Nuclei NEET PYQs (2005–2026) with Solutions

10 solved Nuclei 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 2026EasyQ value of alpha decay

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

   1. AOption A: $3726$
   2. BOption B: $3730$
   3. COption C: $3736$
   4. DOption D: $3740$

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   Correct answer: (A) $3726$

   Explanation

   Mass defect: $\Delta m = 238.050 - (234.043 + 4.003) = 0.004\ \mathrm{u}$. $Q = \Delta m\, c^2 = 0.004 \times 931.5\ \mathrm{MeV} = 3.726\ \mathrm{MeV} = 3726\ \mathrm{keV}$. So the answer is option (A).

2. Question 2 (NEET 2026)

   NEET 2026MediumBohr model, energy levels, hydrogen spectrum

   In the first excited state of hydrogen atom, the energy of its electron is −3.4 eV. The radial distance of the electron from the hydrogen nucleus in this case is approximately: (Take 1 eV = 1.6 × 10⁻¹⁹ J, e = 1.6 × 10⁻¹⁹ C and 1/4πε₀ = 9 × 10⁹ N m²/C²)

   1. AOption A: 2.1 × 10⁻⁹ m
   2. BOption B: 2.1 × 10⁻⁸ m
   3. COption C: 2.1 × 10⁻¹⁰ m
   4. DOption D: 2.1 × 10⁻¹¹ m

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   Correct answer: (C) 2.1 × 10⁻¹⁰ m

   Explanation

   For hydrogen atom: Total energy, E = -k e² / (2r) Magnitude: |E| = k e² / (2r) Substitute: 3.4 eV = (9×10⁹ × (1.6×10⁻¹⁹)²)/(2r) Converting eV to joule: 3.4 × 1.6×10⁻¹⁹ = (9×10⁹ × (1.6×10⁻¹⁹)²)/(2r) r = (9×10⁹ × 1.6×10⁻¹⁹)/(2×3.4) r = 2.12 × 10⁻¹⁰ m ≈ 2.1 × 10⁻¹⁰ m Hence option (C) is correct.

3. Question 3 (NEET 2026)

   NEET 2026MediumMass Defect & Binding Energy

   Four statements are given (A is mass number): A. The volume of a nucleus is proportional to A¹ᐟ³. B. The volume of a nucleus is proportional to A. C. The difference in mass of an atom and its nucleus is called the mass defect. D. The difference in mass of a nucleus and its constituents is called the mass defect. Choose the correct answer from the options given below:

   1. AOption A: A and C are true, but B and D are false
   2. BOption B: B and C are true, but A and D are false
   3. COption C: A and D are true, but B and C are false
   4. DOption D: B and D are true, but A and C are false

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   Correct answer: (D) B and D are true, but A and C are false

   Explanation

   Nuclear radius r=r₀A^(1/3). Hence volume V∝r³∝A, so statement B is true and A is false. Mass defect is the difference between the mass of a nucleus and the sum of masses of its constituent nucleons, so D is true and C is false. Hence option (D) is correct.

4. Question 4 (NEET 2026)

   NEET 2026HardNuclear Composition

   An unknown nucleus has a nuclear density of 2.29 × 10¹⁷ kg/m³ and mass of 19.926 × 10⁻²⁷ kg. Its mass number A is approximately: (Take R₀ = 1.2 × 10⁻¹⁵ m, 4π = 12.56)

   1. AOption A: 12
   2. BOption B: 20
   3. COption C: 16
   4. DOption D: 19

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   Correct answer: (A) 12

   Explanation

   Using R = R₀A^(1/3) and volume=(4/3)πR³=M/ρ, solving gives A≈12. Hence option (A) is correct.

5. Question 5 (NEET 2024)

   NEET 2024MediumRadioactivity

   In the nuclear emission stated above, the mass number and atomic number of the product Q respectively, are:

   1. AOption A: 288, 82
   2. BOption B: 286, 81
   3. COption C: 280, 81
   4. DOption D: 286, 80

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   Correct answer: (B) 286, 81

   Explanation

   Initial nucleus: $$^{290}_{82}Y$$ After α-emission: Mass number decreases by 4 and atomic number by 2: $$A = 286, \quad Z = 80$$ After positron emission (e⁺): Atomic number decreases by 1: $$Z = 79$$ After β⁻ emission: Atomic number increases by 1: $$Z = 80$$ After electron emission (e⁻): This corresponds to β⁻ decay, so atomic number increases by 1: $$Z = 81$$ Mass number remains unchanged throughout beta processes. Therefore, product Q has: $$A = 286, \quad Z = 81$$ Hence, option (B) is correct.

