CHAPTER 13: NUCLEI
SAMPLE QUESTION PAPER (ASSIGNMENT-1)
- There are 33 questions in all. All questions are compulsory.
- Section A: 16 questions (12 MCQs and 4 Assertion-Reasoning) of 1 mark each.
- Section B: 5 short answer questions of 2 marks each.
- Section C: 7 short answer questions of 3 marks each.
- Section D: 2 case study-based questions of 4 marks each.
- Section E: 3 long answer questions of 5 marks each.
- Physical Constants: $\displaystyle {m_p = 1.007825\text{ u}}$, $\displaystyle {m_n = 1.008665\text{ u}}$, $\displaystyle {1\text{ u} = 931.5\text{ MeV}/c^2}$, $\displaystyle {R_0 = 1.2\text{ fm} = 1.2 \times 10^{-15}\text{ m}}$, $\displaystyle {N_A = 6.023 \times 10^{23}\text{ mol}^{-1}}$.
SECTION A (16 Marks)
Q.1. Two lithium nuclei in lithium vapor at room temperature do not combine to form a carbon nucleus because:
Explanation: Positively charged lithium nuclei ($\displaystyle {Z=3}$) experience strong mutual electrostatic Coulomb repulsion that prevents them from approaching within the short range of attractive nuclear forces at room temperature thermal energies.
Q.2. The curve of binding energy per nucleon as a function of atomic mass number shows a sharp peak for the helium nucleus ($\displaystyle {^4_2\text{He}}$). This implies that the helium nucleus is:
Explanation: A local maximum peak in the binding energy per nucleon curve indicates exceptionally high binding energy per nucleon ($\displaystyle {\approx 7.07\text{ MeV/nucleon}}$), rendering $\displaystyle {^4_2\text{He}}$ far more tightly bound and stable than adjacent light nuclei.
Q.3. Which of the following statements about nuclear forces is NOT true?
Explanation: Statement (b) is false because the strong nuclear force is the strongest fundamental force in nature, approximately 100 times stronger than the electrostatic Coulomb force within nuclear dimensions ($\displaystyle {r < 1.4\text{ fm}}$).
Q.4. Heavy stable nuclei contain more neutrons than protons ($\displaystyle {N > Z}$). This is primarily because:
Explanation: As proton count $\displaystyle {Z}$ increases, long-range disruptive Coulomb repulsion grows as $\displaystyle {Z^2}$. Additional neutrons provide extra attractive strong nuclear force without adding disruptive electrical repulsion, thereby stabilizing the heavy nucleus.
Q.5. The atomic mass unit ($\displaystyle {1\text{ u}}$) is defined as:
Explanation: By international standard, $\displaystyle {1\text{ u} = \frac{1}{12} \times \text{mass of one } ^{12}_6\text{C} \text{ atom} \approx 1.660539 \times 10^{-27}\text{ kg}}$.
Q.6. The ratio of the radius of an atom ($\displaystyle {\sim 10^{-10}\text{ m}}$) to the radius of its nucleus ($\displaystyle {\sim 10^{-15}\text{ m}}$) is of the order of:
Explanation: $\displaystyle {\frac{R_{\text{atom}}}{R_{\text{nucleus}}} = \frac{10^{-10}\text{ m}}{10^{-15}\text{ m}} = 10^5}$. The atomic radius is 100,000 times larger than the nuclear radius.
Q.7. The number of neutrons in a polonium nucleus $\displaystyle {^{218}_{84}\text{Po}}$ is:
Explanation: Number of neutrons $\displaystyle {N = A - Z = 218 - 84 = 134}$.
Q.8. If the nuclear radius of $\displaystyle {^{27}_{13}\text{Al}}$ is $\displaystyle {3.6\text{ fm}}$, the approximate nuclear radius of $\displaystyle {^{64}_{29}\text{Cu}}$ in $\displaystyle {\text{fm}}$ is:
Explanation: Since $\displaystyle {R \propto A^{1/3}}$, $\displaystyle {\frac{R_{\text{Cu}}}{R_{\text{Al}}} = \left(\frac{64}{27}\right)^{1/3} = \frac{4}{3} \implies R_{\text{Cu}} = 3.6 \times \frac{4}{3} = 4.8\text{ fm}}$.
