \times 10^{-19}\ \mathrm{C}$ and $\frac{1}{4\pi\varepsilon_0} = 9 \times 10^9\ \mathrm{Nm^2C^{-2}}$]

   1. AOption A: $4$
   2. BOption B: $3$
   3. COption C: $2$
   4. DOption D: $1$

   Show answer & explanationHide answer & explanation

   Correct answer: (D) $1$

   Explanation

   The Coulomb attraction provides the centripetal force: $\frac{mv^2}{r} = \frac{1}{4\pi\varepsilon_0}\frac{e^2}{r^2} \Rightarrow r = \frac{e^2}{4\pi\varepsilon_0 m v^2}$. $v^2 = 25.6 \times 10^{10} = 2.56 \times 10^{11}\ \mathrm{m^2s^{-2}}$. $$r = \frac{9 \times 10^9 \times (1.6 \times 10^{-19})^2}{9 \times 10^{-31} \times 2.56 \times 10^{11}} = \frac{9 \times 10^9 \times 2.56 \times 10^{-38}}{9 \times 10^{-31} \times 2.56 \times 10^{11}} = 10^{-9}\ \mathrm{m}$$ So $x = 1$, option (D).

3. Question 3 (NEET 2025)

   NEET 2025HardBohr Model

   A particle of mass $m$ is moving around the origin with a constant force $F$ pulling it towards the origin. If Bohr model is used to describe its motion, the radius $r$ of the $n^{th}$ orbit and the particle’s speed $v$ in the orbit depend on $n$ as

   1. AOption A: $r \propto n^{1/3},\; v \propto n^{1/3}$
   2. BOption B: $r \propto n^{1/3},\; v \propto n^{2/3}$
   3. COption C: $r \propto n^{2/3},\; v \propto n^{1/3}$
   4. DOption D: $r \propto n^{4/3},\; v \propto n^{-1/3}$

   Show answer & explanationHide answer & explanation

   Correct answer: (C) $r \propto n^{2/3},\; v \propto n^{1/3}$

   Explanation

   For circular motion under constant central force: $$\frac{mv^2}{r} = F$$ So, $$v^2 \propto r \Rightarrow v \propto r^{1/2}$$ Using Bohr’s quantization condition: $$mvr = n\hbar$$ Substituting $v \propto r^{1/2}$:$$mr^{3/2} \propto n$$Hence,$$r \propto n^{2/3}$$and therefore,$$v \propto r^{1/2} \propto n^{1/3}$$ Thus, option (C) is correct.

4. Question 4 (NEET 2025)

   NEET 2025MediumHydrogen Spectrum

   Energy and radius of first Bohr orbit of $\mathrm{He}^+$ and $\mathrm{Li}^{2+}$ are given.

   1. AOption A: Option (1)
   2. BOption B: Option (2)
   3. COption C: Option (3)
   4. DOption D: Option (4)

   Show answer & explanationHide answer & explanation

   Correct answer: (A) Option (1)

   Explanation

   For hydrogen-like species: $E_n = -13.6 Z^2\ \mathrm{eV} = -R_H Z^2$ and $r_n = \dfrac{a_0}{Z}$ for the first orbit $(n=1)$. For $\mathrm{Li}^{2+}$ $(Z=3)$: $E = -2.18 \times 10^{-18} \times 9 = -19.62 \times 10^{-18}\ \mathrm{J}$ $r = \dfrac{52.9}{3} = 17.6\ \mathrm{pm}$ For $\mathrm{He}^{+}$ $(Z=2)$: $E = -2.18 \times 10^{-18} \times 4 = -8.72 \times 10^{-18}\ \mathrm{J}$ $r = \dfrac{52.9}{2} = 26.4\ \mathrm{pm}$ Hence, option (A) is correct.

5. Question 5 (NEET 2024)

   NEET 2024MediumHydrogen Spectrum

   Match List I with List II. Choose the correct answer from the options given below:

   1. AOption A: A-IV, B-III, C-I, D-II
   2. BOption B: A-I, B-II, C-III, D-IV
   3. COption C: A-II, B-I, C-IV, D-III
   4. DOption D: A-III, B-IV, C-II, D-I

   Show answer & explanationHide answer & explanation

   Correct answer: (D) A-III, B-IV, C-II, D-I

   Explanation

   These are Balmer series transitions of hydrogen: $$n_2 = 3 \to n_1 = 2 \Rightarrow 656.3\,nm$$ $$n_2 = 4 \to n_1 = 2 \Rightarrow 486.1\,nm$$ $$n_2 = 5 \to n_1 = 2 \Rightarrow 434.1\,nm$$ $$n_2 = 6 \to n_1 = 2 \Rightarrow 410.2\,nm$$ Thus: A → III B → IV C → II D → I Hence, option (D) is correct.

