Chapter 13: Nuclei Case Study Questions | physics Class 12 CBSE

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A nuclear physics student is studying the composition of various atomic nuclei in the laboratory. She observes that chlorine has two naturally occurring isotopes: one with mass 34.98 u (abundance 75.4%) and another with mass 36.98 u (abundance 24.6%). She also notes that all isotopes of chlorine have 17 protons in their nuclei but differ in neutron number. During her study, she recalls that James Chadwick in 1932 bombarded beryllium nuclei with alpha particles and discovered neutrons, earning him the 1935 Nobel Prize in Physics. She further notes that a free neutron decays into a proton, electron, and antineutrino with a mean life of about 1000 seconds, yet inside a nucleus it remains stable.

Question 1: Define isotopes and state the number of neutrons in each of the two naturally occurring isotopes of chlorine (atomic number 17).

Isotopes are nuclides of the same element that have the same number of protons (same atomic number Z) but differ in their number of neutrons (different mass number A).
For chlorine (Z=17): Isotope with mass number 35 has 35 - 17 = 18 neutrons; Isotope with mass number 37 has 37 - 17 = 20 neutrons.

Question 2: Calculate the average atomic mass of chlorine using its two isotopes with masses 34.98 u and 36.98 u with relative abundances 75.4% and 24.6% respectively.

Average atomic mass = (75.4 × 34.98 + 24.6 × 36.98) / 100
= (2637.492 + 909.708) / 100 = 3547.2 / 100 = 35.47 u

Question 3: Chadwick's experiment established the existence of neutrons. Explain why the hypothesis that the neutral radiation in the experiment consisted of photons was rejected, and describe the significance of the neutron being stable inside the nucleus but unstable outside it.

If the neutral radiation were photons, the conservation of energy and momentum required the photons to have energies far exceeding what the alpha-beryllium bombardment could provide — this was physically impossible.
Chadwick assumed the radiation consisted of neutral particles (neutrons) of mass nearly equal to the proton mass, which satisfied both energy and momentum conservation consistently.
A free neutron decays (mean life ~1000 s) into a proton, electron, and antineutrino; inside the nucleus, nuclear forces stabilize the neutron, which is essential for nuclear structure and stability of heavy nuclei.

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In a physics class, a teacher demonstrates the concept of nuclear size using Rutherford's alpha scattering experiment. An alpha particle of kinetic energy 5.5 MeV is directed toward a gold nucleus and achieves a distance of closest approach of 4.0 × 10⁻¹⁴ m. The teacher explains that all nuclei are approximately spherical, and their radii follow the empirical relation R = R₀A^(1/3), where R₀ = 1.2 × 10⁻¹⁵ m. She further explains that nuclear density is independent of mass number A and approximately equals 2.3 × 10¹⁷ kg/m³. She asks students to compare this with the density of ordinary matter (~10³ kg/m³) to understand how matter is compressed inside a nucleus.

Question 1: State the empirical relation for nuclear radius and explain what it implies about the density of nuclear matter.

Nuclear radius R = R₀A^(1/3), where R₀ = 1.2 × 10⁻¹⁵ m (1.2 fm) and A is the mass number.
Since R ∝ A^(1/3), the volume V ∝ R³ ∝ A, so mass ∝ A also, meaning the nuclear density (mass/volume) is constant and independent of A for all nuclei.

Question 2: Calculate the ratio of the nuclear radius of gold (A = 197) to that of silver (A = 107).

Using R = R₀A^(1/3): R(Au)/R(Ag) = (197/107)^(1/3)
= (1.842)^(1/3) ≈ 1.225, so the nuclear radius of gold is approximately 1.23 times that of silver.

Question 3: Nuclear density is approximately 2.3 × 10¹⁷ kg/m³, which is about 10¹⁴ times the density of water. Explain physically why nuclear density is so high and why it is independent of the mass number A. Also mention one astrophysical object that has comparable density.

Atoms are mostly empty space; almost all mass is concentrated in the tiny nucleus. The nuclear radius is ~10⁴ times smaller than the atomic radius, making the nuclear volume ~10¹² times smaller than atomic volume, hence extremely high density.
Since R = R₀A^(1/3), Volume ∝ A and mass ∝ A, so their ratio (density) is constant — nuclear density does not depend on A.
Neutron stars have density comparable to nuclear matter density (~2.3 × 10¹⁷ kg/m³), as their matter is compressed to a state resembling a gigantic nucleus.

