#radius

Radioactive Decay, Nuclear Forces and Stability

What is the nuclear density of a nucleus with mass \( 3.67 \times 10^{-27} \, \text{kg} \) and radius \( 2.0 \times 10^{

**Stability** belt of stability N≈Z for light, N>Z for heavy due to Coulomb, beyond leads to alpha decay, fission, stability requires balance, nuclear force saturated explains constant density and BE/A. Density = (mass/volume) , Volume = (4/3) π R³ . R³ = (2.0 × 10⁻¹⁵)³ = 8.0 × 10⁻⁴⁵ m³ . Volume = (4/3) × 3.14 × 8.0 × 10⁻⁴⁵ ≈ 3.35 × 10⁻⁴⁴ m³ . Density = (3.67 × 10⁻²⁷/3.35 × 10⁻⁴⁴) ≈ 1.10 × 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 u = 931.5

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

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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 \( 8.35 \times 10^{-27} \, \text{kg} \) and radius \( 2.7 \times 10^{

**Nuclear density** nearly constant because R ∝ A^{1/3} so volume ∝ A, mass ∝ A, ratio constant, ~10¹⁷ kg/m³, 10¹⁴ times water density, shows nucleus compact, nuclear force short-range saturated, mass number 16 radius ~3×10⁻¹⁵ m, mass number from radius R=5.4×10⁻¹⁵ m => A=(R/R₀)³=(4.5)³=91. Density = (mass/volume) , Volume = (4/3) π R³ . R³ = (2.7 × 10⁻¹⁵)³ = 1.9683 × 10⁻⁴⁴ m³ . Volume = (4/3) × 3.14 × 1.9683 × 10⁻⁴⁴ ≈ 8.25 × 10⁻⁴⁴ m³ . Density = (8.35 × 10⁻²⁷/8.25 × 10⁻⁴⁴) ≈ 1.01 × 10¹⁷ kg/m³ . Using E_n = -13.6/n² eV, r_n = n² a₀, L = n

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

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Nuclear Size, Density and Structure

What is the nuclear density of a nucleus with mass \( 1.66 \times 10^{-27} \, \text{kg} \) and radius \( 1.5 \times 10^{

**Nuclear radius** R = R₀ A^{1/3}, R₀=1.2×10⁻¹⁵ m, A mass number, for A=16 R=1.2×10⁻¹⁵×2.52=3.02×10⁻¹⁵ m, for A=36 R=1.2×10⁻¹⁵×3.30=3.96×10⁻¹⁵ m, nuclear density ρ = mass/volume = (A×1.66×10⁻²⁷ kg)/(4/3 π R³) ≈2.3×10¹⁷ kg/m³ independent of A, extremely high, mass 3.67×10⁻²⁷ kg radius 2.0×10⁻¹⁵ m gives density =3.67×10⁻²⁷/(4/3 π×8×10⁻⁴⁵)=1.09×10¹⁷ kg/m³. Density = (mass/volume) , Volume = (4/3) π R³ . R³ = (1.5 × 10⁻¹⁵)³ = 3.375 × 10⁻⁴⁵ m³ . Volume = (4/3) × 3.14 × 3.375 × 10⁻⁴⁵ ≈ 1.41 × 10⁻⁴⁴ m³ . Density = (1.66 × 10⁻²⁷/1.41 × 10⁻⁴⁴) ≈ 1.18 × 10¹⁷ kg/m³ . Using E_n = -13.6/n² eV, r_n = n² a₀,

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

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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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Hydrogen Atom Properties - Radius, Speed and Energy

What is the volume of a nucleus with radius \( 3.6 \times 10^{-15} \, \text{m} \)? (Use \( \pi = 3.14 \))

**Angular momentum** L_n = n h/2π, h=6.6×10⁻³⁴ J·s, L₂=2×6.6×10⁻³⁴/2π=2.11×10⁻³⁴ J·s, speed v_n = e²/(2 ε₀ h) ×1/n ≈2.2×10⁶/n m/s, kinetic energy ½ m v² =13.6/n² eV, illustrating Bohr model predictions for hydrogen-like atoms. Volume = (4/3) π R³ . R³ = (3.6 × 10⁻¹⁵)³ = 4.6656 × 10⁻⁴⁴ m³ . Volume = (4/3) × 3.14 × 4.6656 × 10⁻⁴⁴ ≈ 1.95 × 10⁻⁴³ 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 u = 931.5 MeV, evaluation yields 1.95 × 10⁻⁴³ m³, consistent with Bohr model and nuclear binding energy systematics.

Ref: NCERT > Physics Book > Atoms and Nuclei > Hydrogen Atom Properties - Radius, Speed and Energy

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Nuclear Size, Density and Structure

What is the nuclear density of a nucleus with mass \( 2.33 \times 10^{-27} \, \text{kg} \) and radius \( 1.7 \times 10^{

**Nuclear density** nearly constant because R ∝ A^{1/3} so volume ∝ A, mass ∝ A, ratio constant, ~10¹⁷ kg/m³, 10¹⁴ times water density, shows nucleus compact, nuclear force short-range saturated, mass number 16 radius ~3×10⁻¹⁵ m, mass number from radius R=5.4×10⁻¹⁵ m => A=(R/R₀)³=(4.5)³=91. Density = (mass/volume) , Volume = (4/3) π R³ . R³ = (1.7 × 10⁻¹⁵)³ = 4.913 × 10⁻⁴⁵ m³ . Volume = (4/3) × 3.14 × 4.913 × 10⁻⁴⁵ ≈ 2.06 × 10⁻⁴⁴ m³ . Density = (2.33 × 10⁻²⁷/2.06 × 10⁻⁴⁴) ≈ 1.13 × 10¹⁷ kg/m³ . Using E_n = -13.6/n² eV, r_n = n² a₀, L = n

