Nuclei NEET PYQReal NEET & AIPMT 2005–2021 questions, year-tagged, with solutions
By the RankUp team, founded by Dr. Deepak Sahu (MBBS Gold Medalist, NEET PG AIR 1731) · Updated
Every Nuclei question from the NEET & AIPMT papers on this site (2005–2021), newest first. Each is tagged with its paper. Tap an option to check your answer and read the explanation.
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Nuclei previous year questions
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Q1NEET 2021
A radioactive nucleus ᴬZX undergoes spontaneous decay in the sequence ᴬZX → (Z−1)B → (Z−3)C → (Z−2)D, where Z is the atomic number of element X. The possible decay particles in the sequence are :
Not quite — the answer is C.
Z → Z − 1: atomic number falls by 1 with no other change, which is β⁺ emission. Z − 1 → Z − 3: falls by 2, which is α emission. Z − 3 → Z − 2: rises by 1, which is β⁻ emission. So the sequence is β⁺, α, β⁻.
Q2NEET 2021
The half-life of a radioactive nuclide is 100 hours. The fraction of original activity that will remain after 150 hours would be :
Not quite — the answer is B.
150 h is 1.5 half-lives, so the fraction of activity remaining is (1/2)^1.5 = 1/(2√2) ≈ 0.354. Linear interpolation between 1/2 and 1/4 is wrong because decay is exponential.
Q3NEET 2021
A nucleus with mass number 240 breaks into two fragments each of mass number 120, the binding energy per nucleon of unfragmented nuclei is 7.6 MeV while that of fragments is 8.5 MeV. The total gain in the Binding Energy in the process is :
Not quite — the answer is D.
Initial binding energy = 240 × 7.6 = 1824 MeV; final = 2 × 120 × 8.5 = 2040 MeV. Gain = 2040 − 1824 = 216 MeV. 0.9 MeV is only the change per nucleon, not the total.
Q4NEET 2020 · Q142
The energy equivalent of 0.5 g of a substance is :
Not quite — the answer is B.
E = mc² = 0.5×10⁻³ kg × (3×10⁸ m/s)² = 0.5×10⁻³ × 9×10¹⁶ = 4.5×10¹³ J. Forgetting to convert grams to kilograms gives the 4.5×10¹⁶ J trap.
Q5NEET 2020 · Q158
When a uranium isotope ²³⁵₉₂U is bombarded with a neutron, it generates ⁸⁹₃₆Kr, three neutrons and :
Not quite — the answer is A.
Conserving mass number: 235 + 1 = 89 + 3 + A, so A = 144; conserving charge: 92 = 36 + Z, so Z = 56, which is barium-144. The krypton options keep Z = 36 and so violate charge conservation.
Q6NEET 2019 · Q31
α-particle consists of:
Not quite — the answer is A.
An α-particle is a helium-4 nucleus: 2 protons and 2 neutrons, with no electrons (hence its +2e charge). Including 2 electrons would describe a neutral helium atom.
Q7NEET 2018 · Q13
For a radioactive material, half-life is 10 minutes. If initially there are 600 number of nuclei, the time taken (in minutes) for the disintegration of 450 nuclei is
Not quite — the answer is C.
After 450 decay, 150 of 600 remain, i.e. 1/4 = (1/2)². That is two half-lives = 20 minutes. 30 minutes would leave only 75 nuclei.
Q8NEET 2017 (Gujarati paper) · Q136
The energy liberated per nuclear fission is 200 MeV. If 10²⁰ fissions occur per second the amount of power produced will be:
Not quite — the answer is A.
200 MeV = 200 × 10⁶ × 1.6 × 10⁻¹⁹ J = 3.2 × 10⁻¹¹ J. Power = 10²⁰ s⁻¹ × 3.2 × 10⁻¹¹ J = 3.2 × 10⁹ W = 32 × 10⁸ W. Forgetting to convert MeV to joules gives a value like 2 × 10²² W.
Q9NEET 2016 Phase 2 · Q42
The half-life of a radioactive substance is 30 minutes. The time (in minutes) taken between 40% decay and 85% decay of the same radioactive substance is
Not quite — the answer is D.