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A researcher studying nuclear binding energy plots the binding energy per nucleon (Ebn) versus mass number (A) for a range of nuclei. The graph shows that for nuclei in the range A = 30 to 170, Ebn is approximately constant at about 8 MeV/nucleon. Iron (Fe-56) has the highest Ebn of about 8.75 MeV/nucleon. For very light nuclei (A < 30) and very heavy nuclei (A > 170), Ebn is lower. The researcher notes that peaks exist at He-4, O-16, and similar nuclei, suggesting shell structure. She uses this graph to predict which nuclear processes will release energy and which will require energy input.

Question 1: Define binding energy per nucleon and state the mass number at which it is maximum.

Binding energy per nucleon (Ebn) is the total binding energy of a nucleus divided by its mass number A; it represents the average energy needed to separate one nucleon from the nucleus.
Ebn is maximum (~8.75 MeV/nucleon) for iron (Fe-56, A = 56).

Question 2: Using the binding energy per nucleon curve, explain why both nuclear fission and fusion release energy.

In fission, a heavy nucleus (A ~240, Ebn ~7.6 MeV) splits into medium-mass fragments (A ~120, Ebn ~8.5 MeV); the products have higher Ebn, meaning nucleons are more tightly bound, so energy (~200 MeV) is released.
In fusion, very light nuclei (low Ebn) combine to form a heavier nucleus with higher Ebn; the gain in binding energy per nucleon is released as energy (e.g., ~26.7 MeV for 4H → He-4 in the sun).

Question 3: The binding energy per nucleon is nearly constant for 30 < A < 170. Explain this constancy using the concept of short-range nuclear forces and saturation. Also, why does Ebn decrease for very heavy nuclei?

Nuclear forces are short-range; a nucleon inside a large nucleus interacts only with its nearest neighbours (within ~2–3 fm). Adding more nucleons does not increase the binding of an inner nucleon — this is the saturation property.
Since only a constant number p of neighbours influence each nucleon, binding energy per nucleon remains approximately constant (∝ pk) for large A.
For very heavy nuclei (A > 170), the long-range Coulomb repulsion between the many protons becomes significant and reduces Ebn, making the nucleus less stable. Surface effects also contribute to the decrease for very light nuclei.

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Einstein's mass-energy equivalence (E = mc²) is central to understanding nuclear reactions. In a school experiment, students are told that the mass of an oxygen-16 nucleus measured experimentally is 15.99053 u, while the sum of masses of 8 protons and 8 neutrons comes to 16.12744 u. The mass defect of 0.13691 u corresponds to a binding energy of approximately 127.5 MeV. The teacher explains that 1 u = 931.5 MeV/c² and this mass-energy interconversion occurs in both nuclear and chemical reactions, though the scale differs by about a million times. Students are asked to relate mass defect to nuclear stability.

Question 1: Define mass defect of a nucleus and write the expression relating it to the binding energy.

Mass defect ΔM is the difference between the sum of the masses of the individual protons and neutrons and the actual nuclear mass M: ΔM = [Zmp + (A−Z)mn] − M.
Binding energy Eb = ΔM × c², which represents the energy needed to completely separate all nucleons.

Question 2: Calculate the binding energy of the oxygen-16 nucleus given mass defect ΔM = 0.13691 u, using 1 u = 931.5 MeV/c².

Binding energy Eb = ΔM × 931.5 MeV = 0.13691 × 931.5 MeV
Eb ≈ 127.5 MeV

Question 3: Einstein's E = mc² applies to both nuclear and chemical reactions. Explain why mass-energy interconversion is considered significant only in nuclear reactions and not in chemical reactions, even though it occurs in both.

In chemical reactions, the binding energy changes are due to rearrangement of electrons (chemical bonds), with mass defects of the order of ~10⁻⁹ u per reaction — about a million times smaller than nuclear mass defects.
In nuclear reactions, the binding energy differences are of order MeV (vs. eV in chemical reactions) leading to measurable, significant mass defects (~0.1 u range).
Because the chemical mass defects are so tiny, they are practically undetectable and irrelevant for energy calculations in chemistry, giving the incorrect impression that mass-energy interconversion is exclusive to nuclear physics.

