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NEET Atoms PYQs & Practice Questions

10 free NEET Physics questions on Atoms, starting with previous year questions (2016–2026). Tap “Show answer” to check each one. No login needed.

Topics: Bohr Model / Magnetic Moment · Bohr's Model - Atomic Radii · Rutherford's Nuclear Model · Bohr's Model - Energy Levels · Spectral Series · Bohr's Model - Velocity of Electron · de Broglie's Explanation of Bohr's Postulate · Rutherford's Nuclear Model - Distance of Closest Approach · Spectral Lines - Hydrogen Spectrum · Bohr's Model - Number of Spectral Lines · Limitations of Rutherford's Model · Bohr Model - Generalized Central Force

  1. Q1NEET 2026Geiger-Marsden (Alpha 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?

    • A. N(theta) stays roughly flat and high from 0 to 90 degrees, then declines sharply towards 180 degrees
    • B. N(theta) starts very high near 0 degrees and decreases steeply and monotonically as theta increases towards 180 degrees
    • C. N(theta) is low near 0 degrees, rises to a plateau/peak around 90 degrees, then falls back down near 180 degrees
    • D. N(theta) increases roughly linearly from a low value at 0 degrees to a higher value at 180 degrees
    Show answer

    Answer: A — N(theta) stays roughly flat and high from 0 to 90 degrees, then declines sharply towards 180 degrees

    In the Geiger-Marsden (Rutherford gold foil) scattering experiment, the number of scattered alpha particles N(theta) is maximum at small angles (most particles pass through with little deflection) and decreases monotonically and steeply as the scattering angle increases, since large-angle scattering (close encounters with the nucleus) is increasingly rare. This gives a steeply decreasing curve starting high near theta=0.

  2. Q2NEET 2026Bohr Model - Radius of Orbit

    Consider that an electron is revolving in an excited state of a Hydrogen atom with velocity √25.6 × 10⁵ ms⁻¹. The radius of the orbit is x × 10⁻⁹ m. The value of x is: [Take the mass of electron to be 9×10⁻³¹ kg, charge of electron = −1.6×10⁻¹⁹ C and 1/(4πε₀) = 9×10⁹ N m² C⁻²]

    • A. 3
    • B. 2
    • C. 1
    • D. 4
    Show answer

    Answer: B — 2

    For a circular Bohr orbit, the Coulomb force provides centripetal force: ke²/r² = mv²/r, so r = ke²/(mv²). Here v² = 25.6×10¹⁰ = 2.56×10¹¹ (ms⁻¹)². ke² = 9×10⁹ × (1.6×10⁻¹⁹)² = 9×10⁹×2.56×10⁻³⁸ = 2.304×10⁻²⁸. mv² = 9×10⁻³¹×2.56×10¹¹ = 2.304×10⁻¹⁹. r = 2.304×10⁻²⁸/2.304×10⁻¹⁹ = 1×10⁻⁹ m. So x = 1.

  3. Q3NEET 2025Bohr Model / Magnetic Moment

    A model for quantized motion of an electron in a uniform magnetic field B states that the flux passing through the orbit of the electron is n(h/e) where n is an integer, h is Planck's constant and e is the magnitude of electron's charge. According to the model, the magnetic moment of an electron in its lowest energy state will be (m is the mass of the electron)

    • A. he/(πm)
    • B. he/(2πm)
    • C. heB/(πm)
    • D. heB/(2πm)
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    Answer: A — he/(πm)

    Flux φ = nh/e = B·A = B·πR². Magnetic moment M = iA = (e/T)A = (e·eB/(2πm))×πR² ... simplifying using πR² = nh/(eB) gives M = nhe/(2πm); for lowest energy state n=1, so M = he/(2πm).