After 40% decay 60% remains; after 85% decay 15% remains. 60% → 15% is a factor of 4 = 2², i.e. two half-lives = 2 × 30 = 60 minutes. Subtracting the percentages (45%) and treating decay as linear is the trap.
Q10AIPMT 2015
If radius of the ²⁷₁₃Al nucleus is taken to be R_Al, then the radius of ¹²⁵₅₃Te nucleus is nearly:
Not quite — the answer is C.
Nuclear radius R = R₀A^(1/3), so R_Te/R_Al = (125/27)^(1/3) = 5/3. Using atomic numbers instead of mass numbers gives the (53/13)^(1/3) trap.
Q11AIPMT 2014
The Binding energy per nucleon of ⁷₃Li and ⁴₂He nuclei are 5.60 MeV and 7.06 MeV, respectively. In the nuclear reaction ⁷₃Li + ¹₁H → ⁴₂He + ⁴₂He + Q, the value of energy Q released is:
Not quite — the answer is C.
Q = BE of products − BE of reactants = 2 × 4 × 7.06 − 7 × 5.60 = 56.48 − 39.20 = 17.28 ≈ 17.3 MeV (the proton has no binding energy).
Q12AIPMT 2014
A radio isotope 'X' with a half life 1.4 × 10⁹ years decays to 'Y' which is stable. A sample of the rock from a cave was found to contain 'X' and 'Y' in the ratio 1 : 7. The age of the rock is:
Not quite — the answer is B.
X : Y = 1 : 7 means 1/8 of the original X remains, i.e. 3 half-lives. Age = 3 × 1.4 × 10⁹ = 4.2 × 10⁹ years.
Q13NEET 2013
The half-life of a radioactive isotope ‘X’ is 20 years. It decays to another element ‘Y’ which is stable. The two elements ‘X’ and ‘Y’ were found to be in the ratio 1 : 7 in a sample of a given rock. The age of the rock is estimated to be -
Not quite — the answer is B.
X : Y = 1 : 7 means 1/8 of the original X remains. 1/8 = (1/2)³, so three half-lives have passed: 3 × 20 = 60 years. 80 years would leave 1/16.
Q14AIPMT 2012 Mains
The half life of a radioactive nucleus is 50 days. The time interval (t₂ − t₁) between the time t₂ when 2/3 of it has decayed and the time t₁ when 1/3 of it had decayed is
Not quite — the answer is B.
When 1/3 has decayed, 2/3 remains; when 2/3 has decayed, 1/3 remains. The undecayed amount halves between these instants, which takes exactly one half-life, 50 days.
Q15AIPMT 2012 Prelims
If the nuclear radius of ²⁷Al is 3.6 fermi, the approximate nuclear radius of ⁶⁴Cu in fermi is
Not quite — the answer is D.
Nuclear radius R ∝ A^(1/3), so R_Cu = 3.6 × (64/27)^(1/3) = 3.6 × 4/3 = 4.8 fermi. A heavier nucleus must have a larger radius, which rules out the smaller values.
Q16AIPMT 2012 Prelims
A mixture consists of two radioactive materials A₁ and A₂ with half lives of 20 s and 10 s respectively. Initially the mixture has 40 g of A₁ and 160 g of A₂. The amount of the two in the mixture will become equal after
Not quite — the answer is A.
After t seconds, A₁ = 40(½)^(t/20) and A₂ = 160(½)^(t/10). Equating: 4 = 2^(t/10 − t/20) = 2^(t/20), so t/20 = 2 and t = 40 s. Check: both become 10 g at 40 s.
Q17AIPMT 2011 Mains
Two radioactive nuclei P and Q in a given sample decay into a stable nucleus R. At time t = 0, number of P species are 4N₀ and that of Q are N₀. Half-life of P (for conversion to R) is 1 minute whereas that of Q is 2 minutes. Initially there are no nuclei of R present in the sample. When number of nuclei of P and Q are equal, the number of nuclei of R present in the sample would be
Not quite — the answer is D.