6. Question 6 (NEET 2024)

   NEET 2024EasyAtomic Spectra

   Given below are two statements: Statement I: Atoms are electrically neutral as they contain equal number of positive and negative charges. Statement II: Atoms of each element are stable and emit their characteristic spectrum. In the light of the above statements, choose the most appropriate answer from the options given below:

   1. AOption A: Statement I is correct but Statement II is incorrect.
   2. BOption B: Statement I is incorrect but Statement II is correct.
   3. COption C: Both Statement I and Statement II are correct.
   4. DOption D: Both Statement I and Statement II are incorrect.

   Show answer & explanationHide answer & explanation

   Correct answer: (A) Statement I is correct but Statement II is incorrect., (C) Both Statement I and Statement II are correct.

   Explanation

   Statement I is correct because atoms are electrically neutral due to equal numbers of protons and electrons. Statement II is also considered correct in the context of atomic physics because atoms emit characteristic spectra corresponding to transitions between quantized energy levels. Hence, both Statement I and Statement II are correct. The official NEET (UG) 2024 T1 answer key lists Q25 with multiple correct answers: options (A) and (C).

7. Question 7 (NEET 2023)

   NEET 2023MediumBohr Model

   In hydrogen spectrum, the shortest wavelength in the Balmer series is λ. The shortest wavelength in the Brackett series is :

   1. AOption A: 2λ
   2. BOption B: 4λ
   3. COption C: 9λ
   4. DOption D: 16λ

   Show answer & explanationHide answer & explanation

   Correct answer: (B) 4λ

   Explanation

   Using Rydberg formula: 1/λ = R(1/n₁² − 1/n₂²) For shortest wavelength of Balmer series: n₁ = 2, n₂ = ∞ 1/λB = R/4 For shortest wavelength of Brackett series: n₁ = 4, n₂ = ∞ 1/λBr = R/16 Therefore, λBr/λB = 16/4 = 4 Hence, shortest wavelength in Brackett series = 4λ. Therefore, option (B) is correct.

8. Question 8 (NEET 2023)

   NEET 2023EasyBohr Model

   The radius of inner most orbit of hydrogen atom is 5.3 × 10⁻¹¹ m. What is the radius of third allowed orbit of hydrogen atom?

   1. AOption A: 0.53 Å
   2. BOption B: 1.06 Å
   3. COption C: 1.59 Å
   4. DOption D: 4.77 Å

   Show answer & explanationHide answer & explanation

   Correct answer: (D) 4.77 Å

   Explanation

   According to Bohr's model: rn = n^2 r1 For third orbit: r3 = 3^2 × 5.3×10^-11 = 9 × 5.3×10^-11 = 47.7×10^-11 m = 4.77×10^-10 m Since 1 Å = 10^-10 m, r3 = 4.77 Å.

9. Question 9 (NEET 2022)

   NEET 2022MediumBohr Model

   Let T₁ and T₂ be the energy of an electron in the first and second excited states of hydrogen atoms, respectively. According to Bohr's model of an atom, the ratio T₁ : T₂ is:

   1. AOption A: 9 : 4
   2. BOption B: 1 : 4
   3. COption C: 4 : 1
   4. DOption D: 4 : 9

   Show answer & explanationHide answer & explanation

   Correct answer: (A) 9 : 4

   Explanation

   For hydrogen atom: Eₙ ∝ -1/n² First excited state: n = 2 Second excited state: n = 3 T₁:T₂ = (1/2²):(1/3²) = 1/4 : 1/9 = 9 : 4 Hence option (A) is correct.

10. Question 10 (NEET 2022)

    NEET 2022EasyBohr Model

    If radius of second Bohr orbit of the He+ ion is 105.8 pm, what is the radius of third Bohr orbit of of third Bohr orbit of Li²⁺ ion?

    1. AOption A: 158.7 Å
    2. BOption B: 158.7 pm
    3. COption C: 15.87 pm
    4. DOption D: 1.587 pm

    Show answer & explanationHide answer & explanation

    Correct answer: (B) 158.7 pm

    Explanation

    For hydrogen-like species: rₙ ∝ n²/Z Given: For He+ (Z = 2), n = 2: r₂ = 105.8 pm For Li2+ (Z = 3), n = 3: r = 105.8 × (3²/3)/(2²/2) = 105.8 × (3/2) = 158.7 pm Hence option (B) is correct.