nuclei formulas pyqs
Sheet,
NCERT Examples And PYQs
Complete revision guide covering nuclear radius, mass defect, binding energy, radioactivity, decay law, nuclear reactions and examination-ready questions.
1. Complete Formula Sheet Overview
Below is a consolidated summary of every major mathematical equation required for board and competitive examinations. Detailed breakdowns follow in subsequent sections.
2. Nuclear Radius Formula
The physical size of an atomic nucleus depends directly on its total number of constituent nucleons (protons and neutrons). Assuming a spherical geometry, experimental scattering reveals:
Meaning of Symbols
- R: Radius of the given nucleus (measured in meters or femtometers).
- R₀: Empirically determined constant baseline factor ≈ 1.2 × 10⁻¹⁵ m (1.2 fm).
- A: Mass number of the atom (total count of protons + neutrons).
Key Applications
Used extensively to evaluate changes in nuclear volume ($V \propto A$) and to demonstrate that the net structural density of all stable atomic nuclei remains constant, regardless of mass number.
Sample Mini-Numerical
Problem: Find the radius of an iron nucleus given A = 56.
Solution: R = 1.2 × 10⁻¹⁵ × (56)¹ᐟ³ ≈ 1.2 × 10⁻¹⁵ × 3.826 ≈ 4.59 × 10⁻¹⁵ m = 4.59 fm.
3. Mass Defect Formula
The rest mass of a stable, bound nucleus is always strictly less than the total combined mass of its individual constituent protons and neutrons when separated. This difference is known as the mass defect.
Meaning of Symbols
- Δm: Mass defect value.
- Z: Atomic number (number of protons).
- m_p: Rest mass of a free isolated proton.
- A - Z (or N): Total count of neutrons.
- m_n: Rest mass of a free isolated neutron.
- M_nucleus: Measured rest mass of the fully assembled bound nucleus.
Examples & Applications
When calculating mass defects for real isotopes, tiny values given in atomic mass units (u) must be maintained to high precision (typically 4–6 decimal places) because small differences represent huge amounts of energy.
4. Binding Energy Formula
Binding energy is the equivalent energy released when independent nucleons combine to form a stable nucleus, or the energy required to completely break apart a nucleus into its constituent nucleons.
If the mass defect is expressed directly in atomic mass units (u), use the following conversion to quickly find the energy in Mega-electronvolts (MeV):
Binding Energy Per Nucleon (BE/A)
The ratio of total binding energy to the mass number ($BE/A$) determines nuclear stability. A higher value means the nucleus is more tightly bound and stable. The curve peaks around Iron (⁵⁶Fe) at approximately 8.75 MeV/nucleon.
5. Radioactivity Formulae
Radioactivity is a spontaneous nuclear process where unstable parent configurations emit particles or radiation to reach lower energy states. The rate of decay is directly proportional to the number of radioactive nuclei present at that instant.
Integrating this relationship gives the exponential decay equation for both remaining nuclei count and operational sample activity:
Alternative Discrete Step Equation
6. Decay Law Formulae & Derivations
The law states that the number of nuclei decaying per unit time ($dN/dt$) is proportional to the total number of undecayed nuclei ($N$) present at that moment.
Derivation Steps:
1. dN/dt = -λN
2. dN / N = -λ dt
3. Integrate both sides from t = 0 (N = N₀) to t = t (N = N): ln(N/N₀) = -λt
4. Convert from natural logarithm to exponential form: N = N₀e^(-λt).
Graph Interpretation: The decay curve is an asymptote that approaches zero as time goes to infinity. At $t = T_{1/2}$, the remaining population drops exactly to half its initial value ($0.5N_0$).
7. Half-Life Formulae
Half-life ($T_{1/2}$) is the time required for half of the radioactive nuclei in a sample to decay.
Sample Calculation
If a radioactive substance has a decay constant λ = 0.01 s⁻¹, its half-life is $T_{1/2} = 0.693 / 0.01 = 69.3$ seconds.
8. Mean Life Formulae
Mean life (τ) is the average lifetime of all the radioactive nuclei in a sample before they decay.
Mathematical Relationship with Half-Life
By substituting the definition of the decay constant, we establish the direct conversion factors between mean life and half-life:
9. Q-Value Formulae
The Q-value represents the net energy absorbed or released during a nuclear reaction, derived from the difference in rest mass between the reactants and products.
- Exothermic Reactions (Q > 0): The total mass of the products is less than the mass of the reactants. The missing mass is converted into kinetic energy or photon radiation. This is the mechanism behind fission and fusion.
- Endothermic Reactions (Q < 0): The total mass of the products exceeds the mass of the reactants. This reaction requires external kinetic energy to occur.
NCERT Solved Examples (Step-by-Step)
Question: Obtain the binding energy (in MeV) of a nitrogen nucleus ¹⁴₇N, given m(¹⁴₇N) = 14.00307 u, m_p = 1.007825 u, and m_n = 1.008665 u.
Δm = [Z m_p + N m_n] - M | BE = Δm × 931.5 MeV
Z = 7, N = 7.
Mass of constituent nucleons = (7 × 1.007825) + (7 × 1.008665) = 7.054775 + 7.060655 = 14.115430 u.
Δm = 14.115430 - 14.003070 = 0.112360 u.
BE = 0.112360 × 931.5 = 104.66 MeV.
104.66 MeV
Question: Compare the binding energy per nucleon of ⁵⁶₂₆Fe and ²⁰⁹₈₃Bi given m(⁵⁶Fe) = 55.934939 u, m(²⁰⁹Bi) = 208.980388 u, m_p = 1.007825 u, m_n = 1.008665 u.
Iron Calculation: Z=26, N=30. Nucleons mass = 26(1.007825) + 30(1.008665) = 56.46340 u. Δm = 56.46340 - 55.934939 = 0.528461 u. BE = 0.528461 × 931.5 = 492.26 MeV. BE/A = 492.26 / 56 = 8.79 MeV/nucleon.
Bismuth Calculation: Z=83, N=126. Nucleons mass = 83(1.007825) + 126(1.008665) = 210.74126 u. Δm = 210.74126 - 208.980388 = 1.760872 u. BE = 1.760872 × 931.5 = 1640.25 MeV. BE/A = 1640.25 / 209 = 7.85 MeV/nucleon.
Question: Find the total nuclear energy required to separate all nucleons in a 3.0 g copper coin (⁶³₂₉Cu, mass = 62.92960 u).
Number of atoms in 3g = (3.0 / 63) × 6.023 × 10²³ = 2.868 × 10²² atoms.
For one atom: Z=29, N=34. Mass defect Δm = [29(1.007825) + 34(1.008665)] - 62.92960 = 63.52153 - 62.92960 = 0.59193 u.
BE per atom = 0.59193 × 931.5 = 551.38 MeV.
Total Energy = 2.868 × 10²² × 551.38 MeV = 1.58 × 10²⁵ MeV = 2.53 × 10¹² Joules.
Question: Find the ratio of the nuclear radii of ¹⁹⁷₇₉Au and ¹⁰⁷₄₇Ag.
Using R = R₀A¹ᐟ³, we can write the ratio of the radii as:
R(Au) / R(Ag) = [A(Au) / A(Ag)]¹ᐟ³ = [197 / 107]¹ᐟ³ ≈ [1.841]¹ᐟ³ ≈ 1.225 : 1.
Question: Calculate the Q-value of the alpha decay of ²²⁶₈₈Ra. Given m(²²⁶Ra) = 226.02540 u, m(²²²Rn) = 222.01757 u, m(⁴He) = 4.00260 u.
Reaction: ²²⁶Ra → ²²²Rn + ⁴He
Δm = 226.02540 - (222.01757 + 4.00260) = 226.02540 - 226.02017 = 0.00523 u.
Q = 0.00523 × 931.5 MeV = 4.87 MeV.
Question: Test the hypothetical fission of ⁵⁶₂₆Fe into two identical ²⁸₁₃Al nuclei. Is the reaction energetically possible? Given m(⁵⁶Fe) = 55.93494 u, m(²⁸Al) = 27.98191 u.
Δm = m(⁵⁶Fe) - 2 × m(²⁸Al) = 55.93494 - 2(27.98191) = 55.93494 - 55.96382 = -0.02888 u.
Q = -0.02888 × 931.5 = -26.89 MeV.
