Nuclei: CBSE Class 12 Physics

Composition and Size of the Nucleus

At the center of every atom lies the nucleus, a region of immense density containing positively charged protons and electrically neutral neutrons. Collectively, protons and neutrons are called nucleons. The identity of an element is determined by its atomic number (Z), which is the number of protons in its nucleus. The mass number (A) represents the total number of nucleons (protons + neutrons). The number of neutrons (N) is therefore given by N = A - Z. A specific nucleus is represented as ³ᴬXₙ, where X is the chemical symbol.

Atoms with the same number of protons but different numbers of neutrons are called isotopes (e.g., ¹H, ²H, ³H). Atoms with the same mass number but different atomic numbers are isobars (e.g., ³¹⁴C and ³¹⁴N). Atoms with the same number of neutrons are isotones (e.g., ³⁹K₁₉ and ⁴⁰Ca₂₀, both have 21 neutrons).

Experiments show that nuclei are roughly spherical. The radius (R) of a nucleus is empirically found to be proportional to the cube root of its mass number (A). This relationship is given by the formula: R = R₀A¹/³, where R₀ is a constant approximately equal to 1.2 × 10⁻¹⁵ m (or 1.2 femtometers). A fascinating consequence of this relation is that the nuclear density is nearly constant for all nuclei, regardless of their size. It's incredibly high, about 2.3 × 10¹⁷ kg/m³, which is trillions of times denser than water! This means a sugar cube made of nuclear matter would weigh as much as a mountain.

Key Concepts: Mass-Energy and Nuclear Binding Energy

Mass-Energy Equivalence

Proposed by Albert Einstein, this fundamental principle states that mass and energy are interconvertible. The relationship is given by the iconic equation E = mc², where E is energy, m is mass, and c is the speed of light in vacuum (≈ 3 × 10⁸ m/s). This explains how a small amount of mass can be converted into a huge amount of energy in nuclear reactions.

Mass Defect (Δm)

The mass of a stable nucleus is always slightly less than the sum of the masses of its constituent protons and neutrons. This difference in mass is called the mass defect. Δm = [Z mₚ + (A-Z) mₙ] - Mₙᵤₙ, where mₚ is the mass of a proton, mₙ is the mass of a neutron, and Mₙᵤₙ is the actual mass of the nucleus.

Nuclear Binding Energy (B.E.)

The binding energy of a nucleus is the energy equivalent of its mass defect. It represents the energy that would be required to break the nucleus apart into its individual protons and neutrons. It is calculated as B.E. = Δm * c². A higher binding energy per nucleon indicates a more stable nucleus.

Binding Energy per Nucleon (B.E./A)

This is the average energy required to remove one nucleon from the nucleus. It is calculated by dividing the total binding energy (B.E.) by the mass number (A). The plot of B.E./A versus A, known as the binding energy curve, shows that nuclei near A=56 (Iron) are the most stable.