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

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Nuclear Size, Density and Structure

Calculate the nuclear density of a nucleus if its mass is \( 1.67 \times 10^{-27} \, \text{kg} \) and radius is \( 1.2 \

**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 = (1.2 × 10⁻¹⁵)³ = 1.728 × 10⁻⁴⁵ m³ . Volume = (4/3) × 3.14 × 1.728 × 10⁻⁴⁵ = 7.24 × 10⁻⁴⁵ m³ . Density = (1.67 × 10⁻²⁷/7.24 × 10⁻⁴⁵) ≈ 2.31 × 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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De Broglie Hypothesis and Quantization in Bohr Model

What is the volume of a nucleus with radius \( 3.0 \times 10^{-15} \, \text{m} \)? (Use \( \pi = 3.14 \))

**Bohr's quantization** angular momentum L = m v r = n h/2π, for n=2 L=2h/2π= h/π=2.11×10⁻³⁴ J·s, for n=5 L=5h/2π, de Broglie λ = h/p, p= m v, for first orbit v=2.2×10⁶ m/s, λ= h/(m v)=6.6×10⁻³⁴/(9.1×10⁻³¹×2.2×10⁶)=3.3×10⁻¹⁰ m, circumference 2πr=3.33×10⁻¹⁰ m, one wavelength fits for n=1. Volume = (4/3) π R³ . R³ = (3.0 × 10⁻¹⁵)³ = 2.7 × 10⁻⁴⁴ m³ . Volume = (4/3) × 3.14 × 2.7 × 10⁻⁴⁴ ≈ 1.13 × 10⁻⁴³ 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 u = 931.5 MeV, evaluation

Ref: NCERT > Physics Book > Atoms and Nuclei > De Broglie Hypothesis and Quantization in Bohr Model

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Capacitance of Parallel Plate and Spherical Capacitor

A spherical conductor of radius 5 cm has a charge of \( 5 \times 10^{-8} \, \text{C} \). What is the potential at its su

**Parallel plate capacitor** capacitance C = ε₀ A/d, ε₀=8.85×10⁻¹² F/m, A plate area (m²), d separation (m), for air, with dielectric C = K ε₀ A/d. For A=0.08 m², d=0.4 mm=4×10⁻⁴ m, C=8.85×10⁻¹²×0.08/4×10⁻⁴=1.77×10⁻⁹ F=1.77 nF, illustrating small capacitance for cm separation. Potential at the surface: V = (1/4 π ε₀) (Q/R) . V = 9 × 10⁹ × (5 × 10⁻⁸/0.05) = 9 × 10⁹ × 10⁻⁶ = 9000 V . Using V = kQ/r, U = k q₁q₂/r, E = -dV/dr, C = ε₀A/d, 1/C_eq series, C_eq = ΣC parallel, U = ½ C V² and common potential V = Q_total/C_total, result 9000 V follows, reflecting potential-capacitance relations.

Ref: NCERT > Physics Book > Electrostatic Potential and Capacitance > Capacitance of Parallel Plate and Spherical Capacitor

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Magnetic Force on Moving Charge - Lorentz Force and Motion

A proton moves with a speed of \( 1.8 \times 10^6 \, \text{m/s} \) perpendicular to a magnetic field of \( 0.5 \, \text{

**Effect of doubling velocity** on magnetic force F = q v B sinθ is linear increase, F doubles for same θ and B. Electron with charge 1.6×10⁻¹⁹ C, v = 4.5×10⁶ m/s, B = 0.35 T, θ = 90°, F = 1.6×10⁻¹⁹×4.5×10⁶×0.35 = 2.52×10⁻¹³ N, illustrating magnitude for typical lab values. Radius r = (mv/qB) . r = (1.67 × 10⁻²⁷ × 1.8 × 10⁶/1.6 × 10⁻¹⁹ × 0.5) = (3.006 × 10⁻²¹/8 × 10⁻²⁰) = 3.7575 × 10⁻² m = 3.76 cm . Using F = q v B sinθ, F = I l B sinθ, B = μ₀ I/(2π r), B = μ₀

Ref: NCERT > Physics Book > Moving Charge and Magnetism > Magnetic Force on Moving Charge - Lorentz Force and Motion

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Magnetic Field Due to Current - Straight Wire and Circular Loop

A circular coil of 60 turns and radius \( 8 \, \text{cm} \) carries a current of \( 0.9 \, \text{A} \). What is the magn

**Field at centre of circular loop** with N turns is B = μ₀ N I/(2R), R radius (m), direction along axis via right-hand rule, magnitude proportional to N I/R. For R = 0.09 m, N = 45, I = 1.2 A, B = 4π×10⁻⁷×45×1.2/(2×0.09) = 3.77×10⁻⁴ T, showing N enhancement. Magnetic field B = (μ₀ N I/2 R) . B = (4 π × 10⁻⁷ × 60 × 0.9/2 × 0.08) = (21.6 π × 10⁻⁶/0.16) = 1.35 π × 10⁻⁴ ≈ 4.24 × 10⁻⁴ T . Using F = q v B sinθ, F = I l B sinθ, B = μ₀ I/(2π r),

Ref: NCERT > Physics Book > Moving Charge and Magnetism > Magnetic Field Due to Current - Straight Wire and Circular Loop

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