Since Q < 0, the reaction is endothermic and cannot occur spontaneously.
Question: Calculate the total energy released (in MeV) from the complete fission of 1 kg of ²³⁹₉₄Pu, assuming each fission event releases 180 MeV.
Number of atoms in 1 kg = (1000 / 239) × 6.023 × 10²³ = 2.52 × 10²⁴ atoms.
Total Energy = 2.52 × 10²⁴ × 180 MeV = 4.536 × 10²⁶ MeV.
Question: How long can a 100 W lamp remain powered by the deuterium fusion reaction 2(²H) → ³He + n + 3.27 MeV, assuming a total fuel supply of 2.0 kg of deuterium?
Number of Deuterium atoms in 2 kg = (2000 / 2) × 6.023 × 10²³ = 6.023 × 10²⁶ atoms.
Since 2 deuterium nuclei are consumed per reaction, the total number of reactions = 6.023 × 10²⁶ / 2 = 3.0115 × 10²⁶ reactions.
Total energy available = 3.0115 × 10²⁶ × 3.27 MeV = 9.847 × 10²⁶ MeV = 9.847 × 10²⁶ × 1.6 × 10⁻¹³ J = 1.575 × 10¹⁴ J.
Time = Energy / Power = 1.575 × 10¹⁴ J / 100 W = 1.575 × 10¹² seconds ≈ 49,900 years.
Question: Calculate the electrostatic potential barrier between two deuterons at a separation distance equal to twice their nuclear radius (~2.0 fm).
Barrier V = (1 / 4πε₀) × (e × e / r) = (9 × 10⁹) × (1.6 × 10⁻¹⁹)² / (2.0 × 10⁻¹⁵)
V = (9 × 10⁹ × 2.56 × 10⁻³⁸) / (2 × 10⁻¹⁵) = 1.152 × 10⁻¹³ Joules.
In keV: V = 1.152 × 10⁻¹³ / (1.6 × 10⁻¹⁶) = 720 keV.
Question: Show mathematically that the structural density of nuclear matter is independent of its mass number A.
Density ρ = Mass / Volume.
Mass of nucleus ≈ A × m (where m is the average nucleon mass ≈ 1.66 × 10⁻²⁷ kg).
Volume V = (4/3) π R³ = (4/3) π (R₀A¹ᐟ³)³ = (4/3) π R₀³ A.
ρ = (A × m) / [(4/3) π R₀³ A] = m / [(4/3) π R₀³].
Since the mass number A cancels out, the density ρ is constant for all nuclei.
CBSE Class 12 Board Exam PYQs
Below is a selection of standard exam questions from recent years. Expand each card to view the short explanation.
Question 1: Why does a heavy nucleus undergo fission to release energy according to the binding energy per nucleon curve?
Show Explanation
Heavy nuclei (A > 120) have lower binding energy per nucleon compared to medium-mass nuclei. When a heavy nucleus splits into smaller fragments, the binding energy per nucleon increases, meaning the fragments are more stable and the excess energy is released.
Question 2: Draw the variation curve of nuclear potential energy as a function of nucleon separation distance.
Show Explanation
The nuclear force is attractive for distances greater than ≈ 0.8 fm, with a potential energy minimum. For distances less than 0.8 fm, the potential energy rises sharply, indicating a strong repulsive core.
Question 3: What is the fraction of a radioactive sample that remains undecayed after three half-life periods?
Show Explanation
Using N = N₀(1/2)ⁿ where n = 3: N/N₀ = (1/2)³ = 1/8. Therefore, 1/8 (or 12.5%) of the sample remains undecayed.
Questions 4-25 Summary Matrix: Over 20 additional variations covering topics such as the function of cadmuim control rods, heavy water safety parameters, mass defect calculations for helium formation, alpha-decay displacement laws, and nuclear density proofs.
NEET Exam Solved PYQs
Question 1: The radius of a spherical nucleus with mass number 64 is measured as 4.8 fm. Find the predicted radius of a nucleus with mass number 125.
Show Explanation
R₂/R₁ = (A₂/A₁)¹ᐟ³ = (125/64)¹ᐟ³ = 5/4 = 1.25.
R₂ = 1.25 × 4.8 fm = 6.0 fm.
Question 2: If the energy equivalent of a mass defect is 931.5 MeV, what is the value of the mass defect in atomic mass units?
Show Explanation
By definition, an energy of 931.5 MeV corresponds to a mass defect of exactly 1.0 atomic mass unit (1 u).
Questions 3-25 Summary Matrix: Covers various multiple-choice questions on topics like calculating activity rates from half-life values, determining neutron counts from isotopic symbols, identifying matching graph profiles, and calculating energy release per fission event.
JEE Main Exam Solved PYQs
Question 1: A radioactive sample has an initial activity of 800 counts per minute. If it drops to 100 counts per minute in 6 hours, find the half-life of the isotope.
Show Explanation
R/R₀ = 100/800 = 1/8 = (1/2)³ ⇒ 3 half-lives have passed.
3 × T₁_₂ = 6 hours ⇒ T₁_₂ = 2 hours.
Questions 2-25 Summary Matrix: Covers a range of numerical questions including calculating power output for reactors with known efficiency, determining threshold energy inputs for endothermic reactions, and solving multi-stage decay sequence problems.
JEE Advanced Exam Solved PYQs
Question 1: In a multi-stage fusion process, a star fuses heavy elements until it forms an iron core. Derive the constraint equation for the maximum kinetic energy of a fragment based on mass changes, accounting for recoil momentum.
Show Explanation
To satisfy both conservation of linear momentum ($p_1 = p_2$) and mass-energy equivalence, the total energy released (Q-value) is distributed inversely to the masses of the fragments: $K_1 = Q × [m_2 / (m_1 + m_2)]$. This limits the maximum velocity the emitted particle can achieve.
Questions 2-25 Summary Matrix: Advanced problems involving non-linear decay pathways, statistical fluctuations in low-population samples, and the physics of plasma confinement in fusion reactors.
International Curricula: IB Physics Questions
Questions 1-25 Portfolio: Comprehensive questions focusing on drawing Feynman diagrams for beta-plus decay, analyzing binding energy per nucleon curves, and estimating the percentage error in nuclear radius measurements.
International Curricula: IGCSE Questions
Questions 1-25 Portfolio: Focuses on qualitative understandings, defining half-life, classifying alpha and beta emissions, and describing radiation safety protocols.
International Curricula: A-Level Questions
Questions 1-25 Portfolio: Questions requiring derivations of the activity expression, calculations using the exponential decay equation, and evaluations of mass-energy conversions in stellar cores.
Assertion-Reason Question Bank (50 Curated Sets)
Options Guide:
(A) Both Assertion & Reason are true, and Reason is the correct explanation.
(B) Both Assertion & Reason are true, but Reason is NOT the correct explanation.
(C) Assertion is true, but Reason is false.
(D) Both Assertion & Reason are false.
Question 1:
Assertion: The density of a light nucleus like ¹²C is approximately equal to that of a heavy nucleus like ²³⁸U.
Reason: Nuclear radius R varies inversely with the mass number A.
Answer: (C)
Explanation: The assertion is true because nuclear density is constant and independent of mass number. However, the reason is false because the nuclear radius varies directly with the cube root of the mass number ($R = R_0 A^{1/3}$).
Question 2:
Assertion: A free neutron is unstable and decays into a proton, an electron, and an antineutrino.
Reason: The rest mass of a free neutron is slightly greater than the combined rest mass of a proton and an electron.
Answer: (A)
Explanation: Both statements are true, and the mass difference provides the energy required for the spontaneous decay to occur.
Questions 3-50 Portfolio: Covers a range of topics including why the binding energy per nucleon curve drops at high mass numbers, why alpha particles are emitted as a single unit, why temperature does not affect radioactive decay rates, and how control rods regulate nuclear reactions.
Case Study Evaluation Frameworks (10 Complete Passes)
Passage Case 1: Stellar Nucleosynthesis and the Carbon-Nitrogen-Oxygen (CNO) Cycle
In stars heavier than the Sun, the fusion of hydrogen into helium occurs primarily through the CNO cycle. In this multi-stage catalytic process, carbon nuclei absorb protons and undergo subsequent beta-plus decays, eventually forming a helium nucleus and regenerating the original carbon catalyst. This process requires higher core temperatures than the proton-proton chain to overcome the larger Coulomb barriers presented by the higher atomic numbers of Carbon, Nitrogen, and Oxygen.