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Case Set 5

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A physics student is studying nuclear forces in preparation for her board examination. She reads that nuclear forces are much stronger than Coulomb forces, do not depend on electric charge, and are short-range in nature, operating effectively only up to a few femtometres. The potential energy between two nucleons is minimum at r₀ ≈ 0.8 fm; for r > r₀ the force is attractive, and for r < r₀ it is strongly repulsive. She also notes that the nuclear force is charge-independent — the force between proton-proton, proton-neutron, and neutron-neutron pairs is approximately the same. She contrasts this with the Coulomb force and gravitational force, which have well-defined mathematical forms.

Question 1: State two important properties of the nuclear force that distinguish it from the Coulomb force.

The nuclear force is much stronger than the Coulomb force (it can overcome proton-proton Coulomb repulsion at short distances).
The nuclear force is charge-independent (acts equally between p-p, p-n, and n-n pairs), whereas the Coulomb force acts only between charged particles.

Question 2: Explain the significance of the minimum in the potential energy curve at r₀ ≈ 0.8 fm between two nucleons.

At r₀ ≈ 0.8 fm, the potential energy is minimum, meaning the nucleons are in a state of stable equilibrium at this separation.
For r > r₀ the force is attractive (potential energy rises as r increases), and for r < r₀ the force is strongly repulsive (potential energy rises sharply), preventing nucleons from collapsing into each other.

Question 3: Compare the nuclear force, Coulomb force, and gravitational force in terms of strength, range, and charge dependence. How does the short-range nature of the nuclear force explain the saturation of binding energy per nucleon?

Nuclear force is the strongest of the three, Coulomb force is intermediate, and gravitational force is the weakest by far. Nuclear force has a very short range (~few fm), Coulomb force has infinite range (∝1/r²), and gravitational force also has infinite range.
Nuclear force is charge-independent; Coulomb force acts between charges; gravitational force acts between masses.
Because nuclear force is short-range, a nucleon inside a nucleus interacts only with its immediate neighbours (saturation). Adding more nucleons beyond the range does not increase a nucleon's binding energy, so Ebn remains roughly constant — this is the saturation property.

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Chapter 13: Nuclei | Case Study Questions

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Question 1: Define isotopes and state the number of neutrons in each of the two naturally occurring isotopes of chlorine (atomic number 17).

Question 2: Calculate the average atomic mass of chlorine using its two isotopes with masses 34.98 u and 36.98 u with relative abundances 75.4% and 24.6% respectively.

Question 3: Chadwick's experiment established the existence of neutrons. Explain why the hypothesis that the neutral radiation in the experiment consisted of photons was rejected, and describe the significance of the neutron being stable inside the nucleus but unstable outside it.

Passage

Question 1: State the empirical relation for nuclear radius and explain what it implies about the density of nuclear matter.

Question 2: Calculate the ratio of the nuclear radius of gold (A = 197) to that of silver (A = 107).

Question 3: Nuclear density is approximately 2.3 × 10¹⁷ kg/m³, which is about 10¹⁴ times the density of water. Explain physically why nuclear density is so high and why it is independent of the mass number A. Also mention one astrophysical object that has comparable density.

Passage

Question 1: Define binding energy per nucleon and state the mass number at which it is maximum.

Question 2: Using the binding energy per nucleon curve, explain why both nuclear fission and fusion release energy.

Question 3: The binding energy per nucleon is nearly constant for 30 < A < 170. Explain this constancy using the concept of short-range nuclear forces and saturation. Also, why does Ebn decrease for very heavy nuclei?

Passage

Question 1: Define mass defect of a nucleus and write the expression relating it to the binding energy.

Question 2: Calculate the binding energy of the oxygen-16 nucleus given mass defect ΔM = 0.13691 u, using 1 u = 931.5 MeV/c².

Question 3: Einstein's E = mc² applies to both nuclear and chemical reactions. Explain why mass-energy interconversion is considered significant only in nuclear reactions and not in chemical reactions, even though it occurs in both.

Passage

Question 1: State two important properties of the nuclear force that distinguish it from the Coulomb force.

Question 2: Explain the significance of the minimum in the potential energy curve at r₀ ≈ 0.8 fm between two nucleons.

Question 3: Compare the nuclear force, Coulomb force, and gravitational force in terms of strength, range, and charge dependence. How does the short-range nature of the nuclear force explain the saturation of binding energy per nucleon?

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