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NEET Atoms and Nuclei MCQs with answers

254 questions on Atoms and Nuclei for NEET 2027. Below are 15 to try; tap a question to see its answer and explanation.

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15 sample questions

easy

1. Consider the following two statements regarding nuclear binding energy: Statement I: A higher total binding energy of a nucleus always indicates greater stability of that nucleus compared to another. Statement II: The binding energy per nucleon is a direct measure of the nuclear stability against breakup into its constituent nucleons. In the light of the above statements, choose the most appropriate answer from the options given below:

  1. Both Statement I and Statement II are correct
  2. Statement I is correct but Statement II is incorrect
  3. Statement I is incorrect but Statement II is correct
  4. Both Statement I and Statement II are incorrect
Show answer and explanation

Answer: (C) Statement I is incorrect but Statement II is correct

Statement I is incorrect because nuclear stability is governed by the binding energy per nucleon (E_bn), not by the total binding energy; for example, a heavy uranium nucleus has a much larger total binding energy than an iron nucleus, yet iron is far more stable. Statement II is correct as the binding energy per nucleon directly represents the average energy needed to liberate a nucleon from the nucleus.

medium

2. In a nuclear reactor, the prompt neutron lifetime is 1.0 × 10⁻⁴ s. If the effective multiplication factor k becomes 1.002, by what factor will the neutron population increase after 0.10 s (neglecting delayed neutrons)?

  1. e
  2. e⁻²
  3. e⁰˙²
  4. e²
Show answer and explanation

Answer: (D) e²

The rate of growth is given by N(t) = N₀ exp[(k - 1)t / l]. Here (k - 1) = 0.002, t = 0.10 s, and l = 10⁻⁴ s. The exponent is (0.002 × 0.10) / 10⁻⁴ = 2 × 10⁻⁴ / 10⁻⁴ = 2. Thus, the population increases by a factor of e².

hard

3. A beam of α-particles of kinetic energy 5 MeV is incident on a gold foil of thickness t. The number of particles scattered at an angle of 60° is 3600 per minute. If the foil is replaced by another gold foil of thickness 2t, and the incident α-particle energy is increased to 10 MeV, the number of particles scattered at 60° per minute will be:

  1. 900
  2. 3600
  3. 1800
  4. 450
Show answer and explanation

Answer: (C) 1800

The number of scattered particles N is directly proportional to foil thickness t and inversely proportional to K², i.e., N ∝ t / K². Therefore, N₂ / N₁ = (t₂ / t₁) × (K₁ / K₂)² = (2) × (5 / 10)² = 2 × (1/4) = 1/2. Thus, N₂ = 3600 / 2 = 1800 per minute.

medium

4. Consider the following two statements regarding Bohr's model of the hydrogen atom: Statement I: The Balmer series lies entirely in the visible region of the electromagnetic spectrum. Statement II: The shortest wavelength line of any spectral series occurs when an electron transitions from n = ∞ to the lower state defining that series. In light of the above statements, choose the correct answer:

  1. Both Statement I and Statement II are correct.
  2. Statement I is correct but Statement II is incorrect.
  3. Statement I is incorrect but Statement II is correct.
  4. Both Statement I and Statement II are incorrect.
Show answer and explanation

Answer: (C) Statement I is incorrect but Statement II is correct.

Statement I is incorrect because the Balmer series has lines in the visible region as well as the near ultraviolet region (for transitions from high n to n = 2). Statement II is correct because the series limit (n = ∞ to n₁) corresponds to maximum energy and therefore minimum (shortest) wavelength.

easy

5. Consider the following statements regarding the Geiger-Marsden α-particle scattering experiment: Statement I: The α-particles were collimated into a narrow beam by passing them through lead bricks. Statement II: The scattered α-particles produced bright flashes on a screen coated with zinc sulphide, which were observed using a microscope. Choose the correct option:

  1. Both Statement I and Statement II are correct
  2. Statement I is correct but Statement II is incorrect
  3. Statement I is incorrect but Statement II is correct
  4. Both Statement I and Statement II are incorrect
Show answer and explanation

Answer: (A) Both Statement I and Statement II are correct

Both statements are historically and conceptually accurate as described in NCERT. Lead bricks served as the collimator to produce a narrow beam, and a ZnS screen coupled with a travelling microscope detected scintillations.

hard

6. Two spherical nuclei X and Y have mass numbers A_X and A_Y. If the de Broglie wavelength of a thermal neutron scattered off nucleus X is equal to the radius of X, and for nucleus Y it is equal to the radius of Y, then the ratio of their kinetic energies (K_X / K_Y) when their de Broglie wavelengths match their respective radii is:

  1. (A_Y / A_X)^(1/3)
  2. (A_X / A_Y)^(1/3)
  3. (A_X / A_Y)^(2/3)
  4. (A_Y / A_X)^(2/3)
Show answer and explanation

Answer: (D) (A_Y / A_X)^(2/3)

The de Broglie wavelength is λ = h / √(2m K). Since λ = R = R₀ A^(1/3), we have h / √(2m K) = R₀ A^(1/3). Squaring gives 1/K ∝ A^(2/3), which implies K ∝ A^(-2/3). Therefore, K_X / K_Y = (A_Y / A_X)^(2/3).

medium

7. Consider the following statements regarding the Geiger-Marsden experiment: Statement I: The gold foil used in the experiment was extremely thin to ensure that an α-particle suffers at most one collision before emerging. Statement II: Most of the mass of the atom and all its positive charge are concentrated in a tiny volume called the nucleus. Choose the correct option:

  1. Statement I is correct, but Statement II is incorrect.
  2. Statement I is incorrect, but Statement II is correct.
  3. Both Statement I and Statement II are correct.
  4. Both Statement I and Statement II are incorrect.
Show answer and explanation

Answer: (C) Both Statement I and Statement II are correct.