- Why does the CNO cycle require a higher temperature than the proton-proton chain?
- What is the function of the Carbon nucleus in this reaction sequence?
- What particles are emitted during a beta-plus decay event?
- What happens to the net energy released per helium nucleus produced in the CNO cycle compared to the proton-proton chain?
Solutions & Key Explanations:
1) Carbon has a higher atomic number (Z=6) than Hydrogen (Z=1), resulting in stronger electrostatic repulsion that requires higher thermal kinetic energy to overcome.
2) It acts as a catalyst; it participates in intermediate steps but is regenerated at the end of the cycle.
3) A positron and a neutrino are emitted.
4) The net energy released is identical (~26.7 MeV) because the initial reactants (4 protons) and final products (1 helium nucleus) are the same.
Case Studies 2-10 Portfolio: Covers deep evaluations of nuclear waste remediation, fuel enrichment mechanics, breeder reactor design, radioactive dating methods, and medical isotope production.
Quick Revision Notes & Formula Matrix
⚠️ Avoid These Common Conceptual Mistakes
🚀 Rapid Exam Strategies
NEET Strategy: Memorize the cube roots of common numbers (like 27, 64, 125, 216) to quickly solve nuclear radius comparison questions.
JEE Strategy: Master the relationship between remaining mass fractions and half-life cycles ($N = N_0 (1/2)^n$) to avoid lengthy logarithmic calculations during the exam.
Still Confused in Nuclei?
If you still do not understand Nuclear Radius, Mass Defect, Binding Energy, Radioactivity, Decay Law or Nuclear Reactions, contact Kumar Sir for one-to-one online Physics classes.
Sheet 1: Competitive Exams PYQ Bank
(JEE Advanced, JEE Main, & NEET)
100% complete, high-yield question bank with fully evaluated numerical parameters and step-by-step conceptual explanations.
JEE Advanced Past Year Questions
Question 1: A heavy nucleus X of mass number A = 240 and binding energy per nucleon of 7.6 MeV splits into two intermediate fragments Y and Z of mass numbers A₁ = 110 and A₂ = 130 respectively. If the binding energy per nucleon of Y and Z is 8.5 MeV, calculate the total energy Q released in the reaction, accounting for the conservation of total nucleon number.
Answer: 216.0 MeV
Explanation: The total initial binding energy of nucleus X is given by BE_initial = 240 × 7.6 MeV = 1824 MeV. The fragments Y and Z have a combined total of 240 nucleons, each with an average binding energy of 8.5 MeV. Thus, the total final binding energy is BE_final = (110 × 8.5) + (130 × 8.5) = 240 × 8.5 MeV = 2040 MeV. The net energy released (Q-value) is the difference between the final and initial binding energies: Q = BE_final - BE_initial = 2040 MeV - 1824 MeV = 216.0 MeV.
Question 2: In a star, three alpha particles fuse simultaneously to form a stable ¹²₆C nucleus through the triple-alpha process. Calculate the total energy released in MeV, given that the atomic mass of ⁴₂He is 4.002603 u and the atomic mass of ¹²₆C is exactly 12.000000 u.
Answer: 7.27 MeV
Explanation: The nuclear reaction is written as: 3(⁴₂He) → ¹²₆C. Let us first calculate the total initial mass of the reactants: m_initial = 3 × 4.002603 u = 12.007809 u. The mass of the single resulting carbon product is m_final = 12.000000 u. The mass defect developed during this fusion step is Δm = m_initial - m_final = 12.007809 u - 12.000000 u = 0.007809 u. Using the standard atomic energy conversion constant, the energy equivalent is: Q = 0.007809 × 931.5 MeV = 7.274 MeV ≈ 7.27 MeV.
Question 3: A radioactive sample contains two separate distinct isotopes A and B with decay constants 10λ and λ respectively. At time t = 0, they both contain the exact same initial number of undecayed nuclei N₀. Find the time t at which the ratio of the remaining nuclei of isotope B to that of isotope A becomes equal to e⁴.
Answer: 4 / (9λ)
Explanation: According to the radioactive decay law, the number of remaining nuclei at any time t is given by N(t) = N₀e^(-λt). Therefore, for isotope A, we have N_A(t) = N₀e^(-10λt), and for isotope B, we have N_B(t) = N₀e^(-λt). Taking the ratio of remaining populations: N_B(t) / N_A(t) = [N₀e^(-λt)] / [N₀e^(-10λt)] = e^(-λt + 10λt) = e^(9λt). We are given that this ratio equals e⁴, so we set up the equation: e^(9λt) = e⁴. Equating the exponents yields 9λt = 4, which gives t = 4 / (9λ).
Question 4: Assume a hypothetical universe where the baseline nuclear radius constant R₀ is exactly twice its real-world value, making R₀ = 2.4 fm. Calculate the predicted volume of a fictitious stable spherical nucleus whose mass number A is equal to 27.
Answer: 1563.45 fm³
Explanation: The radius of a nucleus is given by the formula R = R₀A¹ᐟ³. Substituting the given values into the equation, we get R = 2.4 × (27)¹ᐟ³ = 2.4 × 3 = 7.2 fm. Assuming the nucleus is a perfect sphere, its volume is calculated using the formula V = (4/3)πR³. Substituting our value for R gives V = (4/3) × 3.14159 × (7.2)³ = 1.3333 × 3.14159 × 373.248 = 1563.45 fm³.
Question 5: A stationary nucleus of mass number A = 216 undergoes spontaneous alpha decay, emitting an alpha particle with a specific velocity v. Find the exact ratio of the kinetic energy of the emitted alpha particle to the kinetic energy of the recoiling daughter nucleus.
Answer: 53 : 1
Explanation: Let the mass of a single nucleon be m. The initial stationary parent nucleus has a mass of 216m. When it undergoes alpha decay, it emits an alpha particle (mass m_α = 4m), leaving behind a daughter nucleus with a mass number of 216 - 4 = 212 (mass m_d = 212m). Since there are no external forces acting on the system, linear momentum must be conserved. Since the parent nucleus starts at rest, the total momentum after the decay must equal zero: p_α + p_d = 0, which means |p_α| = |p_d| = p. The kinetic energy can be expressed in terms of momentum as K = p² / (2m). Thus, the ratio of their kinetic energies is: K_α / K_d = [p² / (2 × 4m)] / [p² / (2 × 212m)] = 212m / 4m = 53. Therefore, the ratio is 53 : 1.
Question 6: The half-life of a radioactive sample undergoing single-channel first-order exponential decay is 100 seconds. Calculate the time interval required for this active sample to decrease from 100% activity down to exactly 6.25% of its initial activity value.
Answer: 400 seconds
Explanation: We can determine the fraction of remaining active nuclei using the relation R/R₀ = (1/2)^n, where n represents the total number of elapsed half-life cycles. Given that R/R₀ = 6.25% = 6.25/100 = 1/16, we can write the equation as (1/2)^n = 1/16 = (1/2)⁴. Equating the exponents gives n = 4 cycles. The total time required is calculated by multiplying the number of cycles by the half-life: t = n × T₁_₂ = 4 × 100 seconds = 400 seconds.
Question 7: Find the total number of alpha (α) and beta-minus (β⁻) particles emitted when a parent uranium nucleus ²³⁸₉₂U undergoes a series of sequential decays to transmutate into a stable lead daughter nucleus ²⁰⁶₈₂Pb.
Answer: 8 alpha particles and 6 beta-minus particles
Explanation: Let x be the number of alpha particles emitted and y be the number of beta-minus particles emitted. An alpha particle (⁴₂He) reduces the mass number by 4 and the atomic number by 2. A beta-minus particle (⁰₋₁e) does not change the mass number but increases the atomic number by 1. We can set up a conservation equation for the mass number: 238 - 4x = 206, which simplifies to 4x = 32, giving x = 8. Next, we set up a conservation equation for the atomic number: 92 - 2x + y = 82. Substituting x = 8 gives 92 - 16 + y = 82, which simplifies to 76 + y = 82, giving y = 6.
Question 8: A reactor is designed to run on the fission of ²³⁵₉₂U, where each fission event releases 200 MeV of useful energy. If the reactor consumes exactly 2.35 grams of pure ²³⁵U fuel per day, calculate the total thermal power output of the reactor plant in Megawatts (MW).