P: 4N₀(1/2)^t and Q: N₀(1/2)^(t/2) with t in minutes. Setting them equal gives 2^(2 − t) = 2^(−t/2), so t = 4 min, when P = Q = N₀/4. R = 5N₀ − N₀/2 = 9N₀/2. 5N₀/2 would count only the decayed P nuclei incorrectly.
Q18AIPMT 2011 Prelims
The power obtained in a reactor using U-235 disintegration is 1000 kW. The mass decay of U-235 per hour is
Not quite — the answer is D.
Energy produced in one hour = 10⁶ W × 3600 s = 3.6 × 10⁹ J. Mass converted = E/c² = 3.6 × 10⁹/(9 × 10¹⁶) = 4 × 10⁻⁸ kg = 40 μg. Forgetting the factor 3600 or using kW as W gives the smaller options.
Q19AIPMT 2011 Prelims
The half life of a radioactive isotope X is 50 years. It decays to another element Y which is stable. The two elements X and Y were found to be in the ratio of 1 : 15 in a sample of a given rock. The age of the rock was estimated to be
Not quite — the answer is C.
If X : Y = 1 : 15, the fraction of X remaining is 1/16 = (1/2)⁴, so four half-lives have passed: 4 × 50 = 200 years. Taking 1/15 as the remaining fraction or counting 3 half-lives gives the wrong ages.
Q20AIPMT 2011 Prelims
Fusion reaction takes place at high temperature because
Not quite — the answer is D.
Fusing nuclei are positively charged and repel each other; only at very high temperatures (~10⁷ K) do they have enough kinetic energy to overcome this Coulomb barrier. Ionisation of atoms or breaking of molecules is not the reason fusion needs high temperature.
Q21AIPMT 2011 Prelims
A nucleus ᵐₙX emits one α-particle and two β⁻-particles. The resulting nucleus is
Not quite — the answer is D.
Alpha emission lowers mass number by 4 and atomic number by 2; each β⁻ raises atomic number by 1 with no mass change. Net: mass number m − 4, atomic number n − 2 + 2 = n, so the product is an isotope of X, ᵐ⁻⁴ₙX.
Q22AIPMT 2010 Mains · Q76
The decay constant of a radio isotope is λ. If A₁ and A₂ are its activities at times t₁ and t₂ respectively, the number of nuclei which have decayed during the time (t₁ − t₂) is
Not quite — the answer is C.
Activity A = λN, so N₁ = A₁/λ and N₂ = A₂/λ. The number decayed between the two instants is N₁ − N₂ = (A₁ − A₂)/λ. The difference A₁ − A₂ alone is a difference of decay rates, not a number of nuclei.
Q23AIPMT 2010 Mains · Q78
The binding energy per nucleon in deuterium and helium nuclei are 1.1 MeV and 7.0 MeV, respectively. When two deuterium nuclei fuse to form a helium nucleus the energy released in the fusion is
Not quite — the answer is A.
Binding energy of two deuterons = 2 × (2 × 1.1) = 4.4 MeV; of helium-4 = 4 × 7.0 = 28.0 MeV. Energy released = 28.0 − 4.4 = 23.6 MeV. 28.0 MeV is just the helium binding energy without subtracting that of the reactants.
Q24AIPMT 2010 Prelims · Q42
The activity of a radioactive sample is measured as N₀ counts per minute at t = 0 and N₀/e counts per minute at t = 5 minutes. The time (in minutes) at which the activity reduces to half its value is
Not quite — the answer is A.
Activity falls to 1/e in 5 min, so λ = 1/5 per minute (mean life 5 min). T½ = ln 2/λ = 5 logₑ2 min ≈ 3.47 min. 5/logₑ2 inverts the relation.
Q25AIPMT 2010 Prelims · Q44
The mass of a ⁷₃Li nucleus is 0.042 u less than the sum of the masses of all its nucleons. The binding energy per nucleon of ⁷₃Li nucleus is nearly
Not quite — the answer is C.
Binding energy = 0.042 × 931 ≈ 39.1 MeV. Per nucleon: 39.1/7 ≈ 5.6 MeV. 39 MeV would be the total, not per nucleon.