Q.9. The energy equivalent of $\displaystyle {2\text{ g}}$ of a substance converted completely into energy is:
Explanation: Using Einstein's equation $\displaystyle {E = m c^2 = (2 \times 10^{-3}\text{ kg}) \times (3 \times 10^8\text{ m/s})^2 = 2 \times 10^{-3} \times 9 \times 10^{16} = 1.8 \times 10^{14}\text{ J} = 18 \times 10^{13}\text{ J}}$.
Q.10. Given mass of $\displaystyle {^{56}_{26}\text{Fe} = 55.934939\text{ u}}$, $\displaystyle {m_p = 1.007825\text{ u}}$, and $\displaystyle {m_n = 1.008665\text{ u}}$. The binding energy per nucleon of $\displaystyle {^{56}_{26}\text{Fe}}$ is approximately:
Explanation: Mass defect $\displaystyle {\Delta m = 26(1.007825) + 30(1.008665) - 55.934939 = 26.20345 + 30.25995 - 55.934939 = 0.528461\text{ u}}$.
Total $\displaystyle {BE = 0.528461 \times 931.5\text{ MeV} \approx 492.26\text{ MeV}}$. Binding energy per nucleon $\displaystyle {BE/A = 492.26 / 56 \approx 8.790\text{ MeV/nucleon}}$.
Q.11. How much mass has to be converted into energy per second to produce electrical power of $\displaystyle {200\text{ MW}}$ for 1 hour?
Explanation: Total energy $\displaystyle {E = P \times t = (200 \times 10^6\text{ W}) \times 3600\text{ s} = 7.2 \times 10^{11}\text{ J}}$.
Mass converted $\displaystyle {m = \frac{E}{c^2} = \frac{7.2 \times 10^{11}}{(3 \times 10^8)^2} = \frac{7.2 \times 10^{11}}{9 \times 10^{16}} = 8 \times 10^{-6}\text{ kg}}$.
Q.12. The fundamental physical nature of the strong nuclear force is:
Explanation: Strong nuclear force is an independent short-range non-central force, entirely charge-independent and non-electromagnetic in nature.
Directions for Q.13 to Q.16:
(a) Both Assertion and Reason are correct and Reason is correct explanation of Assertion.
(b) Both Assertion and Reason are correct but Reason is NOT correct explanation of Assertion.
(c) Assertion is correct, Reason is incorrect.
(d) Both Assertion and Reason are incorrect.
Q.13. Assertion (A): Nuclear density is extremely large compared to ordinary atomic matter.
Reason (R): Nuclear density increases continuously with an increase in mass number $\displaystyle {A}$.
Explanation: Assertion is true ($\displaystyle {\rho_{\text{nucleus}} \sim 2.3 \times 10^{17}\text{ kg/m}^3}$ vs water $\displaystyle {10^3\text{ kg/m}^3}$). Reason is false because nuclear density is completely constant and independent of mass number $\displaystyle {A}$ ($\displaystyle {\rho \propto A/A^1 = \text{constant}}$).
Q.14. Assertion (A): The nuclear force between $\displaystyle {n-n}$, $\displaystyle {p-n}$, and $\displaystyle {p-p}$ is approximately of equal strength.
Reason (R): The strong nuclear force is independent of electric charge.
Q.15. Assertion (A): Stable light nuclei contain approximately equal numbers of protons and neutrons ($\displaystyle {N \approx Z}$).
Reason (R): Nuclear stability is determined by the neutron-to-proton ratio.
Q.16. Assertion (A): Total energy released per unit mass in nuclear fusion is much greater than in nuclear fission.
Reason (R): Thermonuclear fusion serves as the ultimate energy source powering stars like the Sun.
Explanation: Fusion yields $\displaystyle {\sim 6.5\text{ MeV/nucleon}}$ (e.g. $\displaystyle {^1\text{H} \to ^4\text{He}}$) compared to fission's $\displaystyle {\sim 0.85\text{ MeV/nucleon}}$ (e.g. $\displaystyle {^{235}\text{U}}$), making fusion more energetic per unit mass. Both statements are true, but solar energy origin does not explain the nuclear binding curve energetics.