Worked Examples: Calculating Binding Energy

• Example 1: Find the binding energy and binding energy per nucleon of an alpha particle (⁴He nucleus). Given: Mass of proton (mₚ) = 1.007276 u, Mass of neutron (mₙ) = 1.008665 u, Mass of Helium nucleus (Mₙᵤₙ) = 4.001506 u, and 1 u = 931.5 MeV/c². Step 1: Identify the number of protons and neutrons. For a Helium nucleus (⁴He₂), Atomic number Z = 2 (2 protons), Mass number A = 4. Number of neutrons N = A - Z = 4 - 2 = 2. Step 2: Calculate the total mass of the constituent nucleons. Total mass of 2 protons = 2 × mₚ = 2 × 1.007276 u = 2.014552 u. Total mass of 2 neutrons = 2 × mₙ = 2 × 1.008665 u = 2.017330 u. Sum of constituent masses = 2.014552 u + 2.017330 u = 4.031882 u. Step 3: Calculate the mass defect (Δm). Δm = (Sum of constituent masses) - (Mass of Helium nucleus) Δm = 4.031882 u - 4.001506 u = 0.030376 u. Step 4: Calculate the total Binding Energy (B.E.). B.E. = Δm × 931.5 MeV/u B.E. = 0.030376 u × 931.5 MeV/u = 28.297 MeV. Step 5: Calculate the Binding Energy per Nucleon (B.E./A). B.E./A = Total B.E. / Mass Number (A) B.E./A = 28.297 MeV / 4 = 7.074 MeV/nucleon. Final Answer: The binding energy of the alpha particle is 28.297 MeV, and its binding energy per nucleon is 7.074 MeV.
• Example 2: Understanding the Law of Radioactive Decay. A radioactive sample has a half-life of 10 minutes. If the initial number of nuclei is 600, find the number of nuclei remaining after 30 minutes. Step 1: Determine the number of half-lives. The total time elapsed is t = 30 minutes. The half-life is T₁/₂ = 10 minutes. The number of half-lives (n) = Total time / Half-life = t / T₁/₂ n = 30 minutes / 10 minutes = 3. Step 2: Use the formula for remaining nuclei after n half-lives. The number of nuclei remaining (N) after n half-lives is given by the formula: N = N₀ / 2ⁿ, where N₀ is the initial number of nuclei. Step 3: Substitute the values and calculate. N₀ = 600 n = 3 N = 600 / 2³ = 600 / 8 = 75. Alternatively, using step-by-step decay: After 1st half-life (10 min): Nuclei remaining = 600 / 2 = 300. After 2nd half-life (20 min): Nuclei remaining = 300 / 2 = 150. After 3rd half-life (30 min): Nuclei remaining = 150 / 2 = 75. Final Answer: The number of nuclei remaining after 30 minutes is 75.

Exam Traps and Key Pointers

Watch out for these common pitfalls in your exams:

• Units are Crucial: Be very careful with units. Mass is often given in atomic mass units (u). Remember the conversion factor 1 u = 931.5 MeV/c². When calculating binding energy, you can directly use B.E. = Δm (in u) × 931.5 MeV.
• Mass of Nucleus vs. Mass of Atom: Problems sometimes give the mass of the neutral atom, which includes electrons. For precise binding energy calculations, you should use the mass of the nucleus. The mass of the nucleus can be found by subtracting the mass of the electrons from the atomic mass: M_nucleus = M_atom - Z*m_e. However, in many CBSE problems, this difference is negligible and can be ignored if you use atomic masses for both reactants and products, as the electron masses often cancel out.
• Mass Number (A) ≠ Atomic Mass: Mass number (A) is an integer count of nucleons. Atomic mass is the actual measured mass of the atom/nucleus and is never a perfect integer (except for Carbon-12 by definition).
• Radioactive Decay Formula: Don't confuse the two main decay formulas: N = N₀e⁻ˡᵗ and N = N₀ / 2ⁿ. The first uses the decay constant (λ) and time (t), while the second uses the number of half-lives (n). Use the one that fits the given data.

