How Boards Test This Chapter

Nuclei is a compact, formula-driven Board topic worth 4-6 marks. The recurring demands:

  • 1 mark: define mass defect / binding energy / isotopes-isobars-isotones; state a property of the nuclear force; units of activity; why nuclear density is constant.
  • 2 marks: draw the binding-energy-per-nucleon curve and mark its features; show density is independent of A; distinguish fission from fusion; alpha/beta/gamma comparisons; half-life numericals.
  • 3 marks: explain energy release in fission AND fusion from the curve; properties of the nuclear force with the pair-potential graph; Q-value calculations with given masses; the p-p cycle and the Sun's energy.
  • 5 marks: the full binding-energy story — mass defect → EbE_b → the curve → its four conclusions — with a numerical.

The questions below are Board-style previous-year questions with full step-by-step solutions embedded in the explanations. Attempt each before reading its solution. Years are attached only where attribution is certain; otherwise questions are tagged simply [CBSE Board].

The Definitions and Statements Boards Reward (Model Answers)

Mass defect: the difference between the total mass of a nucleus's constituent nucleons (free) and the actual nuclear mass: ΔM=[Zmp+(AZ)mn]M\Delta M = [Zm_p + (A-Z)m_n] - M.

Binding energy: the energy Eb=ΔMc2E_b = \Delta Mc^2 required to separate a nucleus into its free nucleons (equivalently, released when they assemble). Binding energy per nucleon Ebn=Eb/AE_{bn} = E_b/A measures stability.

Nuclear force (three properties): (i) strongest force — dominates Coulomb repulsion inside the nucleus (gravity negligible); (ii) short-range — falls rapidly to zero beyond a few fm, causing saturation and the flat EbnE_{bn} plateau; attractive beyond ~0.8 fm, strongly repulsive within; (iii) charge-independent — n-n, n-p, p-p nuclear forces approximately equal.

Curve features to draw: rise from light nuclei, plateau ~8.0 MeV for 30 < A < 170, maximum ~8.75 MeV at A = 56, fall to 7.6 MeV at A = 238; mark fission (right → middle) and fusion (left → middle) arrows.

Q value: Q = final KE - initial KE = (initial masses - final masses)c². Q > 0: exothermic.

Radioactive decay law [JEE/NEET-retained]: N=N0eλtN = N_0e^{-\lambda t}; T1/2=0.693/λT_{1/2} = 0.693/\lambda; τ=1/λ\tau = 1/\lambda; activity R=λNR = \lambda N (Bq).

[Board Tip] Marks leak at: drawing the curve without axis labels or the A = 56 peak; 'mass defect' stated without the formula; forgetting that BOTH fission and fusion follow from the same curve; and confusing R0R_0 = 1.2 fm (radius constant) with r0r_0 = 0.8 fm (potential minimum).

Board PYQ Set A: Short Answer (1-2 marks)

PYQ 1. Define mass defect and binding energy of a nucleus. How are they related? [CBSE Board]

Solution:

  1. Mass defect: ΔM=[Zmp+(AZ)mn]M\Delta M = [Zm_p + (A-Z)m_n] - M — constituents' mass minus nuclear mass.
  2. Binding energy: the energy needed to break the nucleus into free nucleons.
  3. Relation: Eb=ΔMc2E_b = \Delta Mc^2 (= ΔM×931.5\Delta M \times 931.5 MeV with ΔM\Delta M in u).

PYQ 2. Why is the density of a nucleus independent of its mass number? [CBSE Board]

Solution:

  1. R=R0A1/3R = R_0A^{1/3} → volume =43πR03AA= \frac{4}{3}\pi R_0^3A \propto A.
  2. Mass ≈ A × (nucleon mass) ∝ A.
  3. Density = mass/volume — A cancels: ρ=3m4πR032.3×1017\rho = \dfrac{3m}{4\pi R_0^3} \approx 2.3 \times 10^{17} kg/m³ for every nucleus.

PYQ 3. State two properties of the nuclear force. [CBSE Board]

Solution:

  1. It is short-ranged (a few fm) and saturating — hence the constant binding energy per nucleon.
  2. It is charge-independent — approximately equal for n-n, n-p and p-p pairs (and much stronger than the Coulomb force).