SECTION B (10 Marks)
Q.17. What is the nuclear radius of $\displaystyle {^{125}_{56}\text{Fe}}$ if the nuclear radius of $\displaystyle {^{27}_{13}\text{Al}}$ is $\displaystyle {3.6\text{ fm}}$?
$\displaystyle {\frac{R_{\text{Fe}}}{R_{\text{Al}}} = \left(\frac{A_{\text{Fe}}}{A_{\text{Al}}}\right)^{1/3} = \left(\frac{125}{27}\right)^{1/3} = \frac{5}{3}}$.
$\displaystyle {R_{\text{Fe}} = 3.6 \text{ fm} \times \frac{5}{3} = 6.0\text{ fm}}$.
Q.18. Draw a graph showing the variation of potential energy between a pair of nucleons as a function of their separation distance $\displaystyle {r}$. Mark the regions where potential energy is (i) positive (repulsive) and (ii) negative (attractive).
1. For separation $\displaystyle {r > 0.8\text{ fm}}$, potential energy $\displaystyle {U < 0}$ (negative), indicating an attractive nuclear force.
2. For separation $\displaystyle {r < 0.8\text{ fm}}$, potential energy rises steeply above zero ($\displaystyle {U > 0}$), indicating a strong repulsive nuclear core that prevents nucleons from collapsing into one another.
3. At $\displaystyle {r = r_0 \approx 0.8\text{ fm}}$, potential energy reaches its minimum ($\displaystyle {\approx -100\text{ MeV}}$), where nucleons exist in stable equilibrium.
Q.19. Explain the process of nuclear fusion using the plot of binding energy per nucleon ($\displaystyle {BE/A}$) versus mass number $\displaystyle {A}$.
Q.20. Find the ratio of the radii of two nuclei with mass numbers 27 and 64 respectively.
Q.21. Calculate how many times nuclear matter is denser than liquid water ($\displaystyle {\rho_{\text{water}} = 10^3\text{ kg/m}^3}$). Take nucleon mass $\displaystyle {m_N = 1.67 \times 10^{-27}\text{ kg}}$ and $\displaystyle {R_0 = 1.2 \times 10^{-15}\text{ m}}$.
Nuclear Density: $\displaystyle {\rho_{\text{nuclear}} = \frac{m_N}{\frac{4}{3}\pi R_0^3} = \frac{3 \times (1.67 \times 10^{-27})}{4 \pi \times (1.2 \times 10^{-15})^3} = \frac{5.01 \times 10^{-27}}{4 \times 3.1416 \times 1.728 \times 10^{-45}} \approx 2.31 \times 10^{17}\text{ kg/m}^3}$.
Ratio to water density: $\displaystyle {\frac{\rho_{\text{nuclear}}}{\rho_{\text{water}}} = \frac{2.31 \times 10^{17}}{10^3} = 2.31 \times 10^{14}}$ times denser.
SECTION C (21 Marks)
Q.22. Find the binding energy per nucleon of a $\displaystyle {^{40}_{20}\text{Ca}}$ nucleus. Given mass of $\displaystyle {^{40}_{20}\text{Ca} = 39.962589\text{ u}}$, $\displaystyle {m_n = 1.008665\text{ u}}$, and $\displaystyle {m_p = 1.007825\text{ u}}$.
Mass defect $\displaystyle {\Delta m = [20(1.007825) + 20(1.008665)] - 39.962589 = [20.1565 + 20.1733] - 39.962589 = 40.3298 - 39.962589 = 0.367211\text{ u}}$.
Total Binding Energy: $\displaystyle {BE = 0.367211 \times 931.5\text{ MeV} \approx 341.87\text{ MeV}}$.
Binding Energy per nucleon: $\displaystyle {BE/A = \frac{341.87\text{ MeV}}{40} = 8.547\text{ MeV/nucleon}}$.
Q.23. How long can an electric lamp of $\displaystyle {100\text{ W}}$ be kept glowing by the fusion of $\displaystyle {2\text{ kg}}$ of deuterium ($\displaystyle {^2_1\text{H}}$)? Take fusion reaction $\displaystyle {^2_1\text{H} + ^2_1\text{H} \to ^3_2\text{He} + n + 3.27\text{ MeV}}$.