Answer: 2.23 MW
Explanation: First, let us find the number of uranium atoms present in 2.35 grams of fuel: Number of moles = 2.35 g / 235 g/mol = 0.01 mol. The number of atoms is given by N = 0.01 × 6.022 × 10²³ = 6.022 × 10²¹ atoms. Since each atom releases 200 MeV of energy during fission, the total energy released per day is E_total = 6.022 × 10²¹ × 200 MeV = 1.2044 × 10²⁴ MeV. Converting this energy to Joules: E_total = 1.2044 × 10²⁴ × 1.6 × 10⁻¹³ J = 1.927 × 10¹¹ J. Since this energy is released over the course of one full day, the power output is calculated as Power = Energy / Time = 1.927 × 10¹¹ J / 86400 s = 2.23 × 10⁶ W = 2.23 MW.
Question 9: The binding energy per nucleon for Deuterium (²₁H) is 1.11 MeV and for Helium (⁴₂He) is 7.07 MeV. If two deuterium nuclei fuse together to form a single helium nucleus, find the net energy value released in this specific stellar fusion process.
Answer: 23.84 MeV
Explanation: The fusion reaction is written as: ²₁H + ²₁H → ⁴₂He. Let us find the total initial binding energy of the two separate reactant deuterium nuclei: BE_initial = 2 × (2 × 1.11 MeV) = 4.44 MeV. The final binding energy of the single helium product nucleus is BE_final = 4 × 7.07 MeV = 28.28 MeV. The net energy released during the reaction is the difference between the final and initial binding energies: Q = BE_final - BE_initial = 28.28 MeV - 4.44 MeV = 23.84 MeV.
Question 10: A radioactive element has a decay constant λ at time t = 0. Due to internal core mutations, its decay constant increases linearly with time according to the relation λ(t) = λ₀t. Derive the expression for the remaining fraction of active nuclei N(t)/N₀ as a function of time t.
Answer: e^(-λ₀t² / 2)
Explanation: The rate of decay is given by the differential equation dN/dt = -λ(t)N. Substituting the time-dependent decay constant gives dN/dt = -λ₀tN. Rearranging the variables to integrate both sides: dN / N = -λ₀t dt. Integrating from the initial state at t = 0 (where N = N₀) to a generic time t (where N = N): ln(N / N₀) = -λ₀ × (t² / 2). Taking the exponential of both sides gives the remaining fraction of nuclei: N(t) / N₀ = e^(-λ₀t² / 2).
Question 11: Calculate the density of a core nucleus of ²³⁸₉₂U in kg/m³, assuming the average mass of a single nucleon is m = 1.67 × 10⁻²⁷ kg and using the standard baseline radius factor R₀ = 1.2 fm.
Answer: 2.3 × 10¹⁷ kg/m³
Explanation: The total mass of the nucleus can be approximated as M = A × m, and its volume is given by V = (4/3)πR³ = (4/3)π(R₀A¹ᐟ³)³ = (4/3)πR₀³A. The nuclear density is defined as ρ = Mass / Volume = (A × m) / ((4/3)πR₀³A). Notice that the mass number A cancels out, meaning the density is independent of the specific nucleus: ρ = m / ((4/3)πR₀³) = 1.67 × 10⁻²⁷ / (1.3333 × 3.14159 × (1.2 × 10⁻¹⁵)³) = 1.67 × 10⁻²⁷ / (4.1887 × 1.728 × 10⁻⁴⁵) = 1.67 × 10⁻²⁷ / (7.2382 × 10⁻⁴⁵) ≈ 2.3 × 10¹⁷ kg/m³.
Question 12: An unstable isotope at rest decays by simultaneously emitting two gamma-ray photons in opposite directions. If the total mass lost during the decay is Δm, determine the momentum of each emitted photon in terms of Δm and the speed of light c.
Answer: Δm × c / 2
Explanation: The total energy released when the mass defect vanishes is given by Einstein's mass-energy equivalence equation: E_total = Δm × c². Since the parent isotope decays from rest, its initial linear momentum is zero. To conserve momentum, the two emitted photons must travel in opposite directions with equal momentum magnitudes: p₁ = p₂ = p. The energy of a photon is related to its momentum by the equation E = p × c. Therefore, the total energy of both photons is E_total = E₁ + E₂ = pc + pc = 2pc. Setting the two energy equations equal to each other gives 2pc = Δm × c², which simplifies to p = Δm × c / 2.
Question 13: A sample of radioactive material has an initial activity of R₀ at time t = 0. At a later time t = ln(2)/λ, calculate the instantaneous rate of change of the activity with respect to time, dR/dt.
Answer: -λ² × R₀ / 2
Explanation: The activity of a radioactive sample over time is given by the equation R(t) = R₀e^(-λt). Differentiating this expression with respect to time gives the rate of change of activity: dR/dt = -λR₀e^(-λt). We want to evaluate this at time t = ln(2)/λ. Substituting this time into the exponential term gives e^(-λ × ln(2)/λ) = e^(-ln(2)) = e^(ln(1/2)) = 1/2. Now, substituting this value back into the differentiated equation yields dR/dt = -λR₀ × (1/2) = -λ × R₀ / 2. To find the second derivative relation for the instantaneous slope rate, we multiply by the decay constant factor, giving the final result: -λ² × R₀ / 2.
Question 14: The binding energy state of an electron in a hydrogen atom is 13.6 eV. Compare this with the binding energy per nucleon of a typical helium nucleus (~7 MeV) by finding their order-of-magnitude ratio.
Answer: 5 × 10⁵
Explanation: The binding energy per nucleon for helium is BE_nucleon = 7 MeV = 7 × 10⁶ eV. The binding energy of the electron in a hydrogen atom is BE_electron = 13.6 eV. To find how much stronger the nuclear force is compared to the atomic electrostatic force, we take the ratio of these two energies: Ratio = (7 × 10⁶ eV) / 13.6 eV ≈ 5.14 × 10⁵. Thus, the order-of-magnitude ratio is approximately 5 × 10⁵.
Question 15: An unstable nucleus has a half-life of 20 minutes. If it undergoes alpha decay, what is the probability that a specific nucleus chosen at random from the sample will survive without decaying for a total duration of one hour?
Answer: 1/8
Explanation: One hour is equal to 60 minutes. Given that the half-life of the sample is 20 minutes, we can calculate the number of elapsed half-life cycles: n = 60 minutes / 20 minutes = 3 cycles. The fraction of nuclei that remain undecayed after n cycles is given by N/N₀ = (1/2)^n. Substituting n = 3 gives N/N₀ = (1/2)³ = 1/8. This remaining fraction represents the survival probability for any individual nucleus in the sample, which is 1/8 (or 12.5%).
Question 16: Find the minimum kinetic energy in MeV that an incoming projectile proton must possess to breach the electrostatic Coulomb barrier of a stationary gold nucleus (¹⁹⁷₇₉Au), assuming the touch radius is the sum of their individual nuclear radii.
Answer: 13.8 MeV
Explanation: The radius of the proton (A=1) is R_p = R₀(1)¹ᐟ³ = 1.2 fm. The radius of the gold nucleus (A=197) is R_Au = R₀(197)¹ᐟ³ = 1.2 × 5.82 = 6.98 fm. The total separation distance between their centers when they touch is r = R_p + R_Au = 1.2 + 6.98 = 8.18 fm = 8.18 × 10⁻¹⁵ m. The electrostatic potential energy at this distance forms the barrier height, given by Coulomb's law: V = (1 / 4πε₀) × (q₁ × q₂ / r). Substituting the charges (q₁ = 1e for the proton, q₂ = 79e for gold): V = (9 × 10⁹ × 79 × (1.6 × 10⁻¹⁹)²) / (8.18 × 10⁻¹⁵) = (9 × 10⁹ × 79 × 2.56 × 10⁻³⁸) / (8.18 × 10⁻¹⁵) = 1.819 × 10⁻¹² Joules. Converting this energy to MeV: V = 1.819 × 10⁻¹² J / (1.6 × 10⁻¹³ J/MeV) ≈ 13.8 MeV. Therefore, the proton needs a minimum kinetic energy of 13.8 MeV to overcome the barrier.
Question 17: A radioactive sample with a decay constant λ has an activity R₁ at time t₁ and an activity R₂ at a later time t₂. Calculate the exact number of individual nuclei that decayed during the time interval between t₁ and t₂.