Q26AIPMT 2009 Prelims · Q1
In the nuclear decay given below: (A, Z)X → (A, Z+1)Y → (A−4, Z−1)B* → (A−4, Z−1)B the particles emitted in the sequence are: [notation: (mass number, atomic number)]
Not quite — the answer is D.
X to Y keeps the mass number and raises the atomic number by 1, which is β⁻ emission. Y to B* lowers A by 4 and Z by 2, which is α emission. B* to B changes neither A nor Z, so the excited nucleus de-excites by γ emission, giving the order β, α, γ.
Q27AIPMT 2009 Prelims · Q17
The number of beta particles emitted by a radioactive substance is twice the number of alpha particles emitted by it. The resulting daughter is an:
Not quite — the answer is C.
One α emission lowers Z by 2 and A by 4; two β⁻ emissions raise Z by 2 and leave A unchanged. Net: same Z, A reduced by 4, so the daughter is an isotope of the parent, not an isobar (same mass number).
Q28AIPMT 2008 Prelims · Q44
Two radioactive materials X₁ and X₂ have decay constants 5λ and λ respectively. If initially they have the same number of nuclei, then the ratio of the number of nuclei of X₁ to that of X₂ will be 1/e after a time
Not quite — the answer is D.
N₁/N₂ = e^(−5λt)/e^(−λt) = e^(−4λt). Setting this to 1/e gives 4λt = 1, so t = 1/(4λ). λ and λ/2 cannot be times, since λ has units of inverse time.
Q29AIPMT 2008 Prelims · Q45
Two nuclei have their mass numbers in the ratio of 1 : 3. The ratio of their nuclear densities would be
Not quite — the answer is A.
Nuclear radius R = R₀A^(1/3), so volume is proportional to A, just as mass is. Their ratio, the density, is the same for all nuclei, 1 : 1. The 3^(1/3) form is the ratio of radii, not of densities.
Q30AIPMT 2008 Prelims · Q46
If M(A; Z), Mp and Mn denote the masses of the nucleus ᴬ_ZX, proton and neutron respectively in units of u (1 u = 931.5 MeV/c²) and BE represents its binding energy in MeV, then
Not quite — the answer is B.
A bound nucleus is lighter than its free nucleons by the mass defect, and the mass defect equals BE/c². So M(A, Z) = ZMp + (A − Z)Mn − BE/c². The plus sign would make the nucleus heavier than its parts, and subtracting BE without c² mixes energy with mass units.
Q31AIPMT 2007 Prelims · Q6
Two radioactive substances A and B have decay constants 5λ and λ respectively. At t = 0 they have the same number of nuclei. The ratio of number of nuclei of A to those of B will be (1/e)² after a time interval
Not quite — the answer is C.
N_A/N_B = e^(−5λt)/e^(−λt) = e^(−4λt). Setting this equal to e^(−2) gives 4λt = 2, so t = 1/(2λ). Answers like 2λ have the wrong dimensions for a time.
Q32AIPMT 2007 Prelims · Q8
In a radioactive decay process, the negatively charged emitted β -particles are
Not quite — the answer is A.
In β⁻ decay a neutron inside the nucleus changes into a proton, emitting an electron (and an antineutrino). Electrons do not exist inside the nucleus beforehand, and orbital electrons are not involved.
Q33AIPMT 2007 Prelims · Q14
A nucleus ᴬZX has mass represented by M(A, Z). If Mp and Mn denote the mass of proton and neutron respectively and B.E. the binding energy in MeV, then
Not quite — the answer is A.
Binding energy is the mass defect times c²: B.E. = [ZMp + (A − Z)Mn − M(A, Z)]c². The expression without c² is a mass, not an energy, and the reversed sign would give a negative binding energy.
Q34AIPMT 2007 Prelims · Q16
If the nucleus ²⁷₁₃Al has nuclear radius of about 3.6 fm, then ¹²⁵₃₂Te would have its radius approximately as
Not quite — the answer is D.