$\displaystyle {N = \frac{2000}{2} \times 6.023 \times 10^{23} = 6.023 \times 10^{26} \text{ nuclei}}$.
Since 2 deuterium nuclei release $\displaystyle {3.27\text{ MeV}}$, total fusion energy released:
$\displaystyle {E_{\text{total}} = \frac{N}{2} \times 3.27\text{ MeV} = \frac{6.023 \times 10^{26}}{2} \times 3.27 \times 1.6 \times 10^{-13}\text{ J} \approx 1.575 \times 10^{14}\text{ J}}$.
Time lamp glows: $\displaystyle {t = \frac{E_{\text{total}}}{P} = \frac{1.575 \times 10^{14}\text{ J}}{100\text{ W}} = 1.575 \times 10^{12}\text{ s} \approx 4.99 \times 10^4 \text{ years}}$.
Q.24. Four light nuclei undergo fusion to form a single heavier daughter nucleus with the release of energy. Which of the two—parent or daughter—has a higher binding energy per nucleon? Justify.
Q.25. Distinguish between nuclear fission and nuclear fusion. Explain the energy release in both processes on the basis of the binding energy per nucleon curve.
1. Fission vs Fusion: Fission splits a heavy unstable nucleus ($\displaystyle {A > 230}$) into medium fragments ($\displaystyle {A \approx 120}$). Fusion combines light nuclei ($\displaystyle {A \le 10}$) into a heavier stable nucleus.
2. Fission Basis: Heavy nuclei have lower $\displaystyle {BE/A}$ ($\displaystyle {\sim 7.6\text{ MeV/nucleon}}$) due to Coulomb repulsion. When split into intermediate nuclei with higher $\displaystyle {BE/A}$ ($\displaystyle {\sim 8.5\text{ MeV/nucleon}}$), mass defect appears as released kinetic energy ($\displaystyle {\sim 200\text{ MeV}}$).
3. Fusion Basis: Very light nuclei have low $\displaystyle {BE/A}$ ($\displaystyle {< 3\text{ MeV/nucleon}}$). Combining them shifts the product higher on the curve ($\displaystyle {^4_2\text{He} \approx 7.07\text{ MeV/nucleon}}$), liberating energy.
Q.26. How is the size of a nucleus found experimentally? Derive the relation between nuclear radius and mass number, and prove that nuclear density is independent of mass number $\displaystyle {A}$.
1. Experimental Measurement: Nuclear size is measured by high-energy electron scattering or Rutherford $\alpha$-particle scattering experiments.
2. Radius Formula: Experiments show nuclear volume is proportional to nucleon count: $\displaystyle {V = \frac{4}{3}\pi R^3 \propto A \implies R = R_0 A^{1/3}}$, where $\displaystyle {R_0 \approx 1.2\text{ fm}}$.
3. Density Proof: Nuclear Mass $\displaystyle {M \approx A \cdot m_N}$. Volume $\displaystyle {V = \frac{4}{3}\pi R_0^3 A}$.
$\displaystyle {\rho_{\text{nuclear}} = \frac{M}{V} = \frac{A \cdot m_N}{\frac{4}{3}\pi R_0^3 A} = \frac{3 m_N}{4\pi R_0^3} = \text{constant}}$. Since $\displaystyle {A}$ cancels, nuclear density is identical for all nuclei.
Q.27. Calculate the binding energy of an $\alpha$-particle ($\displaystyle {^4_2\text{He}}$) in $\displaystyle {\text{MeV}}$. Given $\displaystyle {m_p = 1.007825\text{ u}}$, $\displaystyle {m_n = 1.008665\text{ u}}$, mass of $\displaystyle {^4_2\text{He} = 4.002800\text{ u}}$, and $\displaystyle {1\text{ u} = 931.5\text{ MeV/c}^2}$.
$\displaystyle {\Delta m = [2(1.007825) + 2(1.008665)] - 4.002800 = [2.01565 + 2.01733] - 4.002800 = 4.03298 - 4.002800 = 0.03018\text{ u}}$.