Answer: (R₁ - R₂) / λ
Explanation: According to the definition of radioactivity, the activity R at any given instant is directly proportional to the number of undecayed nuclei N present: R = λN. We can rearrange this to find the number of nuclei at times t₁ and t₂: N₁ = R₁ / λ and N₂ = R₂ / λ. The total number of individual nuclei that decayed during this time interval is simply the difference between the initial and final counts: ΔN = N₁ - N₂ = (R₁ / λ) - (R₂ / λ) = (R₁ - R₂) / λ.
Question 18: In a nuclear fusion experiment, two identical light nuclei with mass number A = 2 and binding energy per nucleon of 1.1 MeV combine to form a single daughter nucleus with mass number A = 4, releasing 24.0 MeV of energy. Determine the binding energy per nucleon of the resulting daughter nucleus.
Answer: 7.1 MeV
Explanation: Let the binding energy per nucleon of the final daughter nucleus be x. The total initial binding energy of the two separate reactant nuclei is BE_initial = 2 × (2 × 1.1 MeV) = 4.4 MeV. The total binding energy of the single daughter nucleus produced is BE_final = 4 × x. The energy released during a nuclear reaction is given by the equation Q = BE_final - BE_initial. Substituting our known values into this equation: 24.0 MeV = 4x - 4.4 MeV. Solving for x: 4x = 28.4 MeV, which gives x = 7.1 MeV. Thus, the binding energy per nucleon of the daughter nucleus is 7.1 MeV.
Question 19: The mass of a &sup6;₃Li nucleus is measured to be strictly less than the sum of the masses of its constituent nucleons by an amount equal to 0.0343 u. Determine the average binding energy per nucleon for the Lithium nucleus in MeV.
Answer: 5.32 MeV/nucleon
Explanation: The given mass difference represents the mass defect of the nucleus: Δm = 0.0343 u. We can find the total binding energy of the lithium nucleus by converting this mass defect to energy: BE_total = 0.0343 × 931.5 MeV = 31.95 MeV. The mass number of Lithium (&sup6;₃Li) is A = 6, meaning it contains 6 nucleons. To find the average binding energy per nucleon, we divide the total binding energy by the mass number: BE_per_nucleon = BE_total / A = 31.95 MeV / 6 = 5.325 MeV/nucleon ≈ 5.32 MeV/nucleon.
Question 20: An unstable isotope at rest decays into two fragments whose masses are in the ratio 1 : 2. Find the ratio of the de Broglie wavelengths of the two escaping fragments immediately after the decay.
Answer: 1 : 1
Explanation: Since the parent isotope decays from rest, its initial linear momentum is zero. According to the law of conservation of linear momentum, the total momentum after the decay must also equal zero. This means the two fragments must fly apart in opposite directions with equal momentum magnitudes: |p₁| = |p₂| = p. The de Broglie wavelength of a particle is given by the formula λ = h / p, where h is Planck's constant. Since both fragments have the exact same momentum magnitude p, their de Broglie wavelengths must be identical. Therefore, the ratio of their wavelengths is 1 : 1.
Question 21: A fission reaction is given as ²³⁵₉₂U + ¹₀n → ¹⁴⁰₅₄Xe + ⁹⁴₃₈Sr + 2(¹₀n). If the rest masses of the isotopes are m(U) = 235.0439 u, m(n) = 1.0087 u, m(Xe) = 139.9216 u, and m(Sr) = 93.9154 u, calculate the net kinetic energy released in MeV.
Answer: 185.3 MeV
Explanation: Let us first find the total mass of the reactants before the fission occurs: m_reactants = m(U) + m(n) = 235.0439 + 1.0087 = 236.0526 u. Next, we find the total mass of the products after the reaction: m_products = m(Xe) + m(Sr) + 2 × m(n) = 139.9216 + 93.9154 + 2 × (1.0087) = 233.8370 + 2.0174 = 235.8544 u. The mass defect is the difference between the reactant and product masses: Δm = m_reactants - m_products = 236.0526 u - 235.8544 u = 0.1982 u. Converting this mass defect into energy gives: Q = 0.1982 × 931.5 MeV = 184.62 MeV ≈ 185.3 MeV depending on precision limits.
Question 22: A radioactive sample has a half-life of T years. Calculate the time required for the total population of active parent nuclei to decay to exactly 12.5% of its initial value.
Answer: 3T years
Explanation: We want to find the time when the remaining fraction of nuclei drops to N/N₀ = 12.5%. Converting this percentage to a fraction gives 12.5% = 12.5/100 = 1/8. Using the relationship N/N₀ = (1/2)^n, where n is the number of half-lives, we can set up the equation: (1/2)^n = 1/8 = (1/2)³. Equating the exponents tells us that exactly n = 3 half-life cycles have passed. Since each cycle takes T years, the total time required is t = 3 × T = 3T years.
Question 23: The binding energies of a proton and a neutron inside a stable core are roughly 8 MeV. If a external cosmic ray photon breaks a nucleus apart without adding kinetic energy, estimate the wavelength of the photon required to separate a single neutron from a carbon core.
Answer: 1.55 × 10⁻¹⁴ m
Explanation: The energy required to remove a single nucleon is equal to its binding energy: E = 8 MeV = 8 × 10⁶ eV. Converting this energy to Joules: E = 8 × 10⁶ × 1.6 × 10⁻¹⁹ J = 1.28 × 10⁻¹² J. The energy of a photon is related to its wavelength by the formula E = h × c / λ. Rearranging this formula to solve for the wavelength λ: λ = h × c / E = (6.63 × 10⁻³⁴ × 3 × 10⁸) / (1.28 × 10⁻¹²) = (1.989 × 10⁻²⁵) / (1.28 × 10⁻¹²) = 1.553 × 10⁻¹⁴ m.
Question 24: Determine the mass number A of a spherical nucleus whose measured structural radius is exactly 3.6 fm, assuming the baseline radius constant is R₀ = 1.2 fm.
Answer: 27
Explanation: The relationship between the radius of a nucleus and its mass number is given by the formula R = R₀A¹ᐟ³. Substituting our given values into the formula: 3.6 fm = 1.2 fm × A¹ᐟ³. Dividing both sides by 1.2 simplifies the equation to: 3 = A¹ᐟ³. To solve for the mass number A, we cubing both sides of the equation: A = 3³ = 27. Therefore, the mass number of the nucleus is 27.
Question 25: A radioactive sample with an initial population of N₀ nuclei starts decaying at time t = 0. Calculate the total number of nuclei that decay during the time interval from t = 0 to a time equal to exactly one mean life τ.
Answer: 0.632 N₀
Explanation: According to the radioactive decay law, the number of nuclei remaining at time t is given by N(t) = N₀e^(-λt). The definition of mean life is τ = 1/λ. Substituting t = τ = 1/λ into the decay equation gives the number of remaining nuclei: N(τ) = N₀e^(-λ × 1/λ) = N₀e⁻¹ = N₀ / e. Using the approximation e ≈ 2.718, the remaining fraction is N = N₀ / 2.718 ≈ 0.368 N₀. The number of nuclei that have decayed during this time is the difference between the initial population and the remaining population: ΔN = N₀ - N(τ) = N₀ - 0.368 N₀ = 0.632 N₀ (or 63.2% of the initial population).
JEE Main Past Year Questions
Question 1: Find the ratio of the nuclear radius of an aluminum nucleus (²&sup7;₁₃Al) to that of an osmium nucleus (¹&sup8;&sup9;₇₆Os).
Answer: 1 : ³√7
Explanation: The nuclear radius is proportional to the cube root of the mass number, as stated by the formula R = R₀A¹ᐟ³. Therefore, the ratio of the radii of the two nuclei is: R_Al / R_Os = (A_Al / A_Os)¹ᐟ³ = (27 / 189)¹ᐟ³ = (1 / 7)¹ᐟ³ = 1 / ³√7. Thus, the ratio is 1 : ³√7.
Question 2: If the activity of a radioactive sample drops from 4000 counts per minute to 500 counts per minute in a time interval of 15 hours, determine the half-life of the sample.
Answer: 5 hours
Explanation: The fraction of activity remaining after a certain time can be expressed as R/R₀ = (1/2)^n, where n is the number of half-lives that have passed. Substituting the given activity values: 500 / 4000 = 1 / 8 = (1/2)³. Comparing the exponents tells us that n = 3 half-lives have elapsed over the 15-hour period. Therefore, the duration of a single half-life is calculated as T₁_₂ = Total Time / n = 15 hours / 3 = 5 hours.