R ∝ A^(1/3), so R_Te = 3.6 × (125/27)^(1/3) = 3.6 × 5/3 = 6.0 fm. Scaling the radius in proportion to A instead of its cube root would give a much larger value.
Q35AIPMT 2006 Prelims · Q10
The binding energy of deuteron is 2.2 MeV and that of ⁴₂He is 28 MeV. If two deuterons are fused to form one ⁴₂He then the energy released is :
Not quite — the answer is B.
Energy released = binding energy of products − binding energy of reactants = 28 − 2 × 2.2 = 23.6 MeV. Subtracting the binding energy of only one deuteron gives 25.8 MeV, and adding them gives 32.4 MeV, neither of which applies.
Q36AIPMT 2006 Prelims · Q11
In a radioactive material the activity at time t₁ is R₁ and at a later time t₂, it is R₂. If the decay constant of the material is λ, then :
Not quite — the answer is A.
Activity decays as R = R₀e^(−λt), so R₁ = R₀e^(−λt₁) and R₂ = R₀e^(−λt₂). Dividing, R₁ = R₂e^(−λ(t₁ − t₂)); since t₁ < t₂ the exponent is positive and R₁ > R₂, as expected for the earlier time. The form with e^(λ(t₁ − t₂)) would make the earlier activity smaller.
Q37AIPMT 2006 Prelims · Q32
The radius of germanium (Ge) nuclide is measured to be twice the radius of ⁹₄Be. The number of nucleons in Ge are :
Not quite — the answer is D.
Nuclear radius R = R₀A^(1/3), so doubling R needs A to increase by 2³ = 8. The Ge nucleus therefore has 8 × 9 = 72 nucleons. The other numbers are other germanium isotopes, not the one fixed by this radius.
Q38AIPMT 2005 Prelims · Q9
In the reaction ²₁H + ³₁H → ⁴₂He + ¹₀n, if the binding energies of ²₁H, ³₁H and ⁴₂He are respectively a, b and c (in MeV), then the energy (in MeV) released in this reaction is :
Not quite — the answer is B.
Energy released = total binding energy of products − total binding energy of reactants = c − (a + b). The free neutron has zero binding energy. a + b − c has the sign reversed.
Q39AIPMT 2005 Prelims · Q13
The nuclei of which one of the following pairs of nuclei are isotones ?
Not quite — the answer is A.
Isotones have the same neutron number N = A − Z. Se: 74 − 34 = 40; Ga: 71 − 31 = 40. The Sr pair are isotopes (same Z) and the Mo/Zr pair are isobars (same A).
Q40AIPMT 2005 Prelims · Q23
Fission of nuclei is possible because the binding energy per nucleon in them :
Not quite — the answer is B.
At high mass numbers the binding energy per nucleon falls as A increases, so splitting a heavy nucleus into medium nuclei raises the binding energy per nucleon and releases energy. Increase with A at low mass numbers explains fusion, not fission.
Q41AIPMT 2005 Prelims · Q27
In any fission process the ratio (mass of fission products/mass of parent nucleus) is :
Not quite — the answer is A.
Fission releases energy Q = Δm c², which comes from a loss of mass, so the products are lighter than the parent nucleus and the ratio is less than 1. 'Equal to 1' ignores the mass defect.
How often Nuclei came up
Questions from this chapter in each paper on this site.
- NEET 2021: 3
- NEET 2020: 2
- NEET 2019: 1
- NEET 2018: 1
- NEET 2017 (Gujarati paper): 1
- NEET 2016 Phase 2: 1
- AIPMT 2015: 1
- AIPMT 2014: 2
- NEET 2013: 1
- AIPMT 2012 Mains: 1
- AIPMT 2012 Prelims: 2
- AIPMT 2011 Mains: 1
- AIPMT 2011 Prelims: 4
- AIPMT 2010 Mains: 2
- AIPMT 2010 Prelims: 2
- AIPMT 2009 Prelims: 2
- AIPMT 2008 Prelims: 3
- AIPMT 2007 Prelims: 4
- AIPMT 2006 Prelims: 3
- AIPMT 2005 Prelims: 4
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