A very thin gold foil (~10⁻⁷ m) was used to ensure single-scattering events so that multiple deflections do not occur. The experimental observations directly led to the conclusion that positive charge and mass are concentrated in the nucleus.

easy

8. The mass defect of a nucleus is 0.1 u. Taking 1 u = 931.5 MeV/c², its binding energy is:

  1. 9315 MeV
  2. 93.2 MeV
  3. 0.931 MeV
  4. 46.6 MeV
Show answer and explanation

Answer: (B) 93.2 MeV

BE = Δm × 931.5 = 0.1 × 931.5 = 93.2 MeV.

medium

9. In the fission of ²³⁵₉₂U induced by a slow neutron, one set of products is ⁹⁵₄₀Zr and ¹³⁹₅₂Te. How many free neutrons are emitted in this specific fission reaction?

  1. 2
  2. 1
  3. 3
  4. 4
Show answer and explanation

Answer: (A) 2

The reaction is ²³⁵₉₂U + ¹₀n → ⁹⁵₄₀Zr + ¹³⁹₅₂Te + x ¹₀n. Conservation of mass number gives 235 + 1 = 95 + 139 + x, which gives 236 = 234 + x, yielding x = 2 neutrons.

medium

10. Given that the binding energies per nucleon for ₃Li⁷ and ₂He⁴ are 5.60 MeV and 7.06 MeV, respectively, what is the Q-value of the reaction: ₃Li⁷ + ₁H¹ → 2 (₂He⁴)? (Assume the proton has zero binding energy).

  1. 19.64 MeV
  2. 1.46 MeV
  3. 8.64 MeV
  4. 17.28 MeV
Show answer and explanation

Answer: (D) 17.28 MeV

Q = Total BE(products) - Total BE(reactants). Total BE(reactants) = BE(₃Li⁷) = 7 × 5.60 = 39.20 MeV. Total BE(products) = 2 × [4 × 7.06] = 2 × 28.24 = 56.48 MeV. Therefore, Q = 56.48 - 39.20 = 17.28 MeV.

easy

11. In the fission of ²³⁵₉₂U, the primary reason why fission fragments are unstable and undergo a series of β⁻ decays is that:

  1. they have an excess of protons relative to stable nuclei of similar mass
  2. nuclear binding energy per nucleon decreases for medium mass nuclei
  3. they are produced in highly excited states with excess kinetic energy
  4. they have an excess of neutrons compared to stable nuclei of similar mass
Show answer and explanation

Answer: (D) they have an excess of neutrons compared to stable nuclei of similar mass

Heavy nuclei like ²³⁵U have an N/Z ratio around 1.56, whereas stable medium-mass nuclei (fission fragments) have an N/Z ratio around 1.2 to 1.3. Consequently, fission fragments have a large neutron excess and achieve stability by converting neutrons to protons via successive β⁻ emissions.

medium

12. An α-particle of kinetic energy K is directed head-on towards a stationary nucleus of atomic number Z. If an incident particle with twice the charge of an α-particle and half its kinetic energy is directed head-on towards the same nucleus, the distance of closest approach becomes:

  1. d₀
  2. 4 d₀
  3. 2 d₀
  4. 8 d₀
Show answer and explanation

Answer: (B) 4 d₀

The distance of closest approach is d = (1/(4πε₀)) × (q₁·q₂ / K). For an α-particle, q₁ = 2e, giving d₀ ∝ 2/K. For the new particle, q₁' = 2(2e) = 4e and K' = K/2. Thus d' ∝ 4/(K/2) = 8/K = 4 × (2/K) = 4 d₀.

medium

13. A hypothetical nucleus of mass number A = 236 undergoes symmetric binary fission into two identical fragments of mass number A' = 118 each. If the binding energy per nucleon of the parent nucleus is 7.5 MeV and that of each fragment is 8.5 MeV, the total energy released in this fission process is:

  1. 236 MeV
  2. 118 MeV
  3. 177 MeV
  4. 472 MeV
Show answer and explanation

Answer: (A) 236 MeV

Total initial binding energy = 236 × 7.5 MeV = 1770 MeV. Total final binding energy = 2 × (118 × 8.5 MeV) = 236 × 8.5 MeV = 2006 MeV. Energy released Q = BE_final - BE_initial = 2006 MeV - 1770 MeV = 236 MeV (or Q = 236 × (8.5 - 7.5) MeV = 236 MeV).

medium

14. The ratio of the magnetic dipole moment associated with the orbital motion of an electron in the ground state of hydrogen to that in its first excited state is:

  1. 2 : 1
  2. 1 : 4
  3. 1 : 8
  4. 1 : 2
Show answer and explanation

Answer: (D) 1 : 2

The orbital magnetic dipole moment is μ = (e / 2m) L, where L = nh/(2π). Thus, μ ∝ n. For the ground state, n = 1, and for the first excited state, n = 2. Therefore, μ₁ : μ₂ = 1 : 2.

easy

15. An electron in a He⁺ ion is in an excited state with principal quantum number n = 4. What is the total mechanical energy of the electron in this state?

  1. -3.40 eV
  2. -0.85 eV
  3. -13.6 eV
  4. -1.70 eV
Show answer and explanation

Answer: (A) -3.40 eV

For a hydrogen-like ion, total energy is Eₙ = -13.6 × (Z²/n²) eV. For He⁺, Z = 2 and n = 4, giving E₄ = -13.6 × (2²/4²) = -13.6 × (4/16) = -3.40 eV.

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