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#mass calculation

9 public questions tagged with this topic.

Radioactive Decay, Nuclear Forces and Stability

What is the nuclear density of a nucleus with mass \( 5.01 \times 10^{-27} \, \text{kg} \) and radius \( 2.1 \times 10^{

**Nuclear force** strong, short-range ~1 fm, attractive, charge independent, saturated in large nuclei because each nucleon interacts only with neighbors, not all others, so BE/A saturates ~8 MeV, primary factor limiting stable nuclei size is Coulomb repulsion between protons growing as Z² vs strong force saturating, beyond Z≈83 no stable nuclei, competition between Coulomb and strong. Density = (mass/volume) , Volume = (4/3) π R³ . R³ = (2.1 × 10⁻¹⁵)³ = 9.261 × 10⁻⁴⁵ m³ . Volume = (4/3) × 3.14 × 9.261 × 10⁻⁴⁵ ≈ 3.88 × 10⁻⁴⁴ m³ . Density = (5.01 × 10⁻²⁷/3.88 × 10⁻⁴⁴) ≈ 1.29 × 10¹⁷ kg/m³ .

Ref: NCERT > Physics Book > Atoms and Nuclei > Radioactive Decay, Nuclear Forces and Stability

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Nuclear Size, Density and Structure

What is the nuclear density of a nucleus with mass \( 6.68 \times 10^{-27} \, \text{kg} \) and radius \( 2.4 \times 10^{

**Nuclear size** order 10⁻¹⁵ m femtometer, atomic size 10⁻¹⁰ m, ratio 10⁵, nucleus contains protons and neutrons bound by strong force, density independent of A indicates incompressibility, R₀ determined from electron scattering experiments. Density = (mass/volume) , Volume = (4/3) π R³ . R³ = (2.4 × 10⁻¹⁵)³ = 1.382 × 10⁻⁴⁴ m³ . Volume = (4/3) × 3.14 × 1.382 × 10⁻⁴⁴ ≈ 5.79 × 10⁻⁴⁴ m³ . Density = (6.68 × 10⁻²⁷/5.79 × 10⁻⁴⁴) ≈ 1.15 × 10¹⁷ kg/m³ . Using E_n = -13.6/n² eV, r_n = n² a₀, L = n h/2π, R = R₀ A^¹/³, BE = Δm c² and 1

Ref: NCERT > Physics Book > Atoms and Nuclei > Nuclear Size, Density and Structure

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Potential Energy of Mass Systems

Two masses 2kg and 8kg are separated by 4m. What is the gravitational potential at the midpoint? (G\=6.67×10−11N m2/kg2)

Distance to midpoint: r = 2m. U = −Gm1r−Gm2r. U = −6.67×10−11(22+82). U = −6.67×10−11×(1+4) = −3.335×10−10J/kg. As per NCERT, applying relevant law/formula with correct units and sign convention leads to -3.3 × 10⁻¹⁰ J/kg. This satisfies dimensional consistency and physical conditions given, so option B is scientifically correct.

Ref: Halliday & Resnick, Fundamentals of Physics, 11th ed., Chapter 13: Gravitation.

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Corrosion and Applications of Electrochemistry

A mercury cell produces 0.5 g of Hg (atomic mass 200 g/mol) at the cathode. What mass of Zn (atomic mass 68 g/mol) is ox

Cathode: HgO + H₂O + 2e⁻ → Hg + 2OH⁻ , Moles = (0.5/200) = 0.0025 mol , Charge = 0.0025 × 2 × 96500 = 482.5 C . Anode: Zn → Zn²⁺ + 2e⁻ , Moles Zn = 0.0025 mol , Mass = 0.0025 × 68 = 0.17 g .

Ref: NCERT Class 12 Chemistry > Chapter 2: Electrochemistry > Topic: Corrosion and Applications of Electrochemistry

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Dalton's Atomic Theory and Atomic and Molecular Masses

A mixture of 6 g H₂ and 32 g O₂ is ignited to form H₂O. What is the mass of H₂O produced? (Atomic masses: H = 1, O = 16)

Reaction: 2H₂ + O₂ → 2H₂O. Moles: H₂ = 6/2 = 3 mol, O₂ = 32/32 = 1 mol. 2 mol H₂ need 1 mol O₂; 1 mol O₂ limits, reacts with 2 mol H₂. 2 mol H₂O formed = 2 × 18 = 36 g.

Ref: NCERT Class 11 Chemistry > Chapter 1: Some Basic Concepts of Chemistry > Topic: Dalton's Atomic Theory and Atomic and Molecular Masses

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Dalton's Atomic Theory and Atomic and Molecular Masses

A mixture of 6 g H₂ and 44 g O₂ is ignited to form H₂O. What is the mass of H₂O produced? (Atomic masses: H = 1, O = 16)

Reaction: 2 H₂ + O₂ → 2 H₂O. Moles: H₂ = 6/2 = 3 mol, O₂ = 44/32 = 1.375 mol. 2 mol H₂ need 1 mol O₂; 1.375 mol O₂ limits, reacts with 2.75 mol H₂. 2.75 mol H₂O formed = 2.75 × 18 = 49.5 g.

Ref: NCERT Class 11 Chemistry > Chapter 1: Some Basic Concepts of Chemistry > Topic: Dalton's Atomic Theory and Atomic and Molecular Masses

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