#mass

A particle of mass \( 2.5 \times 10^{-29} \, \text{kg} \) has a de Broglie wavelength of \( 0.5 \, \text{nm} \). What is

**De Broglie wavelength calculations** for electron moving at 2×10⁶ m/s λ= h/(m v)=6.63×10⁻³⁴/(9.11×10⁻³¹×2×10⁶)=3.64×10⁻¹⁰ m=0.364 nm, for mass 5×10⁻³⁰ kg v=2×10⁵ m/s λ=6.63×10⁻³⁴/(1×10⁻²⁴)=6.63×10⁻¹⁰ m, showing λ decreases with increasing m v. p = (h/λ) = (6.63 × 10⁻³⁴/0.5 × 10⁻⁹) = 1.326 × 10⁻²⁴ kg m/s . v = (p/m) = (1.326 × 10⁻²⁴/2.5 × 10⁻²⁹) = 5.304 × 10⁴ m/s . Applying E = h f = h c/λ, p = h/λ, K_max = h f - Φ, Φ = h f₀ = h c/λ₀, λ = h/√(2 m e V) and f_max = e V/h, result 5.304 × 10⁴ m/s follows, reflecting photoelectric and de Broglie relations.

Ref: NCERT > Physics Book > Dual Nature of Matter > De Broglie Wavelength and Matter Waves

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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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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 \( 2.33 \times 10^{-27} \, \text{kg} \) 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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Damped Oscillations

A spring system has \( m = 1.0 \, \text{kg}, k = 400 \, \text{N/m}, A = 6 \, \text{cm} \). What is the potential energy

**Effect of damping** is gradual amplitude reduction while period remains nearly constant for light damping. Mechanical energy decreases as work done against damping force, E(t) = ½ k A(t)² decaying exponentially, and motion ceases without external energy input, distinguishing from ideal undamped SHM. Potential energy: U = (1/2) k x² . k = 400 N/m, x = 0.03 m . U = 0.5 × 400 × (0.03)² = 0.5 × 400 × 0.0009 = 0.18 J . Applying x = A cos(ωt + φ), v = -ωA sin(ωt + φ), a = -ω²x and E = ½kA² = ½mω²A², result 0.18 J follows, reflecting SHM