  4. Q4NEET 2025Bohr Model - Generalized Central Force

    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 nth orbit and the particle's speed v in the orbit depend on n as

    • A. r ∝ n^(1/3); v ∝ n^(1/3)
    • B. r ∝ n^(1/3); v ∝ n^(2/3)
    • C. r ∝ n^(2/3); v ∝ n^(1/3)
    • D. r ∝ n^(4/3); v ∝ n^(−1/3)
    Show answer

    Answer: B — r ∝ n^(1/3); v ∝ n^(2/3)

    F = mv²/r ⟹ v = √(Fr/m). Bohr quantization: mvr = nh ⟹ mFr³ = n²h² (squaring m√(Fr/m)·r = nh) ⟹ r³ ∝ n² ⟹ r ∝ n^(2/3). Substituting into v = √(Fr/m) gives v ∝ √(n^(2/3)) = n^(1/3).

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  5. Q5NEET 2025De-Broglie Wavelength in Bohr Orbit

    De-Broglie wavelength of an electron orbiting in the n = 2 state of hydrogen atom is close to (Given Bohr radius = 0.052 nm)

    • A. 0.067 nm
    • B. 0.67 nm
    • C. 1.67 nm
    • D. 2.67 nm
    Show answer

    Answer: A — 0.067 nm

    rn = n²a0 = 4×0.052 = 0.208 nm. Bohr quantization: 2πrn = nλ, so λ = 2πrn/n = 2π×4×0.052/2 = 2π×2×0.052 ≈ 0.653 ≈ 0.67 nm.

  6. Q6NEET 2024Hydrogen Spectral Lines - Balmer Series Wavelengths (Match List)

    Match List I with List II. List I (Spectral Lines of Hydrogen for transitions from): (A) n2=3 to n1=2 (B) n2=4 to n1=2 (C) n2=5 to n1=2 (D) n2=6 to n1=2 List II (Wavelengths (nm)): (I) 410.2 (II) 434.1 (III) 656.3 (IV) 486.1 Choose the correct answer from the options given below:

    • A. A-IV, B-III, C-I, D-II
    • B. A-I, B-II, C-III, D-IV
    • C. A-II, B-I, C-IV, D-III
    • D. A-III, B-IV, C-II, D-I
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    Answer: C — A-II, B-I, C-IV, D-III

    Using 1/λ = RH(1/n1² - 1/n2²) with RH = 1.09×10⁷ m⁻¹ for the Balmer series (n1=2): n2=3→2 gives λ=656.3nm (III), n2=4→2 gives λ=486.1nm (IV), n2=5→2 gives λ=434.1nm (II), n2=6→2 gives λ=410.2nm (I). So A-III, B-IV, C-II, D-I.

  7. Q7NEET 2023Hydrogen Spectrum - Balmer and Bracket Series

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

    • A. 4λ
    • B. 9λ
    • C. 16λ
    • D. 2λ
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    Answer: —

    For Balmer series shortest wavelength: 1/λ = R/4. For Bracket series shortest wavelength: 1/λ0 = R/16. Dividing: λ0/λ = 16/4 = 4, so λ0 = 4λ.

  8. Q8NEET 2023Bohr Model - Orbit Radius Calculation

    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?

    • A. 1.06 Å
    • B. 1.59 Å
    • C. 4.77 Å
    • D. 0.53 Å
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    Answer: B — 1.59 Å

    Since r ∝ n²/Z, r2 = r1 × (n2/n1)² = 5.3×10^-11 × 9 = 4.77×10^-10 m = 4.77 Å.

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  9. Q9NEET 2022Bohr Model Energy Levels

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

    • A. 4 : 1
    • B. 4 : 9
    • C. 9 : 4
    • D. 1 : 4
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    Answer: B — 4 : 9

    First excited state: n=2, T₁ = -13.6×z²/n² = -13.6/4 eV. Second excited state: n=3, T₂ = -13.6/9 eV. T₁:T₂ = (1/4):(1/9) = 9:4.

  10. Q10NEET 2022Bohr Model - Energy of Excited States

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

    • A. 1:4
    • B. 4:1
    • C. 4:9
    • D. 9:4
    Show answer

    Answer: C — 4:9

    Energy in nth state E_n = -13.6/n^2 eV. First excited state is n=2, second excited state is n=3. T1:T2 = (1/4):(1/9) = 9:4.

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