Binding Energy $\displaystyle {BE = 0.03018\text{ u} \times 931.5\text{ MeV/u} \approx 28.11\text{ MeV}}$ ($\displaystyle {\approx 28.098\text{ MeV}}$).
Q.28. If both proton and neutron numbers are conserved in a nuclear reaction like $\displaystyle {^{12}_6\text{C} + ^{12}_6\text{C} \to ^{20}_{10}\text{Ne} + ^4_2\text{He}}$, explain how mass is converted into energy.
SECTION D: CASE STUDY QUESTIONS (8 Marks)
Q.29. Case Study 1: Fundamental Forces & Electroweak Unification.
There are four fundamental forces in nature: Gravitational, Electromagnetic, Weak Nuclear, and Strong Nuclear force. Strong nuclear force holds nucleons together within short ranges ($\displaystyle {< 1.4\text{ fm}}$). Standard Electroweak theory unified weak nuclear and electromagnetic forces.
(i) What are the four fundamental forces of nature?
(a) Gravitational, Electromagnetic, Nuclear, Tension
(b) Gravitational, Electromagnetic, Weak Nuclear, Strong Nuclear
(c) Friction, Electric, Nuclear, Magnetic
(ii) Which fundamental force is strictly attractive at all distances?
(a) Electric (b) Magnetic (c) Gravitational (d) Strong Nuclear
(iii) Which two fundamental forces were unified by Standard Electroweak Theory?
(a) Weak nuclear and Electromagnetic (b) Strong nuclear and Electromagnetic
(iv) Which forces possess infinite range?
(a) Weak and Strong nuclear (b) Gravitational and Electromagnetic
Q.30. Case Study 2: Nuclear Discovery and Mass Properties.
Rutherford discovered the dense central positive nucleus in 1911. Experiments established nuclear radius $\displaystyle {R = R_0 A^{1/3}}$.
(i) The ratio of the mass of a nucleus to the mass of its entire neutral atom is approximately:
(a) 1 (b) $\displaystyle {10^{-3}}$ (c) $\displaystyle {10^{-5}}$
(ii) If $\displaystyle {R}$ is nuclear radius and $\displaystyle {A}$ is mass number, the graph of $\displaystyle {\log R}$ versus $\displaystyle {\log A}$ is:
(a) A straight line with slope $\displaystyle {1/3}$ (b) A parabola (c) An ellipse
(iii) What is the ratio of masses of Hydrogen ($\displaystyle {^1\text{H}}$), Deuterium ($\displaystyle {^2\text{H}}$), and Tritium ($\displaystyle {^3\text{H}}$)?
(a) $\displaystyle {1 : 2 : 3}$ (b) $\displaystyle {1 : 1 : 1}$ (c) $\displaystyle {1 : 4 : 9}$
(iv) What are isotones?
(i) (a) Approximately 1 ($\displaystyle {> 99.9\%}$ of atomic mass is concentrated in the nucleus).
(ii) (a) A straight line ($\displaystyle {\log R = \log R_0 + \frac{1}{3}\log A}$).
(iii) (a) $\displaystyle {1 : 2 : 3}$.
(iv) Isotones are nuclides belonging to different chemical elements ($\displaystyle {\text{different } Z}$) that possess an identical number of neutrons ($\displaystyle {N = A - Z}$).
SECTION E (15 Marks)
Q.31. Define the $\displaystyle {Q}$-value of a nuclear reaction. When can a nuclear process NOT proceed spontaneously? Explain how mass converts into energy in a nuclear reaction even when proton and neutron counts are strictly conserved.
1. $\displaystyle {Q}$-value Definition: $\displaystyle {Q = (M_{\text{reactants}} - M_{\text{products}}) c^2}$. It represents the total net kinetic energy released or absorbed during a nuclear transformation.
2. Spontaneous Condition: A process cannot proceed spontaneously if $\displaystyle {Q < 0}$ (endothermic reaction). An endothermic reaction requires external kinetic energy input to overcome the mass deficit.