PYQ 4. A nucleus ZAX^{A}_{Z}X emits one alpha and one beta-minus particle. Write the final nuclide. [CBSE Board]

Solution:

  1. Alpha: (A, Z) → (A - 4, Z - 2).
  2. Beta-minus: (A - 4, Z - 2) → (A - 4, Z - 1).
  3. Final: Z1A4Y^{A-4}_{Z-1}Y.

PYQ 5. Draw the binding energy per nucleon versus mass number curve and mark its main features. What conclusions about fission and fusion follow? [CBSE Board]

Solution:

  1. Sketch: steep rise for light A (spikes at 4^4He, 16^{16}O), plateau ~8.0 MeV over 30 < A < 170, peak ~8.75 MeV at A = 56, gentle fall to 7.6 MeV at A = 238. Axes: EbnE_{bn} (MeV) vs A.
  2. Fission: heavy nuclei (right end, lower EbnE_{bn}) splitting into middle-mass fragments (higher EbnE_{bn}) release energy.
  3. Fusion: very light nuclei fusing into heavier ones climb the left flank — energy released.
  4. Both processes move nucleons toward the iron peak — tighter binding, energy out.

PYQ 6. Distinguish between nuclear fission and fusion, giving one example of each. [CBSE Board]

Solution:

  1. Fission: a heavy nucleus splits into intermediate-mass fragments; e.g. n+235U144Ba+89Kr+3nn + {}^{235}U \to {}^{144}Ba + {}^{89}Kr + 3n (~200 MeV). Occurs at ordinary temperatures (neutron-triggered).
  2. Fusion: light nuclei combine into a heavier nucleus; e.g. 2H+2H3He+n^2H + {}^2H \to {}^3He + n (3.27 MeV). Requires ~10710^7-10910^9 K to beat the Coulomb barrier (thermonuclear).
  3. Per kilogram, fusion releases several times more energy; both are curve-downhill processes.

PYQ 7. Why is energy released in BOTH fission and fusion, though they are opposite processes? [CBSE Board]

Solution:

  1. Energy release requires only that products be more tightly bound (higher EbnE_{bn}) than reactants.
  2. The curve peaks in the MIDDLE (A ≈ 56): heavy nuclei get tighter by splitting toward it, light nuclei by fusing toward it.
  3. Opposite directions on the A-axis, same direction on the binding axis — uphill in EbnE_{bn} both ways.

Board PYQ Set B: Standard Numericals (2-3 marks)

PYQ 8. Calculate the binding energy per nucleon of 2656^{56}_{26}Fe. Given m = 55.934939 u, mHm_H = 1.007825 u, mnm_n = 1.008665 u. [CBSE Board]

Solution:

  1. ΔM=[26×1.007825+30×1.008665]55.934939=0.528461\Delta M = [26 \times 1.007825 + 30 \times 1.008665] - 55.934939 = 0.528461 u.
  2. Eb=0.528461×931.5=492.3E_b = 0.528461 \times 931.5 = 492.3 MeV.
  3. Ebn=492.3/56=8.79E_{bn} = 492.3/56 = 8.79 MeV/nucleon — essentially the curve's peak.

PYQ 9. The radii ratio of two nuclei is 2 : 3. Find the ratio of their mass numbers and of their densities. [CBSE Board]

Solution:

  1. AR3A \propto R^3: ratio =8:27= 8 : 27.
  2. Densities: 1 : 1 — nuclear density is A-independent.

PYQ 10. A radioactive sample has half-life 30 s. Find (a) its decay constant, (b) the time for the sample to decay to 1/16 of its initial amount. [CBSE Board]

Solution:

  1. (a) λ=0.69330=0.0231\lambda = \dfrac{0.693}{30} = 0.0231 s⁻¹.
  2. (b) 1/16=(1/2)41/16 = (1/2)^4 → 4 half-lives → t = 120 s.