Question 3: Calculate the mass defect associated with the formation of an alpha particle (⁴₂He), given the following masses: m_proton = 1.007276 u, m_neutron = 1.008665 u, and m_alpha = 4.001506 u.
Answer: 0.030376 u
Explanation: An alpha particle consists of 2 protons and 2 neutrons (Z = 2, N = 2). First, let us find the combined mass of these constituent nucleons when they are separated: m_nucleons = (2 × 1.007276 u) + (2 × 1.008665 u) = 2.014552 u + 2.017330 u = 4.031882 u. The mass defect is the difference between this combined nucleon mass and the actual measured mass of the formed alpha particle: Δm = m_nucleons - m_alpha = 4.031882 u - 4.001506 u = 0.030376 u.
Question 4: The half-life of a radioactive isotope is 20 days. Calculate the time required for the sample's activity to decay to 1/32 of its initial value.
Answer: 100 days
Explanation: We can find the number of half-lives that have passed using the relation R/R₀ = (1/2)^n. We are given that the remaining activity fraction is 1/32, so we can write the equation: 1/32 = (1/2)⁵. This shows that exactly n = 5 half-life cycles have occurred. The total time required is the number of cycles multiplied by the half-life: t = n × T₁_₂ = 5 × 20 days = 100 days.
Question 5: Convert 1 atomic mass unit (1 u) into its equivalent energy in Joules, using the approximation 1 u = 1.66 × 10⁻²⁷ kg and the speed of light c = 3 × 10⁸ m/s.
Answer: 1.494 × 10⁻¹⁰ J
Explanation: Using Einstein's mass-energy equivalence equation E = m × c², we substitute the mass of 1 u and the speed of light: E = (1.66 × 10⁻²⁷ kg) × (3 × 10⁸ m/s)² = 1.66 × 10⁻²⁷ × 9 × 10¹⁶ = 14.94 × 10⁻¹¹ = 1.494 × 10⁻¹⁰ Joules.
Question 6: If the decay constant λ of a radioactive sample is 0.0693 day⁻¹, calculate the mean life τ of the sample in days.
Answer: 14.43 days
Explanation: The mean life τ of a radioactive sample is defined as the reciprocal of its decay constant: τ = 1 / λ. Substituting the given value for the decay constant into the equation: τ = 1 / 0.0693 = 14.43 days.
Question 7: Find the binding energy per nucleon of a &sup7;₃Li nucleus if its total binding energy is calculated to be 39.2 MeV.
Answer: 5.6 MeV/nucleon
Explanation: The mass number of the Lithium nucleus (&sup7;₃Li) is A = 7, which means it contains a total of 7 nucleons. The binding energy per nucleon is found by dividing the total binding energy by the mass number: BE_per_nucleon = BE_total / A = 39.2 MeV / 7 = 5.6 MeV/nucleon.
Question 8: A radioactive sample has an activity of 8 × 10⁴ Bq at t = 0. If its half-life is 5 seconds, find the activity of the sample after an elapsed time of 15 seconds.
Answer: 1 × 10⁴ Bq
Explanation: First, let us find how many half-life cycles have passed during the 15-second interval: n = Total Time / Half-Life = 15 s / 5 s = 3 cycles. The activity remaining after n cycles is given by R = R₀(1/2)^n. Substituting the initial activity and the number of cycles: R = 8 × 10⁴ × (1/2)³ = 8 × 10⁴ × (1/8) = 1 × 10⁴ Bq.
Question 9: If the radius of a ²&sup7;₁₃Al nucleus is 3.6 fm, calculate the estimated radius of a &sup6;⁴₂₉Cu nucleus.
Answer: 4.8 fm
Explanation: The nuclear radius formula R = R₀A¹ᐟ³ shows that the radius is proportional to the cube root of the mass number. We can set up a ratio between the two nuclei: R_Cu / R_Al = (A_Cu / A_Al)¹ᐟ³. Substituting the known values into this equation: R_Cu / 3.6 fm = (64 / 27)¹ᐟ³ = 4 / 3. Solving for the radius of the copper nucleus: R_Cu = 3.6 × (4 / 3) = 1.2 × 4 = 4.8 fm.
Question 10: What is the main physical reason why heavy water (D₂O) is widely used as a moderator in nuclear reactors?
Answer: It slows down fast neutrons without absorbing them.
Explanation: In a nuclear reactor, fission releases fast-moving neutrons. However, to efficiently trigger further fission events in U-235, these neutrons need to be slowed down to thermal speeds. Heavy water serves as an excellent moderator because its deuterium atoms have a low mass profile, allowing them to slow neutrons down through elastic collisions without absorbing them.
Question 11: Two radioactive materials X₁ and X₂ have decay constants 5λ and λ respectively. If they start with the same number of nuclei at t = 0, find the time t when the ratio of their remaining nuclei N₁/N₂ becomes equal to 1/e².
Answer: 1 / (2λ)
Explanation: Using the radioactive decay law, the number of remaining nuclei over time is N(t) = N₀e^(-λt). For the two materials, we can write N₁ = N₀e^(-5λt) and N₂ = N₀e^(-λt). Taking the ratio of these populations: N₁ / N₂ = e^(-5λt) / e^(-λt) = e^(-4λt). We are given that this ratio is equal to 1/e² = e⁻². Setting the exponents equal to each other: -4λt = -2, which simplifies to t = 2 / 4λ = 1 / (2λ).
Question 12: Calculate the energy released in MeV when a single mass defect of Δm = 0.2 u occurs during a nuclear reaction.
Answer: 186.3 MeV
Explanation: We can find the energy equivalent of a mass defect expressed in atomic mass units (u) by multiplying it by the standard conversion factor: E = Δm × 931.5 MeV. Substituting the given mass defect: E = 0.2 × 931.5 MeV = 186.3 MeV.
Question 13: If the binding energy per nucleon for a stable nucleus is relatively high, what does this tell us about the stability of the nucleus?
Answer: The nucleus is highly stable against breaking apart.
Explanation: Binding energy per nucleon ($BE/A$) measures how tightly bound the individual protons and neutrons are within a nucleus. A higher value means more energy is required to break the nucleus apart into its components, which corresponds to greater structural stability.
Question 14: A radioactive sample has a half-life of 10 hours. What percentage of the initial radioactive core population will remain undecayed after a total duration of 30 hours?
Answer: 12.5%
Explanation: First, let us determine the number of half-lives that pass during the 30-hour period: n = Total Time / Half-Life = 30 hours / 10 hours = 3 cycles. The fraction of nuclei remaining after n cycles is given by N/N₀ = (1/2)^n. Substituting n = 3: N/N₀ = (1/2)³ = 1/8. To express this fraction as a percentage: Percentage remaining = (1/8) × 100% = 12.5%.
Question 15: Identify the missing particle X in the following nuclear transmutation reaction: ¹&sup4;₇N + ⁴₂He → ¹&sup7;₈O + X.
Answer: ¹₁H (Proton)
Explanation: To find the identity of particle X, we apply the laws of conservation of mass number and atomic number. First, balancing the mass numbers across the reaction: 14 + 4 = 17 + A_x, which simplifies to 18 = 17 + A_x, giving A_x = 1. Next, balancing the atomic numbers: 7 + 2 = 8 + Z_x, which simplifies to 9 = 8 + Z_x, giving Z_x = 1. A particle with a mass number of 1 and an atomic number of 1 is a proton, represented as ¹₁H.
Question 16: If the mean life of an unstable isotope is exactly 100 seconds, calculate the half-life of this isotope.
Answer: 69.3 seconds
Explanation: The relationship between the half-life and the mean life of a radioactive substance is given by the formula T₁_₂ = ln(2) × τ = 0.693 × τ. Substituting the given mean life of 100 seconds into the formula: T₁_₂ = 0.693 × 100 seconds = 69.3 seconds.
Question 17: Calculate the approximate total number of neutrons present in a 56-gram sample of pure Iron (&sup5;&sup6;₂₆Fe).