3. Mass-Energy Conversion: Nucleon rest masses depend on nuclear binding environment. Product nucleons are more tightly bound with greater binding energy, producing a rest mass defect $\displaystyle {\Delta m}$ that releases kinetic energy $\displaystyle {E = \Delta m \cdot c^2}$.
Q.32. (a) Why is the rest mass of a stable nucleus always less than the sum of rest masses of its constituent free protons and neutrons?
(b) Draw the potential energy curve of two nucleons as a function of separation distance $\displaystyle {r}$ and explain its key features.
Calculate and compare the energy released by (i) fusion of $\displaystyle {1\text{ kg}}$ of hydrogen ($\displaystyle {4\,^1\text{H} \to ^4\text{He} + 26\text{ MeV}}$), and (ii) complete fission of $\displaystyle {1\text{ kg}}$ of $\displaystyle {^{235}_{92}\text{U}}$ ($\displaystyle {200\text{ MeV}}$ per fission event).
(a) When free nucleons coalesce to form a nucleus, short-range attractive nuclear forces do work, causing a decrease in potential energy. This lost binding energy is radiated away, resulting in a rest mass reduction $\displaystyle {\Delta m = BE/c^2}$.
OR Part Calculations:
(i) Hydrogen Fusion ($\displaystyle {1\text{ kg}}$):
Number of nuclei in 1 kg $\displaystyle {= 1000 \times 6.023 \times 10^{23} = 6.023 \times 10^{26} \text{ protons}}$.
Since 4 protons release $\displaystyle {26\text{ MeV} = 4.16 \times 10^{-12}\text{ J}}$:
$\displaystyle {E_{\text{fusion}} = \frac{6.023 \times 10^{26}}{4} \times 4.16 \times 10^{-12}\text{ J} \approx 6.26 \times 10^{14}\text{ J}}$.
(ii) Uranium Fission ($\displaystyle {1\text{ kg}}$):
Number of $\displaystyle {^{235}\text{U}}$ nuclei $\displaystyle {= \frac{1000}{235} \times 6.023 \times 10^{23} = 2.563 \times 10^{24} \text{ nuclei}}$.
$\displaystyle {E_{\text{fission}} = 2.563 \times 10^{24} \times (200 \times 1.6 \times 10^{-13}\text{ J}) \approx 8.20 \times 10^{13}\text{ J}}$.
Comparison: Fusion of 1 kg hydrogen yields roughly 7.6 times more energy than fission of 1 kg Uranium-235.
Q.33. Explain the operation of a controlled nuclear fission reactor. Why are light elements used as moderators, and why are cadmium control rods provided?
1. Moderator Role: Fast neutrons ($\displaystyle {\sim 2\text{ MeV}}$) emitted during fission must be slowed to thermal speeds ($\displaystyle {\sim 0.025\text{ eV}}$) to induce efficient fission in $\displaystyle {^{235}\text{U}}$. Light nuclei (like $\displaystyle {\text{D}_2\text{O}}$ or graphite) undergo elastic collisions with neutrons, transferring maximum kinetic energy per impact without absorbing them.
2. Control Rods Role: Cadmium or boron rods possess huge absorption cross-sections for thermal neutrons. Inserting or withdrawing them controls the multiplication factor $\displaystyle {k \approx 1}$, preventing explosive runaway chain reactions.
ASSIGNMENT – 2 (MCQ PRACTICE)
1. Two nuclei have mass numbers in the ratio $\displaystyle {1 : 8}$. The ratio of their nuclear radii is:
Explanation: $\displaystyle {\frac{R_1}{R_2} = \left(\frac{A_1}{A_2}\right)^{1/3} = \left(\frac{1}{8}\right)^{1/3} = \frac{1}{2}}$.
2. A nucleus with mass number $\displaystyle {A = 220}$ initially at rest emits an $\alpha$-particle. If the $\displaystyle {Q}$-value of the reaction is $\displaystyle {5.5\text{ MeV}}$, the kinetic energy of the emitted $\alpha$-particle is:
Explanation: Conservation of momentum gives $\displaystyle {K_\alpha = Q \left(\frac{A-4}{A}\right) = 5.5 \times \left(\frac{216}{220}\right) = 5.4\text{ MeV}}$.