Practice Questions with Solutions

• Q: The radius of a ¹²⁵₅₂Te nucleus is measured to be 6.0 fm. What would be the expected radius of a ²⁷₁₃Al nucleus? A: Step 1: State the relationship between nuclear radius (R) and mass number (A). The formula is R = R₀A¹/³. Step 2: Set up a ratio to eliminate the constant R₀. Let R₁ and A₁ be for Tellurium (Te) and R₂ and A₂ be for Aluminium (Al). Then, (R₂/R₁) = (R₀A₂¹/³)/(R₀A₁¹/³) = (A₂/A₁)¹/³. Step 3: Substitute the given values. A₁ = 125, R₁ = 6.0 fm, A₂ = 27. R₂ = R₁ (A₂/A₁)¹/³ = 6.0 fm (27/125)¹/³. Step 4: Calculate the cube root. (27/125)¹/³ = 3/5 = 0.6. Step 5: Calculate the final radius. R₂ = 6.0 fm * 0.6 = 3.6 fm. Final answer: The expected radius of the ²⁷₁₃Al nucleus is 3.6 fm.
• Q: Calculate the energy released in the following fission reaction: ²³⁵₉₂U + ¹₀n → ¹⁴¹₅₆Ba + ⁹²₃₆Kr + 3(¹₀n). Given masses: M(²³⁵U) = 235.0439 u, M(¹⁴¹Ba) = 140.9144 u, M(⁹²Kr) = 91.9262 u, M(¹n) = 1.00866 u. A: Step 1: Calculate the total mass of the reactants. Mass_reactants = M(²³⁵U) + M(¹n) = 235.0439 u + 1.00866 u = 236.05256 u. Step 2: Calculate the total mass of the products. Mass_products = M(¹⁴¹Ba) + M(⁹²Kr) + 3 M(¹n) = 140.9144 u + 91.9262 u + 3 1.00866 u Mass_products = 140.9144 + 91.9262 + 3.02598 = 235.86658 u. Step 3: Calculate the mass defect (Δm) for the reaction. Δm = Mass_reactants - Mass_products = 236.05256 u - 235.86658 u = 0.18598 u. Step 4: Calculate the energy released (Q-value) using E = Δm c². Use the conversion 1 u = 931.5 MeV/c². Energy Released (Q) = Δm 931.5 MeV/u = 0.18598 u * 931.5 MeV/u = 173.23 MeV. Final answer: The energy released in the fission reaction is approximately 173.23 MeV.
• Q: A radioactive isotope has a half-life of T years. How long will it take for its activity to reduce to 3.125% of its original activity? A: Step 1: Relate activity to the number of nuclei. Activity (R) is proportional to the number of undecayed nuclei (N), so R/R₀ = N/N₀. The final activity is 3.125% of the initial activity R₀. So, R = 0.03125 R₀, which means N/N₀ = 0.03125. Step 2: Express the remaining fraction as a power of 1/2. 0.03125 = 3125 / 100000 = 1/32. We know that 32 = 2⁵. So, N/N₀ = 1/2⁵. Step 3: Use the formula for radioactive decay in terms of half-lives. The fraction of nuclei remaining after 'n' half-lives is given by N/N₀ = (1/2)ⁿ. Step 4: Compare the expressions to find 'n'. From Step 2 and 3, (1/2)ⁿ = 1/2⁵. Therefore, the number of half-lives, n, is 5. Step 5: Calculate the total time taken. Total time (t) = n Half-life (T). t = 5 * T = 5T. Final answer: It will take 5T years for the activity to reduce to 3.125% of its original value.
• Q: Two nuclei have mass numbers in the ratio 1:8. What is the ratio of their nuclear radii and nuclear densities? A: Step 1: Analyze the ratio of nuclear radii. The formula for nuclear radius is R = R₀A¹/³. Let the mass numbers be A₁ and A₂. We are given A₁/A₂ = 1/8. The ratio of their radii is R₁/R₂ = (R₀A₁¹/³)/(R₀A₂¹/³) = (A₁/A₂)¹/³. Step 2: Calculate the ratio of radii. R₁/R₂ = (1/8)¹/³ = 1/2. So, the ratio of their radii is 1:2. Step 3: Analyze the nuclear density. Nuclear density (ρ) is given by ρ = Mass / Volume = (A m) / (4/3 π R³), where m is the average mass of a nucleon. Step 4: Substitute R = R₀A¹/³ into the density formula. ρ = (A m) / (4/3 π (R₀A¹/³)³) = (A m) / (4/3 π R₀³ A) = (m) / (4/3 π R₀³). Step 5: Conclude the ratio of densities. As you can see, the 'A' term cancels out. The density ρ depends only on constants (m, R₀, π). Therefore, nuclear density is approximately constant for all nuclei. The ratio of their densities is 1:1. Final answer: The ratio of their nuclear radii is 1:2, and the ratio of their nuclear densities is 1:1.

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