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Nuclei: NEET previous year questions

Every Nuclei question from the official NEET papers on this site, with the answer and a worked solution.

The questions come from NEET 2026, Re-NEET 2026 and NEET 2020. Each shows its answer and solution when you open it, so you can attempt it first. For what the chapter covers and how NEET tests it, see the Nuclei guide.

Practise on the CBT screen5 questions, 5 minutes, marked +4 / −1Download PDF3 pages, 0.7 MB, with answers and solutions

NEET 2026 · the full paper

  1. NEET 2026 · Q31

    Four statements are given (AA is mass number). Choose the correct answer from the options given below:

    Statement I: The volume of a nucleus is proportional to A1/3A^{1/3}.
    Statement II: The volume of a nucleus is proportional to AA.
    Statement III: The difference in mass of an atom and its nucleus is called the mass defect.
    Statement IV: The difference in mass of a nucleus and its constituents is called the mass defect.

    1. (1)

      I and III are true, but II and IV are false

    2. (2)

      II and III are true, but I and IV are false

    3. (3)

      I and IV are true, but II and III are false

    4. (4)

      II and IV are true, but I and III are false

    Answer and solution

    Answer: (4)

    Nuclear radius r=r0A1/3r = r_0 A^{1/3}, so volume ∝r3∝A\propto r^3 \propto A (statement II correct, I wrong). The mass defect is the difference between the total mass of the constituent nucleons and the actual mass of the nucleus (statement IV correct, III wrong). Hence only II and IV are true.

  2. NEET 2026 · Q32

    An unknown nucleus has a nuclear density of 2.29×10172.29 \times 10^{17} kg/m3^3 and mass of 19.926×10−2719.926 \times 10^{-27} kg. Its mass number AA is approximately: (Take R0=1.2×10−15R_0 = 1.2 \times 10^{-15} m, 4π=12.564\pi = 12.56)

    1. (1)

      12

    2. (2)

      20

    3. (3)

      16

    4. (4)

      19

    Answer and solution

    Answer: (1)

    Using density = mass/volume with volume =43πR3= \frac{4}{3}\pi R^3 and R=R0A1/3R = R_0 A^{1/3}, we get 43πR03A=mρ\frac{4}{3}\pi R_0^3 A = \frac{m}{\rho}. Solving, A=mρ⋅34πR03=19.926×10−27×32.29×1017×12.56×(1.2×10−15)3≈12A = \frac{m}{\rho}\cdot\frac{3}{4\pi R_0^3} = \frac{19.926\times10^{-27}\times3}{2.29\times10^{17}\times12.56\times(1.2\times10^{-15})^3} \approx 12.

Re-NEET 2026 · the full paper

  1. Re-NEET 2026 · Q7

    Consider the following nuclear reaction 238U→234Th+4He^{238}\mathrm{U} \rightarrow ^{234}\mathrm{Th} + ^{4}\mathrm{He}. Take masses of 238U^{238}\mathrm{U}, 234Th^{234}\mathrm{Th} and 4He^{4}\mathrm{He} as 238.050 u, 234.043 u and 4.003 u, respectively. The Q value for the reaction, in keV, is: [Given: 1 u = 931.5 MeV c−2c^{-2}]

    1. (1)

      3726

    2. (2)

      3730

    3. (3)

      3736

    4. (4)

      3740

    Answer and solution

    Answer: (1)

    The Q value is the mass defect converted to energy: Δm=238.050−(234.043+4.003)=0.004 u\Delta m = 238.050 - (234.043 + 4.003) = 0.004\ \text{u}. Multiplying by the conversion factor, Q=0.004×931.5=3.726 MeV=3726 keVQ = 0.004 \times 931.5 = 3.726\ \text{MeV} = 3726\ \text{keV}. The positive value confirms the decay is spontaneous. NCERT Class 12, Nuclei (mass-energy and nuclear reactions).

NEET 2020 · the full paper

  1. NEET 2020 · Q151

    When a uranium isotope 92235^{235}_{92}U is bombarded with a neutron, it generates 3689^{89}_{36}Kr, three neutrons and

    1. (1)

      4091^{91}_{40}Zr

    2. (2)

      36101^{101}_{36}Kr

    3. (3)

      36103^{103}_{36}Kr

    4. (4)

      56144^{144}_{56}Ba

    Answer and solution

    Answer: (4)

    The nuclear reaction is 92235U+01n→3689Kr+3 01n+56144Ba^{235}_{92}U + ^1_0n → ^{89}_{36}Kr + 3\ ^1_0n + ^{144}_{56}Ba. The mass and charge numbers must be conserved, so the remaining product is 56144Ba^{144}_{56}Ba. Chapter: Nuclei.

  2. NEET 2020 · Q174

    The energy equivalent of 0.5 g of a substance is

    1. (1)

      4.5×10134.5 \times 10^{13} J

    2. (2)

      1.5×10131.5 \times 10^{13} J

    3. (3)

      0.5×10130.5 \times 10^{13} J

    4. (4)

      4.5×10164.5 \times 10^{16} J

    Answer and solution

    Answer: (1)

    E=mc2=0.5×10−3×(3×108)2=4.5×1013E = mc^{2} = 0.5 \times 10^{-3} \times (3 \times 10^{8})^{2} = 4.5 \times 10^{13} J. Chapter: Nuclei.

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Questions

Are these the official NEET questions on Nuclei?

Yes. They are the questions NTA set in each paper, in its wording. The solutions are written by BrainIt.

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