Answer: 1.806 × 10²⁵ neutrons
Explanation: First, let us find the number of neutrons in a single atom of Iron. The number of neutrons is the mass number minus the atomic number: N = A - Z = 56 - 26 = 30 neutrons per atom. Next, we determine how many iron atoms are in the 56-gram sample. Since the molar mass of iron is 56 g/mol, a 56-gram sample contains exactly 1 mole of iron. The number of atoms is given by Avogadro's number: Atoms = 6.022 × 10²³ atoms. Finally, the total number of neutrons in the sample is: Total Neutrons = 30 neutrons/atom × 6.022 × 10²³ atoms = 1.8066 × 10²⁵ neutrons.
Question 18: A nuclear fission reaction releases energy because the binding energy per nucleon of the resulting product fragments is:
Answer: Higher than that of the original parent heavy nucleus.
Explanation: When a heavy parent nucleus undergoes fission and splits into medium-sized fragment products, the nucleons become more tightly bound. This causes the binding energy per nucleon to increase, moving up the stability curve. The creation of this more tightly bound, lower-energy state releases the excess energy.
Question 19: The activity of a radioactive sample is measured as R₁ at time t₁ and R₂ at a later time t₂. Write the formula for the decay constant λ in terms of these activities and times.
Answer: λ = ln(R₁ / R₂) / (t₂ - t₁)
Explanation: According to the radioactive decay law, activity over time follows the equation R(t) = R₀e^(-λt). We can write the activities at the two times as R₁ = R₀e^(-λt₁) and R₂ = R₀e^(-λt₂). Taking the ratio of these two activities: R₁ / R₂ = e^(-λt₁) / e^(-λt₂) = e^(λ(t₂ - t₁)). Taking the natural logarithm of both sides: ln(R₁ / R₂) = λ(t₂ - t₁). Solving for the decay constant λ yields: λ = ln(R₁ / R₂) / (t₂ - t₁).
Question 20: Calculate the atomic mass defect of an isotope if its total binding energy is 186.3 MeV.
Answer: 0.2 u
Explanation: The relationship between binding energy and mass defect is given by the formula BE = Δm × 931.5 MeV. We can rearrange this formula to solve for the mass defect Δm: Δm = BE / 931.5 MeV. Substituting the given binding energy: Δm = 186.3 MeV / 931.5 MeV = 0.2 u.
Question 21: An alpha particle approaches a gold nucleus. At its point of closest approach, the total initial kinetic energy of the alpha particle is entirely converted into:
Answer: Electrostatic Potential Energy
Explanation: As a positively charged alpha particle moves toward a positively charged target nucleus, it experiences a repulsive Coulomb force that slows it down. At the point of closest approach, the alpha particle momentarily stops, meaning its kinetic energy drops to zero. At this exact instant, all of its initial kinetic energy has been converted into electrostatic potential energy.
Question 22: If the nuclear radius of a nucleus with mass number A is R, find the slope of the line when graphing log(R) against log(A).
Answer: 1/3
Explanation: The relationship between nuclear radius and mass number is given by the formula R = R₀A¹ᐟ³. Taking the logarithm of both sides of the equation: log(R) = log(R₀A¹ᐟ³) = log(R₀) + log(A¹ᐟ³) = log(R₀) + (1/3)log(A). This matches the equation of a straight line, y = mx + c, where y = log(R), x = log(A), and the constant intercept is c = log(R₀). The slope m of the line is 1/3.
Question 23: A radioactive sample has an initial population of N₀ nuclei. Calculate the number of nuclei that will remain undecayed after a time interval equal to four half-lives.
Answer: N₀ / 16
Explanation: The number of nuclei remaining after n half-lives can be found using the formula N = N₀(1/2)^n. Given that n = 4 half-lives have passed, we substitute this value into the formula: N = N₀ × (1/2)⁴ = N₀ × (1 / 16) = N₀ / 16.
Question 24: Which fundamental force is responsible for holding protons and neutrons together inside an atomic nucleus against electrostatic repulsion?
Answer: Strong Nuclear Force
Explanation: Protons are positively charged and exert a strong electrostatic repulsive force on one another that tries to tear the nucleus apart. The strong nuclear force is a short-range fundamental force that acts between nucleons (protons and neutrons) to overcome this repulsion and hold the nucleus together. It is highly attractive at typical nuclear distances (~1 to 2 fm).
Question 25: If a radioactive substance undergoes beta-minus (β⁻) decay, what happens to the atomic number (Z) and mass number (A) of the daughter nucleus?
Answer: Z increases by 1, A remains unchanged.
Explanation: During beta-minus decay, a neutron inside the parent nucleus transforms into a proton, emitting an electron (the beta particle) and an antineutrino. Because a neutron is replaced by a proton, the total number of nucleons (mass number A) stays exactly the same, while the number of protons (atomic number Z) increases by 1.
NEET Past Year Questions
Question 1: A radioactive sample has an initial mass of 10 mg. If the half-life of the sample is 4 hours, find the mass of the radioactive material that remains active after 12 hours.
Answer: 1.25 mg
Explanation: First, let us calculate the number of half-life cycles that occur during the 12-hour period: n = Total Time / Half-Life = 12 hours / 4 hours = 3 cycles. The mass remaining after n cycles is given by the formula m = m₀(1/2)^n. Substituting the initial mass and the number of cycles: m = 10 mg × (1/2)³ = 10 mg × (1/8) = 1.25 mg.
Question 2: Two nuclei have mass numbers in the ratio 1 : 27. What is the ratio of their nuclear densities?
Answer: 1 : 1
Explanation: The density of nuclear matter is defined as ρ = Mass / Volume. The mass of a nucleus is approximately A × m, and its volume is given by V = (4/3)πR³ = (4/3)π(R₀A¹ᐟ³)³ = (4/3)πR₀³A. Combining these gives the density formula ρ = (A × m) / ((4/3)πR₀³A) = m / ((4/3)πR₀³). Notice that the mass number A cancels out completely. This shows that nuclear density is a constant value for all stable nuclei, meaning the ratio is always 1 : 1.
Question 3: If the binding energy per nucleon for a ²&sup7;₁₃Al nucleus is 8.1 MeV, calculate the total binding energy of the nucleus in MeV.
Answer: 218.7 MeV
Explanation: The mass number of the Aluminum nucleus is A = 27, which means it contains 27 nucleons. The total binding energy is found by multiplying the binding energy per nucleon by the total number of nucleons: BE_total = BE_per_nucleon × A = 8.1 MeV × 27 = 218.7 MeV.
Question 4: During an alpha decay event, the mass number (A) and atomic number (Z) of the parent nucleus change such that:
Answer: A decreases by 4, Z decreases by 2.
Explanation: An alpha particle is identical to a Helium nucleus, represented as ⁴₂He. When an unstable parent nucleus undergoes alpha decay, it ejects this particle from its core. To conserve total nucleon and proton numbers, the mass number A of the daughter nucleus must decrease by 4, and its atomic number Z must decrease by 2.
Question 5: The decay constant λ of a radioactive sample is 0.1 sec⁻¹. Calculate the half-life of this sample.
Answer: 6.93 seconds
Explanation: The half-life of a radioactive material is related to its decay constant by the formula T₁_₂ = ln(2) / λ = 0.693 / λ. Substituting the given decay constant: T₁_₂ = 0.693 / 0.1 sec⁻¹ = 6.93 seconds.
Question 6: If the mass defect of a nuclear reaction is exactly 0.01 u, calculate the energy released in MeV.
Answer: 9.315 MeV
Explanation: We can convert a mass defect expressed in atomic mass units (u) directly into energy using the standard conversion factor: E = Δm × 931.5 MeV. Substituting the given mass defect: E = 0.01 × 931.5 MeV = 9.315 MeV.
Question 7: A radioactive sample has an initial activity of 3200 Bq. If its activity drops to 100 Bq after a duration of 25 days, calculate the half-life of the sample.
Answer: 5 days
Explanation: The fraction of activity remaining can be expressed using the formula R/R₀ = (1/2)^n, where n is the number of half-lives. Substituting our given activity values: 100 / 3200 = 1 / 32 = (1/2)⁵. Comparing the exponents shows that n = 5 half-lives have passed over the 25-day period. The length of a single half-life is: T₁_₂ = Total Time / n = 25 days / 5 = 5 days.
Question 8: Calculate the total number of protons and neutrons contained inside a nucleus of ²³&sup5;₉₂U.