PYQ 11. Determine the Q value of the reaction 1H+3H2H+2H^{1}H + {}^{3}H \to {}^{2}H + {}^{2}H, given the masses 1.007825, 3.016049, 2.014102 u. Is it exo- or endothermic? [CBSE Board]

Solution:

  1. Q=[1.007825+3.0160492×2.014102]×931.5Q = [1.007825 + 3.016049 - 2 \times 2.014102] \times 931.5.
  2. =0.004330×931.54.03= -0.004330 \times 931.5 \approx -4.03 MeV.
  3. Q < 0 → endothermic: 4.03 MeV must be supplied (NCERT Exercise 13.5i).

PYQ 12. How much energy is released when 1 g of matter is fully converted? Compare with the daily output (~101410^{14} J) of a large power station. [CBSE Board]

Solution:

  1. E=mc2=103×9×1016=9×1013E = mc^2 = 10^{-3} \times 9 \times 10^{16} = 9 \times 10^{13} J.
  2. Comparable to the power station's full day — one gram equals a day of grid-scale generation.

Board PYQ Set C: Long-Answer Patterns (3-5 marks)

PYQ 13. Explain, with the binding-energy curve, how the constancy of EbnE_{bn} over 30 < A < 170 follows from the short range of the nuclear force. [CBSE Board]

Solution:

  1. A nucleon in a large nucleus interacts only with neighbours within the force's few-fm range — at most p of them.
  2. Its binding ≈ pk, independent of total A; adding remote nucleons changes nothing for it.
  3. Interior nucleons dominate, so the average EbnE_{bn} ≈ pk — the plateau. This neighbour-only behaviour is the saturation property.

PYQ 14. Describe the proton-proton cycle. Show that its net effect releases 26.7 MeV, and explain why such fusion requires very high temperature. [CBSE Board]

Solution:

  1. Steps: (i) p+pd+e++νp + p \to d + e^+ + \nu (0.42 MeV); (ii) e++e2γe^+ + e^- \to 2\gamma (1.02 MeV); (iii) d+p3He+γd + p \to {}^3He + \gamma (5.49 MeV); (iv) 3He+3He4He+2p^3He + {}^3He \to {}^4He + 2p (12.86 MeV).
  2. Net (2i + 2ii + 2iii + iv): 4p+2e4He+2ν+6γ4p + 2e^- \to {}^4He + 2\nu + 6\gamma; energy = 0.84 + 2.04 + 10.98 + 12.86 = 26.7 MeV.
  3. Temperature: the positive nuclei must beat a ~400 keV Coulomb barrier; only at ~10710^7-10910^9 K do enough particles have such energies (thermonuclear fusion). In the Sun, tail-of-distribution protons do the burning.

PYQ 15. With a labelled diagram of the potential energy of a nucleon pair versus separation, describe the nature of the nuclear force at different distances. [CBSE Board]

Solution:

  1. Sketch: U(r) dipping to a minimum at r00.8r_0 \approx 0.8 fm, rising steeply for r < r0r_0, rising gently toward zero for r > r0r_0.
  2. r > 0.8 fm: force attractive (potential increases with r), dying off beyond a few fm.
  3. r < 0.8 fm: strongly repulsive (hard core) — prevents collapse, fixes nuclear density.
  4. r = 0.8 fm: equilibrium separation — the potential minimum.

PYQ 16. (a) State the law of radioactive decay and derive N=N0eλtN = N_0e^{-\lambda t}. (b) Define half-life and obtain T1/2=0.693/λT_{1/2} = 0.693/\lambda. [CBSE Board]

Solution:

  1. (a) Law: the decay rate is proportional to the number of undecayed nuclei: dNdt=λN\dfrac{dN}{dt} = -\lambda N.
  2. Separate and integrate: dNN=λdt\int \dfrac{dN}{N} = -\lambda\int dtlnNN0=λt\ln\dfrac{N}{N_0} = -\lambda tN=N0eλtN = N_0e^{-\lambda t}. ∎
  3. (b) Half-life: the time for N to fall to N0/2N_0/2. Setting N0/2=N0eλT1/2N_0/2 = N_0e^{-\lambda T_{1/2}}: T1/2=ln2λ=0.693λT_{1/2} = \dfrac{\ln 2}{\lambda} = \dfrac{0.693}{\lambda}. ∎