3. The mass defect of a $\displaystyle {^7_3\text{Li}}$ nucleus is $\displaystyle {0.042\text{ u}}$. The binding energy per nucleon of $\displaystyle {^7_3\text{Li}}$ is approximately:
Explanation: Total $\displaystyle {BE = 0.042 \times 931.5 \approx 39.12\text{ MeV}}$. Binding energy per nucleon $\displaystyle {BE/A = 39.12 / 7 \approx 5.59\text{ MeV/nucleon} \approx 5.6\text{ MeV/nucleon}}$.
4. The binding energy per nucleon of deuterium ($\displaystyle {^2_1\text{H}}$) and helium ($\displaystyle {^4_2\text{He}}$) are $\displaystyle {1.1\text{ MeV}}$ and $\displaystyle {7.0\text{ MeV}}$ respectively. When two deuterium nuclei fuse to form a helium nucleus, the energy released is:
Explanation: Initial $\displaystyle {BE = 2 \times (2 \times 1.1) = 4.4\text{ MeV}}$. Final $\displaystyle {BE = 4 \times 7.0 = 28.0\text{ MeV}}$.
Released energy $\displaystyle {Q = 28.0 - 4.4 = 23.6\text{ MeV}}$.
ASSIGNMENT – 3 (ASSERTION & REASON)
1. Assertion (A): Neutrons penetrate matter far more readily than protons.
Reason (R): Neutrons are slightly more massive than protons.
Explanation: Neutrons penetrate matter easily because they carry zero electrical charge and suffer no Coulomb repulsion; mass difference is irrelevant to penetration.
2. Assertion (A): The net attractive force between two protons ($\displaystyle {f_{pp}}$) is slightly smaller than between two neutrons ($\displaystyle {f_{nn}}$).
Reason (R): Electrostatic Coulomb repulsion between protons reduces the net attractive force.
3. Assertion (A): Rest mass alone is not conserved in nuclear reactions, but total mass-energy is conserved.
Reason (R): Mass and energy interconvert according to Einstein's relation $\displaystyle {E = m c^2}$.
ASSIGNMENT – 4 (CASE STUDY QUESTIONS)
Case Study: Binding Energy Curve & Nuclear Stability
The binding energy per nucleon ($\displaystyle {BE/A}$) curve peaks at $\displaystyle {A \approx 56}$ ($\displaystyle {^{56}\text{Fe}}$) with $\displaystyle {\sim 8.8\text{ MeV/nucleon}}$ and falls off for both light ($\displaystyle {A < 30}$) and heavy ($\displaystyle {A > 170}$) nuclei.
(i) Which nucleus possesses the maximum binding energy per nucleon?
(a) $\displaystyle {^1\text{H}}$ (b) $\displaystyle {^{56}\text{Fe}}$ (c) $\displaystyle {^{235}\text{U}}$
(ii) When a heavy nucleus undergoes fission, total binding energy:
(a) Decreases (b) Remains constant (c) Increases
(iii) What is mass defect? Write its formula.
(i) (b) $\displaystyle {^{56}\text{Fe}}$ ($\displaystyle {8.75 - 8.8\text{ MeV/nucleon}}$).
(ii) (c) Increases (product fragments sit higher on the curve and are more tightly bound).
(iii) Mass defect $\displaystyle {\Delta m}$ is the difference between the sum of rest masses of constituent nucleons and the actual rest mass of the nucleus: $\displaystyle {\Delta m = [Z m_p + (A-Z) m_n] - M_{\text{nucleus}}}$.
ASSIGNMENT – 5 (FORMULAE & CONCEPTUAL QUESTIONS)
CORE FORMULAE SUMMARY
$\displaystyle {\Delta m = [Z m_p + (A-Z) m_n] - M_{\text{nucleus}}}$
$\displaystyle {BE = \Delta m \cdot c^2 = \Delta m (\text{in u}) \times 931.5\text{ MeV}}$
CONCEPTUAL SHORT QUESTIONS
1. Why is nuclear density constant across all atomic nuclei regardless of mass number $\displaystyle {A}$?
2. Why does a nuclear chain reaction sometimes die out in a small block of enriched Uranium?
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