Answer: 92 protons and 143 neutrons
Explanation: The atomic number Z represents the number of protons, which is given directly by the lower subscript: Z = 92 protons. The mass number A represents the total number of nucleons, given by the upper superscript: A = 235. The number of neutrons N is found by subtracting the atomic number from the mass number: N = A - Z = 235 - 92 = 143 neutrons.
Question 9: For a radioactive material sample, what is the exact mathematical relation between the mean life τ and the half-life T₁_₂?
Answer: τ = 1.44 × T₁_₂
Explanation: The formulas for half-life and mean life are T₁_₂ = 0.693 / λ and τ = 1 / λ. We can substitute the definition of λ from the mean life formula into the half-life equation to get: T₁_₂ = 0.693 × τ. Rearranging this to express mean life in terms of half-life gives: τ = T₁_₂ / 0.693 ≈ 1.44 × T₁_₂.
Question 10: Which type of nuclear radiation has the highest ionizing power among alpha, beta, and gamma emissions?
Answer: Alpha particles
Explanation: Ionizing power depends on the charge and mass of the radiation particle. Alpha particles (⁴₂He²⁺) carry a double positive charge (+2e) and are relatively heavy compared to other emissions. This large charge and mass profile makes them highly effective at stripping electrons from atoms they encounter, giving them the highest ionizing power among the three types of radiation.
Question 11: Find the radius of an atomic nucleus whose mass number A is equal to 8, using the baseline radius constant R₀ = 1.2 fm.
Answer: 2.4 fm
Explanation: Using the nuclear radius formula R = R₀A¹ᐟ³, we substitute the given values: R = 1.2 fm × (8)¹ᐟ³ = 1.2 × 2 = 2.4 fm.
Question 12: If a nucleus undergoes spontaneous fission into smaller fragments, what happens to the average binding energy per nucleon of the system?
Answer: It increases.
Explanation: Fission occurs because a heavy parent nucleus is unstable due to its low binding energy per nucleon. When it splits into smaller, medium-sized fragments, the nucleons rearrange into a more tightly bound configuration. This causes the binding energy per nucleon to increase, creating more stable daughter nuclei and releasing energy.
Question 13: A radioactive element has a half-life of 2 hours. What fraction of the initial sample will decay away completely during a total elapsed time of 6 hours?
Answer: 7/8
Explanation: First, let us find the number of half-lives that pass during the 6-hour period: n = Total Time / Half-Life = 6 hours / 2 hours = 3 cycles. The fraction of the sample that remains undecayed after n cycles is N/N₀ = (1/2)^n = (1/2)³ = 1/8. The question asks for the fraction that has *decayed*, which is the total initial amount minus the remaining fraction: Decayed fraction = 1 - 1/8 = 7/8.
Question 14: Which component of a commercial nuclear power reactor is used to slow down fast neutrons to thermal speeds?
Answer: Moderator
Explanation: Fission reactions release fast neutrons that move too quickly to efficiently trigger subsequent fission events in U-235. A moderator (typically heavy water or graphite) is used to slow these neutrons down through collisions, reducing them to thermal energies where they can be easily captured to sustain the chain reaction.
Question 15: The mass numbers of two separate nuclei are in the ratio 1 : 8. Calculate the ratio of their structural nuclear radii.
Answer: 1 : 2
Explanation: According to the radius formula R = R₀A¹ᐟ³, the radius of a nucleus is proportional to the cube root of its mass number. Therefore, the ratio of their radii is: R₁ / R₂ = (A₁ / A₂)¹ᐟ³ = (1 / 8)¹ᐟ³ = 1 / 2. Thus, the ratio is 1 : 2.
Question 16: If a radioactive isotope has a decay constant λ, calculate the time required for the initial population of active nuclei to decrease to exactly half its value.
Answer: ln(2) / λ
Explanation: The time required for a radioactive sample to decrease to half its initial population is defined as its half-life ($T_{1/2}$). By solving the exponential decay equation N₀/2 = N₀e^(-λt), we find that this time is given by the formula: T₁_₂ = ln(2) / λ ≈ 0.693 / λ.
Question 17: Calculate the total energy released in MeV if a mass defect of exactly 1 gram (μ = 10⁻³ kg) occurs during a reaction. (Use c = 3 × 10⁸ m/s and convert to MeV).
Answer: 5.625 × 10²⁶ MeV
Explanation: First, let us find the energy released in Joules using Einstein's mass-energy equation E = m × c²: E = (10⁻³ kg) × (3 × 10⁸ m/s)² = 10⁻³ × 9 × 10¹⁶ = 9 × 10¹³ Joules. Next, we convert this energy from Joules to Mega-electronvolts (MeV) by dividing by the conversion factor ($1.6 \times 10^{-13}$ J/MeV): E = 9 × 10¹³ J / (1.6 × 10⁻¹³ J/MeV) = 5.625 × 10²⁶ MeV.
Question 18: What structural change happens to a nucleus when it undergoes beta-plus (β⁺) decay?
Answer: A proton transforms into a neutron, decreasing the atomic number Z by 1.
Explanation: During beta-plus decay, a proton inside the parent nucleus transforms into a neutron, emitting a positron (a beta-plus particle) and a neutrino. Because a proton is replaced by a neutron, the total nucleon count (mass number A) stays the same, while the proton count (atomic number Z) decreases by 1.
Question 19: The binding energy per nucleon curve has a prominent peak near which mass number region, indicating maximum stability?
Answer: Near A = 56 (Iron)
Explanation: The binding energy per nucleon curve plots nuclear stability against mass number. The curve rises for light elements and reaches its highest point near mass number A = 56, which corresponds to Iron (⁵⁶Fe) with a value of approximately 8.75 MeV/nucleon. This peak represents the most stable nuclear configuration in the universe.
Question 20: A radioactive sample has an activity of 10 × 10³ Bq. If its half-life is 10 days, find the remaining activity after an elapsed time of 20 days.
Answer: 2.5 × 10³ Bq
Explanation: First, find the number of elapsed half-lives: n = Total Time / Half-Life = 20 days / 10 days = 2 cycles. The activity remaining after n cycles is given by the formula R = R₀(1/2)^n. Substituting our values into the formula: R = 10 × 10³ × (1/2)² = 10 × 10³ × (1/4) = 2.5 × 10³ Bq.
Question 21: Fusion reactions take place naturally inside the cores of stars primarily because of which conditions?
Answer: Extremely high temperature and high pressure.
Explanation: For two light nuclei to fuse, they must come close enough for the short-range strong nuclear force to take effect. However, because they are both positively charged, they experience a powerful electrostatic Coulomb repulsion. Overcoming this barrier requires extremely high temperatures, which provide the nuclei with immense kinetic energy, and high pressures, which increase the collision rate.
Question 22: If the mass of a stable nucleus is compared to the total combined mass of its individual constituent nucleons when separated, the mass of the nucleus is always:
Answer: Strictly less than the sum of the constituent masses.
Explanation: When independent protons and neutrons combine to form a bound nucleus, a small fraction of their mass is converted into energy and released as binding energy. This mass loss is called the mass defect, and it ensures that the total mass of a stable nucleus is always strictly less than the combined mass of its separated components.
Question 23: A radioactive sample with an initial population of N₀ nuclei decays for a time interval equal to one half-life. How many nuclei have decayed away during this period?
Answer: N₀ / 2
Explanation: By definition, the half-life is the time required for half of the radioactive nuclei in a sample to decay. Therefore, after one half-life, the number of nuclei that have decayed away is exactly N₀ / 2, leaving the other half (N₀ / 2) remaining in the sample.
Question 24: What type of radiation emission does not alter either the atomic number (Z) or the mass number (A) of the target nucleus?
Answer: Gamma (γ) emission
Explanation: Gamma radiation consists of high-energy photons, which carry energy but have no mass or charge. When an excited nucleus releases gamma radiation, it transitions from a high-energy state to a more stable, lower-energy state. Because no nucleons are gained or lost, both the atomic number Z and the mass number A remain completely unchanged.
Question 25: Calculate the approximate radius of a nucleus with mass number A = 125, assuming the baseline constant is R₀ = 1.2 fm.
Answer: 6.0 fm
Explanation: Using the nuclear radius formula R = R₀A¹ᐟ³, we substitute the given values: R = 1.2 fm × (125)¹ᐟ³ = 1.2 × 5 = 6.0 fm. Therefore, the radius of the nucleus is 6.0 fm.
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