6. Question 6 (NEET 2023)

   NEET 2023MediumRadioactivity

   The half life of a radioactive substance is 20 minutes. In how much time, the activity of the substance drops to (1/16)th of its initial value?

   1. AOption A: 20 minutes
   2. BOption B: 40 minutes
   3. COption C: 60 minutes
   4. DOption D: 80 minutes

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   Correct answer: (D) 80 minutes

   Explanation

   Activity is proportional to the number of undecayed nuclei. (1/2)^n = 1/16 = (1/2)^4 So, n = 4 half lives. Time = 4 × 20 = 80 minutes. Hence, option (D) is correct.

7. Question 7 (NEET 2022)

   NEET 2022MediumBeta Decay Nuclear Reactions

   In the given nuclear reaction, the element X is: ²²₁₁Na → X + e⁺ + ν

   1. AOption A: ²²₁₂Mg
   2. BOption B: ²³₁₁Na
   3. COption C: ²⁰₁₀Ne
   4. DOption D: ²²₁₀Ne

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   Correct answer: (D) ²²₁₀Ne

   Explanation

   In β⁺ decay, atomic number decreases by 1 while mass number remains unchanged. Hence ²²₁₁Na → ²²₁₀Ne + e⁺ + ν.

8. Question 8 (NEET 2022)

   NEET 2022EasyNuclear Composition

   A nucleus of mass number 189 splits into two nuclei having mass numbers 125 and 64. The ratio of radius of two daughter nuclei respectively is:

   1. AOption A: 25 : 16
   2. BOption B: 1 : 1
   3. COption C: 4 : 5
   4. DOption D: 5 : 4

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   Correct answer: (D) 5 : 4

   Explanation

   Nuclear radius: R ∝ A^(1/3) Thus: R₁/R₂ = (125/64)^(1/3) = 5/4 Hence the ratio is 5 : 4. Therefore option (D) is correct.

9. Question 9 (NEET 2021)

   NEET 2021MediumRadioactivity

   A radioactive nucleus $^{A}_{Z}X$ undergoes spontaneous decay in the sequence: where Z is the atomic number of element X The possible decay particles in the sequence are:

   1. AOption A: $\beta^- ,\ \alpha ,\ \beta^+$
   2. BOption B: $\alpha ,\ \beta^- ,\ \beta^+$
   3. COption C: $\alpha ,\ \beta^+ ,\ \beta^-$
   4. DOption D: $\beta^+ ,\ \alpha ,\ \beta^-$

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   Correct answer: (D) $\beta^+ ,\ \alpha ,\ \beta^-$

   Explanation

   Step 1: $$^{A}_{Z}X \rightarrow \ ^{A}_{Z-1}B$$ Atomic number decreases by 1 while mass number remains unchanged. This corresponds to **$\beta^+$decay (positron emission)**.$$p \rightarrow n + e^+ + \nu$$Step 2:$$^{A}_{Z-1}B \rightarrow \ ^{A-4}_{Z-3}C$$Mass number decreases by 4 and atomic number decreases by 2. This corresponds to **$\alpha$decay**.$$^{4}_{2}He$$is emitted. Step 3:$$^{A-4}_{Z-3}C \rightarrow \ ^{A-4}_{Z-2}D$$Atomic number increases by 1 while mass number remains unchanged. This corresponds to **$\beta^-$decay**.$$n \rightarrow p + e^- + \bar{\nu}$$Therefore, the correct sequence is:$$\beta^+ ,\ \alpha ,\ \beta^-$$ Hence, **Option D** is correct.

10. Question 10 (NEET 2021)

    NEET 2021EasyNuclear Energy

    A nucleus with mass number $240$ breaks into two fragments each of mass number $120$. The binding energy per nucleon of the unfragmented nucleus is $7.6\,\text{MeV}$, while that of each fragment is $8.5\,\text{MeV}$. The total gain in binding energy in the process is:

    1. AOption A: $216\,\text{MeV}$
    2. BOption B: $0.9\,\text{MeV}$
    3. COption C: $9.4\,\text{MeV}$
    4. DOption D: $804\,\text{MeV}$

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    Correct answer: (A) $216\,\text{MeV}$

    Explanation

    Initial total binding energy of the nucleus: $$BE_i = 240 \times 7.6 = 1824\,\text{MeV}$$ Final total binding energy of the two fragments: $$BE_f = 2 \times (120 \times 8.5) = 2040\,\text{MeV}$$ Gain in binding energy: $$\Delta BE = BE_f - BE_i$$ $$\Delta BE = 2040 - 1824 = 216\,\text{MeV}$$ Hence, the correct answer is **Option A**.